Wavefunctions as gravitational waves

This is the paper I always wanted to write. It is there now, and I think it is good – and that‘s an understatement. 🙂 It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.

It’s great stuff. If you have been following my blog – and many of you have – you will want to digest this. 🙂

Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass – which is, of course, the dimension of acceleration (m/s2) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit charge) by the new N/kg = m/s2 dimension.

The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a physical normalization condition. Also, Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.

While the approach offers a physical interpretation of the wavefunction, the author argues that the core of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does not explain the particle nature of matter.

Introduction

This is not another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an interpretation of wave mechanics. As such, we do not challenge the complementarity principle: the physical interpretation of the wavefunction that is offered here explains the wave nature of matter only. It explains diffraction and interference of amplitudes but it does not explain why a particle will hit the detector not as a wave but as a particle. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.

The basic ideas in this paper stem from a simple observation: the geometric similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (E and B) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (ψ = a·ei∙θ = a∙cosθ – a∙sinθ) differ by 90 degrees (π/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?

We show the answer is positive and remarkably straightforward.  If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?

The similarity between the energy of a (one-dimensional) linear oscillator (E = m·a2·ω2/2) and Einstein’s relativistic energy equation E = m∙c2 inspires us to interpret the energy as a two-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.

As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]

Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as real vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]

Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einstein’s basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some physical explanation for the calculated probabilities.[7]

We will, therefore, start with Einstein’s relativistic energy equation (E = mc2) and wonder what it could possibly tell us. 

I. Energy as a two-dimensional oscillation of mass

The structural similarity between the relativistic energy formula, the formula for the total energy of an oscillator, and the kinetic energy of a moving body, is striking:

  1. E = mc2
  2. E = mω2/2
  3. E = mv2/2

In these formulas, ω, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω2/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in two dimensions, so it stores an amount of energy that is equal to E = 2·m·ω2/2 = m·ω2?

That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring. Hence, we can describe it by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs.

Figure 1: Oscillations in two dimensionsV-2 engine

If we assume there is no friction, we have a perpetuum mobile here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]

At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to a, then the motion of the piston (or the mass on a spring) will be described by x = a·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our t = 0 point, and ω is the natural angular frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by x = a·cos(ω·t), while the motion of the other is given by x = a·cos(ω·t–π/2) = a·sin(ω·t).

The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:

  1. K.E. = T = m·v2/2 = (1/2)·m·ω2·a2·sin2(ω·t + Δ)
  2. P.E. = U = k·x2/2 = (1/2)·k·a2·cos2(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω2. Hence, the total energy is equal to:

E = T + U = (1/2)· m·ω2·a2·[sin2(ω·t + Δ) + cos2(ω·t + Δ)] = m·a2·ω2/2

To facilitate the calculations, we will briefly assume k = m·ω2 and a are equal to 1. The motion of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin2θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:

d(sin2θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ

Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin2(θ−π /2), and how it changes – as a function of θ – will be equal to:

2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ

We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = ma2ω2.

We have a great metaphor here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent half of the total energy of our particle? Should we think of the c in our E = mc2 formula as an angular velocity?

These are sensible questions. Let us explore them. 

II. The wavefunction as a two-dimensional oscillation

The elementary wavefunction is written as:

ψ = a·ei[E·t − px]/ħa·ei[E·t − px]/ħ = a·cos(px E∙t/ħ) + i·a·sin(px E∙t/ħ)

When considering a particle at rest (p = 0) this reduces to:

ψ = a·ei∙E·t/ħ = a·cos(E∙t/ħ) + i·a·sin(E∙t/ħ) = a·cos(E∙t/ħ) i·a·sin(E∙t/ħ) 

Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates clockwise with time, while the mathematical convention for measuring the phase angle (ϕ) is counter-clockwise.

Figure 2: Euler’s formula760px-eulers_formula

If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and px/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – will either freeze x or, else, t. Alternatively, one can google web animations varying both. The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the x-direction, then the oscillations are along the y– and z-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine  – the real and imaginary part of our wavefunction – appear to give some spin to the whole. I will come back to this.

Figure 3: Geometric representation of the wavefunction5d_euler_f

Hence, if we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary part of the wavefunction may then describe how energy propagates through space over time.

Let us consider, once again, a particle at rest. Hence, p = 0 and the (elementary) wavefunction reduces to ψ = a·ei∙E·t/ħ. Hence, the angular velocity of both oscillations, at some point x, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = mc2.

Can we, somehow, relate this to the m·a2·ω2 energy formula for our V-2 perpetuum mobile? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the square of the amplitude: E µ a2. We may, therefore, think that the a2 factor in the E = m·a2·ω2 energy will surely be relevant as well.

However, here is a complication: an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ak, and their own ωi = -Ei/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both ai as well as Ei will matter.

What is Ei? Ei varies around some average E, which we can associate with some average mass m: m = E/c2. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:F1We can re-write this as:F2What is the meaning of this equation? We may look at it as some sort of physical normalization condition when building up the Fourier sum. Of course, we should relate this to the mathematical normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy densities, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma: what is mass?

Before we do so, let us quickly calculate the value of c2ħ2: it is about 1´1051 N2∙m4. Let us also do a dimensional analysis: the physical dimensions of the E = m·a2·ω2 equation make sense if we express m in kg, a in m, and ω in rad/s. We then get: [E] = kg∙m2/s2 = (N∙s2/m)∙m2/s2 = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N3∙m5. 

III. What is mass?

We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new aether theory is, of course, not an option, but then what is it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einstein’s energy equation once again. If we want to look at mass, we should re-write it as m = E/c2:

[m] = [E/c2] = J/(m/s)2 = N·m∙s2/m2 = N·s2/m = kg

This is not very helpful. It only reminds us of Newton’s definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the absolute nature of the speed of light. Einstein’s E = mc2 equation implies we can write the ratio between the energy and the mass of any particle is always the same, so we can write, for example:F3This reminds us of the ω2= C1/L or ω2 = k/m of harmonic oscillators once again.[13] The key difference is that the ω2= C1/L and ω2 = k/m formulas introduce two or more degrees of freedom.[14] In contrast, c2= E/m for any particle, always. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in one physical space only: our spacetime. Hence, the speed of light c emerges here as the defining property of spacetime – the resonant frequency, so to speak. We have no further degrees of freedom here.

 

The Planck-Einstein relation (for photons) and the de Broglie equation (for matter-particles) have an interesting feature: both imply that the energy of the oscillation is proportional to the frequency, with Planck’s constant as the constant of proportionality. Now, for one-dimensional oscillations – think of a guitar string, for example – we know the energy will be proportional to the square of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us two waves for the price of one, so to speak, each carrying half of the total energy of the oscillation but, as a result, we get a proportionality between E and f instead of between E and f2.

However, such reflections do not answer the fundamental question we started out with: what is mass? At this point, it is hard to go beyond the circular definition that is implied by Einstein’s formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and c emerges us as the property of spacetime that defines how exactly.

When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real number that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, condense into elementary particles? That is what the Higgs mechanism is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.

What we can do, however, is look at the wave equation again (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation. 

IV. Schrödinger’s equation as an energy diffusion equation

The interpretation of Schrödinger’s equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:

“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”[17]

Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/meff)·∇2ψ(x, t)

The ubiquitous diffusion equation in physics is:

∂φ(x, t)/∂t = D·∇2φ(x, t)

The structural similarity is obvious. The key difference between both equations is that the wave equation gives us two equations for the price of one. Indeed, because ψ is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:

  1. Re(∂ψ/∂t) = −(1/2)·(ħ/meffIm(∇2ψ)
  2. Im(∂ψ/∂t) = (1/2)·(ħ/meffRe(∇2ψ)

These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):

  1. B/∂t = –∇×E
  2. E/∂t = c2∇×B

The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.

Figure 4: Propagation mechanismspropagation

The Laplacian operator (∇2), when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m2). In this case, it is operating on ψ(x, t), so what is the dimension of our wavefunction ψ(x, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/meff) factor:

  1. As a mathematical constant of proportionality, it will quantify the relationship between both derivatives (i.e. the time derivative and the Laplacian);
  2. As a physical constant, it will ensure the physical dimensions on both sides of the equation are compatible.

Now, the ħ/meff factor is expressed in (N·m·s)/(N· s2/m) = m2/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s1 while, as mentioned above, the dimension of ∇2ψ is m2. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?

At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of Schrödinger’s equation. One may argue, effectively, that its argument, (px – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.

To this, we may object that ħ may be looked as a mathematical scaling constant only. If we do that, then the argument of ψ will, effectively, be expressed in action units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?

We may have a closer look at Maxwell’s equations for inspiration here. The electric field vector is expressed in newton (the unit of force) per unit of charge (coulomb). Now, there is something interesting here. The physical dimension of the magnetic field is N/C divided by m/s.[19] We may write B as the following vector cross-product: B = (1/c)∙ex×E, with ex the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)∙ex× operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90° degrees. Hence, we may boldly write: B = (1/c)∙ex×E = (1/c)∙iE. This allows us to also geometrically interpret Schrödinger’s equation in the way we interpreted it above (see Figure 3).[20]

Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newton’s and Coulomb’s force laws:F4Hence, if the electric field vector E is expressed in force per unit charge (N/C), then we may want to think of associating the real part of our wavefunction with a force per unit mass (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 N·s2/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s2/m)= m/s2

What is this: m/s2? Is that the dimension of the a·cosθ term in the a·eiθ a·cosθ − i·a·sinθ wavefunction?

My answer is: why not? Think of it: m/s2 is the physical dimension of acceleration: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for any particle – matter-particles or particles with zero rest mass (photons) – and the associated wave equation (which has to be the same for all, as the spacetime we live in is one) are mutually consistent.

In this regard, we should think of how we would model a gravitational wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.

V. Energy densities and flows

Pursuing the geometric equivalence between the equations for an electromagnetic wave and Schrödinger’s equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:F5E and B are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:F6Needless to say, the ∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’s source or sink at a given point. To be precise, the divergence gives us the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. In this case, it gives us the volume density of the flux of S.

We can analyze the dimensions of the equation for the energy density as follows:

  1. E is measured in newton per coulomb, so [EE] = [E2] = N2/C2.
  2. B is measured in (N/C)/(m/s), so we get [BB] = [B2] = (N2/C2)·(s2/m2). However, the dimension of our c2 factor is (m2/s2) and so we’re also left with N2/C2.
  3. The ϵ0 is the electric constant, aka as the vacuum permittivity. As a physical constant, it should ensure the dimensions on both sides of the equation work out, and they do: [ε0] = C2/(N·m2) and, therefore, if we multiply that with N2/C2, we find that is expressed in J/m3.[21]

Replacing the newton per coulomb unit (N/C) by the newton per kg unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute ϵ0 for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – then the probabilities should be proportional to them.

Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting B for (1/c)∙iE or for −(1/c)∙iE gives us the following result:F7Zero!? An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously, as shown below.[22] This is because their phase is the same.

Figure 5: Electromagnetic wave: E and BEM field

Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between a·cosθ and a·sinθ, which gives a different picture of the propagation of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:F8We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !

|ψ|2  = |a·ei∙E·t/ħ|2 = a2 = u

This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (rest) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.

As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:

“Why is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.” (Feynman, Lectures, III-4-1)

The physical interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’: the physical dimension of the oscillation is just very different. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more. 

VI. Group and phase velocity of the matter-wave

The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively travels through space and time. But what is traveling, exactly? It is the pulse – or the signal – only: the phase velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the group velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves with our particle.

Here, we should also reiterate that we did not answer the question as to what is oscillating up and down and/or sideways: we only associated a physical dimension with the components of the wavefunction – newton per kg (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (newton per coulomb, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated energy densities and a Poynting vector for energy dissipation. In addition, we showed that Schrödinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:

ψ = a·ei[E·t − px]/ħa·ei[E·t − px]/ħ = a·cos(px/ħ − E∙t/ħ) + i·a·sin(px/ħ − E∙t/ħ)

The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−v∙t) wavefunction will always describe some wave that is traveling in the positive x-direction (with the wave velocity), while an F(x+v∙t) wavefunction will travel in the negative x-direction. For a geometric interpretation of the wavefunction in three dimensions, we need to agree on how to define i or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving counterclockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, p = 0, and the wavefunction reduces to:

ψ = a·ei∙E·t/ħ = a·cos(−E∙t/ħ) + i·a·sin(−E0∙t/ħ) = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)

E0 is, of course, the rest mass of our particle and, now that we are here, we should probably wonder whose time we are talking about: is it our time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t is, effectively, the proper time so perhaps we should write it as t0. It does not matter. You can see what we expect to see: E0/ħ pops up as the natural frequency of our matter-particle: (E0/ħ)∙t = ω∙t. Remembering the ω = 2π·f = 2π/T and T = 1/formulas, we can associate a period and a frequency with this wave, using the ω = 2π·f = 2π/T. Noting that ħ = h/2π, we find the following:

T = 2π·(ħ/E0) = h/E0 ⇔ = E0/h = m0c2/h

This is interesting, because we can look at the period as a natural unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (vg) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by vp = λ·= (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k = p/ħ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is not at rest? We write:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·vg) = (m·c2)/(m·vg) = c2/vg

This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with as a simple scaling constant. Indeed, the graph below shows the shape of the function does not change with the value of c, and we may also re-write the relation above as:

vp/= βp = c/vp = 1/βg = 1/(c/vp)

Figure 6: Reciprocal relation between phase and group velocitygraph

We can also write the mentioned relationship as vp·vg = c2, which reminds us of the relationship between the electric and magnetic constant (1/ε0)·(1/μ0) = c2. This is interesting in light of the fact we can re-write this as (c·ε0)·(c·μ0) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]

Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k = p/ħ. We would probably like to think it implies the elementary wavefunction should always be associated with some momentum, because the concept of zero momentum clearly leads to weird math: something times zero cannot be equal to c2! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then neither Δx nor Δp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually being at some specific point in time and in space does not make sense: it has to move. It tells us that our concept of dimensionless points in time and space are mathematical notions only. Actual particles – including photons – are always a bit spread out, so to speak, and – importantly – they have to move.

For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = m·c = m·c2/= E/c. Using the relationship above, we get:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = c ⇒ vg = c2/vp = c2/c = c

This is good: we started out with some reflections on the matter-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle having to move, we should remind ourselves, once again, of the fact that an actual particle is always localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·ei[E·t − px]/ħ or, for a particle at rest, the ψ = a·ei∙E·t/ħ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own ωi = −Ei/ħ. Indeed, in section II, we showed that each of these wavefunctions will contribute some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both ai as well as Ei matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.

To calculate a meaningful group velocity, we must assume the vg = ∂ωi/∂ki = ∂(Ei/ħ)/∂(pi/ħ) = ∂(Ei)/∂(pi) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate ωi as a function of ki here, or Ei as a function of pi. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇2ψ wave equation and, hence, re-write it as the following pair of two equations:

  1. Re(∂ψ/∂t) = −[ħ/(2meff)]·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·[ħ/(2meff)]·cos(kx − ωt)
  2. Im(∂ψ/∂t) = [ħ/(2meff)]·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·[ħ/(2meff)]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k2/(2meff)

Of course, we need to think about the subscripts now: we have ωi, ki, but… What about meff or, dropping the subscript, m? Do we write it as mi? If so, what is it? Well… It is the equivalent mass of Ei obviously, and so we get it from the mass-energy equivalence relation: mi = Ei/c2. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: Ei varies around some average energy E and, therefore, the Uncertainty Principle kicks in. 

VII. Explaining spin

The elementary wavefunction vector – i.e. the vector sum of the real and imaginary component – rotates around the x-axis, which gives us the direction of propagation of the wave (see Figure 3). Its magnitude remains constant. In contrast, the magnitude of the electromagnetic vector – defined as the vector sum of the electric and magnetic field vectors – oscillates between zero and some maximum (see Figure 5).

We already mentioned that the rotation of the wavefunction vector appears to give some spin to the particle. Of course, a circularly polarized wave would also appear to have spin (think of the E and B vectors rotating around the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the equivalent mass of its energy may be thought of as rotating as well. But so here we are looking at a matter-wave.

The basic idea is the following: if we look at ψ = a·ei∙E·t/ħ as some real vector – as a two-dimensional oscillation of mass, to be precise – then we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.

Figure 7: Torque and angular momentum vectorsTorque_animation

A torque on some mass about a fixed axis gives it angular momentum, which we can write as the vector cross-product L = r×p or, perhaps easier for our purposes here as the product of an angular velocity (ω) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:

L = I·ω

Note we can write L and ω in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the period of the matter-wave is equal to T = 2π·(ħ/E0). Hence, the angular velocity must be equal to:

ω = 2π/[2π·(ħ/E0)] = E0

We also know the distance r, so that is the magnitude of r in the Lr×p vector cross-product: it is just a, so that is the magnitude of ψ = a·ei∙E·t/ħ. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = m·v. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = r·ω. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (ω), the angular mass (I). Here we need to make some assumption about the mass (or energy) distribution. Now, it may or may not sense to assume the energy in the oscillation – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·r2/2. If we keep the analysis non-relativistic, then m = m0. Of course, the energy-mass equivalence tells us that m0 = E0/c2. Hence, this is what we get:

L = I·ω = (m0·r2/2)·(E0/ħ) = (1/2)·a2·(E0/c2)·(E0/ħ) = a2·E02/(2·ħ·c2)

Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m2·J2 = m2·N2·m2 in the numerator and N·m·s·m2/s2 in the denominator. Hence, the dimensions work out: we get N·m·s as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the action dimension, of course, and that cannot be a coincidence. Also note that the E = mc2 equation allows us to re-write it as:

L = a2·E02/(2·ħ·c2)

Of course, in quantum mechanics, we associate spin with the magnetic moment of a charged particle, not with its mass as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in N·m·s units, and which can only take on one of two possible values: J = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·a2·m02/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10−14 N∙m, and a… What value should we take for a?

We have an obvious trio of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.

Let us start with the Bohr radius, so that is about 0.×10−10 N∙m. We get L = a2·E02/(2·ħ·c2) = 9.9×10−31 N∙m∙s. Now that is about 1.88×104 times ħ/2. That is a huge factor. The Bohr radius cannot be right: we are not looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)×10−34 joule in energy. So our electron should pack about 1.24×10−20 oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626×10−34 joule for E0 and the Bohr radius is equal to 6.49×10−59 N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10−20), we get about 8.01×10−51 N∙m∙s, so that is a totally different number.

The classical electron radius is about 2.818×10−15 m. We get an L that is equal to about 2.81×10−39 N∙m∙s, so now it is a tiny fraction of ħ/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631×10−12 m.

This gives us an L of 2.08×10−33 N∙m∙s, which is only 20 times ħ. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we take for a so as to ensure L = a2·E02/(2·ħ·c2) = ħ/2? Let us write it out:F9

In fact, this is the formula for the so-called reduced Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for a (you can calculate it: it is about 3.8616×10−33 m), we get what we should find:F10

This is a rather spectacular result, and one that would – a priori – support the interpretation of the wavefunction that is being suggested in this paper. 

VIII. The boson-fermion dichotomy

Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an actual particle is localized in space and that it can, therefore, not be represented by the elementary wavefunction ψ = a·ei[E·t − px]/ħ or, for a particle at rest, the ψ = a·ei∙E·t/ħ function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude ai, and their own ωi = −Ei/ħ. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.

Think of the apparent right-handedness of the elementary wavefunction: surely, Nature can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: J = +ħ/2 or −ħ/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they either consist of (elementary) right-handed waves or, else, of (elementary) left-handed waves. An elementary right-handed wave would be written as:

ψ(θi= ai·(cosθi + i·sinθi)

In contrast, an elementary left-handed wave would be written as:

ψ(θi= ai·(cosθii·sinθi)

How does that work out with the E0·t argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but Nature surely does not care how we count time: counting like 1, 2, 3, etcetera or like −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:

ψ = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)

If we count time like −1, −2, −3, etcetera then we write it as:

 ψ = a·cos(E0∙t/ħ) − i·a·sin(E0∙t/ħ)= a·cos(E0∙t/ħ) + i·a·sin(E0∙t/ħ)

Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have either positive or negative quantum-mechanical spin (+ħ/2 or −ħ/2). It is the usual thing: we have two mathematical possibilities here, and so we must have two physical situations that correspond to it.

It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynman’s Lecture on it (Feynman, III-4), which is confusing and – I would dare to say – even inconsistent: how are photons or electrons supposed to know that they need to interfere with a positive or a negative sign? They are not supposed to know anything: knowledge is part of our interpretation of whatever it is that is going on there.

Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different physical dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually carry charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called vacuum – and the rest mass of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:

(vphase·c)·(vgroup·c) = 1 ⇔ vp·vg = c2

The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any charge. They should, therefore, not have any magnetic moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass as well as charge.[26] 

IX. Concluding remarks

There are, of course, other ways to look at the matter – literally. For example, we can imagine two-dimensional oscillations as circular rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.

Figure 8: Two-dimensional circular movementoscillation-of-a-ball

The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the reality of the wavefunction. Stating that it is a mathematical construct only without physical significance amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.

The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) charge unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.

The interpretation that is offered here has some advantages over other explanations, as it explains the how of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does not explain its particle nature: while we think of the energy as being spread out, we will still observe electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of ‘hook’ the whole blob of energy, so to speak?

The interpretation of the wavefunction that is offered here does not explain this. Hence, the complementarity principle of the Copenhagen interpretation of the wavefunction surely remains relevant.

Appendix 1: The de Broglie relations and energy

The 1/2 factor in Schrödinger’s equation is related to the concept of the effective mass (meff). It is easy to make the wrong calculations. For example, when playing with the famous de Broglie relations – aka as the matter-wave equations – one may be tempted to derive the following energy concept:

  1. E = h·f and p = h/λ. Therefore, f = E/h and λ = p/h.
  2. v = λ = (E/h)∙(p/h) = E/p
  3. p = m·v. Therefore, E = v·p = m·v2

E = m·v2? This resembles the E = mc2 equation and, therefore, one may be enthused by the discovery, especially because the m·v2 also pops up when working with the Least Action Principle in classical mechanics, which states that the path that is followed by a particle will minimize the following integral:F11Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the sum of the kinetic and the potential energy is zero throughout the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE − PE = m·v2.[27]

However, that is classical mechanics and, therefore, not so relevant in the context of the de Broglie equations, and the apparent paradox should be solved by distinguishing between the group and the phase velocity of the matter wave.

Appendix 2: The concept of the effective mass

The effective mass – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’s equation written as:F12This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:

∂ψ(x, t)/∂t = i·(1/2)·(ħ/meff)·∇2ψ(x, t)

We just moved the i·ħ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute a·ei∙[E·t − p∙x]/ħ for ψ), this implies the following:

a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ = −i·(ħ/2meffa·(p22 ei∙[E·t − p∙x]/ħ 

⇔ E = p2/(2meff) ⇔ meff = m∙(v/c)2/2 = m∙β2/2

It is an ugly formula: it resembles the kinetic energy formula (K.E. = m∙v2/2) but it is, in fact, something completely different. The β2/2 factor ensures the effective mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define meff as two times the old meff (hence, meffNEW = 2∙meffOLD), as a result of which the formula will look somewhat better:

meff = m∙(v/c)2 = m∙β2

We know β varies between 0 and 1 and, therefore, meff will vary between 0 and m. Feynman drops the subscript, and just writes meff as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a stationary electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).

In the context of the derivation of the electron orbitals, we do have the potential energy term – which is the equivalent of a source term in a diffusion equation – and that may explain why the above-mentioned meff = m∙(v/c)2 = m∙β2 formula does not apply.

References

This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynman’s Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.

Notes

[1] Of course, an actual particle is localized in space and can, therefore, not be represented by the elementary wavefunction ψ = a·ei∙θa·ei[E·t − px]/ħ = a·(cosθ i·a·sinθ). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude ak and its own argument θk = (Ek∙t – pkx)/ħ. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.

[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s2), thereby facilitating a direct interpretation in terms of Newton’s force law.

[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the author’s perpetuum mobile may be replaced by springs.

[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.

[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.

[6] For example, when using Schrödinger’s equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)

[7] This sentiment is usually summed up in the apocryphal quote: “God does not play dice.”The actual quote comes out of one of Einstein’s private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)

[8] Of course, both are different velocities: ω is an angular velocity, while v is a linear velocity: ω is measured in radians per second, while v is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = m·a2∙ω2/2. The additional factor (a) is the (maximum) amplitude of the oscillator.

[9] We also have a 1/2 factor in the E = mv2/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As for the exclusion of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy mirror each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = mv2. Appendix 1 provides some notes on that.

[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.

[11] It is interesting to note that we may look at the energy in the rotating flywheel as potential energy because it is energy that is associated with motion, albeit circular motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. rotational inertia (I) and angular velocity ω. The kinetic energy of a rotating object is then given by K.E. = (1/2)·I·ω2.

[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only approximately, but you can easily imagine the idealized limit situation.

[13] The ω2= 1/LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as ω2= C1/L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.

[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the inertia, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as γm and as R = γL respectively.

[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (Lectures, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 108, which means the wave train will last about 10–8 seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/e). For example, for sodium light, the radiation will last about 3.2×10–8 seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×1012 oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom.

[16] This is a general result and is reflected in the K.E. = T = (1/2)·m·ω2·a2·sin2(ω·t + Δ) and the P.E. = U = k·x2/2 = (1/2)· m·ω2·a2·cos2(ω·t + Δ) formulas for the linear oscillator.

[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the local conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.

[18] The meff is the effective mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write meff = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + i∙b and c + i∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = i∙(ħ/meff)∙∇2ψ equation amounts to writing something like this: a + i∙b = i∙(c + i∙d). Now, remembering that i2 = −1, you can easily figure out that i∙(c + i∙d) = i∙c + i2∙d = − d + i∙c.

[19] The dimension of B is usually written as N/(m∙A), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 A∙s and, hence, 1 N/(m∙A) = 1 (N/C)/(m/s).     

[20] Of course, multiplication with i amounts to a counterclockwise rotation. Hence, multiplication by –i also amounts to a rotation by 90 degrees, but clockwise. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)∙iE, we assume we are looking in the negative x-direction. If we are looking in the positive x-direction, we should write: B = -(1/c)∙iE. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.

[21] In fact, when multiplying C2/(N·m2) with N2/C2, we get N/m2, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (joule per unit volume) can also be measured in newton (force per unit area.

[22] The illustration shows a linearly polarized wave, but the obtained result is general.

[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinθ = cos(θ−π /2).

[24] I must thank a physics blogger for re-writing the 1/(ε0·μ0) = c2 equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).

[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90° difference in phase.

[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.

[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.

[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).

The Imaginary Energy Space

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link.

Original post:

Intriguing title, isn’t it? You’ll think this is going to be highly speculative and you’re right. In fact, I could also have written: the imaginary action space, or the imaginary momentum space. Whatever. It all works ! It’s an imaginary space – but a very real one, because it holds energy, or momentum, or a combination of both, i.e. action. 🙂

So the title is either going to deter you or, else, encourage you to read on. I hope it’s the latter. 🙂

In my post on Richard Feynman’s exposé on how Schrödinger got his famous wave equation, I noted an ambiguity in how he deals with the energy concept. I wrote that piece in February, and we are now May. In-between, I looked at Schrödinger’s equation from various perspectives, as evidenced from the many posts that followed that February post, which I summarized on my Deep Blue page, where I note the following:

  1. The argument of the wavefunction (i.e. θ = ωt – kx = [E·t – p·x]/ħ) is just the proper time of the object that’s being represented by the wavefunction (which, in most cases, is an elementary particle—an electron, for example).
  2. The 1/2 factor in Schrödinger’s equation (∂ψ/∂t = i·(ħ/2m)·∇2ψ) doesn’t make all that much sense, so we should just drop it. Writing ∂ψ/∂t = i·(m/ħ)∇2ψ (i.e. Schrödinger’s equation without the 1/2 factor) does away with the mentioned ambiguities and, more importantly, avoids obvious contradictions.

Both remarks are rather unusual—especially the second one. In fact, if you’re not shocked by what I wrote above (Schrödinger got something wrong!), then stop reading—because then you’re likely not to understand a thing of what follows. 🙂 In any case, I thought it would be good to follow up by devoting a separate post to this matter.

The argument of the wavefunction as the proper time

Frankly, it took me quite a while to see that the argument of the wavefunction is nothing but the t’ = (t − v∙x)/√(1−v2)] formula that we know from the Lorentz transformation of spacetime. Let me quickly give you the formulas (just substitute the for v):

relativity

In fact, let me be precise: the argument of the wavefunction also has the particle’s rest mass m0 in it. That mass factor (m0) appears in it as a general scaling factor, so it determines the density of the wavefunction both in time as well as in space. Let me jot it down:

ψ(x, t) = a·ei·(mv·t − p∙x) = a·ei·[(m0/√(1−v2))·t − (m0·v/√(1−v2))∙x] = a·ei·m0·(t − v∙x)/√(1−v2)

Huh? Yes. Let me show you how we get from θ = ωt – kx = [E·t – p·x]/ħ to θ = mv·t − p∙x. It’s really easy. We first need to choose our units such that the speed of light and Planck’s constant are numerically equal to one, so we write: = 1 and ħ = 1. So now the 1/ħ factor no longer appears.

[Let me note something here: using natural units does not do away with the dimensions: the dimensions of whatever is there remain what they are. For example, energy remains what it is, and so that’s force over distance: 1 joule = 1 newton·meter (1 J = 1 N·m. Likewise, momentum remains what it is: force times time (or mass times velocity). Finally, the dimension of the quantum of action doesn’t disappear either: it remains the product of force, distance and time (N·m·s). So you should distinguish between the numerical value of our variables and their dimension. Always! That’s where physics is different from algebra: the equations actually mean something!]

Now, because we’re working in natural units, the numerical value of both and cwill be equal to 1. It’s obvious, then, that Einstein’s mass-energy equivalence relation reduces from E = mvc2 to E = mv. You can work out the rest yourself – noting that p = mv·v and mv = m0/√(1−v2). Done! For a more intuitive explanation, I refer you to the above-mentioned page.

So that’s for the wavefunction. Let’s now look at Schrödinger’s wave equation, i.e. that differential equation of which our wavefunction is a solution. In my introduction, I bluntly said there was something wrong with it: that 1/2 factor shouldn’t be there. Why not?

What’s wrong with Schrödinger’s equation?

When deriving his famous equation, Schrödinger uses the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v2/2, and that’s why – after all the complicated turns – that 1/2 factor is there. There are many reasons why that factor doesn’t make sense. Let me sum up a few.

[I] The most important reason is that de Broglie made it quite clear that the energy concept in his equations for the temporal and spatial frequency for the wavefunction – i.e. the ω = E/ħ and k = p/ħ relations – is the total energy, including rest energy (m0), kinetic energy (m·v2/2) and any potential energy (V). In fact, if we just multiply the two de Broglie (aka as matter-wave equations) and use the old-fashioned v = λ relation (so we write E as E = ω·ħ = (2π·f)·(h/2π) = f·h, and p as p = k·ħ = (2π/λ)·(h/2π) = h/λ and, therefore, we have = E/h and p = h/p), we find that the energy concept that’s implicit in the two matter-wave equations is equal to E = m∙v2, as shown below:

  1. f·λ = (E/h)·(h/p) = E/p
  2. v = λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v2

Huh? E = m∙v2? Yes. Not E = m∙c2 or m·v2/2 or whatever else you might be thinking of. In fact, this E = m∙v2 formula makes a lot of sense in light of the two following points.

Skeptical note: You may – and actually should – wonder whether we can use that v = λ relation for a wave like this, i.e. a wave with both a real (cos(-θ)) as well as an imaginary component (i·sin(-θ). It’s a deep question, and I’ll come back to it later. But… Yes. It’s the right question to ask. 😦

[II] Newton told us that force is mass time acceleration. Newton’s law is still valid in Einstein’s world. The only difference between Newton’s and Einstein’s world is that, since Einstein, we should treat the mass factor as a variable as well. We write: F = mv·a = mv·= [m0/√(1−v2)]·a. This formula gives us the definition of the newton as a force unit: 1 N = 1 kg·(m/s)/s = 1 kg·m/s2. [Note that the 1/√(1−v2) factor – i.e. the Lorentz factor (γ) – has no dimension, because is measured as a relative velocity here, i.e. as a fraction between 0 and 1.]

Now, you’ll agree the definition of energy as a force over some distance is valid in Einstein’s world as well. Hence, if 1 joule is 1 N·m, then 1 J is also equal to 1 (kg·m/s2)·m = 1 kg·(m2/s2), so this also reflects the E = m∙v2 concept. [I can hear you mutter: that kg factor refers to the rest mass, no? No. It doesn’t. The kg is just a measure of inertia: as a unit, it applies to both mas well as mv. Full stop.]

Very skeptical note: You will say this doesn’t prove anything – because this argument just shows the dimensional analysis for both equations (i.e. E = m∙v2 and E = m∙c2) is OK. Hmm… Yes. You’re right. 🙂 But the next point will surely convince you! 🙂

[III] The third argument is the most intricate and the most beautiful at the same time—not because it’s simple (like the arguments above) but because it gives us an interpretation of what’s going on here. It’s fairly easy to verify that Schrödinger’s equation, ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation (including the 1/2 factor to which I object), is equivalent to the following set of two equations:

  1. Re(∂ψ/∂t) = −(ħ/2m)·Im(∇2ψ)
  2. Im(∂ψ/∂t) = (ħ/2m)·Re(∇2ψ)

[In case you don’t see it immediately, note that two complex numbers a + i·b and c + i·d are equal if, and only if, their real and imaginary parts are the same. However, here we have something like this: a + i·b = i·(c + i·d) = i·c + i2·d = − d + i·c (remember i= −1).]

Now, before we proceed (i.e. before I show you what’s wrong here with that 1/2 factor), let us look at the dimensions first. For that, we’d better analyze the complete Schrödinger equation so as to make sure we’re not doing anything stupid here by looking at one aspect of the equation only. The complete equation, in its original form, is:

schrodinger 5

Notice that, to simplify the analysis above, I had moved the and the ħ on the left-hand side to the right-hand side (note that 1/= −i, so −(ħ2/2m)/(i·ħ) = ħ/2m). Now, the ħfactor on the right-hand side is expressed in J2·s2. Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)2 units. So our ħ2/2m coefficient is expressed in (J2·s2)/[J/(m/s)2] = J·m2. Now we multiply that by that Laplacian operating on some scalar, which yields some quantity per square meter. So the whole right-hand side becomes some amount expressed in joule, i.e. the unit of energy! Interesting, isn’t it?

On the left-hand side, we have i and ħ. We shouldn’t worry about the imaginary unit because we can treat that as just another number, albeit a very special number (because its square is minus 1). However, in this equation, it’s like a mathematical constant and you can think of it as something like π or e. [Think of the magical formula: eiπ = i2 = −1.] In contrast, ħ is a physical constant, and so that constant comes with some dimension and, therefore, we cannot just do what we want. [I’ll show, later, that even moving it to the other side of the equation comes with interpretation problems, so be careful with physical constants, as they really mean something!] In this case, its dimension is the action dimension: J·s = N·m·s, so that’s force times distance times time. So we multiply that with a time derivative and we get joule once again (N·m·s/s = N·m = J), so that’s the unit of energy. So it works out: we have joule units both left and right in Schrödinger’s equation. Nice! Yes. But what does it mean? 🙂

Well… You know that we can – and should – think of Schrödinger’s equation as a diffusion equation – just like a heat diffusion equation, for example – but then one describing the diffusion of a probability amplitude. [In case you are not familiar with this interpretation, please do check my post on it, or my Deep Blue page.] But then we didn’t describe the mechanism in very much detail, so let me try to do that now and, in the process, finally explain the problem with the 1/2 factor.

The missing energy

There are various ways to explain the problem. One of them involves calculating group and phase velocities of the elementary wavefunction satisfying Schrödinger’s equation but that’s a more complicated approach and I’ve done that elsewhere, so just click the reference if you prefer the more complicated stuff. I find it easier to just use those two equations above:

  1. Re(∂ψ/∂t) = −(ħ/2m)·Im(∇2ψ)
  2. Im(∂ψ/∂t) = (ħ/2m)·Re(∇2ψ)

The argument is the following: if our elementary wavefunction is equal to ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt), then it’s easy to proof that this pair of conditions is fulfilled if, and only if, ω = k2·(ħ/2m). [Note that I am omitting the normalization coefficient in front of the wavefunction: you can put it back in if you want. The argument here is valid, with or without normalization coefficients.] Easy? Yes. Check it out. The time derivative on the left-hand side is equal to:

∂ψ/∂t = −iω·iei(kx − ωt) = ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

And the second-order derivative on the right-hand side is equal to:

2ψ = ∂2ψ/∂x= i·k2·ei(kx − ωt) = k2·cos(kx − ωt) + i·k2·sin(kx − ωt)

So the two equations above are equivalent to writing:

  1. Re(∂ψB/∂t) =   −(ħ/2m)·Im(∇2ψB) ⇔ ω·cos(kx − ωt) = k2·(ħ/2m)·cos(kx − ωt)
  2. Im(∂ψB/∂t) = (ħ/2m)·Re(∇2ψB) ⇔ ω·sin(kx − ωt) = k2·(ħ/2m)·sin(kx − ωt)

So both conditions are fulfilled if, and only if, ω = k2·(ħ/2m). You’ll say: so what? Well… We have a contradiction here—something that doesn’t make sense. Indeed, the second of the two de Broglie equations (always look at them as a pair) tells us that k = p/ħ, so we can re-write the ω = k2·(ħ/2m) condition as:

ω/k = vp = k2·(ħ/2m)/k = k·ħ/(2m) = (p/ħ)·(ħ/2m) = p/2m ⇔ p = 2m

You’ll say: so what? Well… Stop reading, I’d say. That p = 2m doesn’t make sense—at all! Nope! In fact, if you thought that the E = m·v2  is weird—which, I hope, is no longer the case by now—then… Well… This p = 2m equation is much weirder. In fact, it’s plain nonsense: this condition makes no sense whatsoever. The only way out is to remove the 1/2 factor, and to re-write the Schrödinger equation as I wrote it, i.e. with an ħ/m coefficient only, rather than an (1/2)·(ħ/m) coefficient.

Huh? Yes.

As mentioned above, I could do those group and phase velocity calculations to show you what rubbish that 1/2 factor leads to – and I’ll do that eventually – but let me first find yet another way to present the same paradox. Let’s simplify our life by choosing our units such that = ħ = 1, so we’re using so-called natural units rather than our SI units. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on.] Our mass-energy equivalence then becomes: E = m·c= m·1= m. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on. So we’d still measure energy and mass in different but equivalent units. Hence, the equality sign should not make you think mass and energy are actually the same: energy is energy (i.e. force times distance), while mass is mass (i.e. a measure of inertia). I am saying this because it’s important, and because it took me a while to make these rather subtle distinctions.]

Let’s now go one step further and imagine a hypothetical particle with zero rest mass, so m0 = 0. Hence, all its energy is kinetic and so we write: K.E. = mv·v/2. Now, because this particle has zero rest mass, the slightest acceleration will make it travel at the speed of light. In fact, we would expect it to travel at the speed, so mv = mc and, according to the mass-energy equivalence relation, its total energy is, effectively, E = mv = mc. However, we just said its total energy is kinetic energy only. Hence, its total energy must be equal to E = K.E. = mc·c/2 = mc/2. So we’ve got only half the energy we need. Where’s the other half? Where’s the missing energy? Quid est veritas? Is its energy E = mc or E = mc/2?

It’s just a paradox, of course, but one we have to solve. Of course, we may just say we trust Einstein’s E = m·c2 formula more than the kinetic energy formula, but that answer is not very scientific. 🙂 We’ve got a problem here and, in order to solve it, I’ve come to the following conclusion: just because of its sheer existence, our zero-mass particle must have some hidden energy, and that hidden energy is also equal to E = m·c2/2. Hence, the kinetic and the hidden energy add up to E = m·c2 and all is alright.

Huh? Hidden energy? I must be joking, right?

Well… No. Let me explain. Oh. And just in case you wonder why I bother to try to imagine zero-mass particles. Let me tell you: it’s the first step towards finding a wavefunction for a photon and, secondly, you’ll see it just amounts to modeling the propagation mechanism of energy itself. 🙂

The hidden energy as imaginary energy

I am tempted to refer to the missing energy as imaginary energy, because it’s linked to the imaginary part of the wavefunction. However, it’s anything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it. And read on!]

Back to that factor 1/2. As mentioned above, it also pops up when calculating the group and the phase velocity of the wavefunction. In fact, let me show you that calculation now. [Sorry. Just hang in there.] It goes like this.

The de Broglie relations tell us that the k and the ω in the ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. Let’s now think of that zero-mass particle once more, so we assume all of its energy is kinetic: no rest energy, no potential! So… If we now use the kinetic energy formula E = m·v2/2 – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p2/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction as:

vg = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p2/2m]/∂p = 2p/2m = p/m = v

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.]

Fine. Now the phase velocity. For the phase velocity of our ei(kx − ωt) wavefunction, we find:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = (p2/2m)/p = p/2m = v/2

So that’s only half of v: it’s the 1/2 factor once more! Strange, isn’t it? Why would we get a different value for the phase velocity here? It’s not like we have two different frequencies here, do we? Well… No. You may also note that the phase velocity turns out to be smaller than the group velocity (as mentioned, it’s only half of the group velocity), which is quite exceptional as well! So… Well… What’s the matter here? We’ve got a problem!

What’s going on here? We have only one wave here—one frequency and, hence, only one k and ω. However, on the other hand, it’s also true that the ei(kx − ωt) wavefunction gives us two functions for the price of one—one real and one imaginary: ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt). So the question here is: are we adding waves, or are we not? It’s a deep question. If we’re adding waves, we may get different group and phase velocities, but if we’re not, then… Well… Then the group and phase velocity of our wave should be the same, right? The answer is: we are and we aren’t. It all depends on what you mean by ‘adding’ waves. I know you don’t like that answer, but that’s the way it is, really. 🙂

Let me make a small digression here that will make you feel even more confused. You know – or you should know – that the sine and the cosine function are the same except for a phase difference of 90 degrees: sinθ = cos(θ + π/2). Now, at the same time, multiplying something with amounts to a rotation by 90 degrees, as shown below.

Hence, in order to sort of visualize what our ei(kx − ωt) function really looks like, we may want to super-impose the two graphs and think of something like this:

vision

You’ll have to admit that, when you see this, our formulas for the group or phase velocity, or our v = λ relation, do no longer make much sense, do they? 🙂

Having said that, that 1/2 factor is and remains puzzling, and there must be some logical reason for it. For example, it also pops up in the Uncertainty Relations:

Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2

So we have ħ/2 in both, not ħ. Why do we need to divide the quantum of action here? How do we solve all these paradoxes? It’s easy to see how: the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙v2 formula rather than the kinetic energy E = m∙v2/2. But then… What energy formula is the correct one: E = m∙v2 or m∙c2? Einstein’s formula is always right, isn’t it? It must be, so let me postpone the discussion a bit by looking at a limit situation. If v = c, then we don’t need to make a choice, obviously. 🙂 So let’s look at that limit situation first. So we’re discussing our zero-mass particle once again, assuming it travels at the speed of light. What do we get?

Well… Measuring time and distance in natural units, so c = 1, we have:

E = m∙c2 = m and p = m∙c = m, so we get: E = m = p

Waw ! E = m = p ! What a weird combination, isn’t it? Well… Yes. But it’s fully OK. [You tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units and, hence, the numerical value of c and c2 is 1. Just figure it out for yourself.] The point to note is that the E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our ei(kx − ωt) wavefunction, we get:

vg = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (remember: the 1 stands for c here, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! Isn’t that great? My ‘hidden energy theory’ makes a lot of sense.:-)

However, if there’s hidden energy, we still need to show where it’s hidden. 🙂 Now that question is linked to the propagation mechanism that’s described by those two equations, which now – leaving the 1/2 factor out, simplify to:

  1. Re(∂ψ/∂t) = −(ħ/m)·Im(∇2ψ)
  2. Im(∂ψ/∂t) = (ħ/m)·Re(∇2ψ)

Propagation mechanism? Yes. That’s what we’re talking about here: the propagation mechanism of energy. Huh? Yes. Let me explain in another separate section, so as to improve readability. Before I do, however, let me add another note—for the skeptics among you. 🙂

Indeed, the skeptics among you may wonder whether our zero-mass particle wavefunction makes any sense at all, and they should do so for the following reason: if x = 0 at t = 0, and it’s traveling at the speed of light, then x(t) = t. Always. So if E = m = p, the argument of our wavefunction becomes E·t – p·x = E·t – E·t = 0! So what’s that? The proper time of our zero-mass particle is zero—always and everywhere!?

Well… Yes. That’s why our zero-mass particle – as a point-like object – does not really exist. What we’re talking about is energy itself, and its propagation mechanism. 🙂

While I am sure that, by now, you’re very tired of my rambling, I beg you to read on. Frankly, if you got as far as you have, then you should really be able to work yourself through the rest of this post. 🙂 And I am sure that – if anything – you’ll find it stimulating! 🙂

The imaginary energy space

Look at the propagation mechanism for the electromagnetic wave in free space, which (for = 1) is represented by the following two equations:

  1. B/∂t = –∇×E
  2. E/∂t = ∇×B

[In case you wonder, these are Maxwell’s equations for free space, so we have no stationary nor moving charges around.] See how similar this is to the two equations above? In fact, in my Deep Blue page, I use these two equations to derive the quantum-mechanical wavefunction for the photon (which is not the same as that hypothetical zero-mass particle I introduced above), but I won’t bother you with that here. Just note the so-called curl operator in the two equations above (∇×) can be related to the Laplacian we’ve used so far (∇2). It’s not the same thing, though: for starters, the curl operator operates on a vector quantity, while the Laplacian operates on a scalar (including complex scalars). But don’t get distracted now. Let’s look at the revised Schrödinger’s equation, i.e. the one without the 1/2 factor:

∂ψ/∂t = i·(ħ/m)·∇2ψ

On the left-hand side, we have a time derivative, so that’s a flow per second. On the right-hand side we have the Laplacian and the i·ħ/m factor. Now, written like this, Schrödinger’s equation really looks exactly the same as the general diffusion equation, which is written as: ∂φ/∂t = D·∇2φ, except for the imaginary unit, which makes it clear we’re getting two equations for the price of one here, rather than one only! 🙂 The point is: we may now look at that ħ/m factor as a diffusion constant, because it does exactly the same thing as the diffusion constant D in the diffusion equation ∂φ/∂t = D·∇2φ, i.e:

  1. As a constant of proportionality, it quantifies the relationship between both derivatives.
  2. As a physical constant, it ensures the dimensions on both sides of the equation are compatible.

So the diffusion constant for  Schrödinger’s equation is ħ/m. What is its dimension? That’s easy: (N·m·s)/(N·s2/m) = m2/s. [Remember: 1 N = 1 kg·m/s2.] But then we multiply it with the Laplacian, so that’s something expressed per square meter, so we get something per second on both sides.

Of course, you wonder: what per second? Not sure. That’s hard to say. Let’s continue with our analogy with the heat diffusion equation so as to try to get a better understanding of what’s being written here. Let me give you that heat diffusion equation here. Assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T  – we can write it as:

heat diffusion 2

So that’s structurally similar to Schrödinger’s equation, and to the two equivalent equations we jotted down above. So we’ve got T (temperature) in the role of ψ here—or, to be precise, in the role of ψ ‘s real and imaginary part respectively. So what’s temperature? From the kinetic theory of gases, we know that temperature is not just a scalar: temperature measures the mean (kinetic) energy of the molecules in the gas. That’s why we can confidently state that the heat diffusion equation models an energy flow, both in space as well as in time.

Let me make the point by doing the dimensional analysis for that heat diffusion equation. The time derivative on the left-hand side (∂T/∂t) is expressed in K/s (Kelvin per second). Weird, isn’t it? What’s a Kelvin per second? Well… Think of a Kelvin as some very small amount of energy in some equally small amount of space—think of the space that one molecule needs, and its (mean) energy—and then it all makes sense, doesn’t it?

However, in case you find that a bit difficult, just work out the dimensions of all the other constants and variables. The constant in front (k) makes sense of it. That coefficient (k) is the (volume) heat capacity of the substance, which is expressed in J/(m3·K). So the dimension of the whole thing on the left-hand side (k·∂T/∂t) is J/(m3·s), so that’s energy (J) per cubic meter (m3) and per second (s). Nice, isn’t it? What about the right-hand side? On the right-hand side we have the Laplacian operator  – i.e. ∇= ·, with ∇ = (∂/∂x,  ∂/∂y,  ∂/∂z) – operating on T. The Laplacian operator, when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m2). In this case, it’s operating on T, so the dimension of ∇2T is K/m2. Again, that doesn’t tell us very much (what’s the meaning of a Kelvin per square meter?) but we multiply it by the thermal conductivity (κ), whose dimension is W/(m·K) = J/(m·s·K). Hence, the dimension of the product is  the same as the left-hand side: J/(m3·s). So that’s OK again, as energy (J) per cubic meter (m3) and per second (s) is definitely something we can associate with an energy flow.

In fact, we can play with this. We can bring k from the left- to the right-hand side of the equation, for example. The dimension of κ/k is m2/s (check it!), and multiplying that by K/m(i.e. the dimension of ∇2T) gives us some quantity expressed in Kelvin per second, and so that’s the same dimension as that of ∂T/∂t. Done! 

In fact, we’ve got two different ways of writing Schrödinger’s diffusion equation. We can write it as ∂ψ/∂t = i·(ħ/m)·∇2ψ or, else, we can write it as ħ·∂ψ/∂t = i·(ħ2/m)·∇2ψ. Does it matter? I don’t think it does. The dimensions come out OK in both cases. However, interestingly, if we do a dimensional analysis of the ħ·∂ψ/∂t = i·(ħ2/m)·∇2ψ equation, we get joule on both sides. Interesting, isn’t it? The key question, of course, is: what is it that is flowing here?

I don’t have a very convincing answer to that, but the answer I have is interesting—I think. 🙂 Think of the following: we can multiply Schrödinger’s equation with whatever we want, and then we get all kinds of flows. For example, if we multiply both sides with 1/(m2·s) or 1/(m3·s), we get a equation expressing the energy conservation law, indeed! [And you may want to think about the minus sign of the  right-hand side of Schrödinger’s equation now, because it makes much more sense now!]

We could also multiply both sides with s, so then we get J·s on both sides, i.e. the dimension of physical action (J·s = N·m·s). So then the equation expresses the conservation of actionHuh? Yes. Let me re-phrase that: then it expresses the conservation of angular momentum—as you’ll surely remember that the dimension of action and angular momentum are the same. 🙂

And then we can divide both sides by m, so then we get N·s on both sides, so that’s momentum. So then Schrödinger’s equation embodies the momentum conservation law.

Isn’t it just wonderfulSchrödinger’s equation packs all of the conservation laws!:-) The only catch is that it flows back and forth from the real to the imaginary space, using that propagation mechanism as described in those two equations.

Now that is really interesting, because it does provide an explanation – as fuzzy as it may seem – for all those weird concepts one encounters when studying physics, such as the tunneling effect, which amounts to energy flowing from the imaginary space to the real space and, then, inevitably, flowing back. It also allows for borrowing time from the imaginary space. Hmm… Interesting! [I know I still need to make these points much more formally, but… Well… You kinda get what I mean, don’t you?]

To conclude, let me re-baptize my real and imaginary ‘space’ by referring to them to what they really are: a real and imaginary energy space respectively. Although… Now that I think of it: it could also be real and imaginary momentum space, or a real and imaginary action space. Hmm… The latter term may be the best. 🙂

Isn’t this all great? I mean… I could go on and on—but I’ll stop here, so you can freewheel around yourself. For  example, you may wonder how similar that energy propagation mechanism actually is as compared to the propagation mechanism of the electromagnetic wave? The answer is: very similar. You can check how similar in one of my posts on the photon wavefunction or, if you’d want a more general argument, check my Deep Blue page. Have fun exploring! 🙂

So… Well… That’s it, folks. I hope you enjoyed this post—if only because I really enjoyed writing it. 🙂

[…]

OK. You’re right. I still haven’t answered the fundamental question.

So what about  the 1/2 factor?

What about that 1/2 factor? Did Schrödinger miss it? Well… Think about it for yourself. First, I’d encourage you to further explore that weird graph with the real and imaginary part of the wavefunction. I copied it below, but with an added 45º line—yes, the green diagonal. To make it somewhat more real, imagine you’re the zero-mass point-like particle moving along that line, and we observe you from our inertial frame of reference, using equivalent time and distance units.

spacetime travel

So we’ve got that cosine (cosθ) varying as you travel, and we’ve also got the i·sinθ part of the wavefunction going while you’re zipping through spacetime. Now, THINK of it: the phase velocity of the cosine bit (i.e. the red graph) contributes as much to your lightning speed as the i·sinθ bit, doesn’t it? Should we apply Pythagoras’ basic r2 = x2 + yTheorem here? Yes: the velocity vector along the green diagonal is going to be the sum of the velocity vectors along the horizontal and vertical axes. So… That’s great.

Yes. It is. However, we still have a problem here: it’s the velocity vectors that add up—not their magnitudes. Indeed, if we denote the velocity vector along the green diagonal as u, then we can calculate its magnitude as:

u = √u2 = √[(v/2)2 + (v/2)2] = √[2·(v2/4) = √[v2/2] = v/√2 ≈ 0.7·v

So, as mentioned, we’re adding the vectors, but not their magnitudes. We’re somewhat better off than we were in terms of showing that the phase velocity of those sine and cosine velocities add up—somehow, that is—but… Well… We’re not quite there.

Fortunately, Einstein saves us once again. Remember we’re actually transforming our reference frame when working with the wavefunction? Well… Look at the diagram below (for which I  thank the author)

special relativity

In fact, let me insert an animated illustration, which shows what happens when the velocity goes up and down from (close to) −c to +c and back again.  It’s beautiful, and I must credit the author here too. It sort of speaks for itself, but please do click the link as the accompanying text is quite illuminating. 🙂

Animated_Lorentz_Transformation

The point is: for our zero-mass particle, the x’ and t’ axis will rotate into the diagonal itself which, as I mentioned a couple of times already, represents the speed of light and, therefore, our zero-mass particle traveling at c. It’s obvious that we’re now adding two vectors that point in the same direction and, hence, their magnitudes just add without any square root factor. So, instead of u = √[(v/2)2 + (v/2)2], we just have v/2 + v/2 = v! Done! We solved the phase velocity paradox! 🙂

So… I still haven’t answered that question. Should that 1/2 factor in Schrödinger’s equation be there or not? The answer is, obviously: yes. It should be there. And as for Schrödinger using the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v2/2… Well… What other mass concept would he use? I probably got a bit confused with Feynman’s exposé – especially this notion of ‘choosing the zero point for the energy’ – but then I should probably just re-visit the thing and adjust the language here and there. But the formula is correct.

Thinking it all through, the ħ/2m constant in Schrödinger’s equation should be thought of as the reciprocal of m/(ħ/2). So what we’re doing basically is measuring the mass of our object in units of ħ/2, rather than units of ħ. That makes perfect sense, if only because it’s ħ/2, rather than ħthe factor that appears in the Uncertainty Relations Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2. In fact, in my post on the wavefunction of the zero-mass particle, I noted its elementary wavefunction should use the m = E = p = ħ/2 values, so it becomes ψ(x, t) = a·ei∙[(ħ/2)∙t − (ħ/2)∙x]/ħ = a·ei∙[t − x]/2.

Isn’t that just nice? 🙂 I need to stop here, however, because it looks like this post is becoming a book. Oh—and note that nothing what I wrote above discredits my ‘hidden energy’ theory. On the contrary, it confirms it. In fact, the nice thing about those illustrations above is that it associates the imaginary component of our wavefunction with travel in time, while the real component is associated with travel in space. That makes our theory quite complete: the ‘hidden’ energy is the energy that moves time forward. The only thing I need to do is to connect it to that idea of action expressing itself in time or in space, cf. what I wrote on my Deep Blue page: we can look at the dimension of Planck’s constant, or at the concept of action in general, in two very different ways—from two different perspectives, so to speak:

  1. [Planck’s constant] = [action] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
  2. [Planck’s constant] = [action] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

Hmm… I need to combine that with the idea of the quantum vacuum, i.e. the mathematical space that’s associated with time and distance becoming countable variables…. In any case. Next time. 🙂

Before I sign off, however, let’s quickly check if our a·ei∙[t − x]/2 wavefunction solves the Schrödinger equation:

  • ∂ψ/∂t = −a·ei∙[t − x]/2·(i/2)
  • 2ψ = ∂2[a·ei∙[t − x]/2]/∂x=  ∂[a·ei∙[t − x]/2·(i/2)]/∂x = −a·ei∙[t − x]/2·(1/4)

So the ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation becomes:

a·ei∙[t − x]/2·(i/2) = −i·(ħ/[2·(ħ/2)])·a·ei∙[t − x]/2·(1/4)

⇔ 1/2 = 1/4 !?

The damn 1/2 factor. Schrödinger wants it in his wave equation, but not in the wavefunction—apparently! So what if we take the m = E = p = ħ solution? We get:

  • ∂ψ/∂t = −a·i·ei∙[t − x]
  • 2ψ = ∂2[a·ei∙[t − x]]/∂x=  ∂[a·i·ei∙[t − x]]/∂x = −a·ei∙[t − x]

So the ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation now becomes:

a·i·ei∙[t − x] = −i·(ħ/[2·ħ])·a·ei∙[t − x]

⇔ 1 = 1/2 !?

We’re still in trouble! So… Was Schrödinger wrong after all? There’s no difficulty whatsoever with the ∂ψ/∂t = i·(ħ/m)·∇2ψ equation:

  • a·ei∙[t − x]/2·(i/2) = −i·[ħ/(ħ/2)]·a·ei∙[t − x]/2·(1/4) ⇔ 1 = 1
  • a·i·ei∙[t − x] = −i·(ħ/ħ)·a·ei∙[t − x] ⇔ 1 = 1

What these equations might tell us is that we should measure mass, energy and momentum in terms of ħ (and not in terms of ħ/2) but that the fundamental uncertainty is ± ħ/2. That solves it all. So the magnitude of the uncertainty is ħ but it separates not 0 and ± 1, but −ħ/2 and −ħ/2. Or, more generally, the following series:

…, −7ħ/2, −5ħ/2, −3ħ/2, −ħ/2, +ħ/2, +3ħ/2,+5ħ/2, +7ħ/2,…

Why are we not surprised? The series represent the energy values that a spin one-half particle can possibly have, and ordinary matter – i.e. all fermions – is composed of spin one-half particles.

To  conclude this post, let’s see if we can get any indication on the energy concepts that Schrödinger’s revised wave equation implies. We’ll do so by just calculating the derivatives in the ∂ψ/∂t = i·(ħ/m)·∇2ψ equation (i.e. the equation without the 1/2 factor). Let’s also not assume we’re measuring stuff in natural units, so our wavefunction is just what it is: a·ei·[E·t − p∙x]/ħ. The derivatives now become:

  • ∂ψ/∂t = −a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ
  • 2ψ = ∂2[a·ei∙[E·t − p∙x]/ħ]/∂x=  ∂[a·i·(p/ħ)·ei∙[E·t − p∙x]/ħ]/∂x = −a·(p22ei∙[E·t − p∙x]/ħ

So the ∂ψ/∂t = i·(ħ/m)·∇2ψ = i·(1/m)·∇2ψ equation now becomes:

a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ = −i·(ħ/m)·a·(p22ei∙[E·t − p∙x]/ħ  ⇔ E = p2/m = m·v2

It all works like a charm. Note that we do not assume stuff like E = m = p here. It’s all quite general. Also note that the E = p2/m closely resembles the kinetic energy formula one often sees: K.E. = m·v2/2 = m·m·v2/(2m) = p2/(2m). We just don’t have the 1/2 factor in our E = p2/m formula, which is great—because we don’t want it! :-) Of course, if you’d add the 1/2 factor in Schrödinger’s equation again, you’d get it back in your energy formula, which would just be that old kinetic energy formula which gave us all these contradictions and ambiguities. 😦

Finally, and just to make sure: let me add that, when we wrote that E = m = p – like we did above – we mean their numerical values are the same. Their dimensions remain what they are, of course. Just to make sure you get that subtle point, we’ll do a quick dimensional analysis of that E = p2/m formula:

[E] = [p2/m] ⇔ N·m = N2·s2/kg = N2·s2/[N·m/s2] = N·m = joule (J)

So… Well… It’s all perfect. 🙂

Post scriptum: I revised my Deep Blue page after writing this post, and I think that a number of the ideas that I express above are presented more consistently and coherently there. In any case, the missing energy theory makes sense. Think of it: any oscillator involves both kinetic as well as potential energy, and they both add up to twice the average kinetic (or potential) energy. So why not here? When everything is said and done, our elementary wavefunction does describe an oscillator. 🙂

N-state systems

On the 10th of December, last year, I wrote that my next post would generalize the results we got for two-state systems. That didn’t happen: I didn’t write the ‘next post’—not till now, that is. No. Instead, I started digging—as you can see from all the posts in-between this one and the 10 December piece. And you may also want to take a look at my new Essentials page. 🙂 In any case, it is now time to get back to Feynman’s Lectures on quantum mechanics. Remember where we are: halfway, really. The first half was all about stuff that doesn’t move in space. The second half, i.e. all that we’re going to study now, is about… Well… You guessed it. 🙂 That’s going to be about stuff that does move in space. To see how that works, we first need to generalize the two-state model to an N-state model. Let’s do it.

You’ll remember that, in quantum mechanics, we describe stuff by saying it’s in some state which, as long as we don’t measure in what state exactly, is written as some linear combination of a set of base states. [And please do think about what I highlight here: some state, measureexactly. It all matters. Think about it!] The coefficients in that linear combination are complex-valued functions, which we referred to as wavefunctions, or (probability) amplitudes. To make a long story short, we wrote:

eq1

These Ci coefficients are a shorthand for 〈 i | ψ(t) 〉 amplitudes. As such, they give us the amplitude of the system to be in state i as a function of time. Their dynamics (i.e. the way they evolve in time) are governed by the Hamiltonian equations, i.e.:

Eq2

The Hij coefficients in this set of equations are organized in the Hamiltonian matrix, which Feynman refers to as the energy matrix, because these coefficients do represent energies indeed. So we applied all of this to two-state systems and, hence, things should not be too hard now, because it’s all the same, except that we have N base states now, instead of just two.

So we have a N×N matrix whose diagonal elements Hij are real numbers. The non-diagonal elements may be complex numbers but, if they are, the following rule applies: Hij* = Hji. [In case you wonder: that’s got to do with the fact that we can write any final 〈χ| or 〈φ| state as the conjugate transpose of the initial |χ〉 or |φ〉 state, so we can write: 〈χ| = |χ〉*, or 〈φ| = |φ〉*.]

As usual, the trick is to find those N Ci(t) functions: we do so by solving that set of N equations, assuming we know those Hamiltonian coefficients. [As you may suspect, the real challenge is to determine the Hamiltonian, which we assume to be given here. But… Well… You first need to learn how to model stuff. Once you get your degree, you’ll be paid to actually solve problems using those models. 🙂 ] We know the complex exponential is a functional form that usually does that trick. Hence, generalizing the results from our analysis of two-state systems once more, the following general solution is suggested:

Ci(t) = ai·ei·(E/ħ)·t 

Note that we introduce only one E variable here, but N ai coefficients, which may be real- or complex-valued. Indeed, my examples – see my previous posts – often involved real coefficients, but that’s not necessarily the case. Think of the C2(t) = i·e(i/ħ)·E0·t·sin[(A/ħ)·t] function describing one of the two base state amplitudes for the ammonia molecule—for example. 🙂

Now, that proposed general solution allows us to calculate the derivatives in our Hamiltonian equations (i.e. the d[Ci(t)]/dt functions) as follows:

d[Ci(t)]/dt = −i·(E/ħ)·ai·ei·(E/ħ)·t 

You can now double-check that the set of equations reduces to the following:

Eq4

Please do write it out: because we have one E only, the ei·(E/ħ)·t factor is common to all terms, and so we can cancel it. The other stuff is plain arithmetic: i·i = i2 = 1, and the ħ constants cancel out too. So there we are: we’ve got a very simple set of N equations here, with N unknowns (i.e. these a1, a2,…, aN coefficients, to be specific). We can re-write this system as:

Eq5

The δij here is the Kronecker delta, of course (it’s one for i = j and zero for j), and we are now looking at a homogeneous system of equations here, i.e. a set of linear equations in which all the constant terms are zero. You should remember it from your high school math course. To be specific, you’d write it as Ax = 0, with A the coefficient matrix. The trivial solution is the zero solution, of course: all a1, a2,…, aN coefficients are zero. But we don’t want the trivial solution. Now, as Feynman points out – tongue-in-cheek, really – we actually have to be lucky to have a non-trivial solution. Indeed, you may or may not remember that the zero solution was actually the only solution if the determinant of the coefficient matrix was not equal to zero. So we only had a non-trivial solution if the determinant of A was equal to zero, i.e. if Det[A] = 0. So A has to be some so-called singular matrix. You’ll also remember that, in that case, we got an infinite number of solutions, to which we could apply the so-called superposition principle: if x and y are two solutions to the homogeneous set of equations Ax = 0, then any linear combination of x and y is also a solution. I wrote an addendum to this post (just scroll down and you’ll find it), which explains what systems of linear equations are all about, so I’ll refer you to that in case you’d need more detail here. I need to continue our story here. The bottom line is: the [Hij–δijE] matrix needs to be singular for the system to have meaningful solutions, so we will only have a non-trivial solution for those values of E for which

Det[Hij–δijE] = 0

Let’s spell it out. The condition above is the same as writing:

Eq7

So far, so good. What’s next? Well… The formula for the determinant is the following:

det physicists

That looks like a monster, and it is, but, in essence, what we’ve got here is an expression for the determinant in terms of the permutations of the matrix elements. This is not a math course so I’ll just refer you Wikipedia for a detailed explanation of this formula for the determinant. The bottom line is: if we write it all out, then Det[Hij–δijE] is just an Nth order polynomial in E. In other words: it’s just a sum of products with powers of E up to EN, and so our Det[Hij–δijE] = 0 condition amounts to equating it with zero.

In general, we’ll have N roots, but – sorry you need to remember so much from your high school math classes here – some of them may be multiple roots (i.e. two or more roots may be equal). We’ll call those roots—you guessed it:

EI, EII,…, En,…, EN

Note I am following Feynman’s exposé, and so he uses n, rather than k, to denote the nth Roman numeral (as opposed to Latin numerals). Now, I know your brain is near the melting point… But… Well… We’re not done yet. Just hang on. For each of these values E = EI, EII,…, En,…, EN, we have an associated set of solutions ai. As Feynman puts it: you get a set which belongs to En. In order to not forget that, for each En, we’re talking a set of N coefficients ai (= 1, 2,…, N), we denote that set not by ai(n) but by ai(n). So that’s why we use boldface for our index n: it’s special—and not only because it denotes a Roman numeral! It’s just one of Feynman’s many meaningful conventions.

Now remember that Ci(t) = ai·ei·(E/ħ)·t formula. For each set of ai(n) coefficients, we’ll have a set of Ci(n) functions which, naturally, we can write as:

Ci(n) = ai(nei·(En/ħ)·t

So far, so good. We have N ai(n) coefficients and N Ci(n) functions. That’s easy enough to understand. Now we’ll define also define a set of N new vectors,  which we’ll write as |n〉, and which we’ll refer to as the state vectors that describe the configuration of the definite energy states En (n = I, II,… N). [Just breathe right now: I’ll (try to) explain this in a moment.] Moreover, we’ll write our set of coefficients ai(n) as 〈i|n〉. Again, the boldface n reminds us we’re talking a set of N complex numbers here. So we re-write that set of N Ci(n) functions as follows:

Ci(n) = 〈i|n〉·ei·(En/ħ)·t

We can expand this as follows:

Ci(n) = 〈 i | ψn(t) 〉 = 〈 i | 〉·ei·(En/ħ)·t

which, of course, implies that:

| ψn(t) 〉 = |n〉·ei·(En/ħ)·t

So now you may understand Feynman’s description of those |n〉 vectors somewhat better. As he puts it:

“The |n〉 vectors – of which there are N – are the state vectors that describe the configuration of the definite energy states En (n = I, II,… N), but have the time dependence factored out.”

Hmm… I know. This stuff is hard to swallow, but we’re not done yet: if your brain hasn’t melted yet, it may do so now. You’ll remember we talked about eigenvalues and eigenvectors in our post on the math behind the quantum-mechanical model of our ammonia molecule. Well… We can generalize the results we got there:

  1. The energies EI, EII,…, En,…, EN are the eigenvalues of the Hamiltonian matrix H.
  2. The state vectors |n〉 that are associated with each energy En, i.e. the set of vectors |n〉, are the corresponding eigenstates.

So… Well… That’s it! We’re done! This is all there is to it. I know it’s a lot but… Well… We’ve got a general description of N-state systems here, and so that’s great!

Let me make some concluding remarks though.

First, note the following property: if we let the Hamiltonian matrix act on one of those state vectors |n〉, the result is just En times the same state. We write:

Eq-12

We’re writing nothing new here really: it’s just a consequence of the definition of eigenstates and eigenvalues. The more interesting thing is the following. When describing our two-state systems, we saw we could use the states that we associated with the Eand EII as a new base set. The same is true for N-state systems: the state vectors |n〉 can also be used as a base set. Of course, for that to be the case, all of the states must be orthogonal, meaning that for any two of them, say |n〉 and |m〉, the following equation must hold:

n|m〉 = 0

Feynman shows this will be true automatically if all the energies are different. If they’re not – i.e. if our polynomial in E would accidentally have two (or more) roots with the same energy – then things are more complicated. However, as Feynman points out, this problem can be solved by ‘cooking up’ two new states that do have the same energy but are also orthogonal. I’ll refer you to him for the detail, as well as for the proof of that 〈n|m〉 = 0 equation.

Finally, you should also note that – because of the homogeneity principle – it’s possible to multiply the N ai(n) coefficients by a suitable factor so that all the states are normalized, by which we mean:

n|n〉 = 1

Well… We’re done! For today, at least! 🙂

Addendum on Systems of Linear Equations

It’s probably good to briefly remind you of your high school math class on systems of linear equations. First note the difference between homogeneous and non-homogeneous equations. Non-homogeneous equations have a non-zero constant term. The following three equations are an example of a non-homogeneous set of equations:

  • 3x + 2y − z = 1
  • 2x − 2y + 4z = −2
  • −x + y/2 − z = 0

We have a point solution here: (x, y, z) = (1, −2, −2). The geometry of the situation is something like this:

Secretsharing_3-point

One of the equations may be a linear combination of the two others. In that case, that equation can be removed without affecting the solution set. For the three-dimensional case, we get a line solution, as illustrated below.  Intersecting_Planes_2

Homogeneous and non-homogeneous sets of linear equations are closely related. If we write a homogeneous set as Ax = 0, then a non-homogeneous set of equations can be written as Ax = b. They are related. More in particular, the solution set for Ax = b is going to be a translation of the solution set for Ax = 0. We can write that more formally as follows:

If p is any specific solution to the linear system Ax = b, then the entire solution set can be described as {p + v|v is any solution to Ax = 0}

The solution set for a homogeneous system is a linear subspace. In the example above, which had three variables and, hence, for which the vector space was three-dimensional, there were three possibilities: a point, line or plane solution. All are (linear) subspaces—although you’d want to drop the term ‘linear’ for the point solution, of course. 🙂 Formally, a subspace is defined as follows: if V is a vector space, then W is a subspace if and only if:

  1. The zero vector (i.e. 0) is in W.
  2. If x is an element of W, then any scalar multiple ax will be an element of W too (this is often referred to as the property of homogeneity).
  3. If x and y are elements of W, then the sum of x and y (i.e. x + y) will be an element of W too (this is referred to as the property of additivity).

As you can see, the superposition principle actually combines the properties of homogeneity and additivity: if x and y are solutions, then any linear combination of them will be a solution too.

The solution set for a non-homogeneous system of equations is referred to as a flat. It’s a subset too, so it’s like a subspace, except that it need not pass through the origin. Again, the flats in two-dimensional space are points and lines, while in three-dimensional space we have points, lines and planes. In general, we’ll have flats, and subspaces, of every dimension from 0 to n−1 in n-dimensional space.

OK. That’s clear enough, but what is all that talk about eigenstates and eigenvalues about? Mathematically, we define eigenvectors, aka as characteristic vectors, as follows:

  • The non-zero vector v is an eigenvector of a square matrix A if Av is a scalar multiple of v, i.e. Av = λv.
  • The associated scalar λ is known as the eigenvalue (or characteristic value) associated with the eigenvector v.

Now, in physics, we talk states, rather than vectors—although our states are vectors, of course. So we’ll call them eigenstates, rather than eigenvectors. But the principle is the same, really. Now, I won’t copy what you can find elsewhere—especially not in an addendum to a post, like this one. So let me just refer you elswhere. Paul’s Online Math Notes, for example, are quite good on this—especially in the context of solving a set of differential equations, which is what we are doing here. And you can also find a more general treatment in the Wikipedia article on eigenvalues and eigenstates which, while being general, highlights their particular use in quantum math.

The Hamiltonian revisited

I want to come back to something I mentioned in a previous post: when looking at that formula for those Uij amplitudes—which I’ll jot down once more:

Uij(t + Δt, t) = δij + ΔUij(t + Δt, t) = δij + Kij(t)·Δt ⇔ Uij(t + Δt, t) = δij − (i/ħ)·Hij(t)·Δt

—I noted that it resembles the general y(t + Δt) = y(t) + Δy = y(t) + (dy/dt)·Δt formula. So we can look at our Kij(t) function as being equal to the time derivative of the Uij(t + Δt, t) function. I want to re-visit that here, as it triggers a whole range of questions, which may or may not help to understand quantum math somewhat more intuitively.  Let’s quickly sum up what we’ve learned so far: it’s basically all about quantum-mechanical stuff that does not move in space. Hence, the x in our wavefunction ψ(x, t) is some fixed point in space and, therefore, our elementary wavefunction—which we wrote as:

ψ(x, t) = a·ei·θ a·ei·(ω·t − k∙x) = a·ei·[(E/ħ)·t − (p/ħ)∙x]

—reduces to ψ(t) = a·ei·ω·t = a·ei·[(E/ħ)·t.

Unlike what you might think, we’re not equating x with zero here. No. It’s the p = m·v factor that becomes zero, because our reference frame is that of the system that we’re looking at, so its velocity is zero: it doesn’t move in our reference frame. That immediately answers an obvious question: does our wavefunction look any different when choosing another reference frame? The answer is obviously: yes! It surely matters if the system moves or not, and it also matters how fast it moves, because it changes the energy and momentum values from E and p to some E’ and p’. However, we’ll not consider such complications here: that’s the realm of relativistic quantum mechanics. Let’s start with the simplest of situations.

A simple two-state system

One of the simplest examples of a quantum-mechanical system that does not move in space, is the textbook example of the ammonia molecule. The picture was as simple as the one below: an ammonia molecule consists of one nitrogen atom and three hydrogen atoms, and the nitrogen atom could be ‘up’ or ‘down’ with regard to the motion of the NH3 molecule around its axis of symmetry, as shown below.

Capture

It’s important to note that this ‘up’ or ‘down’ direction is, once again, defined with respect to the reference frame of the system itself. The motion of the molecule around its axis of symmetry is referred to as its spin—a term that’s used in a variety of contexts and, therefore, is annoyingly ambiguous. When we use the term ‘spin’ (up or down) to describe an electron state, for example, we’d associate it with the direction of its magnetic moment. Such magnetic moment arises from the fact that, for all practical purposes, we can think of an electron as a spinning electric charge. Now, while our ammonia molecule is electrically neutral, as a whole, the two states are actually associated with opposite electric dipole moments, as illustrated below. Hence, when we’d apply an electric field (denoted as ε) below, the two states are effectively associated with different energy levels, which we wrote as E0 ± εμ.

ammonia

But we’re getting ahead of ourselves here. Let’s revert to the system in free space, i.e. without an electromagnetic force field—or, what amounts to saying the same, without potential. Now, the ammonia molecule is a quantum-mechanical system, and so there is some amplitude for the nitrogen atom to tunnel through the plane of hydrogens. I told you before that this is the key to understanding quantum mechanics really: there is an energy barrier there and, classically, the nitrogen atom should not sneak across. But it does. It’s like it can borrow some energy – which we denote by A – to penetrate the energy barrier.

In quantum mechanics, the dynamics of this system are modeled using a set of two differential equations. These differential equations are really the equivalent of Newton’s classical Law of Motion (I am referring to the F = m·(dv/dt) = m·a equation here) in quantum mechanics, so I’ll have to explain them—which is not so easy as explaining Newton’s Law, because we’re talking complex-valued functions, but… Well… Let me first insert the solution of that set of differential equations:

graph

This graph shows how the probability of the nitrogen atom (or the ammonia molecule itself) being in state 1 (i.e. ‘up’) or, else, in state 2 (i.e. ‘down’), varies sinusoidally in time. Let me also give you the equations for the amplitudes to be in state 1 or 2 respectively:

  1. C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
  2. C2(t) = 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t = i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

So the P1(t) and P2(t) probabilities above are just the absolute square of these C1(t) and C2(t) functions. So as to help you understand what’s going on here, let me quickly insert the following technical remarks:

  • In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e eiθ eiθ reduces to 2·cosθ, while eiθ − eiθ reduces to 2·i·sinθ.
  • As for how to take the absolute square… Well… I shouldn’t be explaining that here, but you should be able to work that out remembering that (i) |a·b·c|2 = |a|2·|b|2·|c|2; (ii) |eiθ|2 = |e−iθ|= 12 = 1 (for any value of θ); and (iii) |i|2 = 1.
  • As for the periodicity of both probability functions, note that the period of the squared sine and cosine functions is equal to π. Hence, the argument of our sine and cosine function will be equal to 0, π, 2π, 3π etcetera if (A/ħ)·t = 0, π, 2π, 3π etcetera, i.e. if t = 0·ħ/A, π·ħ/A, 2π·ħ/A, 3π·ħ/A etcetera. So that’s why we measure time in units of ħ/A above.

The graph above is actually tricky to interpret, as it assumes that we know in what state the molecule starts out with at t = 0. This assumption is tricky because we usually do not know that: we have to make some observation which, curiously enough, will always yield one of the two states—nothing in-between. Or, else, we can use a state selector—an inhomogeneous electric field which will separate the ammonia molecules according to their state. It’s a weird thing really, and it summarizes all of the ‘craziness’ of quantum-mechanics: as long as we don’t measure anything – by applying that force field – our molecule is in some kind of abstract state, which mixes the two base states. But when we do make the measurement, always along some specific direction (which we usually take to be the z-direction in our reference frame), we’ll always find the molecule is either ‘up’ or, else, ‘down’. We never measure it as something in-between. Personally, I like to think the measurement apparatus – I am talking the electric field here – causes the nitrogen atom to sort of ‘snap into place’. However, physicists use more precise language here: they would say that the electric field does result in the two positions having very different energy levels (E0 + εμ and E0 – εμ, to be precise) and that, as a result, the amplitude for the nitrogen atom to flip back and forth has little effect. Now how do we model that?

The Hamiltonian equations

I shouldn’t be using the term above, as it usually refers to a set of differential equations describing classical systems. However, I’ll also use it for the quantum-mechanical analog, which amounts to the following for our simple two-state example above:

Hamiltonian maser

Don’t panic. We’ll explain. The equations above are all the same but use different formats: the first block writes them as a set of equations, while the second uses the matrix notation, which involves the use of that rather infamous Hamiltonian matrix, which we denote by H = [Hij]. Now, we’ve postponed a lot of technical stuff, so… Well… We can’t avoid it any longer. Let’s look at those Hamiltonian coefficients Hij first. Where do they come from?

You’ll remember we thought of time as some kind of apparatus, with particles entering in some initial state φ and coming out in some final state χ. Both are to be described in terms of our base states. To be precise, we associated the (complex) coefficients C1 and C2 with |φ〉 and D1 and D2 with |χ〉. However, the χ state is a final state, so we have to write it as 〈χ| = |χ〉† (read: chi dagger). The dagger symbol tells us we need to take the conjugate transpose of |χ〉, so the column vector becomes a row vector, and its coefficients are the complex conjugate of D1 and D2, which we denote as D1* and D2*. We combined this with Dirac’s bra-ket notation for the amplitude to go from one base state to another, as a function in time (or a function of time, I should say):

Uij(t + Δt, t) = 〈i|U(t + Δt, t)|j〉

This allowed us to write the following matrix equation:

U coefficients

To see what it means, you should write it all out:

〈χ|U(t + Δt, t)|φ〉 = D1*·(U11(t + Δt, t)·C1 + U12(t + Δt, t)·C2) + D2*·(U21(t + Δt, t)·C1 + U22(t + Δt, t)·C2)

= D1*·U11(t + Δt, t)·C+ D1*·U12(t + Δt, t)·C+ D2*·U21(t + Δt, t)·C+ D2*·U22(t + Δt, t)·C2

It’s a horrendous expression, but it’s a complex-valued amplitude or, quite simply, a complex number. So this is not nonsensical. We can now take the next step, and that’s to go from those Uij amplitudes to the Hij amplitudes of the Hamiltonian matrix. The key is to consider the following: if Δt goes to zero, nothing happens, so we write: Uij = 〈i|U|j〉 → 〈i|j〉 = δij for Δt → 0, with δij = 1 if i = j, and δij = 0 if i ≠ j. We then assume that, for small t, those Uij amplitudes should differ from δij (i.e. from 1 or 0) by amounts that are proportional to Δt. So we write:

Uij(t + Δt, t) = δij + ΔUij(t + Δt, t) = δij + Kij(t)·Δt

We then equated those Kij(t) factors with − (i/ħ)·Hij(t), and we were done: Uij(t + Δt, t) = δij − (i/ħ)·Hij(t)·Δt. […] Well… I show you how we get those differential equations in a moment. Let’s pause here for a while to see what’s going on really. You’ll probably remember how one can mathematically ‘construct’ the complex exponential eiθ by using the linear approximation eiε = 1 + iε near θ = 0 and for infinitesimally small values of ε. In case you forgot, we basically used the definition of the derivative of the real exponential eε for ε going to zero:

FormulaSo we’ve got something similar here for U11(t + Δt, t) = 1 − i·[H11(t)/ħ]·Δt and U22(t + Δt, t) = 1 − i·[H22(t)/ħ]·Δt. Just replace the ε in eiε = 1 + iε by ε = − (E0/ħ)·Δt. Indeed, we know that H11 = H22 = E0, and E0/ħ is, of course, just the energy measured in (reduced) Planck units, i.e. in its natural unit. Hence, if our ammonia molecule is in one of the two base states, we start at θ = 0 and then we just start moving on the unit circle, clockwise, because of the minus sign in eiθ. Let’s write it out:

U11(t + Δt, t) = 1 − i·[H11(t)/ħ]·Δt = 1 − i·[E0/ħ]·Δt and

U22(t + Δt, t) = 1 − i·[H22(t)/ħ]·Δt = 1 − i·[E0/ħ]·Δt

But what about U12 and U21? Is there a similar interpretation? Let’s write those equations down and think about them:

U12(t + Δt, t) = 0 − i·[H12(t)/ħ]·Δt = 0 + i·[A/ħ]·Δt and

U21(t + Δt, t) = 0 − i·[H21(t)/ħ]·Δt = 0 + i·[A/ħ]·Δt

We can visualize this as follows:

circle

Let’s remind ourselves of the definition of the derivative of a function by looking at the illustration below:izvodThe f(x0) value in this illustration corresponds to the Uij(t, t), obviously. So now things make somewhat more sense: U11(t, t) = U11(t, t) = 1, obviously, and U12(t, t) = U21(t, t) = 0. We then add the ΔUij(t + Δt, t) to Uij(t, t). Hence, we can, and probably should, think of those Kij(t) coefficients as the derivative of the Uij(t, t) functions with respect to time. So we can write something like this:

H and U

These derivatives are pure imaginary numbers. That does not mean that the Uij(t + Δt, t) functions are purely imaginary: U11(t + Δt, t) and U22(t + Δt, t) can be approximated by 1 − i·[E0/ħ]·Δt for small Δt, so they do have a real part. In contrast, U12(t + Δt, t) and U21(t + Δt, t) are, effectively, purely imaginary (for small Δt, that is).

I can’t help thinking these formulas reflect a deep and beautiful geometry, but its meaning escapes me so far. 😦 When everything is said and done, none of the reflections above makes things somewhat more intuitive: these wavefunctions remain as mysterious as ever.

I keep staring at those P1(t) and P2(t) functions, and the C1(t) and C2(t) functions that ‘generate’ them, so to speak. They’re not independent, obviously. In fact, they’re exactly the same, except for a phase difference, which corresponds to the phase difference between the sine and cosine. So it’s all one reality, really: all can be described in one single functional form, so to speak. I hope things become more obvious as I move forward. :-/

Post scriptum: I promised I’d show you how to get those differential equations but… Well… I’ve done that in other posts, so I’ll refer you to one of those. Sorry for not repeating myself. 🙂

The math behind the maser

As I skipped the mathematical arguments in my previous post so as to focus on the essential results only, I thought it would be good to complement that post by looking at the math once again, so as to ensure we understand what it is that we’re doing. So let’s do that now. We start with the easy situation: free space.

The two-state system in free space

We started with an ammonia molecule in free space, i.e. we assumed there were no external force fields, like a gravitational or an electromagnetic force field. Hence, the picture was as simple as the one below: the nitrogen atom could be ‘up’ or ‘down’ with regard to its spin around its axis of symmetry.

Capture

It’s important to note that this ‘up’ or ‘down’ direction is defined in regard to the molecule itself, i.e. not in regard to some external reference frame. In other words, the reference frame is that of the molecule itself. For example, if I flip the illustration above – like below – then we’re still talking the same states, i.e. the molecule is still in state 1 in the image on the left-hand side and it’s still in state 2 in the image on the right-hand side. 

Capture

We then modeled the uncertainty about its state by associating two different energy levels with the molecule: E0 + A and E− A. The idea is that the nitrogen atom needs to tunnel through a potential barrier to get to the other side of the plane of the hydrogens, and that requires energy. At the same time, we’ll show the two energy levels are effectively associated with an ‘up’ or ‘down’ direction of the electric dipole moment of the molecule. So that resembles the two spin states of an electron, which we associated with the +ħ/2 and −ħ/2 energies respectively. So if E0 would be zero (we can always take another reference point, remember?), then we’ve got the same thing: two energy levels that are separated by some definite amount: that amount is 2A for the ammonia molecule, and ħ when we’re talking quantum-mechanical spin. I should make a last note here, before I move on: note that these energies only make sense in the presence of some external field, because the + and − signs in the E0 + A and E− A and +ħ/2 and −ħ/2 expressions make sense only with regard to some external direction defining what’s ‘up’ and what’s ‘down’ really. But I am getting ahead of myself here. Let’s go back to free space: no external fields, so what’s ‘up’ or ‘down’ is completely random here. 🙂

Now, we also know an energy level can be associated with a complex-valued wavefunction, or an amplitude as we call it. To be precise, we can associate it with the generic a·e−(i/ħ)·(E·t − px) expression which you know so well by now. Of course,  as the reference frame is that of the molecule itself, its momentum is zero, so the px term in the a·e−(i/ħ)·(E·t − px) expression vanishes and the wavefunction reduces to a·ei·ω·t a·e−(i/ħ)·E·t, with ω = E/ħ. In other words, the energy level determines the temporal frequency, or the temporal variation (as opposed to the spatial frequency or variation), of the amplitude.

We then had to find the amplitudes C1(t) = 〈 1 | ψ 〉 and C2(t) =〈 2 | ψ 〉, so that’s the amplitude to be in state 1 or state 2 respectively. In my post on the Hamiltonian, I explained why the dynamics of a situation like this can be represented by the following set of differential equations:

Hamiltonian

As mentioned, the Cand C2 functions evolve in time, and so we should write them as C= C1(t) and C= C2(t) respectively. In fact, our Hamiltonian coefficients may also evolve in time, which is why it may be very difficult to solve those differential equations! However, as I’ll show below, one usually assumes they are constant, and then one makes informed guesses about them so as to find a solution that makes sense.

Now, I should remind you here of something you surely know: if Cand Care solutions to this set of differential equations, then the superposition principle tells us that any linear combination a·C1 + b·Cwill also be a solution. So we need one or more extra conditions, usually some starting condition, which we can combine with a normalization condition, so we can get some unique solution that makes sense.

The Hij coefficients are referred to as Hamiltonian coefficients and, as shown in the mentioned post, the H11 and H22 coefficients are related to the amplitude of the molecule staying in state 1 and state 2 respectively, while the H12 and H21 coefficients are related to the amplitude of the molecule going from state 1 to state 2 and vice versa. Because of the perfect symmetry of the situation here, it’s easy to see that H11 should equal H22 , and that H12 and H21 should also be equal to each other. Indeed, Nature doesn’t care what we call state 1 or 2 here: as mentioned above, we did not define the ‘up’ and ‘down’ direction with respect to some external direction in space, so the molecule can have any orientation and, hence, switching the i an j indices should not make any difference. So that’s one clue, at least, that we can use to solve those equations: the perfect symmetry of the situation and, hence, the perfect symmetry of the Hamiltonian coefficients—in this case, at least!

The other clue is to think about the solution if we’d not have two states but one state only. In that case, we’d need to solve iħ·[dC1(t)/dt] = H11·C1(t). That’s simple enough, because you’ll remember that the exponential function is its own derivative. To be precise, we write: d(a·eiωt)/dt = a·d(eiωt)/dt = a·iω·eiωt, and please note that can be any complex number: we’re not necessarily talking a real number here! In fact, we’re likely to talk complex coefficients, and we multiply with some other complex number (iω) anyway here! So if we write iħ·[dC1/dt] = H11·C1 as dC1/dt = −(i/ħ)·H11·C1 (remember: i−1 = 1/i = −i), then it’s easy to see that the Ca·e–(i/ħ)·H11·t function is the general solution for this differential equation. Let me write it out for you, just to make sure:

dC1/dt = d[a·e–(i/ħ)H11t]/dt = a·d[e–(i/ħ)H11t]/dt = –a·(i/ħ)·H11·e–(i/ħ)H11t

= –(i/ħ)·H11·a·e–(i/ħ)H11= −(i/ħ)·H11·C1

Of course, that reminds us of our generic wavefunction a·e−(i/ħ)·E0·t wavefunction: we only need to equate H11 with E0 and we’re done! Hence, in a one-state system, the Hamiltonian coefficient is, quite simply, equal to the energy of the system. In fact, that’s a result can be generalized, as we’ll see below, and so that’s why Feynman says the Hamiltonian ought to be called the energy matrix.

In fact, we actually may have two states that are entirely uncoupled, i.e. a system in which there is no dependence of C1 on Cand vice versa. In that case, the two equations reduce to:

iħ·[dC1/dt] = H11·C1 and iħ·[dC2/dt] = H22·C2

These do not form a coupled system and, hence, their solutions are independent:

C1(t) = a·e–(i/ħ)·H11·t and C2(t) = b·e–(i/ħ)·H22·t 

The symmetry of the situation suggests we should equate a and b, and then the normalization condition says that the probabilities have to add up to one, so |C1(t)|+ |C2(t)|= 1, so we’ll find that = 1/√2.

OK. That’s simple enough, and this story has become quite long, so we should wrap it up. The two ‘clues’ – about symmetry and about the Hamiltonian coefficients being energy levels – lead Feynman to suggest that the Hamiltonian matrix for this particular case should be equal to:

H-matrix

Why? Well… It’s just one of Feynman’s clever guesses, and it yields probability functions that makes sense, i.e. they actually describe something real. That’s all. 🙂 I am only half-joking, because it’s a trial-and-error process indeed and, as I’ll explain in a separate section in this post, one needs to be aware of the various approximations involved when doing this stuff. So let’s be explicit about the reasoning here:

  1. We know that H11 = H22 = Eif the two states would be identical. In other words, if we’d have only one state, rather than two – i.e. if H12 and H21 would be zero – then we’d just plug that in. So that’s what Feynman does. So that’s what we do here too! 🙂
  2. However, H12 and H21 are not zero, of course, and so assume there’s some amplitude to go from one position to the other by tunneling through the energy barrier and flipping to the other side. Now, we need to assign some value to that amplitude and so we’ll just assume that the energy that’s needed for the nitrogen atom to tunnel through the energy barrier and flip to the other side is equal to A. So we equate H12 and H21 with −A.

Of course, you’ll wonder: why minus A? Why wouldn’t we try H12 = H21 = A? Well… I could say that a particle usually loses potential energy as it moves from one place to another, but… Well… Think about it. Once it’s through, it’s through, isn’t it? And so then the energy is just Eagain. Indeed, if there’s no external field, the + or − sign is quite arbitrary. So what do we choose? The answer is: when considering our molecule in free space, it doesn’t matter. Using +A or −A yields the same probabilities. Indeed, let me give you the amplitudes we get for H11 = H22 = Eand H12 and H21 = −A:

  1. C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
  2. C2(t) = 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t = i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

[In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e eiθ eiθ reduces to 2·cosθ, while eiθ − eiθ reduces to 2·i·sinθ.]

Now, it’s easy to see that, if we’d have used +A rather than −A, we would have gotten something very similar:

  • C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E+ A)·t + (1/2)·e(i/ħ)·(E− A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
  • C2(t) = 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E+ A)·t – (1/2)·e(i/ħ)·(E− A)·t = −i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

So we get a minus sign in front of our C2(t) function, because cos(α) = cos(–α) but sin(α) = −sin(α). However, the associated probabilities are exactly the same. For both, we get the same P1(t) and P2(t) functions:

  • P1(t) = |C1(t)|2 = cos2[(A/ħ)·t]
  • P2(t) = |C2(t)|= sin2[(A/ħ)·t]

[Remember: the absolute square of and −is |i|= +√12 = +1 and |i|2 = (−1)2|i|= +1 respectively, so the i and −i in the two C2(t) formulas disappear.]

You’ll remember the graph:

graph

Of course, you’ll say: that plus or minus sign in front of C2(t) should matter somehow, doesn’t it? Well… Think about it. Taking the absolute square of some complex number – or some complex function , in this case! – amounts to multiplying it with its complex conjugate. Because the complex conjugate of a product is the product of the complex conjugates, it’s easy to see what happens: the e(i/ħ)·E0·t factor in C1(t) = e(i/ħ)·E0·t·cos[(A/ħ)·t] and C2(t) = ±i·e(i/ħ)·E0·t·sin[(A/ħ)·t] gets multiplied by e+(i/ħ)·E0·t and, hence, doesn’t matter: e(i/ħ)·E0·t·e+(i/ħ)·E0·t = e0 = 1. The cosine factor in C1(t) = e(i/ħ)·E0·t·cos[(A/ħ)·t] is real, and so its complex conjugate is the same. Now, the ±i·sin[(A/ħ)·t] factor in C2(t) = ±i·e(i/ħ)·E0·t·sin[(A/ħ)·t] is a pure imaginary number, and so its complex conjugate is its opposite. For some reason, we’ll find similar solutions for all of the situations we’ll describe below: the factor determining the probability will either be real or, else, a pure imaginary number. Hence, from a math point of view, it really doesn’t matter if we take +A or −A for  or  real factor for those H12 and H21 coefficients. We just need to be consistent in our choice, and I must assume that, in order to be consistent, Feynman likes to think of our nitrogen atom borrowing some energy from the system and, hence, temporarily reducing its energy by an amount that’s equal to −A. If you have a better interpretation, please do let me know! 🙂

OK. We’re done with this section… Except… Well… I have to show you how we got those C1(t) and C1(t) functions, no? Let me copy Feynman here:

solutionNote that the ‘trick’ involving the addition and subtraction of the differential equations is a trick we’ll use quite often, so please do have a look at it. As for the value of the a and b coefficients – which, as you can see, we’ve equated to 1 in our solutions for C1(t) and C1(t) – we get those because of the following starting condition: we assume that at t = 0, the molecule will be in state 1. Hence, we assume C1(0) = 1 and C2(0) = 0. In other words: we assume that we start out on that P1(t) curve in that graph with the probability functions above, so the C1(0) = 1 and C2(0) = 0 starting condition is equivalent to P1(0) = 1 and P1(0) = 0. Plugging that in gives us a/2 + b/2 = 1 and a/2 − b/2 = 0, which is possible only if a = b = 1.

Of course, you’ll say: what if we’d choose to start out with state 2, so our starting condition is P1(0) = 0 and P1(0) = 1? Then a = 1 and b = −1, and we get the solution we got when equating H12 and H21 with +A, rather than with −A. So you can think about that symmetry once again: when we’re in free space, then it’s quite arbitrary what we call ‘up’ or ‘down’.

So… Well… That’s all great. I should, perhaps, just add one more note, and that’s on that A/ħ value. We calculated it in the previous post, because we wanted to actually calculate the period of those P1(t) and P2(t) functions. Because we’re talking the square of a cosine and a sine respectively, the period is equal to π, rather than 2π, so we wrote: (A/ħ)·T = π ⇔ T = π·ħ/A. Now, the separation between the two energy levels E+ A and E− A, so that’s 2A, has been measured as being equal, more or less, to 2A ≈ 10−4 eV.

How does one measure that? As mentioned above, I’ll show you, in a moment, that, when applying some external field, the plus and minus sign do matter, and the separation between those two energy levels E+ A and E− A will effectively represent something physical. More in particular, we’ll have transitions from one energy level to another and that corresponds to electromagnetic radiation being emitted or absorbed, and so there’s a relation between the energy and the frequency of that radiation. To be precise, we can write 2A = h·f0. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, which corresponds to microwave radiation with a wavelength of λ = c/f0 = 1.26 cm. Hence, 2·A ≈ 25×109 Hz times 4×10−15 eV·s = 10−4 eV, indeed, and, therefore, we can write: T = π·ħ/A ≈ 3.14 × 6.6×10−16 eV·s divided by 0.5×10−4 eV, so that’s 40×10−12 seconds = 40 picoseconds. That’s 40 trillionths of a seconds. So that’s very short, and surely much shorter than the time that’s associated with, say, a freely emitting sodium atom, which is of the order of 3.2×10−8 seconds. You may think that makes sense, because the photon energy is so much lower: a sodium light photon is associated with an energy equal to E = h·f = 500×1012 Hz times 4×10−15  eV·s = 2 eV, so that’s 20,000 times 10−4 eV.

There’s a funny thing, however. An oscillation of a frequency of 500 tera-hertz that lasts 3.2×10−8 seconds is equivalent to 500×1012 Hz times 3.2×10−8 s ≈ 16 million cycles. However, an oscillation of a frequency of 23.97 giga-hertz that only lasts 40×10−12 seconds is equivalent to 23.97×109 Hz times 40×10−12 s ≈ 1000×10−3 = 1 ! One cycle only? We’re surely not talking resonance here!

So… Well… I am just flagging it here. We’ll have to do some more thinking about that later. [I’ve added an addendum that may or may not help us in this regard. :-)]

The two-state system in a field

As mentioned above, when there is no external force field, we define the ‘up’ or ‘down’ direction of the nitrogen atom was defined with regard to its its spin around its axis of symmetry, so with regard to the molecule itself. However, when we apply an external electromagnetic field, as shown below, we do have some external reference frame.

Now, the external reference frame – i.e. the physics of the situation, really – may make it more convenient to define the whole system using another set of base states, which we’ll refer to as I and II, rather than 1 and 2. Indeed, you’ve seen the picture below: it shows a state selector, or a filter as we called it. In this case, there’s a filtering according to whether our ammonia molecule is in state I or, alternatively, state II. It’s like a Stern-Gerlach apparatus splitting an electron beam according to the spin state of the electrons, which is ‘up’ or ‘down’ too, but in a totally different way than our ammonia molecule. Indeed, the ‘up’ and ‘down’ spin of an electron has to do with its magnetic moment and its angular momentum. However, there are a lot of similarities here, and so you may want to compare the two situations indeed, i.e. the electron beam in an inhomogeneous magnetic field versus the ammonia beam in an inhomogeneous electric field.

electric field

Now, when reading Feynman, as he walks us through the relevant Lecture on all of this, you get the impression that it’s the I and II states only that have some kind of physical or geometric interpretation. That’s not the case. Of course, the diagram of the state selector above makes it very obvious that these new I and II base states make very much sense in regard to the orientation of the field, i.e. with regard to external space, rather than with respect to the position of our nitrogen atom vis-á-vis the hydrogens. But… Well… Look at the image below: the direction of the field (which we denote by ε because we’ve been using the E for energy) obviously matters when defining the old ‘up’ and ‘down’ states of our nitrogen atom too!

In other words, our previous | 1 〉 and | 2 〉 base states acquire a new meaning too: it obviously matters whether or not the electric dipole moment of the molecule is in the same or, conversely, in the opposite direction of the field. To be precise, the presence of the electromagnetic field suddenly gives the energy levels that we’d associate with these two states a very different physical interpretation.

ammonia

Indeed, from the illustration above, it’s easy to see that the electric dipole moment of this particular molecule in state 1 is in the opposite direction and, therefore, temporarily ignoring the amplitude to flip over (so we do not think of A for just a brief little moment), the energy that we’d associate with state 1 would be equal to E+ με. Likewise, the energy we’d associate with state 2 is equal to E− με.  Indeed, you’ll remember that the (potential) energy of an electric dipole is equal to the vector dot product of the electric dipole moment μ and the field vector ε, but with a minus sign in front so as to get the sign for the energy righ. So the energy is equal to −μ·ε = −|μ|·|ε|·cosθ, with θ the angle between both vectors. Now, the illustration above makes it clear that state 1 and 2 are defined for θ = π and θ = 0 respectively. [And, yes! Please do note that state 1 is the highest energy level, because it’s associated with the highest potential energy: the electric dipole moment μ of our ammonia molecule will – obviously! – want to align itself with the electric field ε ! Just think of what it would imply to turn the molecule in the field!]

Therefore, using the same hunches as the ones we used in the free space example, Feynman suggests that, when some external electric field is involved, we should use the following Hamiltonian matrix:

H-matrix 2

So we’ll need to solve a similar set of differential equations with this Hamiltonian now. We’ll do that later and, as mentioned above, it will be more convenient to switch to another set of base states, or another ‘representation’ as it’s referred to. But… Well… Let’s not get too much ahead of ourselves: I’ll say something about that before we’ll start solving the thing, but let’s first look at that Hamiltonian once more.

When I say that Feynman uses the same clues here, then… Well.. That’s true and not true. You should note that the diagonal elements in the Hamiltonian above are not the same: E+ με ≠ E+ με. So we’ve lost that symmetry of free space which, from a math point of view, was reflected in those identical H11 = H22 = Ecoefficients.

That should be obvious from what I write above: state 1 and state 2 are no longer those 1 and 2 states we described when looking at the molecule in free space. Indeed, the | 1 〉 and | 2 〉 states are still ‘up’ or ‘down’, but the illustration above also makes it clear we’re defining state 1 and state 2 not only with respect to the molecule’s spin around its own axis of symmetry but also vis-á-vis some direction in space. To be precise, we’re defining state 1 and state 2 here with respect to the direction of the electric field ε. Now that makes a really big difference in terms of interpreting what’s going on.

In fact, the ‘splitting’ of the energy levels because of that amplitude A is now something physical too, i.e. something that goes beyond just modeling the uncertainty involved. In fact, we’ll find it convenient to distinguish two new energy levels, which we’ll write as E= E+ A and EII = E− A respectively. They are, of course, related to those new base states | I 〉 and | II 〉 that we’ll want to use. So the E+ A and E− A energy levels themselves will acquire some physical meaning, and especially the separation between them, i.e. the value of 2A. Indeed, E= E+ A and EII = E− A will effectively represent an ‘upper’ and a ‘lower’ energy level respectively.

But, again, I am getting ahead of myself. Let’s first, as part of working towards a solution for our equations, look at what happens if and when we’d switch to another representation indeed.

Switching to another representation

Let me remind you of what I wrote in my post on quantum math in this regard. The actual state of our ammonia molecule – or any quantum-mechanical system really – is always to be described in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C1 + | 2 〉 C2

You’ll say: why? Our molecule is obviously always in either state 1 or state 2, isn’t it? Well… Yes and no. That’s the mystery of quantum mechanics: it is and it isn’t. As long as we don’t measure it, there is an amplitude for it to be in state 1 and an amplitude for it to be in state 2. So we can only make sense of its state by actually calculating 〈 1 | φ 〉 and 〈 2 | φ 〉 which, unsurprisingly are equal to 〈 1 | φ 〉 = 〈 1 | 1 〉 C1 + 〈 1 | 2 〉 C2  = C1(t) and 〈 2 | φ 〉 = 〈 2 | 1 〉 C1 + 〈 2 | 2 〉 C2  = C2(t) respectively, and so these two functions give us the probabilities P1(t) and  P2(t) respectively. So that’s Schrödinger’s cat really: the cat is dead or alive, but we don’t know until we open the box, and we only have a probability function – so we can say that it’s probably dead or probably alive, depending on the odds – as long as we do not open the box. It’s as simple as that.

Now, the ‘dead’ and ‘alive’ condition are, obviously, the ‘base states’ in Schrödinger’s rather famous example, and we can write them as | DEAD 〉 and | ALIVE 〉 you’d agree it would be difficult to find another representation. For example, it doesn’t make much sense to say that we’ve rotated the two base states over 90 degrees and we now have two new states equal to (1/√2)·| DEAD 〉 – (1/√2)·| ALIVE 〉 and (1/√2)·| DEAD 〉 + (1/√2)·| ALIVE 〉 respectively. There’s no direction in space in regard to which we’re defining those two base states: dead is dead, and alive is alive.

The situation really resembles our ammonia molecule in free space: there’s no external reference against which to define the base states. However, as soon as some external field is involved, we do have a direction in space and, as mentioned above, our base states are now defined with respect to a particular orientation in space. That implies two things. The first is that we should no longer say that our molecule will always be in either state 1 or state 2. There’s no reason for it to be perfectly aligned with or against the field. Its orientation can be anything really, and so its state is likely to be some combination of those two pure base states | 1 〉 and | 2 〉.

The second thing is that we may choose another set of base states, and specify the very same state in terms of the new base states. So, assuming we choose some other set of base states | I 〉 and | II 〉, we can write the very same state | φ 〉 = | 1 〉 C1 + | 2 〉 Cas:

| φ 〉 = | I 〉 CI + | II 〉 CII

It’s really like what you learned about vectors in high school: one can go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. It’s just that, just like in high school, we need some direction in regard to which we define our rotation or our translation.

For state vectors, I showed how a rotation of base states worked in one of my posts on two-state systems. To be specific, we had the following relation between the two representations:

matrix

The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as:

transformation

The y-axis? What y-axis? What state filtering apparatus? Just relax. Think about what you’ve learned already. The orientations are shown below: the S apparatus separates ‘up’ and ‘down’ states along the z-axis, while the T-apparatus does so along an axis that is tilted, about the y-axis, over an angle equal to α, or φ, as it’s written in the table above.

tilted

Of course, we don’t really introduce an apparatus at this or that angle. We just introduced an electromagnetic field, which re-defined our | 1 〉 and | 2 〉 base states and, therefore, through the rotational transformation matrix, also defines our | I 〉 and | II 〉 base states.

[…] You may have lost me by now, and so then you’ll want to skip to the next section. That’s fine. Just remember that the representations in terms of | I 〉 and | II 〉 base states or in terms of | 1 〉 and | 2 〉 base states are mathematically equivalent. Having said that, if you’re reading this post, and you want to understand it, truly (because you want to truly understand quantum mechanics), then you should try to stick with me here. 🙂 Indeed, there’s a zillion things you could think about right now, but you should stick to the math now. Using that transformation matrix, we can relate the Cand CII coefficients in the | φ 〉 = | I 〉 CI + | II 〉 CII expression to the Cand CII coefficients in the | φ 〉 = | 1 〉 C1 + | 2 〉 C2 expression. Indeed, we wrote:

  • C= 〈 I | ψ 〉 = (1/√2)·(C1 − C2)
  • CII = 〈 II | ψ 〉 = (1/√2)·(C1 + C2)

That’s exactly the same as writing:

transformation

OK. […] Waw! You just took a huge leap, because we can now compare the two sets of differential equations:

set of equations

They’re mathematically equivalent, but the mathematical behavior of the functions involved is very different. Indeed, unlike the C1(t) and C2(t) amplitudes, we find that the CI(t) and CII(t) amplitudes are stationary, i.e. the associated probabilities – which we find by taking the absolute square of the amplitudes, as usual – do not vary in time. To be precise, if you write it all out and simplify, you’ll find that the CI(t) and CII(t) amplitudes are equal to:

  • CI(t) = 〈 I | ψ 〉 = (1/√2)·(C1 − C2) = (1/√2)·e(i/ħ)·(E0+ A)·t = (1/√2)·e(i/ħ)·EI·t
  • CII(t) = 〈 II | ψ 〉 = (1/√2)·(C1 + C2) = (1/√2)·e(i/ħ)·(E0− A)·t = (1/√2)·e(i/ħ)·EII·t

As the absolute square of the exponential is equal to one, the associated probabilities, i.e. |CI(t)|2 and |CII(t)|2, are, quite simply, equal to |1/√2|2 = 1/2. Now, it is very tempting to say that this means that our ammonia molecule has an equal chance to be in state I or state II. In fact, while I may have said something like that in my previous posts, that’s not how one should interpret this. The chance of our molecule being exactly in state I or state II, or in state 1 or state 2 is varying with time, with the probability being ‘dumped’ from one state to the other all of the time.

I mean… The electric dipole moment can point in any direction, really. So saying that our molecule has a 50/50 chance of being in state 1 or state 2 makes no sense. Likewise, saying that our molecule has a 50/50 chance of being in state I or state II makes no sense either. Indeed, the state of our molecule is specified by the | φ 〉 = | I 〉 CI + | II 〉 CII = | 1 〉 C1 + | 2 〉 Cequations, and neither of these two expressions is a stationary state. They mix two frequencies, because they mix two energy levels.

Having said that, we’re talking quantum mechanics here and, therefore, an external inhomogeneous electric field will effectively split the ammonia molecules according to their state. The situation is really like what a Stern-Gerlach apparatus does to a beam of electrons: it will split the beam according to the electron’s spin, which is either ‘up’ or, else, ‘down’, as shown in the graph below:

diagram 2

The graph for our ammonia molecule, shown below, is very similar. The vertical axis measures the same: energy. And the horizontal axis measures με, which increases with the strength of the electric field ε. So we see a similar ‘splitting’ of the energy of the molecule in an external electric field.

graph new

How should we explain this? It is very tempting to think that the presence of an external force field causes the electrons, or the ammonia molecule, to ‘snap into’ one of the two possible states, which are referred to as state I and state II respectively in the illustration of the ammonia state selector below. But… Well… Here we’re entering the murky waters of actually interpreting quantum mechanics, for which (a) we have no time, and (b) we are not qualified. So you should just believe, or take for granted, what’s being shown here: an inhomogeneous electric field will split our ammonia beam according to their state, which we define as I and II respectively, and which are associated with the energy E0+ A and E0− A  respectively.

electric field

As mentioned above, you should note that these two states are stationary. The Hamiltonian equations which, as they always do, describe the dynamics of this system, imply that the amplitude to go from state I to state II, or vice versa, is zero. To make sure you ‘get’ that, I reproduce the associated Hamiltonian matrix once again:

H-matrix I and II

Of course, that will change when we start our analysis of what’s happening in the maser. Indeed, we will have some non-zero HI,II and HII,I amplitudes in the resonant cavity of our ammonia maser, in which we’ll have an oscillating electric field and, as a result, induced transitions from state I to II and vice versa. However, that’s for later. While I’ll quickly insert the full picture diagram below, you should, for the moment, just think about those two stationary states and those two zeroes. 🙂

maser diagram

Capito? If not… Well… Start reading this post again, I’d say. 🙂

Intermezzo: on approximations

At this point, I need to say a few things about all of the approximations involved, because it can be quite confusing indeed. So let’s take a closer look at those energy levels and the related Hamiltonian coefficients. In fact, in his LecturesFeynman shows us that we can always have a general solution for the Hamiltonian equations describing a two-state system whenever we have constant Hamiltonian coefficients. That general solution – which, mind you, is derived assuming Hamiltonian coefficients that do not depend on time – can always be written in terms of two stationary base states, i.e. states with a definite energy and, hence, a constant probability. The equations, and the two definite energy levels are:

Hamiltonian

solution3

That yields the following values for the energy levels for the stationary states:

solution x

Now, that’s very different from the E= E0+ A and EII = E0− A energy levels for those stationary states we had defined in the previous section: those stationary states had no square root, and no μ2ε2, in their energy. In fact, that sort of answers the question: if there’s no external field, then that μ2ε2 factor is zero, and the square root in the expression becomes ±√A= ±A. So then we’re back to our E= E0+ A and EII = E0− A formulas. The whole point, however, is that we will actually have an electric field in that cavity. Moreover, it’s going to be a field that varies in time, which we’ll write:

field

Now, part of the confusion in Feynman’s approach is that he constantly switches between representing the system in terms of the I and II base states and the 1 and 2 base states respectively. For a good understanding, we should compare with our original representation of the dynamics in free space, for which the Hamiltonian was the following one:

H-matrix

That matrix can easily be related to the new one we’re going to have to solve, which is equal to:

H-matrix 2

The interpretation is easy if we look at that illustration again:

ammonia

If the direction of the electric dipole moment is opposite to the direction ε, then the associated energy is equal to −μ·ε = −μ·ε = −|μ|·|ε|·cosθ = −μ·ε·cos(π) = +με. Conversely, for state 2, we find −μ·ε·cos(0) = −με for the energy that’s associated with the dipole moment. You can and should think about the physics involved here, because they make sense! Thinking of amplitudes, you should note that the +με and −με terms effectively change the H11 and H22 coefficients, so they change the amplitude to stay in state 1 or state 2 respectively. That, of course, will have an impact on the associated probabilities, and so that’s why we’re talking of induced transitions now.

Having said that, the Hamiltonian matrix above keeps the −A for H12 and H21, so the matrix captures spontaneous transitions too!

Still… You may wonder why Feynman doesn’t use those Eand EII formulas with the square root because that would give us some exact solution, wouldn’t it? The answer to that question is: maybe it would, but would you know how to solve those equations? We’ll have a varying field, remember? So our Hamiltonian H11 and H22 coefficients will no longer be constant, but time-dependent. As you’re going to see, it takes Feynman three pages to solve the whole thing using the +με and −με approximation. So just imagine how complicated it would be using that square root expression! [By the way, do have a look at those asymptotic curves in that illustration showing the splitting of energy levels above, so you see how that approximation looks like.]

So that’s the real answer: we need to simplify somehow, so as to get any solutions at all!

Of course, it’s all quite confusing because, after Feynman first notes that, for strong fields, the A2 in that square root is small as compared to μ2ε2, thereby justifying the use of the simplified E= E0+ με = H11 and EII = E0− με = H22 coefficients, he continues and bluntly uses the very same square root expression to explain how that state selector works, saying that the electric field in the state selector will be rather weak and, hence, that με will be much smaller than A, so one can use the following approximation for the square root in the expressions above:

square root sum of squaresThe energy expressions then reduce to:energy 2

And then we can calculate the force on the molecules as:

force

So the electric field in the state selector is weak, but the electric field in the cavity is supposed to be strong, and so… Well… That’s it, really. The bottom line is that we’ve a beam of ammonia molecules that are all in state I, and it’s what happens with that beam then, that is being described by our new set of differential equations:

new

Solving the equations

As all molecules in our ammonia beam are described in terms of the | I 〉 and | II 〉 base states – as evidenced by the fact that we say all molecules that enter the cavity are state I – we need to switch to that representation. We do that by using that transformation above, so we write:

  • C= 〈 I | ψ 〉 = (1/√2)·(C1 − C2)
  • CII = 〈 II | ψ 〉 = (1/√2)·(C1 + C2)

Keeping these ‘definitions’ of Cand CII in mind, you should then add the two differential equations, divide the result by the square root of 2, and you should get the following new equation:

Eq1

Please! Do it and verify the result! You want to learn something here, no? 🙂

Likewise, subtracting the two differential equations, we get:

Eq2

We can re-write this as:set new

Now, the problem is that the Hamiltonian constants here are not constant. To be precise, the electric field ε varies in time. We wrote:

field

So HI,II  and HII,I, which are equal to με, are not constant: we’ve got Hamiltonian coefficients that are a function of time themselves. […] So… Well… We just need to get on with it and try to finally solve this thing. Let me just copy Feynman as he grinds through this:

F1

This is only the first step in the process. Feynman just takes two trial functions, which are really similar to the very general Ca·e–(i/ħ)·H11·t function we presented when only one equation was involved, or – if you prefer a set of two equations – those CI(t) = a·e(i/ħ)·EI·t and CI(t) = b·e(i/ħ)·EII·equations above. The difference is that the coefficients in front, i.e. γI and γII are not some (complex) constant, but functions of time themselves. The next step in the derivation is as follows:

F2

One needs to do a bit of gymnastics here as well to follow what’s going on, but please do check and you’ll see it works. Feynman derives another set of differential equations here, and they specify these γI = γI(t) and γII = γII(t) functions. These equations are written in terms of the frequency of the field, i.e. ω, and the resonant frequency ω0, which we mentioned above when calculating that 23.79 GHz frequency from the 2A = h·f0 equation. So ω0 is the same molecular resonance frequency but expressed as an angular frequency, so ω0 = f0/2π = ħ/2A. He then proceeds to simplify, using assumptions one should check. He then continues:

F3

That gives us what we presented in the previous post:

F4

So… Well… What to say? I explained those probability functions in my previous post, indeed. We’ve got two probabilities here:

  • P= cos2[(με0/ħ)·t]
  • PII = sin2[(με0/ħ)·t]

So that’s just like the P=  cos2[(A/ħ)·t] and P= sin2[(A/ħ)·t] probabilities we found for spontaneous transitions. But so here we are talking induced transitions.

As you can see, the frequency and, hence, the period, depend on the strength, or magnitude, of the electric field, i.e. the εconstant in the ε = 2ε0cos(ω·t) expression. The natural unit for measuring time would be the period once again, which we can easily calculate as (με0/ħ)·T = π ⇔ T = π·ħ/με0.

Now, we had that T = (π·ħ)/(2A) expression above, which allowed us to calculate the period of the spontaneous transition frequency, which we found was like 40 picoseconds, i.e. 40×10−12 seconds. Now, the T = (π·ħ)/(2με0) is very similar, it allows us to calculate the expected, average, or mean time for an induced transition. In fact, if we write Tinduced = (π·ħ)/(2με0) and Tspontaneous = (π·ħ)/(2A), then we can take ratio to find:

Tinduced/Tspontaneous = [(π·ħ)/(2με0)]/[(π·ħ)/(2A)] = A/με0

This A/με0 ratio is greater than one, so Tinduced/Tspontaneous is greater than one, which, in turn, means that the presence of our electric field – which, let me remind you, dances to the beat of the resonant frequency – causes a slower transition than we would have had if the oscillating electric field were not present.

But – Hey! – that’s the wrong comparison! Remember all molecules enter in a stationary state, as they’ve been selected so as to ensure they’re in state I. So there is no such thing as a spontaneous transition frequency here! They’re all polarized, so to speak, and they would remain that way if there was no field in the cavity. So if there was no oscillating electric field, they would never transition. Nothing would happen! Well… In terms of our particular set of base states, of course! Why? Well… Look at the Hamiltonian coefficients HI,II = HII,I = με: these coefficients are zero if ε is zero. So… Well… That says it all.

So that‘s what it’s all about: induced emission and, as I explained in my previous post, because all molecules enter in state I, i.e. the upper energy state, literally, they all ‘dump’ a net amount of energy equal to 2A into the cavity at the occasion of their first transition. The molecules then keep dancing, of course, and so they absorb and emit the same amount as they go through the cavity, but… Well… We’ve got a net contribution here, which is not only enough to maintain the cavity oscillations, but actually also provides a small excess of power that can be drawn from the cavity as microwave radiation of the same frequency.

As Feynman notes, an exact description of what actually happens requires an understanding of the quantum mechanics of the field in the cavity, i.e. quantum field theory, which I haven’t studied yet. But… Well… That’s for later, I guess. 🙂

Post scriptum: The sheer length of this post shows we’re not doing something that’s easy here. Frankly, I feel the whole analysis is still quite obscure, in the sense that – despite looking at this thing again and again – it’s hard to sort of interpret what’s going on, in a physical sense that is. But perhaps one shouldn’t try that. I’ve quoted Feynman’s view on how easy or how difficult it is to ‘understand’ quantum mechanics a couple of times already, so let me do it once more:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and human intuition applies to large objects.”

So… Well… I’ll grind through the remaining Lectures now – I am halfway through Volume III now – and then re-visit all of this. Despite Feynman’s warning, I want to understand it the way I like to, even if I don’t quite know what way that is right now. 🙂

Addendum: As for those cycles and periods, I noted a couple of times already that the Planck-Einstein equation E = h·f  can usefully be re-written as E/= h, as it gives a physical interpretation to the value of the Planck constant. In fact, I said h is the energy that’s associated with one cycle, regardless of the frequency of the radiation involved. Indeed, the energy of a photon divided by the number of cycles per second, should give us the energy per cycle, no?

Well… Yes and no. Planck’s constant h and the frequency are both expressed referencing the time unit. However, if we say that a sodium atom emits one photon only as its electron transitions from a higher energy level to a lower one, and if we say that involves a decay time of the order of 3.2×10−8 seconds, then what we’re saying really is that a sodium light photon will ‘pack’ like 16 million cycles, which is what we get when we multiply the number of cycles per second (i.e. the mentioned frequency of 500×1012 Hz) by the decay time (i.e. 3.2×10−8 seconds): (500×1012 Hz)·(3.2×10−8 s) = 16 ×10cycles, indeed. So the energy per cycle is 2.068 eV (i.e. the photon energy) divided by 16×106, so that’s 0.129×10−6 eV. Unsurprisingly, that’s what we get when we we divide h by 3.2×10−8 s: (4.13567×10−15)/(3.2×10−8 s) = 1.29×10−7 eV. We’re just putting some values in to the E/(T) = h/T equation here.

The logic for that 2A = h·f0 is the same. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, so the photon energy is (23.97×109 Hz)·(4.13567×10−15 eV·s) ≈ 1×10−4 eV. Now, we calculated the transition period T as T = π·ħ/A ≈ (π·6.626×10−16 eV·s)/(0.5×10−4 eV) ≈ 41.6×10−12 seconds. Now, an oscillation of a frequency of 23.97 giga-hertz that only lasts 41.6×10−12 seconds is an oscillation of one cycle only. The consequence is that, when we continue this style of reasoning, we’d have a photon that packs all of its energy into one cycle!

Let’s think about what this implies in terms of the density in space. The wavelength of our microwave radiation is 1.25×10−2 m, so we’ve got a ‘density’ of 1×10−4 eV/1.25×10−2 m = 0.8×10−2 eV/m = 0.008 eV/m. The wavelength of our sodium light is 0.6×10−6 m, so we get a ‘density’ of 1.29×10−7 eV/0.6×10−6 m = 2.15×10−1 eV/m = 0.215 eV/m. So the energy ‘density’ of our sodium light is 26.875 times that of our microwave radiation. 🙂

Frankly, I am not quite sure if calculations like this make much sense. In fact, when talking about energy densities, I should review my posts on the Poynting vector. However, they may help you think things through. 🙂

Quantum math revisited

It’s probably good to review the concepts we’ve learned so far. Let’s start with the foundation of all of our math, i.e. the concept of the state, or the state vector. [The difference between the two concepts is subtle but real. I’ll come back to it.]

State vectors and base states

We used Dirac’s bra-ket notation to denote a state vector, in general, as | ψ 〉. The obvious question is: what is this thing? We called it a vector because we use it like a vector: we multiply it with some number, and then add it to some other vector. So that’s just what you did in high school, when you learned about real vector spaces. In this regard, it is good to remind you of the definition of a vector space. To put it simply, it is is a collection of objects called vectors, which may be added together, and multiplied by numbers. So we have two things here: the ‘objects’, and the ‘numbers’. That’s why we’d say that we have some vector space over a field of numbers. [The term ‘field’ just refers to an algebraic structure, so we can add and multiply and what have you.] Of course, what it means to ‘add’ two ‘objects’, and what it means to ‘multiply’ an object with a number, depends on the type of objects and, unsurprisingly, the type of numbers.

Huh? The type of number?! A number is a number, no?

No, hombre, no! We’ve got natural numbers, rational numbers, real numbers, complex numbers—and you’ve probably heard of quaternions too – and, hence, ‘multiplying’ a ‘number’ with ‘something else’ can mean very different things. At the same time, the general idea is the general idea, so that’s the same, indeed. 🙂 When using real numbers and the kind of vectors you are used to (i.e. Euclidean vectors), then the multiplication amounts to a re-scaling of the vector, and so that’s why a real number is often referred to as a scalar. At the same time, anything that can be used to multiply a vector is often referred to as a scalar in math so… Well… Terminology is often quite confusing. In fact, I’ll give you some more examples of confusing terminology in a moment. But let’s first look at our ‘objects’ here, i.e. our ‘vectors’.

I did a post on Euclidean and non-Euclidean vector spaces two years ago, when I started this blog, but state vectors are obviously very different ‘objects’. They don’t resemble the vectors we’re used to. We’re used to so-called polar vectors, aka as real vectors, like the position vector (x or r), or the momentum vector (p = m·v), or the electric field vector (E). We are also familiar with the so-called pseudo-vectors, aka as axial vectors, like angular momentum (L = r×p), or the magnetic dipole moment. [Unlike what you might think, not all vector cross products yield a pseudo-vector. For example, the cross-product of a polar and an axial vector yields a polar vector.] But here we are talking some very different ‘object’. In math, we say that state vectors are elements in a Hilbert space. So a Hilbert space is a vector space but… Well… With special vectors. 🙂

The key to understanding why we’d refer to states as state vectors is the fact that, just like Euclidean vectors, we can uniquely specify any element in a Hilbert space with respect to a set of base states. So it’s really like using Cartesian coordinates in a two- or three-dimensional Euclidean space. The analogy is complete because, even in the absence of a geometrical interpretation, we’ll require those base states to be orthonormal. Let me be explicit on that by reminding you of your high-school classes on vector analysis: you’d choose a set of orthonormal base vectors e1e2, and e3, and you’d write any vector A as:

A = (Ax, Ay, Az) = Ax·e1 + Ay·e2 + Az·e3 with ei·ej = 1 if i = j, and ei·ej = 0 if i ≠ j

The ei·ej = 1 if i = j and ei·ej = 0 if i ≠ j condition expresses the orthonormality condition: the base vectors need to be orthogonal unit vectors. We wrote it as ei·ej = δij using the Kronecker delta ij = 1 if i = j and 0 if i ≠ j). Now, base states in quantum mechanics do not necessarily have a geometrical interpretation. Indeed, although one often can actually associate them with some position or direction in space, the condition of orthonormality applies in the mathematical sense of the word only. Denoting the base states by i = 1, 2,… – or by Roman numerals, like I and II – so as to distinguish them from the Greek ψ or φ symbols we use to denote state vectors in general, we write the orthonormality condition as follows:

〈 i | j 〉 = δij, with δij = δji is equal to 1 if i = j, and zero if i ≠ j

Now, you may grumble and say: that 〈 i | j 〉 bra-ket does not resemble the ei·ej product. Well… It does and it doesn’t. I’ll show why in a moment. First note how we uniquely specify state vectors in general in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C1 + | 2 〉 C2

Or, if we chose some other set of base states | I 〉 and | II 〉, we’ll write:

| φ 〉 = | I 〉 CI + | II 〉 CII

You should note that the | 1 〉 C1 term in the | φ 〉 = | 1 〉 C1 + | 2 〉 C2 sum is really like the Ax·e1 product in the A = Ax·e1 + Ay·e2 + Az·e3 expression. In fact, you may actually write it as C1·| 1 〉, or just reverse the order and write C1| 1 〉. However, that’s not common practice and so I won’t do that, except occasionally. So you should look at | 1 〉 C1 as a product indeed: it’s the product of a base state and a complex number, so it’s really like m·v, or whatever other product of some scalar and some vector, except that we’ve got a complex scalar here. […] Yes, I know the term ‘complex scalar’ doesn’t make sense, but I hope you know what I mean. 🙂

More generally, we write:

F1

Writing our state vector | ψ 〉, | φ 〉 or | χ 〉 like this also defines these coefficients or coordinates Ci. Unlike our state vectors, or our base states, Cis an actual number. It has to be, of course: it’s the complex number that makes sense of the whole expression. To be precise, Cis an amplitude, or a wavefunction, i.e. a function depending on both space and time. In our previous posts, we limited the analysis to amplitudes varying in time only, and we’ll continue to do so for a while. However, at some point, you’ll get the full picture.

Now, what about the supposed similarity between the 〈 i | j〉 bra-ket and the ei·ej product? Let me invoke what Feynman, tongue-in-cheek as usual, refers to as the Great Law of Quantum Mechanics:

F2

You get this by taking | ψ 〉 out of the | ψ 〉 = ∑| i 〉〈 i | ψ 〉 expression. And, no, don’t say: what nonsense! Because… Well… Dirac’s notation really is that simple and powerful! You just have to read it from right to left. There’s an order to the symbols, unlike what you’re used to in math, because you’re used to operations that are commutative. But I need to move on. The upshot is that we can specify our base states in terms of the base states too. For example, if we have only two base states, let’s say I and II, then we can write:

| I 〉 = ∑| i 〉〈 i | 1 〉 = 1·| I 〉 + 0·| II 〉 and | II 〉 = ∑| i 〉〈 i | II 〉 = 0·| 1 〉 + 0·| II 〉

We can write this using a matrix notation:

M1Now that is silly, you’ll say. What’s the use of this? It doesn’t tell us anything new, and it also does not show us why we should think of the 〈 i | j 〉 bra-ket and the ei·ej product as being similar! Well… Yes and no. Let me show you something else. Let’s assume we’ve got some state χ and φ, which we specify in terms of our chosen set of base states as | χ 〉 = ∑ | i 〉 Di and | φ 〉 = ∑ | i 〉 Ci respectively. Now, from our post on quantum math, you’ll remember that 〈 χ | i 〉 and 〈 i | χ 〉 are each other’s complex conjugates, so we know that 〈 χ | i 〉 = 〈 i | χ 〉* = Di*. So if we have all Ci = 〈 i | φ 〉 and all Di = 〈 i | χ 〉, i.e. the ‘components’ of both states in terms of our base states, then we can calculate 〈 χ | φ 〉 – i.e. the amplitude to go from some state φ to some state χ as:

〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉 =∑ Di*C= ∑ Di*〈 i | φ 〉

We can now scrap | φ 〉 in this expression – yes, it’s the power of Dirac’s notation once more! – so we get:

b

Now, we can re-write this using a matrix notation:

M3

[I assumed that we have three base states now, so as to make the example somewhat less obvious. Please note we can never leave one of the base states out when specifying a state vector, so it’s not like the previous example was not complete. I’ll switch from two-state to three-state systems and back again all the time, so as to show the analysis is pretty general. To visualize things, think of the ammonia molecule as an example of a two-state system versus the spin of a proton or an electron as a three-state system, respectively. OK. Let’s get back to the lesson.]

You’ll say: so what? Well… Look at this:

M4

I just combined the notations for 〈 I | and | III 〉. Can you now see the similarity between the the 〈 i | j〉 bra-ket and the ei·ej product? It really is the same: you just need to respect the subtleties in regard to writing the 〈 i | and | j 〉 vector, or the eand ej vectors, as a row vector or a column vector respectively.

It doesn’t stop here, of course. When learning about vectors in high school, we also learned that we could go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. We showed how a rotation worked in one of our post on two-state systems, where we wrote:

matrix

So we’ve got that transformation matrix, which, of course, isn’t random. To be precise, we got the matrix equation above (note that we’re back to two states only, so as to simplify) because we defined the Cand CII coefficients in the | φ 〉 = | I 〉 CI + | II 〉 CII = | 1 〉 C1 + | 2 〉 C2 expression as follows:

  • C= 〈 I | ψ 〉 = (1/√2)·(C1 − C2)
  • CII = 〈 II | ψ 〉 = (1/√2)·(C1 + C2)

The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as:

transformation

I promised I’d say something more about confusing terminology so let me do that here. We call a set of base states a ‘representation‘, and writing a state vector in terms of a set of base states is often referred to as a ‘projection‘ of that state into the base set. Again, we can see it’s sort of a mathematical projection, rather than a geometrical one. But it makes sense. In any case, that’s enough on state vectors and base states.

Let me wrap it up by inserting one more matrix equation, which you should be able to reconstruct yourself:

M6

The only thing we’re doing here is to substitute 〈 χ | and | φ 〉 for ∑ Dj*〈 j | and ∑ | i 〉 Ci respectively. All the rest follows. Finally, I promised I’d tell you the difference between a state and a state vector. It’s subtle and, in practice, the two concepts refer to the same. However, we write a state as a state, like ψ or, if it’s a base state, like I, or ‘up’, or whatever. When we say a state vector, then we think of a set of numbers. It may be a row vector, like the 〈 χ | row vector with the Di* coefficients, or a column vector, like the | φ 〉 column vector with the Di* coefficients. But so if we say vector, then we think of a one-dimensional array of numbers, while the state itself is… Well… The state. So that’s some reality in physics. So you might define the state vector as the set of numbers that describes the state. While the difference is subtle, it’s important. It’s also important to note that the 〈 χ | and | χ 〉 state vectors are different too. The former appears as the final state in an amplitude, while the latter describes the starting condition. The former is referred to as a bra in the 〈 χ | φ 〉 bra-ket, while the latter is a ket in the 〈 φ | χ 〉 = 〈 χ | φ 〉* amplitude. 〈 χ | is a row vector equal to ∑ Di*〈 i |, while | χ 〉 = ∑ D| χ 〉. So it’s quite different. More in general, we’d define bras and kets as row and column vectors respectively, so we write:

M14

That makes it clear that a bra next to a ket is to be understood as a matrix multiplication. From what I wrote, it is also obvious that the conjugate transpose (which is also known as the Hermitian conjugate) of a bra is the corresponding ket and vice versa, so we write:

M15

Let me formally define the conjugate or Hermitian transpose here: the conjugate transpose of an m-by-n matrix A with complex elements is the n-by-m matrix A† obtained from A by taking the transpose (so we write the rows as columns and vice versa) and then taking the complex conjugate of each element (i.e. we switch the sign of the imaginary part of the complex number). A† is read as ‘A dagger’, but mathematicians will usually denote it by A*. In fact, there are a lot of equivalent notations, as we can write:

M16

OK. That’s it on this.

One more thing, perhaps. We’ll often have states, or base states, that make sense, in a physical sense, that is. But it’s not always the case: we’ll sometimes use base states that may not represent some situation we’re likely to encounter, but that make sense mathematically. We gave the example of the ‘mathematical’ | I 〉 and | II 〉 base states, versus the ‘physical’ | 1 〉 and | 2 〉 base states, in our post on the ammonia molecule, so I won’t say more about this here. Do keep it in mind though. Sometimes it may feel like nothing makes sense, physically, but it usually does mathematically and, therefore, all usually comes out alright in the end. 🙂 To be precise, what we did there, was to choose base states with a unambiguous, i.e. a definite, energy level. That made our calculations much easier, and the end result was the same, indeed!

So… Well… I’ll let this sink in, and move on to the next topic.

The Hamiltonian operator

In my post on the post on the Hamiltonian, I explained that those Ci and Di coefficients are usually a function of time, and how they can be determined. To be precise, they’re determined by a set of differential equations (i.e. equations involving a function and the derivative of that function) which we wrote as:

H6

If we have two base states only, then this set of equations can be written as:

set - two-base

Two equations and two functions – C= C1(t) and C= C2(t) – so we should be able to solve this thing, right? Well… No. We don’t know those Hij coefficients. As I explained in that post, they also evolve in time, so we should write them as Hij(t) instead of Hij tout court, and so it messes the whole thing up. We have two equations and six functions really. Of course, there’s always a way out, but I won’t dwell on that here—not now at least. What I want to do here is look at the Hamiltonian as an operator.

We introduced operators – but not very rigorously – when explaining the Hamiltonian. We did so by ‘expanding’ our 〈 χ | φ 〉 amplitude as follows. We’d say the amplitude to find a ‘thing’ – like a particle, for example, or some system, of particles or other things – in some state χ at the time t = t2, when it was in some state φ state at the time t = t1 was equal to:

H1

Now, a formula like this only makes sense because we’re ‘abstracting away’ from the base states, which we need to describe any state. Hence, to actually describe what’s going on, we have to choose some representation and expand this expression as follows:

H2

That looks pretty monstrous, so we should write it all out. Using the matrix notation I introduced above, we can do that – let’s take a practical example with three base states once again – as follows:

Matrix U

Now, this still looks pretty monstrous, but just think of it. We’re just applying that ‘Great Law of Quantum Physics’ here, i.e. | = ∑ | i 〉〈 i | over all base states i. To be precise, we apply it to an 〈 χ | A | φ 〉 expression, and we do so twice, so we get:

A1

Nothing more, nothing less. 🙂 Now, the idea of an operator is the result of being creative: we just drop the 〈 χ | state from the expression above to write:

C1

Yes. I know. That’s a lot to swallow, but you’ll see it makes sense because of the Great Law of Quantum Mechanics:

C2

Just think about it and continue reading when you’re ready. 🙂 The upshot is: we now think of the particle entering some ‘apparatus’ A in the state ϕ and coming out of A in some state ψ or, looking at A as an operator, we can generalize this. As Feynman puts it:

“The symbol A is neither an amplitude, nor a vector; it is a new kind of thing called an operator. It is something which “operates on” a state to produce a new state.”

Back to our Hamiltonian. Let’s go through the same process of ‘abstraction’. Let’s first re-write that ‘Hamiltonian equation’ as follows:

M7

The Hij(t) are amplitudes indeed, and we can represent them in a 〈 i | Hij(t) | j 〉 matrix indeed! Now let’s take the first step in our ‘abstraction process’: let’s scrap the 〈 i | bit. We get:

M8

We can, of course, also abstract away from the | j 〉 bit, so we get:

M9

Look at this! The right-hand side of this expression is exactly the same as that A | χ 〉 format we presented when introducing the concept of an operator. [In fact, when I say you should ‘abstract away’ from the | j 〉 bit, then you should think of the ‘Great Law’ and that matrix notation above.] So H is an operator and, therefore, it’s something which operates on a state to produce a new state.

OK. Clear enough. But what’s that ‘state’ on the left-hand side? I’ll just paraphrase Feynman here, who says we should think of it as follows: “The time derivative of the state vector |ψ〉 times i is equal to what you get by (1) operating with the Hamiltonian operator H on each base state, (2) multiplying by the amplitude that ψ is in the state j (i.e. 〈j|ψ〉), and (1) summing over all j.” Alternatively, you can also say: “The time derivative, times iħ, of a state |ψ〉 is equal to what you get if you operate on it with the Hamiltonian.” Of course, that’s true for any state, so we can ‘abstract away’ the |ψ〉 bit too and, putting a little hat (^) over the operator to remind ourselves that it’s an operator (rather than just any matrix), we get the Hamiltonian operator equation:

M10

Now, that’s all nice and great, but the key question, of course, is: what can you do with this? Well… It turns out his Hamiltonian operators is useful to calculate lots of stuff. In the first place, of course, it’s a useful operator in the context of those differential equations describing the dynamics of a quantum-mechanical system. When everything is said and done, those equations are the equivalent, in quantum physics, of the law of motion in classical physics. [And I am not joking here.]

In addition, the Hamiltonian operator also has other uses. The one I should really mention here is that you can calculate the average or expected value (EV[X]) of the energy  of a state ψ (i.e. any state, really) by first operating on | ψ 〉 with the Hamiltonian, and then multiplying 〈 ψ | with the result. That sounds a bit complicated, but you’ll understand it when seeing the mathematical expression, which we can write as:

M13

The formula is pretty straightforward. [If you don’t think so, then just write it all out using the matrix notation.] But you may wonder how it works exactly… Well… Sorry. I don’t want to copy all of Feynman here, so I’ll refer you to him on this. In fact, the proof of this formula is actually very straightforward, and so you should be able to get through it with the math you got here. You may even understand Feynman’s illustration of it for the ‘special case’ when base states are, indeed, those mathematically convenient base states with definite energy levels.

Have fun with it! 🙂

Post scriptum on Hilbert spaces:

As mentioned above, our state vectors are actually functions. To be specific, they are wavefunctions, i.e. periodic functions, evolving in space and time, so we usually write them as ψ = ψ(x, t). Our ‘Hilbert space’, i.e. our collection of state vectors, is, therefore, often referred to as a function space. So it’s a set of functions. At the same time, it is a vector space too, because we have those addition and multiplication operations, so our function space has the algebraic structure of a vector space. As you can imagine, there are some mathematical conditions for a space or a set of objects to ‘qualify’ as a Hilbert space, and the epithet itself comes with a lot of interesting properties. One of them is completeness, which is a property that allows us to jot down those differential equations that describe the dynamics of a quantum-mechanical system. However, as you can find whatever you’d need or want to know about those mathematical properties on the Web, I won’t get into it. The important thing here is to understand the concept of a Hilbert space intuitively. I hope this post has helped you in that regard, at least. 🙂

Re-visiting uncertainty…

I re-visited the Uncertainty Principle a couple of times already, but here I really want to get at the bottom of the thing? What’s uncertain? The energy? The time? The wavefunction itself? These questions are not easily answered, and I need to warn you: you won’t get too much wiser when you’re finished reading this. I just felt like freewheeling a bit. [Note that the first part of this post repeats what you’ll find on the Occam page, or my post on Occam’s Razor. But these post do not analyze uncertainty, which is what I will be trying to do here.]

Let’s first think about the wavefunction itself. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows. In my previous post, I said I like to think it travels with the particle, but then doesn’t make much sense either. It’s like a fundamental property of the particle. Like the color of an apple. But where is that color? In the apple, in the light it reflects, in the retina of our eye, or is it in our brain? If you know a thing or two about how perception actually works, you’ll tend to agree the quality of color is not in the apple. When everything is said and done, the wavefunction is a mental construct: when learning physics, we start to think of a particle as a wavefunction, but they are two separate things: the particle is reality, the wavefunction is imaginary.

But that’s not what I want to talk about here. It’s about that uncertainty. Where is the uncertainty? You’ll say: you just said it was in our brain. No. I didn’t say that. It’s not that simple. Let’s look at the basic assumptions of quantum physics:

  1. Quantum physics assumes there’s always some randomness in Nature and, hence, we can measure probabilities only. We’ve got randomness in classical mechanics too, but this is different. This is an assumption about how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
  2. However, as Shakespeare has one of his characters say: there is a method in the madness, and the pioneers– I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered that method: all probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ. The generic functional form of that wavefunction is:

Ψ = Ψ(x, t) = a·eiθ = a·ei(ω·t − k ∙x) = a·ei·[(E/ħ)·t − (p/ħ)∙x]

That should be obvious by now, as I’ve written more than a dozens of posts on this. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. It’s not that complex exponential. That eiφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below. [It’s a sine and a cosine, but they’re the same function: there’s just a phase difference of 90 degrees.] sine

No. To understand the wavefunction, we need to understand those de Broglie relations, ω = E/ħ and k = p/ħ, and then, as mentioned, we need to understand the Uncertainty Principle. We need to understand where it comes from. Let’s try to go as far as we can by making a few remarks:

  • Adding or subtracting two terms in math, (E/ħ)·t − (p/ħ)∙x, implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (energy, or work, is force over distance, remember?) or electronvolts (1 eV = 1.6×10−19 J = 1.6×10−19 N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying E/ħ and p/ħ by t and x respectively gives us a dimensionless number indeed.
  • It’s also an invariant number, which means we’ll always get the same value for it, regardless of our frame of reference. As mentioned above, that’s because the four-vector product pμxμ = E·t − px is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
  • Now, Planck’s quantum of action h, or ħ – h and ħ only differ in their dimension: h is measured in cycles per second, while ħ is measured in radians per second: both assume we can at least measure one cycle – is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon. [In case you want to see the logic of this, see my post on the physical constants c, h and α.]
  • It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c2/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v  with the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it’s in a ‘field-free’ space, so our particle is in a region of uniform potential.] So we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, and the related kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would again yield the same t − β·x argument.

The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:

Φ(φ) = a·eiφ = a·ei(t − β·x)

This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.

Let’s now go back to that Φ(φ) = a·eiφ = a·ei(t − β·x) wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.

So… Well… We’ve pretty much explained what quantum physics is all about here. You just need to get used to that complex exponential: eiφ = cos(−φ) + i·sin(−φ) = cos(φ) −i·sin(φ). It would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

However, the analysis above does not include uncertainty. That’s as fundamental to quantum physics as de Broglie‘s equations, so let’s think about that now.

Introducing uncertainty

Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E± σE, and p = p± σp. Huh? No ΔE or ΔE? Well… It’s the same, really, but I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.

The simplest model is that of a two-state system, because it involves two energy levels only: E = E± A, with A some constant. Large or small, it doesn’t matter. All is relative anyway. 🙂 We explained the basics of the two-state system using the example of an ammonia molecule, i.e. an NHmolecule, so it consists on one nitrogen and three hydrogen atoms. We had two base states in this system: ‘up’ or ‘down’, which we denoted as base state | 1 〉 and base state | 2 〉 respectively. This ‘up’ and ‘down’ had nothing to do with the classical or quantum-mechanical notion of spin, which is related to the magnetic moment. No. It’s much simpler than that: the nitrogen atom could be either beneath or, else, above the plane of the hydrogens, as shown below, with ‘beneath’ and ‘above’ being defined in regard to the molecule’s direction of rotation around its axis of symmetry.

Capture

In any case, for the details, I’ll refer you to the post(s) on it. Here I just want to mention the result. We wrote the amplitude to find the molecule in either one of these two states as:

  • C= 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t
  • C= 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t

That gave us the following probabilities:

graph

If our molecule can be in two states only, and it starts off in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase.

Now, the point you should note is that we get these time-dependent probabilities only because we’re introducing two different energy levels: E+ A and E− A. [Note they separated by an amount equal to 2·A, as I’ll use that information later.] If we’d have one energy level only – which amounts to saying that we know it, and that it’s something definite then we’d just have one wavefunction, which we’d write as:

a·eiθ = a·e−(i/ħ)·(E0·t − p·x) = a·e−(i/ħ)·(E0·t)·e(i/ħ)·(p·x)

Note that we can always split our wavefunction in a ‘time’ and a ‘space’ part, which is quite convenient. In fact, because our ammonia molecule stays where it is, it has no momentum: p = 0. Therefore, its wavefunction reduces to:

a·eiθ = a·e−(i/ħ)·(E0·t)

As simple as it can be. 🙂 The point is that a wavefunction like this, i.e. a wavefunction that’s defined by a definite energy, will always yield a constant and equal probability, both in time as well as in space. That’s just the math of it: |a·eiθ|= a2. Always! If you want to know why, you should think of Euler’s formula and Pythagoras’ Theorem: cos2θ +sin2θ = 1. Always! 🙂

That constant probability is annoying, because our nitrogen atom never ‘flips’, and we know it actually does, thereby overcoming a energy barrier: it’s a phenomenon that’s referred to as ‘tunneling’, and it’s real! The probabilities in that graph above are real! Also, if our wavefunction would represent some moving particle, it would imply that the probability to find it somewhere in space is the same all over space, which implies our particle is everywhere and nowhere at the same time, really.

So, in quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or potentially infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. In our two-state system, n = 2, obviously! So we’ve two energy levels only and so our composite wave consists of two component waves only.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

File-Wave_group

[The animation comes from Wikipedia, and shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

So… OK. That should be clear enough. Let’s now apply these thoughts to our ‘reduced’ wavefunction

Φ(φ) = a·eiφ = a·ei(t − β·x)

Thinking about uncertainty

Frankly, I tried to fool you above. If the functional form of the wavefunction is a·e−(i/ħ)·(E·t − p·x), then we can measure E and p in whatever unit we want, including h or ħ, but we cannot re-scale the argument of the function, i.e. the phase θ, without changing the functional form itself. I explained that in that post for my kids on wavefunctions:, in which I explained we may represent the same electromagnetic wave by two different functional forms:

 F(ct−x) = G(t−x/c)

So F and G represent the same wave, but they are different wavefunctions. In this regard, you should note that the argument of F is expressed in distance units, as we multiply t with the speed of light (so it’s like our time unit is 299,792,458 m now), while the argument of G is expressed in time units, as we divide x by the distance traveled in one second). But F and G are different functional forms. Just do an example and take a simple sine function: you’ll agree that sin(θ) ≠ sin(θ/c) for all values of θ, except 0. Re-scaling changes the frequency, or the wavelength, and it does so quite drastically in this case. 🙂 Likewise, you can see that a·ei(φ/E) = [a·eiφ]1/E, so that’s a very different function. In short, we were a bit too adventurous above. Now, while we can drop the 1/ħ in the a·e−(i/ħ)·(E·t − p·x) function when measuring energy and momentum in units that are numerically equal to ħ, we’ll just revert to our original wavefunction for the time being, which equals

Ψ(θ) = a·eiθ = a·ei·[(E/ħ)·t − (p/ħ)·x]

Let’s now introduce uncertainty once again. The simplest situation is that we have two closely spaced energy levels. In theory, the difference between the two can be as small as ħ, so we’d write: E = E± ħ/2. [Remember what I said about the ± A: it means the difference is 2A.] However, we can generalize this and write: E = E± n·ħ/2, with n = 1, 2, 3,… This does not imply any greater uncertainty – we still have two states only – but just a larger difference between the two energy levels.

Let’s also simplify by looking at the ‘time part’ of our equation only, i.e. a·ei·(E/ħ)·t. It doesn’t mean we don’t care about the ‘space part’: it just means that we’re only looking at how our function varies in time and so we just ‘fix’ or ‘freeze’ x. Now, the uncertainty is in the energy really but, from a mathematical point of view, we’ve got an uncertainty in the argument of our wavefunction, really. This uncertainty in the argument is, obviously, equal to:

(E/ħ)·t = [(E± n·ħ/2)/ħ]·t = (E0/ħ ± n/2)·t = (E0/ħ)·t ± (n/2)·t

So we can write:

a·ei·(E/ħ)·t = a·ei·[(E0/ħ)·t ± (1/2)·t] = a·ei·[(E0/ħ)·t]·ei·[±(n/2)·t]

This is valid for any value of t. What the expression says is that, from a mathematical point of view, introducing uncertainty about the energy is equivalent to introducing uncertainty about the wavefunction itself. It may be equal to a·ei·[(E0/ħ)·t]·ei·(n/2)·t, but it may also be equal to a·ei·[(E0/ħ)·t]·ei·(n/2)·t. The phases of the ei·t/2 and ei·t/2 factors are separated by a distance equal to t.

So… Well…

[…]

Hmm… I am stuck. How is this going to lead me to the ΔE·Δt = ħ/2 principle? To anyone out there: can you help? 🙂

[…]

The thing is: you won’t get the Uncertainty Principle by staring at that formula above. It’s a bit more complicated. The idea is that we have some distribution of the observables, like energy and momentum, and that implies some distribution of the associated frequencies, i.e. ω for E, and k for p. The Wikipedia article on the Uncertainty Principle gives you a formal derivation of the Uncertainty Principle, using the so-called Kennard formulation of it. You can have a look, but it involves a lot of formalism—which is what I wanted to avoid here!

I hope you get the idea though. It’s like statistics. First, we assume we know the population, and then we describe that population using all kinds of summary statistics. But then we reverse the situation: we don’t know the population but we do have sample information, which we also describe using all kinds of summary statistics. Then, based on what we find for the sample, we calculate the estimated statistics for the population itself, like the mean value and the standard deviation, to name the most important ones. So it’s a bit the same here, except that, in quantum mechanics, there may not be any real value underneath: the mean and the standard deviation represent something fuzzy, rather than something precise.

Hmm… I’ll leave you with these thoughts. We’ll develop them further as we will be digging into all much deeper over the coming weeks. 🙂

Post scriptum: I know you expect something more from me, so… Well… Think about the following. If we have some uncertainty about the energy E, we’ll have some uncertainty about the momentum p according to that β = p/E. [By the way, please think about this relationship: it says, all other things being equal (such as the inertia, i.e. the mass, of our particle), that more energy will all go into more momentum. More specifically, note that ∂p/∂p = β according to this equation. In fact, if we include the mass of our particle, i.e. its inertia, as potential energy, then we might say that (1−β)·E is the potential energy of our particle, as opposed to its kinetic energy.] So let’s try to think about that.

Let’s denote the uncertainty about the energy as ΔE. As should be obvious from the discussion above, it can be anything: it can mean two separate energy levels E = E± A, or a potentially infinite set of values. However, even if the set is infinite, we know the various energy levels need to be separated by ħ, at least. So if the set is infinite, it’s going to be a countable infinite set, like the set of natural numbers, or the set of integers. But let’s stick to our example of two values E = E± A only, with A = ħ so E + ΔE = E± ħ and, therefore, ΔE = ± ħ. That implies Δp = Δ(β·E) = β·ΔE = ± β·ħ.

Hmm… This is a bit fishy, isn’t it? We said we’d measure the momentum in units of ħ, but so here we say the uncertainty in the momentum can actually be a fraction of ħ. […] Well… Yes. Now, the momentum is the product of the mass, as measured by the inertia of our particle to accelerations or decelerations, and its velocity. If we assume the inertia of our particle, or its mass, to be constant – so we say it’s a property of the object that is not subject to uncertainty, which, I admit, is a rather dicey assumption (if all other measurable properties of the particle are subject to uncertainty, then why not its mass?) – then we can also write: Δp = Δ(m·v) = Δ(m·β) = m·Δβ. [Note that we’re not only assuming that the mass is not subject to uncertainty, but also that the velocity is non-relativistic. If not, we couldn’t treat the particle’s mass as a constant.] But let’s be specific here: what we’re saying is that, if ΔE = ± ħ, then Δv = Δβ will be equal to Δβ = Δp/m = ± (β/m)·ħ. The point to note is that we’re no longer sure about the velocity of our particle. Its (relative) velocity is now:

β ± Δβ = β ± (β/m)·ħ

But, because velocity is the ratio of distance over time, this introduces an uncertainty about time and distance. Indeed, if its velocity is β ± (β/m)·ħ, then, over some time T, it will travel some distance X = [β ± (β/m)·ħ]·T. Likewise, it we have some distance X, then our particle will need a time equal to T = X/[β ± (β/m)·ħ].

You’ll wonder what I am trying to say because… Well… If we’d just measure X and T precisely, then all the uncertainty is gone and we know if the energy is E+ ħ or E− ħ. Well… Yes and no. The uncertainty is fundamental – at least that’s what’s quantum physicists believe – so our uncertainty about the time and the distance we’re measuring is equally fundamental: we can have either of the two values X = [β ± (β/m)·ħ] T = X/[β ± (β/m)·ħ], whenever or wherever we measure. So we have a ΔX and ΔT that are equal to ± [(β/m)·ħ]·T and X/[± (β/m)·ħ] respectively. We can relate this to ΔE and Δp:

  • ΔX = (1/m)·T·Δp
  • ΔT = X/[(β/m)·ΔE]

You’ll grumble: this still doesn’t give us the Uncertainty Principle in its canonical form. Not at all, really. I know… I need to do some more thinking here. But I feel I am getting somewhere. 🙂 Let me know if you see where, and if you think you can get any further. 🙂

The thing is: you’ll have to read a bit more about Fourier transforms and why and how variables like time and energy, or position and momentum, are so-called conjugate variables. As you can see, energy and time, and position and momentum, are obviously linked through the E·t and p·products in the E0·t − p·x sum. That says a lot, and it helps us to understand, in a more intuitive way, why the ΔE·Δt and Δp·Δx products should obey the relation they are obeying, i.e. the Uncertainty Principle, which we write as ΔE·Δt ≥ ħ/2 and Δp·Δx ≥ ħ/2. But so proving involves more than just staring at that Ψ(θ) = a·eiθ = a·ei·[(E/ħ)·t − (p/ħ)·x] relation.

Having said, it helps to think about how that E·t − p·x sum works. For example, think about two particles, a and b, with different velocity and mass, but with the same momentum, so p= pb ⇔ ma·v= ma·v⇔ ma/v= mb/va. The spatial frequency of the wavefunction  would be the same for both but the temporal frequency would be different, because their energy incorporates the rest mass and, hence, because m≠ mb, we also know that E≠ Eb. So… It all works out but, yes, I admit it’s all very strange, and it takes a long time and a lot of reflection to advance our understanding.

Occam’s Razor

The analysis of a two-state system (i.e. the rather famous example of an ammonia molecule ‘flipping’ its spin direction from ‘up’ to ‘down’, or vice versa) in my previous post is a good opportunity to think about Occam’s Razor once more. What are we doing? What does the math tell us?

In the example we chose, we didn’t need to worry about space. It was all about time: an evolving state over time. We also knew the answers we wanted to get: if there is some probability for the system to ‘flip’ from one state to another, we know it will, at some point in time. We also want probabilities to add up to one, so we knew the graph below had to be the result we would find: if our molecule can be in two states only, and it starts of in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase, which is what is depicted below.

graph

However, the graph above is only a Platonic idea: we don’t bother to actually verify what state the molecule is in. If we did, we’d have to ‘re-set’ our t = 0 point, and start all over again. The wavefunction would collapse, as they say, because we’ve made a measurement. However, having said that, yes, in the physicist’s Platonic world of ideas, the probability functions above make perfect sense. They are beautiful. You should note, for example, that P1 (i.e. the probability to be in state 1) and P2 (i.e. the probability to be in state 2) add up to 1 all of the time, so we don’t need to integrate over a cycle or something: so it’s all perfect!

These probability functions are based on ideas that are even more Platonic: interfering amplitudes. Let me explain.

Quantum physics is based on the idea that these probabilities are determined by some wavefunction, a complex-valued amplitude that varies in time and space. It’s a two-dimensional thing, and then it’s not. It’s two-dimensional because it combines a sine and cosine, i.e. a real and an imaginary part, but the argument of the sine and the cosine is the same, and the sine and cosine are the same function, except for a phase shift equal to π. We write:

a·eiθ = cos(θ) – sin(−θ) = cosθ – sinθ

The minus sign is there because it turns out that Nature measures angles, i.e. our phase, clockwise, rather than counterclockwise, so that’s not as per our mathematical convention. But that’s a minor detail, really. [It should give you some food for thought, though.] For the rest, the related graph is as simple as the formula:

graph sin and cos

Now, the phase of this wavefunction is written as θ = (ω·t − k ∙x). Hence, ω determines how this wavefunction varies in time, and the wavevector k tells us how this wave varies in space. The young Frenchman Comte Louis de Broglie noted the mathematical similarity between the ω·t − k ∙x expression and Einstein’s four-vector product pμxμ = E·t − px, which remains invariant under a Lorentz transformation. He also understood that the Planck-Einstein relation E = ħ·ω actually defines the energy unit and, therefore, that any frequency, any oscillation really, in space or in time, is to be expressed in terms of ħ.

[To be precise, the fundamental quantum of energy is h = ħ·2π, because that’s the energy of one cycle. To illustrate the point, think of the Planck-Einstein relation. It gives us the energy of a photon with frequency f: Eγ = h·f. If we re-write this equation as Eγ/f = h, and we do a dimensional analysis, we get: h = Eγ/f ⇔ 6.626×10−34 joule·second [x joule]/[cycles per second] ⇔ h = 6.626×10−34 joule per cycle. It’s only because we are expressing ω and k as angular frequencies (i.e. in radians per second or per meter, rather than in cycles per second or per meter) that we have to think of ħ = h/2π rather than h.]

Louis de Broglie connected the dots between some other equations too. He was fully familiar with the equations determining the phase and group velocity of composite waves, or a wavetrain that actually might represent a wavicle traveling through spacetime. In short, he boldly equated ω with ω = E/ħ and k with k = p/ħ, and all came out alright. It made perfect sense!

I’ve written enough about this. What I want to write about here is how this also makes for the situation on hand: a simple two-state system that depends on time only. So its phase is θ = ω·t = E0/ħ. What’s E0? It is the total energy of the system, including the equivalent energy of the particle’s rest mass and any potential energy that may be there because of the presence of one or the other force field. What about kinetic energy? Well… We said it: in this case, there is no translational or linear momentum, so p = 0. So our Platonic wavefunction reduces to:

a·eiθ = ae(i/ħ)·(E0·t)

Great! […] But… Well… No! The problem with this wavefunction is that it yields a constant probability. To be precise, when we take the absolute square of this wavefunction – which is what we do when calculating a probability from a wavefunction − we get P = a2, always. The ‘normalization’ condition (so that’s the condition that probabilities have to add up to one) implies that P1 = P2 = a2 = 1/2. Makes sense, you’ll say, but the problem is that this doesn’t reflect reality: these probabilities do not evolve over time and, hence, our ammonia molecule never ‘flips’ its spin direction from ‘up’ to ‘down’, or vice versa. In short, our wavefunction does not explain reality.

The problem is not unlike the problem we’d had with a similar function relating the momentum and the position of a particle. You’ll remember it: we wrote it as a·eiθ = ae(i/ħ)·(p·x). [Note that we can write a·eiθ = a·e−(i/ħ)·(E0·t − p·x) = a·e−(i/ħ)·(E0·t)·e(i/ħ)·(p·x), so we can always split our wavefunction in a ‘time’ and a ‘space’ part.] But then we found that this wavefunction also yielded a constant and equal probability all over space, which implies our particle is everywhere (and, therefore, nowhere, really).

In quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to n energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

File-Wave_group

[The animation also shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

This begs the following question: what’s the uncertainty really? Is it an uncertainty in the energy, or is it an uncertainty in the wavefunction? I mean: we have a function relating the energy to a frequency. Introducing some uncertainty about the energy is mathematically equivalent to introducing uncertainty about the frequency. Of course, the answer is: the uncertainty is in both, so it’s in the frequency and in the energy and both are related through the wavefunction. So… Well… Yes. In some way, we’re chasing our own tail. 🙂

However, the trick does the job, and perfectly so. Let me summarize what we did in the previous post: we had the ammonia molecule, i.e. an NH3 molecule, with the nitrogen ‘flipping’ across the hydrogens from time to time, as illustrated below:

dipole

This ‘flip’ requires energy, which is why we associate two energy levels with the molecule, rather than just one. We wrote these two energy levels as E+ A and E− A. That assumption solved all of our problems. [Note that we don’t specify what the energy barrier really consists of: moving the center of mass obviously requires some energy, but it is likely that a ‘flip’ also involves overcoming some electrostatic forces, as shown by the reversal of the electric dipole moment in the illustration above.] To be specific, it gave us the following wavefunctions for the amplitude to be in the ‘up’ or ‘1’ state versus the ‘down’ or ‘2’ state respectivelly:

  • C= (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t
  • C= (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t

Both are composite waves. To be precise, they are the sum of two component waves with a temporal frequency equal to ω= (E− A)/ħ and ω= (E+ A)/ħ respectively. [As for the minus sign in front of the second term in the wave equation for C2, −1 = e±iπ, so + (1/2)·e(i/ħ)·(E+ A)·t and – (1/2)·e(i/ħ)·(E+ A)·t are the same wavefunction: they only differ because their relative phase is shifted by ±π.] So the so-called base states of the molecule themselves are associated with two different energy levels: it’s not like one state has more energy than the other.

You’ll say: so what?

Well… Nothing. That’s it really. That’s all I wanted to say here. The absolute square of those two wavefunctions gives us those time-dependent probabilities above, i.e. the graph we started this post with. So… Well… Done!

You’ll say: where’s the ‘envelope’? Oh! Yes! Let me tell you. The C1(t) and C2(t) equations can be re-written as:

C2

Now, remembering our rules for adding and subtracting complex conjugates (eiθ + e–iθ = 2cosθ and eiθ − e–iθ = 2sinθ), we can re-write this as:

C3

So there we are! We’ve got wave equations whose temporal variation is basically defined by Ebut, on top of that, we have an envelope here: the cos(A·t/ħ) and sin(A·t/ħ) factor respectively. So their magnitude is no longer time-independent: both the phase as well as the amplitude now vary with time. The associated probabilities are the ones we plotted:

  • |C1(t)|= cos2[(A/ħ)·t], and
  • |C2(t)|= sin2[(A/ħ)·t].

So, to summarize it all once more, allowing the nitrogen atom to push its way through the three hydrogens, so as to flip to the other side, thereby breaking the energy barrier, is equivalent to associating two energy levels to the ammonia molecule as a whole, thereby introducing some uncertainty, or indefiniteness as to its energy, and that, in turn, gives us the amplitudes and probabilities that we’ve just calculated. [And you may want to note here that the probabilities “sloshing back and forth”, or “dumping into each other” – as Feynman puts it – is the result of the varying magnitudes of our amplitudes, so that’s the ‘envelope’ effect. It’s only because the magnitudes vary in time that their absolute square, i.e. the associated probability, varies too.

So… Well… That’s it. I think this and all of the previous posts served as a nice introduction to quantum physics. More in particular, I hope this post made you appreciate the mathematical framework is not as horrendous as it often seems to be.

When thinking about it, it’s actually all quite straightforward, and it surely respects Occam’s principle of parsimony in philosophical and scientific thought, also know as Occam’s Razor: “When trying to explain something, it is vain to do with more what can be done with less.” So the math we need is the math we need, really: nothing more, nothing less. As I’ve said a couple of times already, Occam would have loved the math behind QM: the physics call for the math, and the math becomes the physics.

That’s what makes it beautiful. 🙂

Post scriptum:

One might think that the addition of a term in the argument in itself would lead to a beat note and, hence, a varying probability but, no! We may look at e(i/ħ)·(E+ A)·t as a product of two amplitudes:

e(i/ħ)·(E+ A)·t e(i/ħ)·E0·t·e(i/ħ)·A·t

But, when writing this all out, one just gets a cos(α·t+β·t)–sin(α·t+β·t), whose absolute square |cos(α·t+β·t)–sin(α·t+β·t)|= 1. However, writing e(i/ħ)·(E+ A)·t as a product of two amplitudes in itself is interesting. We multiply amplitudes when an event consists of two sub-events. For example, the amplitude for some particle to go from s to x via some point a is written as:

x | s 〉via a = 〈 x | a 〉〈 a | s 〉

Having said that, the graph of the product is uninteresting: the real and imaginary part of the wavefunction are a simple sine and cosine function, and their absolute square is constant, as shown below. graph

Adding two waves with very different frequencies – A is a fraction of E– gives a much more interesting pattern, like the one below, which shows an eiαt+eiβt = cos(αt)−i·sin(αt)+cos(βt)−i·sin(βt) = cos(αt)+cos(βt)−i·[sin(αt)+sin(βt)] pattern for α = 1 and β = 0.1.

graph 2

That doesn’t look a beat note, does it? The graphs below, which use 0.5 and 0.01 for β respectively, are not typical beat notes either.

 graph 3graph 4

We get our typical ‘beat note’ only when we’re looking at a wave traveling in space, so then we involve the space variable again, and the relations that come with in, i.e. a phase velocity v= ω/k  = (E/ħ)/(p/ħ) = E/p = c2/v (read: all component waves travel at the same speed), and a group velocity v= dω/dk = v (read: the composite wave or wavetrain travels at the classical speed of our particle, so it travels with the particle, so to speak). That’s what’s I’ve shown numerous times already, but I’ll insert one more animation here, just to make sure you see what we’re talking about. [Credit for the animation goes to another site, one on acoustics, actually!]

beats

So what’s left? Nothing much. The only thing you may want to do is to continue thinking about that wavefunction. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows, really, but I like to think it does travel with the particle. So it’s like a fundamental property of the particle. We need it every time when we try to measure something: its position, its momentum, its spin (i.e. angular momentum) or, in the example of our ammonia molecule, its orientation in space. So the funny thing is that, in quantum mechanics,

  1. We can measure probabilities only, so there’s always some randomness. That’s how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
  2. But then, as Polonius famously put, there is a method in this madness, and the pioneers – I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered. All probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ:

θ = (ω·t − k ∙x) = (E/ħ)·t − (p/ħ)·x

That should be obvious by now, as I’ve written dozens of posts on this by now. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. But let’s try to go as far as we can by making a few remarks:

  •  Adding two terms in math implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (force over distance, remember?) or electronvolts (1 eV = 1.6×10−19 J = 1.6×10−19 N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying it by t and x respectively gives us a dimensionless number indeed.
  • It’s also an invariant number, which means we’ll always get the same value for it. As mentioned above, that’s because the four-vector product pμxμ = E·t − px is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
  • Now, Planck’s quantum of action h or ħ (they only differ in their dimension: h is measured in cycles per second and ħ is measured in radians per second) is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon.
  • It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c2/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v  with the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it is present in a ‘field-free’ space, so our particle is in a region of uniform potential.] You see what I am going to do now: we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Now that’s the argument of a wavefunction with the argument expressed in distance units. Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, i.e. in kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would then yield the same t − β·x argument.

The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:

Φ(φ) = a·eiφ = a·ei(t − β·x)

This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.

Let’s now go back to that Φ(φ) = a·eiφ = a·ei(t − β·x) wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.

Of course, the analysis above does not include uncertainty. Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E± σE, and p = p± σp. [I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.] But, including that, we’ve pretty much explained what quantum physics is about here.

You just need to get used to that complex exponential: eiφ = cos(−φ) + i·sin(−φ) = cos(φ) − i·sin(φ). Of course, it would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

So… Well… Just accept that eiφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below.

 sine

And then you need to think of stuff like this (the animation is taken from Wikipedia), but then with a projection of the sine of those phasors too. It’s all great fun, so I’ll let you play with it now. 🙂

Sumafasores

The Hamiltonian for a two-state system: the ammonia example

Ammonia, i.e. NH3, is a colorless gas with a strong smell. Its serves as a precursor in the production of fertilizer, but we also know it as a cleaning product, ammonium hydroxide, which is NH3 dissolved in water. It has a lot of other uses too. For example, its use in this post, is to illustrate a two-state system. 🙂 We’ll apply everything we learned in our previous posts and, as I  mentioned when finishing the last of those rather mathematical pieces, I think the example really feels like a reward after all of the tough work on all of those abstract concepts – like that Hamiltonian matrix indeed – so I hope you enjoy it. So… Here we go!

The geometry of the NH3 molecule can be described by thinking of it as a trigonal pyramid, with the nitrogen atom (N) at its apex, and the three hydrogen atoms (H) at the base, as illustrated below. [Feynman’s illustration is slightly misleading, though, because it may give the impression that the hydrogen atoms are bonded together somehow. That’s not the case: the hydrogen atoms share their electron with the nitrogen, thereby completing the outer shell of both atoms. This is referred to as a covalent bond. You may want to look it up, but it is of no particular relevance to what follows here.]

Capture

Here, we will only worry about the spin of the molecule about its axis of symmetry, as shown above, which is either in one direction or in the other, obviously. So we’ll discuss the molecule as a two-state system. So we don’t care about its translational (i.e. linear) momentum, its internal vibrations, or whatever else that might be going on. It is one of those situations illustrating that the spin vector, i.e. the vector representing angular momentum, is an axial vector: the first state, which is denoted by | 1 〉 is not the mirror image of state | 2 〉. In fact, there is a more sophisticated version of the illustration above, which usefully reminds us of the physics involved.

dipoleIt should be noted, however, that we don’t need to specify what the energy barrier really consists of: moving the center of mass obviously requires some energy, but it is likely that a ‘flip’ also involves overcoming some electrostatic forces, as shown by the reversal of the electric dipole moment in the illustration above. In fact, the illustration may confuse you, because we’re usually thinking about some net electric charge that’s spinning, and so the angular momentum results in a magnetic dipole moment, that’s either ‘up’ or ‘down’, and it’s usually also denoted by the very same μ symbol that’s used below. As I explained in my post on angular momentum and the magnetic moment, it’s related to the angular momentum J through the so-called g-number. In the illustration above, however, the μ symbol is used to denote an electric dipole moment, so that’s different. Don’t rack your brain over it: just accept there’s an energy barrier, and it requires energy to get through it. Don’t worry about its details!

Indeed, in quantum mechanics, we abstract away from such nitty-gritty, and so we just say that we have base states | i 〉 here, with i equal to 1 or 2. One or the other. Now, in our post on quantum math, we introduced what Feynman only half-jokingly refers to as the Great Law of Quantum Physics: | = ∑ | i 〉〈 i | over all base states i. It basically means that we should always describe our initial and end states in terms of base states. Applying that principle to the state of our ammonia molecule, which we’ll denote by | ψ 〉, we can write:

f1

You may – in fact, you should – mechanically apply that | = ∑ | i 〉〈 i | substitution to | ψ 〉 to get what you get here, but you should also think about what you’re writing. It’s not an easy thing to interpret, but it may help you to think of the similarity of the formula above with the description of a vector in terms of its base vectors, which we write as A = Ax·e+ Ay·e2 + Az·e3. Just substitute the Acoefficients for Ci and the ebase vectors for the | i 〉 base states, and you may understand this formula somewhat better. It also explains why the | ψ 〉 state is often referred to as the | ψ 〉 state vector: unlike our  A = ∑ Ai·esum of base vectors, our | 1 〉 C1 + | 2 〉 Csum does not have any geometrical interpretation but… Well… Not all ‘vectors’ in math have a geometric interpretation, and so this is a case in point.

It may also help you to think of the time-dependency. Indeed, this formula makes a lot more sense when realizing that the state of our ammonia molecule, and those coefficients Ci, depend on time, so we write: ψ = ψ(t) and C= Ci(t). Hence, if we would know, for sure, that our molecule is always in state | 1 〉, then C1 = 1 and C2 = 0, and we’d write: | ψ 〉 = | 1 〉 = | 1 〉 1 + | 2 〉 0. [I am always tempted to insert a little dot (·), and change the order of the factors, so as to show we’re talking some kind of product indeed – so I am tempted to write | ψ 〉 = C1·| 1 〉 C1 + C2·| 2 〉 C2, but I note that’s not done conventionally, so I won’t do it either.]  

Why this time dependency? It’s because we’ll allow for the possibility of the nitrogen to push its way through the pyramid – through the three hydrogens, really – and flip to the other side. It’s unlikely, because it requires a lot of energy to get half-way through (we’ve got what we referred to as an energy barrier here), but it may happen and, as we’ll see shortly, it results in us having to think of the the ammonia molecule as having two separate energy levels, rather than just one. We’ll denote those energy levels as E0 ± A. However, I am getting ahead of myself here, so let me get back to the main story.

To fully understand the story, you should really read my previous post on the Hamiltonian, which explains how those Ci coefficients, as a function of time, can be determined. They’re determined by a set of differential equations (i.e. equations involving a function and the derivative of that function) which we wrote as:

H6

 If we have two base states only – which is the case here – then this set of equations is:

set - two-base

Two equations and two functions – C= C1(t) and C= C2(t) – so we should be able to solve this thing, right? Well… No. We don’t know those Hij coefficients. As I explained in my previous post, they also evolve in time, so we should write them as Hij(t) instead of Hij tout court, and so it messes the whole thing up. We have two equations and six functions really. There is no way we can solve this! So how do we get out of this mess?

Well… By trial and error, I guess. 🙂 Let us just assume the molecule would behave nicely—which we know it doesn’t, but so let’s push the ‘classical’ analysis as far as we can, so we might get some clues as to how to solve this problem. In fact, our analysis isn’t ‘classical’ at all, because we’re still talking amplitudes here! However, you’ll agree the ‘simple’ solution would be that our ammonia molecule doesn’t ‘tunnel’. It just stays in the same spin direction forever. Then H12 and H21 must be zero (think of the U12(t + Δt, t) and U21(t + Δt, t) functions) and H11 and H22 are equal to… Well… I’d love to say they’re equal to 1 but… Well… You should go through my previous posts: these Hamiltonian coefficients are related to probabilities but… Well… Same-same but different, as they say in Asia. 🙂 They’re amplitudes, which are things you use to calculate probabilities. But calculating probabilities involve normalization and other stuff, like allowing for interference of amplitudes, and so… Well… To make a long story short, if our ammonia molecule would stay in the same spin direction forever, then H11 and H22  are not one but some constant. In any case, the point is that they would not change in time (so H11(t) = H11  and H22(t ) = H22), and, therefore, our two equations would reduce to:

S1

So the coefficients are now proper coefficients, in the sense that they’ve got some definite value, and so we have two equations and two functions only now, and so we can solve this. Indeed, remembering all of the stuff we wrote on the magic of exponential functions (more in particular, remembering that d[ex]/dx), we can understand the proposed solution:

S2

As Feynman notes: “These are just the amplitudes for stationary states with the energies E= H11 and E= H22.” Now let’s think about that. Indeed, I find the term ‘stationary’ state quite confusing, as it’s ill-defined. In this context, it basically means that we have a wavefunction that is determined by (i) a definite (i.e. unambiguous, or precise) energy level and (ii) that there is no spatial variation. Let me refer you to my post on the basics of quantum math here. We often use a sort of ‘Platonic’ example of the wavefunction indeed:

a·ei·θ ei·(ω·t − k ∙x) = a·e(i/ħ)·(E·t − px)

So that’s a wavefunction assuming the particle we’re looking at has some well-defined energy E and some equally well-defined momentum p. Now, that’s kind of ‘Platonic’ indeed, because it’s more like an idea, rather than something real. Indeed, a wavefunction like that means that the particle is everywhere and nowhere, really—because its wavefunction is spread out all of over space. Of course, we may think of the ‘space’ as some kind of confined space, like a box, and then we can think of this particle as being ‘somewhere’ in that box, and then we look at the temporal variation of this function only – which is what we’re doing now: we don’t consider the space variable x at all. So then the equation reduces to a·e–(i/ħ)·(E·t), and so… Well… Yes. We do find that our Hamiltonian coefficient Hii is like the energy of the | i 〉 state of our NH3 molecule, so we write: H11 = E1, and H22 = E2, and the ‘wavefunctions’ of our Cand Ccoefficients can be written as:

  • Ca·e(i/ħ)·(H11·t) a·e(i/ħ)·(E1·t), with H11 = E1, and
  • C= a·e(i/ħ)·(H22·t) a·e(i/ħ)·(E2·t), with H22 = E2.

But can we interpret Cand  Cas proper amplitudes? They are just coefficients in these equations, aren’t they? Well… Yes and no. From what we wrote in previous posts, you should remember that these Ccoefficients are equal to 〈 i | ψ 〉, so they are the amplitude to find our ammonia molecule in one state or the other.

Back to Feynman now. He adds, logically but brilliantly:

We note, however, that for the ammonia molecule the two states |1〉 and |2〉 have a definite symmetry. If nature is at all reasonable, the matrix elements H11 and H22 must be equal. We’ll call them both E0, because they correspond to the energy the states would have if H11 and H22 were zero.”

So our Cand Camplitudes then reduce to:

  • C〈 1 | ψ 〉 = a·e(i/ħ)·(E0·t)
  • C=〈 2 | ψ 〉 = a·e(i/ħ)·(E0·t)

We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

  • |〈 1 | ψ 〉|= |a·e(i/ħ)·(E0·t)|a
  • |〈 2 | ψ 〉|= |a·e(i/ħ)·(E0·t)|a

Now, the probabilities have to add up to 1, so a+ a= 1 and, therefore, the probability to be in either in state 1 or state 2 is 0.5, which is what we’d expect.

Note: At this point, it is probably good to get back to our | ψ 〉 = | 1 〉 C1 + | 2 〉 Cequation, so as to try to understand what it really says. Substituting the a·e(i/ħ)·(E0·t) expression for C1 and C2 yields:

| ψ 〉 = | 1 〉 a·e(i/ħ)·(E0·t) + | 2 〉 a·e(i/ħ)·(E0·t) = [| 1 〉 + | 2 〉] a·e(i/ħ)·(E0·t)

Now, what is this saying, really? In our previous post, we explained this is an ‘open’ equation, so it actually doesn’t mean all that much: we need to ‘close’ or ‘complete’ it by adding a ‘bra’, i.e. a state like 〈 χ |, so we get a 〈 χ | ψ〉 type of amplitude that we can actually do something with. Now, in this case, our final 〈 χ | state is either 〈 1 | or 〈 2 |, so we write:

  • 〈 1 | ψ 〉 = [〈 1 | 1 〉 + 〈 1 | 2 〉]·a·e(i/ħ)·(E0·t) = [1 + 0]·a·e(i/ħ)·(E0·t)· = a·e(i/ħ)·(E0·t)
  • 〈 2 | ψ 〉 = [〈 2 | 1 〉 + 〈 2 | 2 〉]·a·e(i/ħ)·(E0·t) = [0 + 1]·a·e(i/ħ)·(E0·t)· = a·e(i/ħ)·(E0·t)

Note that I finally added the multiplication dot (·) because we’re talking proper amplitudes now and, therefore, we’ve got a proper product too: we multiply one complex number with another. We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

  • |〈 1 | ψ 〉|= |a·e(i/ħ)·(E0·t)|a
  • |〈 2 | ψ 〉|= |a·e(i/ħ)·(E0·t)|a

Unsurprisingly, we find the same thing: these probabilities have to add up to 1, so a+ a= 1 and, therefore, the probability to be in state 1 or state 2 is 0.5. So the notation and the logic behind makes perfect sense. But let me get back to the lesson now.

The point is: the true meaning of a ‘stationary’ state here, is that we have non-fluctuating probabilities. So they are and remain equal to some constant, i.e. 1/2 in this case. This implies that the state of the molecule does not change: there is no way to go from state 1 to state 2 and vice versa. Indeed, if we know the molecule is in state 1, it will stay in that state. [Think about what normalization of probabilities means when we’re looking at one state only.]

You should note that these non-varying probabilities are related to the fact that the amplitudes have a non-varying magnitude. The phase of these amplitudes varies in time, of course, but their magnitude is and remains aalways. The amplitude is not being ‘enveloped’ by another curve, so to speak.

OK. That should be clear enough. Sorry I spent so much time on this, but this stuff on ‘stationary’ states comes back again and again and so I just wanted to clear that up as much as I can. Let’s get back to the story.

So we know that, what we’re describing above, is not what ammonia does really. As Feynman puts it: “The equations [i.e. the Cand Cequations above] don’t tell us what what ammonia really does. It turns out that it is possible for the nitrogen to push its way through the three hydrogens and flip to the other side. It is quite difficult; to get half-way through requires a lot of energy. How can it get through if it hasn’t got enough energy? There is some amplitude that it will penetrate the energy barrier. It is possible in quantum mechanics to sneak quickly across a region which is illegal energetically. There is, therefore, some [small] amplitude that a molecule which starts in |1〉 will get to the state |2. The coefficients H12 and H21 are not really zero.”

He adds: “Again, by symmetry, they should both be the same—at least in magnitude. In fact, we already know that, in general, Hij must be equal to the complex conjugate of Hji.”

His next step, then, is to interpreted as either a stroke of genius or, else, as unexplained. 🙂 He invokes the symmetry of the situation to boldly state that H12 is some real negative number, which he denotes as −A, which – because it’s a real number (so the imaginary part is zero) – must be equal to its complex conjugate H21. So then Feynman does this fantastic jump in logic. First, he keeps using the E0 value for H11 and H22, motivating that as follows: “If nature is at all reasonable, the matrix elements H11 and H22 must be equal, and we’ll call them both E0, because they correspond to the energy the states would have if H11 and H22 were zero.” Second, he uses that minus A value for H12 and H21. In short, the two equations and six functions are now reduced to:

equations

Solving these equations is rather boring. Feynman does it as follows:

solution

Now, what does these equations actually mean? It depends on those a and b coefficients. Looking at the solutions, the most obvious question to ask is: what if a or b are zero? If b is zero, then the second terms in both equations is zero, and so C1 and C2 are exactly the same: two amplitudes with the same temporal frequency ω = (E− A)/ħ. If a is zero, then C1 and C2 are the same too, but with opposite sign: two amplitudes with the same temporal frequency ω = (E+ A)/ħ. Squaring them – in both cases (i.e. for a = 0 or b = 0) – yields, once again, an equal and constant probability for the spin of the ammonia molecule to in the ‘up’ or ‘down’ or ‘down’. To be precise, we We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

  • For b = 0: |〈 1 | ψ 〉|= |(a/2)·e(i/ħ)·(E− A)·t|a2/4 = |〈 2 | ψ 〉|
  • For a = 0: |〈 1 | ψ 〉|=|(b/2)·e(i/ħ)·(E+ A)·t|= b2/4 = |〈 2 | ψ 〉|(the minus sign in front of b/2 is squared away)

So we get two stationary states now. Why two instead of one? Well… You need to use your imagination a bit here. They actually reflect each other: they’re the same as the one stationary state we found when assuming our nitrogen atom could not ‘flip’ from one position to the other. It’s just that the introduction of that possibility now results in a sort of ‘doublet’ of energy levels. But so we shouldn’t waste our time on this, as we want to analyze the general case, for which the probabilities to be in state 1 or state 2 do vary in time. So that’s when a and b are non-zero.

To analyze it all, we may want to start with equating t to zero. We then get:

C1

This leads us to conclude that a = b = 1, so our equations for C1(t) and C2(t) can now be written as:

C2

Remembering our rules for adding and subtracting complex conjugates (eiθ + e–iθ = 2cosθ and eiθ − e–iθ = 2sinθ), we can re-write this as:

C3

Now these amplitudes are much more interesting. Their temporal variation is defined by Ebut, on top of that, we have an envelope here: the cos(A·t/ħ) and sin(A·t/ħ) factor respectively. So their magnitude is no longer time-independent: both the phase as well as the amplitude now vary with time. What’s going on here becomes quite obvious when calculating and plotting the associated probabilities, which are

  • |C1(t)|= cos2(A·t/ħ), and
  • |C2(t)|= sin2(A·t/ħ)

respectively (note that the absolute square of i is equal to 1, not −1). The graph of these functions is depicted below.

graph

As Feynman puts it: “The probability sloshes back and forth.” Indeed, the way to think about this is that, if our ammonia molecule is in state 1, then it will not stay in that state. In fact, one can be sure the nitrogen atom is going to flip at some point in time, with the probabilities being defined by that fluctuating probability density function above. Indeed, as time goes by, the probability to be in state 2 increases, until it will effectively be in state 2. And then the cycle reverses.

Our | ψ 〉 = | 1 〉 C1 + | 2 〉 Cequation is a lot more interesting now, as we do have a proper mix of pure states now: we never really know in what state our molecule will be, as we have these ‘oscillating’ probabilities now, which we should interpret carefully.

The point to note is that the a = 0 and b = 0 solutions came with precise temporal frequencies: (E− A)/ħ and (E0 + A)/ħ respectively, which correspond to two separate energy levels: E− A and E0 + A respectively, with |A| = H12 = H21. So everything is related to everything once again: allowing the nitrogen atom to push its way through the three hydrogens, so as to flip to the other side, thereby breaking the energy barrier, is equivalent to associating two energy levels to the ammonia molecule as a whole, thereby introducing some uncertainty, or indefiniteness as to its energy, and that, in turn, gives us the amplitudes and probabilities that we’ve just calculated.

Note that the probabilities “sloshing back and forth”, or “dumping into each other” – as Feynman puts it – is the result of the varying magnitudes of our amplitudes, going up and down and, therefore, their absolute square varies too.

So… Well… That’s it as an introduction to a two-state system. There’s more to come. Ammonia is used in the ammonia maser. Now that is something that’s interesting to analyze—both from a classical as well as from a quantum-mechanical perspective. Feynman devotes a full chapter to it, so I’d say… Well… Have a look. 🙂

Post scriptum: I must assume this analysis of the NH3 molecule, with the nitrogen ‘flipping’ across the hydrogens, triggers a lot of questions, so let me try to answer some. Let me first insert the illustration once more, so you don’t have to scroll up:

dipole

The first thing that you should note is that the ‘flip’ involves a change in the center of mass position. So that requires energy, which is why we associate two different energy levels with the molecule: E+ A and E− A. However, as mentioned above, we don’t care about the nitty-gritty here: the energy barrier is likely to combine a number of factors, including electrostatic forces, as evidenced by the flip in the electric dipole moment, which is what the μ symbol here represents! Just note that the two energy levels are separated by an amount that’s equal to 2·A, rather than A and that, once again, it becomes obvious now why Feynman would prefer the Hamiltonian to be called the ‘energy matrix’, as its coefficients do represent specific energy levels, or differences between them! Now, that assumption yielded the following wavefunctions for C= 〈 1 | ψ 〉 and C= 〈 2 | ψ 〉:

  • C= 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t
  • C= 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t

Both are composite waves. To be precise, they are the sum of two component waves with a temporal frequency equal to ω= (E− A)/ħ and ω= (E+ A)/ħ respectively. [As for the minus sign in front of the second term in the wave equation for C2, −1 = e±iπ, so + (1/2)·e(i/ħ)·(E+ A)·t and – (1/2)·e(i/ħ)·(E+ A)·t are the same wavefunction: they only differ because their relative phase is shifted by ±π.]

Now, writing things this way, rather than in terms of probabilities, makes it clear that the two base states of the molecule themselves are associated with two different energy levels, so it is not like one state has more energy than the other. It’s just that the possibility of going from one state to the other requires an uncertainty about the energy, which is reflected by the energy doublet  E± A in the wavefunction of the base states. Now, if the wavefunction of the base states incorporates that energy doublet, then it is obvious that the state of the ammonia molecule, at any point in time, will also incorporate that energy doublet.

This triggers the following remark: what’s the uncertainty really? Is it an uncertainty in the energy, or is it an uncertainty in the wavefunction? I mean: we have a function relating the energy to a frequency. Introducing some uncertainty about the energy is mathematically equivalent to introducing uncertainty about the frequency. Think of it: two energy levels implies two frequencies, and vice versa. More in general, introducing n energy levels, or some continuous range of energy levels ΔE, amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. Of course, the answer is: the uncertainty is in both, so it’s in the frequency and in the energy and both are related through the wavefunction. So… In a way, we’re chasing our own tail.

Having said that, the energy may be uncertain, but it is real. It’s there, as evidenced by the fact that the ammonia molecule behaves like an atomic oscillator: we can excite it in exactly the same way as we can excite an electron inside an atom, i.e. by shining light on it. The only difference is the photon energies: to cause a transition in an atom, we use photons in the optical or ultraviolet range, and they give us the same radiation back. To cause a transition in an ammonia molecule, we only need photons with energies in the microwave range. Here, I should quickly remind you of the frequencies and energies involved. visible light is radiation in the 400–800 terahertz range and, using the E = h·f equation, we can calculate the associated energies of a photon as 1.6 to 3.2 eV. Microwave radiation – as produced in your microwave oven – is typically in the range of 1 to 2.5 gigahertz, and the associated photon energy is 4 to 10 millionths of an eV. Having illustrated the difference in terms of the energies involved, I should add that masers and lasers are based on the same physical principle: LASER and MASER stand for Light/Micro-wave Amplification by Stimulated Emission of Radiation, respectively.

So… How shall I phrase this? There’s uncertainty, but the way we are modeling that uncertainty matters. So yes, the uncertainty in the frequency of our wavefunction and the uncertainty in the energy are mathematically equivalent, but the wavefunction has a meaning that goes much beyond that. [You may want to reflect on that yourself.]

Finally, another question you may have is why would Feynman take minus A (i.e. −A) for H12 and H21. Frankly, my first thought on this was that it should have something to do with the original equation for these Hamiltonian coefficients, which also has a minus sign: Uij(t + Δt, t) = δij + Kij(t)·Δt = δij − (i/ħ)·Hij(t)·Δt. For i ≠ j, this reduces to:

Uij(t + Δt, t) = + Kij(t)·Δt = − (i/ħ)·Hij(t)·Δt

However, the answer is: it really doesn’t matter. One could write: H12 and H21 = +A, and we’d find the same equations. We’d just switch the indices 1 and 2, and the coefficients a and b. But we get the same solutions. You can figure that out yourself. Have fun with it !

Oh ! And please do let me know if some of the stuff above would trigger other questions. I am not sure if I’ll be able to answer them, but I’ll surely try, and good question always help to ensure we sort of ‘get’ this stuff in a more intuitive way. Indeed, when everything is said and done, the goal of this blog is not simply re-produce stuff, but to truly ‘get’ it, as good as we can. 🙂

Quantum math: states as vectors, and apparatuses as operators

I actually wanted to write about the Hamiltonian matrix. However, I realize that, before I can serve the plat de résistance, we need to review or introduce some more concepts and ideas. It all revolves around the same theme: working with states is like working with vectors, but so you need to know how exactly. Let’s go for it. 🙂

In my previous posts, I repeatedly said that a set of base states is like a coordinate system. A coordinate system allows us to describe (i.e. uniquely identify) vectors in an n-dimensional space: we associate a vector with a set of real numbers, like x, y and z, for example. Likewise, we can describe any state in terms of a set of complex numbers – amplitudes, really – once we’ve chosen a set of base states. We referred to this set of base states as a ‘representation’. For example, if our set of base states is +S, 0S and −S, then any state φ can be defined by the amplitudes C+ = 〈 +S | φ 〉, C0 = 〈 0S | φ 〉, and C = 〈 −S | φ 〉.

We have to choose some representation (but we are free to choose which one) because, as I demonstrated when doing a practical example (see my description of muon decay in my post on how to work with amplitudes), we’ll usually want to calculate something like the amplitude to go from one state to another – which we denoted as 〈 χ | φ 〉 – and we’ll do that by breaking it up. To be precise, we’ll write that amplitude 〈 χ | φ 〉  – i.e. the amplitude to go from state φ to state χ (you have to read this thing from right to left, like Hebrew or Arab) – as the following sum:

sum

So that’s a sum over a complete set of base states (that’s why I write all i under the summation symbol ∑). We discussed this rule in our presentation of the ‘Laws’ of quantum math.

Now we can play with this. As χ can be defined in terms of the chosen set of base states too, it’s handy to know that 〈 χ | i 〉 and 〈 i | χ 〉 are each other’s complex conjugates – we write this as: 〈 χ | i 〉 = 〈 i | χ 〉* – so if we have one, we have the other (we can also write: 〈 i | χ 〉* = 〈 χ | i 〉). In other words, if we have all Ci = 〈 i | φ 〉 and all Di = 〈 i | χ 〉, i.e. the ‘components’ of both states in terms of our base states, then we can calculate 〈 χ | φ 〉 as:

〈 χ | φ 〉 = ∑ Di*Ci = ∑〈 χ | i 〉〈 i | φ 〉,

provided we make sure we do the summation over a complete set of base states. For example, if we’re looking at the angular momentum of a spin-1/2 particle, like an electron or a proton, then we’ll have two base states, +ħ/2 and +ħ/2, so then we’ll have only two terms in our sum, but the spin number (j) of a cobalt nucleus is 7/2, so if we’d be looking at the angular momentum of a cobalt nucleus, we’ll have eight (2·j + 1) base states and, hence, eight terms when doing the sum. So it’s very much like working with vectors, indeed, and that’s why states are often referred to as state vectors. So now you know that term too. 🙂

However, the similarities run even deeper, and we’ll explore all of them in this post. You may or may not remember that your math teacher actually also defined ordinary vectors in three-dimensional space in terms of base vectors ei, defined as: e= [1, 0, 0], e= [0, 1, 0] and e= [0, 0, 1]. You may also remember that the units along the x, y and z-axis didn’t have to be the same – we could, for example, measure in cm along the x-axis, but in inches along the z-axis, even if that’s not very convenient to calculate stuff – but that it was very important to ensure that the base vectors were a set of orthogonal vectors. In any case, we’d chose our set of orthogonal base vectors and write all of our vectors as:

A = Ax·e1 + Ay·e+ Az·e3

That’s simple enough. In fact, one might say that the equation above actually defines coordinates. However, there’s another way of defining them. We can write Ax, Ay, and Az as vector dot products, aka scalar vector products (as opposed to cross products, or vector products tout court). Check it:

A= A·e1, A= A·e2, and A= A·e3.

This actually allows us to re-write the vector dot product A·B in a way you’ve probably haven’t seen before. Indeed, you’d usually calculate A·B as |A|∙|B|·cosθ = A∙B·cosθ (A and B is the magnitude of the vectors A and B respectively) or, quite simply, as AxB+ AyB+ AzBz. However, using the dot products above, we can now also write it as:

equation 2

We deliberately wrote B·A instead of Abecause, while the mathematical similarity with the

〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉

equation is obvious, B·A = A·B but 〈 χ | φ 〉 ≠ 〈 φ | χ 〉. Indeed, 〈 χ | φ 〉 and 〈 φ | χ 〉 are complex conjugates – so 〈 χ | φ 〉 = 〈 φ | χ 〉* – but they’re not equal. So we’ll have to watch the order when working with those amplitudes. That’s because we’re working with complex numbers instead of real numbers. Indeed, it’s only because the A·B dot product involves real numbers, whose complex conjugate is the same, that we have that commutativity in the real vector space. Apart from that – so apart from having to carefully check the order of our products – the correspondence is complete.

Let me mention another similarity here. As mentioned above, our base vectors ei had to be orthogonal. We can write this condition as:

ei·ej = δij, with δij = 0 if i ≠ j, and 1 if i = j.

Now, our first quantum-mechanical rule says the same:

〈 i | j 〉 = δij, with δij = 0 if i ≠ j, and 1 if i = j.

So our set of base states also has to be ‘orthogonal’, which is the term you’ll find in physics textbooks, although – as evidenced from our discussion on the base states for measuring angular momentum – one should not try to give any geometrical interpretation here: +ħ/2 and +ħ/2 (so that’s spin ‘up’ and ‘down’ respectively) are not ‘orthogonal’ in any geometric sense, indeed. It’s just that pure states, i.e. base states, are separate, which we write as: 〈 ‘up’ | ‘down’ 〉 = 〈 ‘down’ | ‘up’ 〉 = 0 and 〈 ‘up’ | ‘up’ 〉 = 〈 ‘down’ | ‘down’ 〉 = 1. It just means they are just different base states, and so it’s one or the other. For our +S, 0S and −S example, we’d have nine such amplitudes, and we can organize them in a little matrix:

def base statesIn fact, just like we defined the base vectors ei as e= [1, 0, 0], e= [0, 1, 0] and e= [0, 0, 1] respectively, we may say that the matrix above, which states exactly the same as the 〈 i | j 〉 = δij rule, can serve as a definition of what base states actually are. [Having said that, it’s obvious we like to believe that base states are more than just mathematical constructs: we’re talking reality here. The angular momentum as measured in the x-, y- or z-direction, or in whatever direction, is more than just a number.]

OK. You get this. In fact, you’re probably getting impatient because this is too simple for you. So let’s take another step. We showed that the 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | χ 〉 and B·= ∑(B·ei)(ei·A) are structurally equivalent – from a mathematical point of view, that is – but B and A are separate vectors, while 〈 χ | φ 〉 is just a complex number. Right?

Well… No. We can actually analyze the bra and the ket in the 〈 χ | φ 〉 bra-ket as separate pieces too. Moreover, we’ll show they are actually state vectors too, even if the bra, i.e. 〈 χ |, and the ket, i.e. | φ 〉, are ‘unfinished pieces’, so to speak. Let’s be bold. Let’s just cut the 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | χ 〉 by writing:

bra and ket

Huh? 

Yes. That’s the power of Dirac’s bra-ket notation: we can just drop symbols left or right. It’s quite incredible. But, of course, the question is: so what does this actually mean? Well… Don’t rack your brain. I’ll tell you. We define | φ 〉 as a state vector because we define | i 〉 as a (base) state vector. Look at it this way: we wrote the 〈 +S | φ 〉, 〈 0S | φ 〉 and 〈 −S | φ 〉 amplitudes as C+, C0, C, respectively, so we can write the equation above as:

a

So we’ve got a sum of products here, and it’s just like A = Ax·e+ Ay·e2 + Az·e3. Just substitute the Acoefficients for Ci and the ebase vectors for the | i 〉 base states. We get:

| φ 〉 = |+S〉 C+ + |0S〉 C0  + |+S〉 C

Of course, you’ll wonder what those terms mean: what does it mean to ‘multiply’ C+ (remember: C+  is some complex number) by |+S〉? Be patient. Just wait. You’ll understand when we do some examples, so when you start working with this stuff. You’ll see it all makes sense—later. 🙂

Of course, we’ll have a similar equation for | χ 〉, and so if we write 〈 χ | i 〉 as Di, then we can write | χ 〉 = ∑ | i 〉〈 χ | i 〉 as | χ 〉 = ∑ | i 〉 Di.

So what? Again: be patient. We know that 〈 χ | i 〉 = 〈 i | χ 〉*, so our second equation above becomes:

b

You’ll have two questions now. The first is the same as the one above: what does it mean to ‘multiply’, let’s say, D0* (i.e. the complex conjugate of D0, so if D= a + ib, then D0* = a − ib) with 〈0S|? The answer is the same: be patient. 🙂 Your second question is: why do I use another symbol for the index here? Why j instead of i? Well… We’ll have to re-combine stuff, so it’s better to keep things separate by using another symbol for the same index. 🙂

In fact, let’s re-combine stuff right now, in exactly the same way as we took it apart: we just write the two things right next to each other. We get the following:

c

What? Is that it? So we went through all of this hocus-pocus just to find the same equation as we started out with?

Yes. I had to take you through this so you get used to juggling all those symbols, because that’s what we’ll do in the next post. Just think about it and give yourself some time. I know you’ve probably never ever handled such exercise in symbols before – I haven’t, for sure! – but it all makes sense: we cut and paste. It’s all great! 🙂 [Oh… In case you wonder about the transition from the sum involving i and j to the sum involving i only, think about the Kronecker expression: 〈 j | i 〉 = δij, with δij = 0 if i ≠ j, and 1 if i = j, so most of the terms are zero.]

To summarize the whole discussion, note that the expression above is completely analogous with the B·= BxA+ ByA+ BzAformula. The only difference is that we’re talking complex numbers here, so we need to watch out. We have to watch the order of stuff, and we can’t use the Dnumbers themselves: we have to use their complex conjugates Di*. But, for the rest, we’re all set! 🙂 If we’ve got a set of base states, then we can define any state in terms of a set of ‘coordinates’ or ‘coefficients’ – i.e. the Ci or Di numbers for the φ or χ example above – and we can then calculate the amplitude to go from one state to another as:

d

In case you’d get confused, just take the original equation:

sum

The two equations are fully equivalent.

[…]

So we just went through all of the shit above so as to show that structural similarity with vector spaces?

Yes. It’s important. You just need to remember that we may have two, three, four, five,… or even an infinite number of base states depending on the situation we’re looking at, and what we’re trying to measure. I am sorry I had to take you through all of this. However, there’s more to come, and so you need this baggage. We’ll take the next step now, and that is to introduce the concept of an operator.

Look at the middle term in that expression above—let me copy it:

c

We’ve got three terms in that double sum (a double sum is a sum involving two indices, which is what we have here: i and j). When we have two indices like that, one thinks of matrices. That’s easy to do here, because we represented that 〈 i | j 〉 = δij equation as a matrix too! To be precise, we presented it as the identity matrix, and a simple substitution allows us to re-write our equation above as:

matrix

I must assume you’re shaking your head in disbelief now: we’ve expanded a simple amplitude into a product of three matrices now. Couldn’t we just stick to that sum, i.e that vector dot product ∑ Di*Ci? What’s next? Well… I am afraid there’s a lot more to come. :-/ For starters, we’ll take that idea of ‘putting something in the middle’ to the next level by going back to our Stern-Gerlach filters and whatever other apparatus we can think of. Let’s assume that, instead of some filter S or T, we’ve got something more complex now, which we’ll denote by A. [Don’t confuse it with our vectors: we’re talking an apparatus now, so you should imagine some beam of particles, polarized or not, entering it, going through, and coming out.]

We’ll stick to the symbols we used already, and so we’ll just assume a particle enters into the apparatus in some state φ, and that it comes out in some state χ. Continuing the example of spin-one particles, and assuming our beam has not been filtered – so, using lingo, we’d say it’s unpolarized – we’d say there’s a probability of 1/3 for being either in the ‘plus’, ‘zero’, or ‘minus’ state with respect to whatever representation we’d happen to be working with, and the related amplitudes would be 1/√3. In other words, we’d say that φ is defined by C+ = 〈 +S | φ 〉, C0 = 〈 0S | φ 〉, and C = 〈 −S | φ 〉, with C+ = C0 = C− = 1/√3. In fact, using that | φ 〉 = |+S〉 C+ + |0S〉 C0  + |+S〉 C− expression we invented above, we’d write: | φ 〉 = (1/√3)|+S〉 + (1/√3)|0S〉 C0  + (1/√3)|+S〉 C or, using ‘matrices’—just a row and a column, really:

matrix 2

However, you don’t need to worry about that now. The new big thing is the following expression:

〈 χ | A | φ〉

It looks simple enough: φ to A to χ. Right? Well… Yes and no. The question is: what do you do with this? How would we take its complex conjugate, for example? And if we know how to do that, would it be equal to 〈 φ | A | χ〉?

You guessed it: we’ll have to take it apart, but how? We’ll do this using another fantastic abstraction. Remember how we took Dirac’s 〈 χ | φ 〉 bra-ket apart by writing | φ 〉 = ∑ | i 〉〈 i | φ 〉? We just dropped the 〈 χ left and right in our 〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉 expression. We can go one step further now, and drop the φ 〉 left and right in our | φ 〉 = ∑ | i 〉〈 i | φ 〉 expression. We get the following wonderful thing:

| = ∑ | i 〉〈 i | over all base states i

With characteristic humor, Feynman calls this ‘The Great Law of Quantum Mechanics’ and, frankly, there’s actually more than one grain of truth in this. 🙂

Now, if we apply this ‘Great Law’ to our 〈 χ | A | φ〉 expression – we should apply it twice, actually – we get:

A1

As Feynman points out, it’s easy to add another apparatus in series. We just write:

B1

Just put a | bar between B and A and apply the same trick. The | bar is really like a factor 1 in multiplication. However, that’s all great fun but it doesn’t solve our problem. Our ‘Great Law’ allows us to sort of ‘resolve’ our apparatus A in terms of base states, as we now have 〈 i | A | j 〉 in the middle, rather than 〈 χ | A | φ〉 but, again, how do we work with that?

Well… The answer will surprise you. Rather than trying to break this thing up, we’ll say that the apparatus A is actually being described, or defined, by the nine 〈 i | A | j 〉 amplitudes. [There are nine for this example, but four only for the example involving spin-1/2 particles, of course.] We’ll call those amplitudes, quite simply, the matrix of amplitudes, and we’ll often denote it by Aij.

Now, I wanted to talk about operators here. The idea of an operator comes up when we’re creative again, and when we drop the 〈 χ | state from the 〈 χ | A | φ〉 expression. We write:

C1

So now we think of the particle entering the ‘apparatus’ A in the state ϕ and coming out of A in some state ψ (‘psi’). We can generalize this and think of it as an ‘operator’, which Feynman intuitively defines as follows:

The symbol A is neither an amplitude, nor a vector; it is a new kind of thing called an operator. It is something which “operates on” a state to produce a new state.”

But… Wait a minute! | ψ 〉 is not the same as 〈 χ |. Why can we do that substitution? We can only do it because any state ψ and χ are related through that other ‘Law’ of quantum math:

C2

Combining the two shows our ‘definition’ of an operator is OK. We should just note that it’s an ‘open’ equation until it is completed with a ‘bra’, i.e. a state like 〈 χ |, so as to give the 〈 χ | ψ〉 = 〈 χ | A | φ〉 type of amplitude that actually means something. In practical terms, that means our operator or our apparatus doesn’t mean much as long as we don’t measure what comes out, so then we choose some set of base states, i.e. a representation, which allows us to describe the final state, i.e. 〈 χ |.

[…]

Well… Folks, that’s it. I know this was mighty abstract, but the next posts should bring things back to earth again. I realize it’s only by working examples and doing exercises that one can get some kind of ‘feel’ for this kind of stuff, so that’s what we’ll have to go through now. 🙂

Quantum math: transformations

We’ve come a very long way. Now we’re ready for the Big Stuff. We’ll look at the rules for transforming amplitudes from one ‘base’ to ‘another’. [In quantum mechanics, however, we’ll talk about a ‘representation’, rather than a ‘base’, as we’ll reserve the latter term for a ‘base’ state.] In addition, we’ll look at how physicists model how amplitudes evolve over time using the so-called Hamiltonian matrix. So let’s go for it.

Transformations: how should we think about them?

In my previous post, I presented the following hypothetical set-up: we have an S-filter and a T-filter in series, but the T-filter at the angle α with respect to the first. In case you forgot: these ‘filters’ are modified Stern-Gerlach apparatuses, designed to split a particle beam according to the angular momentum in the direction of the gradient of the magnetic field, in which we may place masks to filter out one or more states.

tilted

The idea is illustrated in the hypothetical example below. The unpolarized beam goes through S, but we have masks blocking all particles with zero or negative spin in the z-direction, i.e. with respect to S. Hence, all particles entering the T-filter are in the +S state. Now, we assume the set-up of the T-filter is such that it filters out all particles with positive or negative spin. Hence, only particles with zero spin go through. So we’ve got something like this:

3

However, we need to be careful as what we are saying here. The T-apparatus is tilted, so the gradient of the magnetic field is different. To be precise, it’s got the same tilt as the T-filter itself (α). Hence, it will be filtering out all particles with positive or negative spin with respect to T. So, unlike what you might think at first, some fraction of the particles in the +S state will get through the T-filter, and come out in the 0T state. In fact, we know how many, because we have formulas for situations like this. To be precise, in this case, we should apply the following formula:

〈 0T | +S 〉 =  −(1/√2)·sinα

This is a real-valued amplitude. As usual, we get the following probability by taking the absolute square, so P = |−(1/√2)·sinα|= (1/2)·sin2α, which gives us the following graph of P:

graph probability

The probability varies between 0  (for α = 0 or π) and 1/2 = 0.5 (for α = π/2 or 3π/2). Now, this graph may or may not make sense to you, so you should think about it. You’ll admit it makes sense to find P = 0 for α = 0, but what about the non-zero values?

Think about what this would mean in classical terms: we’ve got a beam of particles whose angular momentum is ‘up’ in the z-direction. To be precise, this means that Jz = +ħ. [Angular momentum and the quantum of action have the same dimension: the joule·second.] So that’s the maximum value out of the three permitted values, which are +ħ, 0 and –ħ. Note that the particles here must be bosons. So you may think we’re talking photons, in practice but… Well… No. As I’ll explain in a later post, the photon is a spin-one particle but it’s quite particular, because it has no ‘zero spin’-state. Don’t worry about it here – but it’s really quite remarkable. So, instead of thinking of a photon, you should think of some composite matter particle obeying Bose-Einstein statistics. These are not so rare as you may think: all matter-particles that contain an even number of fermions – like elementary particles – have integer spin – but… Well… Their spin number is usually zero – not one. So… Well… Feynman’s particle here is somewhat theoretical – but it doesn’t matter. Let’s move on. 🙂

Let’s look at another transformation formula. More in particular, let’s look at the formula we (should) get for 〈 0T | −S 〉 as a function of α. So we change the set-up of the S-filter to ensure all particles entering T have negative spin. The formula is:

〈 0T | −S 〉 =  +(1/√2)·sinα

That gives the same probabilities: |+(1/√2)·sinα|= (1/2)·sin2α. Adding |〈 0T | +S 〉|2 and |〈 0T | −S 〉|gives us a total probability equal to sin2α, which is equal to 1 if α = π/2 or 3π/2. We may be tempted to interpret this as follows: if a particle is in the +S or −S state before entering the T-apparatus, and the T-apparatus is tilted at an angle α = π/2 or 3π/2 with respect to the S-apparatus, then this particle will come out of the T-apparatus in the 0T-state. No ambiguity here: P = 1.

Is this strange? Well… Let’s think about what it means to tilt the T-apparatus. You’ll have to admit that, if the apparatus is tilted at the angle π/2 or 3π/2, it’s going to measure the angular momentum in the x-direction. [The y-direction is the common axis of both apparatuses here.] So… Well… It’s pretty plausible, isn’t it? If all of the angular momentum is in the positive or negative z-direction, then it’s not going to have any angular momentum in the x-direction, right? And not having any angular momentum in the x-direction effectively corresponds to being in the 0T-state, right?

Oh ! Is it that easy?

Well… No! Not at all! The reasoning above shows how easy it is to be led astray. We forgot to normalize. Remember, if we integrate the probability density function over its domain, i.e. α ∈ [0, 2π], then we have to get one, as all probabilities have to add up to one. The definite integral of (1/2)·sin2α over [0, 2π] is equal to π/2 (the definite integral of the sine or cosine squared over a full cycle is equal to π), so we need to multiply this function by 2/π to get the actual probability density function, i.e. (1/π)·sin2α. It’s got the same shape, obviously, but it gives us maximum probabilities equal to 1/π ≈ 0.32 for α = π/2 or 3π/2, instead of 1/2 = 0.5.

Likewise, the sin2α function we got when adding |〈 0T | +S 〉|2 and |〈 0T | −S 〉|should also be normalized. One really needs to keep one’s wits about oneself here. What we’re saying here is that we have a particle that is either in the +S or the −S state, so let’s say that the chance is 50/50 to be in either of the two states. We then have these probabilities |〈 0T | +S 〉|2 and |〈 0T | −S 〉|2, which we calculated as (1/π)·sin2α. So the total combined probability is equal to 0.5·(1/π)·sin2α + 0.5·(1/π)·sin2α = (1/π)·sin2α. So we’re now weighing the two (1/π)·sin2α functions – and it doesn’t matter if the weights are 50/50 or 75/25 or whatever, as long as the two weights add up to one. The bottom line is: we get the same (1/π)·sin2α function for P, and the same maximum probability 1/π ≈ 0.32 for α = π/2 or 3π/2.

So we don’t get unity: P ≠ 1 for α = π/2 or 3π/2. Why not? Think about it. The classical analysis made sense, didn’t it? If the angular momentum is all in the z-direction (or in one of the two z-directions, I should say), then we cannot have any of it in the x-direction, can it? Well… The surprising answer is: yes, we can. The remarkable thing is that, in quantum physics, we actually never have all of the angular momentum in one direction. As I explained in my post on spin and angular momentum, the classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. In quantum physics, we find that the magnitude of a vector quantity, like angular momentum, or the related magnetic moment, is generally not equal to the maximum value of the component of that quantity in any direction. The general rule is that the maximum value of any component of J in whatever direction – i.e. +ħ in the example we’re discussing here – is smaller than the magnitude of J – which I calculated in the mentioned post as |J| = J = +√2·ħ ≈ 1.414·ħ, so that’s almost 1.5 times ħ! So it’s quite a bit smaller! The upshot is that we cannot associate any precise and unambiguous direction with quantities like the angular momentum J or the magnetic moment μ. So the answer is: the angular momentum can never be all in the z-direction, so we can always have some of it in the x-direction, and so that explains the amplitudes and probabilities we’re having here.

Huh?

Yep. I know. We never seem to get out of this ‘weirdness’, but then that’s how quantum physics is like. Feynman warned us upfront:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and of human intuition applies to large objects. We know how large objects will act, but things on a small scale just do not act that way. So we have to learn about them in a sort of abstract or imaginative fashion and not by connection with our direct experience.”

As I see it, quantum physics is about explaining all sorts of weird stuff, like electron interference and tunneling and what have you, so it shouldn’t surprise us that the framework is as weird as the stuff it’s trying to explain. 🙂 So… Well… All we can do is to try to go along with it, isn’t it? And so that’s what we’ll do here. 🙂

Transformations: the formulas

We need to distinguish various cases here. The first case is the case explained above: the T-apparatus shares the same y-axis – along which the particles move – but it’s tilted. To be precise, we should say that it’s rotated about the common y-axis by the angle α. That implies we can relate the x’, y’, z’ coordinate system of T to the x, y, z coordinate system of S through the following equations: z′ cosα sinα, x′ cosα − sinα, and y′ y. Then the transformation amplitudes are:

transformation 1

We used the formula for 〈 0T | +S 〉 and 〈 0T | −S 〉 above, and you can play with the formulas above by imagining the related set-up of the S and T filters, such as the one below:

1

If you do your homework (just check what formula and what set-up this corresponds to), you should find the following graph for the amplitude and the probability as a function of α: the graph is zero for α = π, but is non-zero everywhere else. As with the other example, you should think about this. It makes sense—sort of, that is. 🙂

probability 2

OK. Next case. Now we’re going to rotate the T-apparatus around the z-axis by some angle β. To illustrate what we’re doing here, we need to take a ‘top view’ of our apparatus, as shown below, which shows a rotation over 90°. More in general, for any angle β, the coordinate transformation is given by z′ z, x′ cosβ sinβ, y′ cosβ − sinβ. [So it’s quite similar to case 1: we’re only rotating the thing in a different plane.]

rotation z

The transformation amplitudes are now given by:

transformation 2

As you can see, we get complex-valued transformation amplitudes, unlike our first case, which yielded real-valued transformation amplitudes. That’s just the way it is. Nobody says transformation amplitudes have to be real-valued. On the contrary, one would expect them to be complex numbers. 🙂 Having said that, the combined set of transformation formulas is, obviously, rather remarkable. The amplitude to go from the +S state to, say, the 0T state is zero. Also, when our particle has zero spin when coming out of S, it will always have zero spin when and if it goes through T. In fact, the absolute value of those e±iβ functions is also equal to one, so they are also associated with probabilities that are equal to one: |e±iβ|2 = 12 =  1. So… Well… Those formulas are simple and weird at the same time, aren’t they? They sure give us plenty of stuff to think about, I’d say.

So what’s next? Well… Not all that much. We’re sort of done, really. Indeed, it’s just a property of space that we can get any rotation of T by combining the two rotations above. As I only want to introduce the basic concepts here, I’ll refer you to Feynman for the details of how exactly that’s being done. [He illustrates it for spin-1/2 particles in particular.] I’ll just wrap up here by generalizing our results from base states to any state.

Transformations: generalization

We mentioned a couple of times already that the base states are like a particular coordinate system: we will usually describe a state in terms of base states indeed. More in particular, choosing S as our representation, we’ll say:

The state φ is defined by the three numbers:

C+ = 〈 +S | φ 〉,

C0 = 〈 0S | φ 〉,

C = 〈 −S | φ 〉.

Now, the very same state can, of course, also be described in the ‘T system’, so then our numbers – i.e. the ‘components’ of φ – would be equal to:

C’+ = 〈 +T | φ 〉, C’0 = 〈 0T | φ 〉, and C’ = 〈 −T | φ 〉.

So how can we go from the unprimed ‘coordinates’ to the primed ones? The trick is to use the second of the three quantum math ‘Laws’ which I introduced in my previous post:

Capture

Just replace χ in [II] by +T, 0T and/or –T. More in general, if we denote +T, 0T or –T by jT, we can re-write this ‘Law’ as:

transformation 3

So the 〈 jT | iS 〉 amplitudes are those nine transformation amplitudes. Now, we can represent those nine amplitudes in a nice three-by-three matrix and, yes, we’ll call that matrix the transformation matrix. So now you know what that is.

To conclude, I should note that it’s only because we’re talking spin-one particles here that we have three base states here and, hence, three ‘components’, which we denoted by C+, C and C0, which transform the way they do when going from one representation to another, and so that is very much like what vectors do when we move to a different coordinate system, which is why spin-one particles are often referred to as ‘vector particles. [I am just mentioning this in case you’d come across the term and wonder why they’re being called that way. Now you know.] In fact, if we have three base states, in respect to whatever representation, and we define some state φ in terms of them, then we can always re-define that state in terms of the following ‘special’ set of components:

special set

The set is ‘special’ because one can show (you can do that yourself that by using those transformation laws) that these components transform exactly the way as x, y, z transform to x, y, z. But so I’ll leave at this.

[…]

Oh… What about the Hamiltonian? Well… I’ll save that for my next posts, as my posts have become longer and longer, and so it’s probably a good idea to separate them out. 🙂

Post scriptum: transformations for spin-1/2 particles

You should actually really check out that chapter of Feynman. The transformation matrices for spin-1/2 particles look different because… Well… Because there’s only two base states for spin-1/2 particles. It’s a pretty technical chapter, but then spin-1/2 particles are the ones that make up the world. 🙂

Quantum math: the rules – all of them! :-)

In my previous post, I made no compromise, and used all of the rules one needs to calculate quantum-mechanical stuff:

Capture

However, I didn’t explain them. These rules look simple enough, but let’s analyze them now. They’re simple and not at the same time, indeed.

[I] The first equation uses the Kronecker delta, which sounds fancy but it’s just a simple shorthand: δij = δji is equal to 1 if i = j, and zero if i ≠ j, with and j representing base states. Equation (I) basically says that base states are all different. For example, the angular momentum in the x-direction of a spin-1/2 particle – think of an electron or a proton – is either +ħ/2 or −ħ/2, not something in-between, or some mixture. So 〈 +x | +x 〉 = 〈 −x | −x 〉 = 1 and 〈 +x | −x 〉 = 〈 −x | +x 〉 = 0.

We’re talking base states here, of course. Base states are like a coordinate system: we settle on an x-, y- and z-axis, and a unit, and any point is defined in terms of an x-, y– and z-number. It’s the same here, except we’re talking ‘points’ in four-dimensional spacetime. To be precise, we’re talking constructs evolving in spacetime. To be even more precise, we’re talking amplitudes with a temporal as well as a spatial frequency, which we’ll often represent as:

ei·θ ei·(ω·t − k ∙x) = a·e(i/ħ)·(E·t − px)

The coefficient in front (a) is just a normalization constant, ensuring all probabilities add up to one. It may not be a constant, actually: perhaps it just ensure our amplitude stays within some kind of envelope, as illustrated below.

Photon wave

As for the ω = E/ħ and k = p/ħ identities, these are the de Broglie equations for a matter-wave, which the young Comte jotted down as part of his 1924 PhD thesis. He was inspired by the fact that the E·t − px factor is an invariant four-vector product (E·t − px = pμxμ) in relativity theory, and noted the striking similarity with the argument of any wave function in space and time (ω·t − k ∙x) and, hence, couldn’t resist equating both. Louis de Broglie was inspired, of course, by the solution to the blackbody radiation problem, which Max Planck and Einstein had convincingly solved by accepting that the ω = E/ħ equation holds for photons. As he wrote it:

“When I conceived the first basic ideas of wave mechanics in 1923–24, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905.” (Louis de Broglie, quoted in Wikipedia)

Looking back, you’d of course want the phase of a wavefunction to be some invariant quantity, and the examples we gave our previous post illustrate how one would expect energy and momentum to impact its temporal and spatial frequency. But I am digressing. Let’s look at the second equation. However, before we move on, note that minus sign in the exponent of our wavefunction: a·ei·θ. The phase turns counter-clockwise. That’s just the way it is. I’ll come back to this.

[II] The φ and χ symbols do not necessarily represent base states. In fact, Feynman illustrates this law using a variety of examples including both polarized as well as unpolarized beams, or ‘filtered’ as well as ‘unfiltered’ states, as he calls it in the context of the Stern-Gerlach apparatuses he uses to explain what’s going on. Let me summarize his argument here.

I discussed the Stern-Gerlach experiment in my post on spin and angular momentum, but the Wikipedia article on it is very good too. The principle is illustrated below: a inhomogeneous magnetic field – note the direction of the gradient ∇B = (∂B/∂x, ∂B/∂y, ∂B/∂z) – will split a beam of spin-one particles into three beams. [Matter-particles with spin one are rather rare (Lithium-6 is an example), but three states (rather than two only, as we’d have when analyzing spin-1/2 particles, such as electrons or protons) allow for more play in the analysis. 🙂 In any case, the analysis is easily generalized.]

stern-gerlach simple

The splitting of the beam is based, of course, on the quantized angular momentum in the z-direction (i.e. the direction of the gradient): its value is either ħ, 0, or −ħ. We’ll denote these base states as +, 0 or −, and we should note they are defined in regard to an apparatus with a specific orientation. If we call this apparatus S, then we can denote these base states as +S, 0S and −S respectively.

The interesting thing in Feynman’s analysis is the imagined modified Stern-Gerlach apparatus, which – I am using Feynman‘s words here 🙂 –  “puts Humpty Dumpty back together.” It looks a bit monstruous, but it’s easy enough to understand. Quoting Feynman once more: “It consists of a sequence of three high-gradient magnets. The first one (on the left) is just the usual Stern-Gerlach magnet and splits the incoming beam of spin-one particles into three separate beams. The second magnet has the same cross section as the first, but is twice as long and the polarity of its magnetic field is opposite the field in magnet 1. The second magnet pushes in the opposite direction on the atomic magnets and bends their paths back toward the axis, as shown in the trajectories drawn in the lower part of the figure. The third magnet is just like the first, and brings the three beams back together again, so that leaves the exit hole along the axis.”

stern-gerlach modified

Now, we can use this apparatus as a filter by inserting blocking masks, as illustrated below.

filter

But let’s get back to the lesson. What about the second ‘Law’ of quantum math? Well… You need to be able to imagine all kinds of situations now. The rather simple set-up below is one of them: we’ve got two of these apparatuses in series now, S and T, with T tilted at the angle α with respect to the first.

tilted

I know: you’re getting impatient. What about it? Well… We’re finally ready now. Let’s suppose we’ve got three apparatuses in series, with the first and the last one having the very same orientation, and the one in the middle being tilted. We’ll denote them by S, T and S’ respectively. We’ll also use masks: we’ll block the 0 and − state in the S-filter, like in that illustration above. In addition, we’ll block the + and − state in the T apparatus and, finally, the 0 and − state in the S’ apparatus. Now try to imagine what happens: how many particles will get through?

[…]

Just try to think about it. Make some drawing or something. Please!  

[…]

OK… The answer is shown below. Despite the filtering in S, the +S particles that come out do have an amplitude to go through the 0T-filter, and so the number of atoms that come out will be some fraction (α) of the number of atoms (N) that came out of the +S-filter. Likewise, some other fraction (β) will make it through the +S’-filter, so we end up with βαN particles.

ratio 2

Now, I am sure that, if you’d tried to guess the answer yourself, you’d have said zero rather than βαN but, thinking about it, it makes sense: it’s not because we’ve got some angular momentum in one direction that we have none in the other. When everything is said and done, we’re talking components of the total angular momentum here, don’t we? Well… Yes and no. Let’s remove the masks from T. What do we get?

[…]

Come on: what’s your guess? N?

[…] You’re right. It’s N. Perfect. It’s what’s shown below.

ratio 3

Now, that should boost your confidence. Let’s try the next scenario. We block the 0 and − state in the S-filter once again, and the + and − state in the T apparatus, so the first two apparatuses are the same as in our first example. But let’s change the S’ apparatus: let’s close the + and − state there now. Now try to imagine what happens: how many particles will get through?

[…]

Come on! You think it’s a trap, isn’t it? It’s not. It’s perfectly similar: we’ve got some other fraction here, which we’ll write as γαN, as shown below.

ratio 1Next scenario: S has the 0 and − gate closed once more, and T is fully open, so it has no masks. But, this time, we set S’ so it filters the 0-state with respect to it. What do we get? Come on! Think! Please!

[…]

The answer is zero, as shown below.

ratio 4

Does that make sense to you? Yes? Great! Because many think it’s weird: they think the T apparatus must ‘re-orient’ the angular momentum of the particles. It doesn’t: if the filter is wide open, then “no information is lost”, as Feynman puts it. Still… Have a look at it. It looks like we’re opening ‘more channels’ in the last example: the S and S’ filter are the same, indeed, and T is fully open, while it selected for 0-state particles before. But no particles come through now, while with the 0-channel, we had γαN.

Hmm… It actually is kinda weird, won’t you agree? Sorry I had to talk about this, but it will make you appreciate that second ‘Law’ now: we can always insert a ‘wide-open’ filter and, hence, split the beams into a complete set of base states − with respect to the filter, that is − and bring them back together provided our filter does not produce any unequal disturbances on the three beams. In short, the passage through the wide-open filter should not result in a change of the amplitudes. Again, as Feynman puts it: the wide-open filter should really put Humpty-Dumpty back together again. If it does, we can effectively apply our ‘Law’:

second law

For an example, I’ll refer you to my previous post. This brings me to the third and final ‘Law’.

[III] The amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ:

〈 χ | φ 〉 = 〈 φ | χ 〉*

This is probably the weirdest ‘Law’ of all, even if I should say, straight from the start, we can actually derive it from the second ‘Law’, and the fact that all probabilities have to add up to one. Indeed, a probability is the absolute square of an amplitude and, as we know, the absolute square of a complex number is also equal to the product of itself and its complex conjugate:

|z|= |z|·|z| = z·z*

[You should go through the trouble of reviewing the difference between the square and the absolute square of a complex number. Just write z as a + ib and calculate (a + ib)= a2 + 2ab+ b2 , as opposed to |z|= a2 + b2. Also check what it means when writing z as r·eiθ = r·(cosθ + i·sinθ).]

Let’s applying the probability rule to a two-filter set-up, i.e. the situation with the S and the tilted T filter which we described above, and let’s assume we’ve got a pure beam of +S particles entering the wide-open T filter, so our particles can come out in either of the three base states with respect to T. We can then write:

〈 +T | +S 〉+ 〈 0T | +S 〉+ 〈 −T | +S 〉= 1

⇔ 〈 +T | +S 〉〈 +T | +S 〉* + 〈 0T | +S 〉〈 0T | +S 〉* + 〈 −T | +S 〉〈 −T | +S 〉* = 1

Of course, we’ve got two other such equations if we start with a 0S or a −S state. Now, we take the 〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉 ‘Law’, and substitute χ and φ for +S, and all states for the base states with regard to T. We get:

〈 +S | +S 〉 = 1 = 〈 +S | +T 〉〈 +T | +S 〉 + 〈 +S | 0T 〉〈 0T | +S 〉 + 〈 +S | –T 〉〈 −T | +S 〉

These equations are consistent only if:

〈 +S | +T 〉 = 〈 +T | +S 〉*,

〈 +S | 0T 〉 = 〈 0T | +S 〉*,

〈 +S | −T 〉 = 〈 −T | +S 〉*,

which is what we wanted to prove. One can then generalize to any state φ and χ. However, proving the result is one thing. Understanding it is something else. One can write down a number of strange consequences, which all point to Feynman‘s rather enigmatic comment on this ‘Law’: “If this Law were not true, probability would not be ‘conserved’, and particles would get ‘lost’.” So what does that mean? Well… You may want to think about the following, perhaps. It’s obvious that we can write:

|〈 φ | χ 〉|= 〈 φ | χ 〉〈 φ | χ 〉* = 〈 χ | φ 〉*〈 χ | φ 〉 = |〈 χ | φ 〉|2

This says that the probability to go from the φ-state to the χ-state  is the same as the probability to go from the χ-state to the φ-state.

Now, when we’re talking base states, that’s rather obvious, because the probabilities involved are either 0 or 1. However, if we substitute for +S and −T, or some more complicated states, then it’s a different thing. My guts instinct tells me this third ‘Law’ – which, as mentioned, can be derived from the other ‘Laws’ – reflects the principle of reversibility in spacetime, which you may also interpret as a causality principle, in the sense that, in theory at least (i.e. not thinking about entropy and/or statistical mechanics), we can reverse what’s happening: we can go back in spacetime.

In this regard, we should also remember that the complex conjugate of a complex number in polar form, i.e. a complex number written as r·eiθ, is equal to r·eiθ, so the argument in the exponent gets a minus sign. Think about what this means for our a·ei·θ ei·(ω·t − k ∙x) = a·e(i/ħ)·(E·t − pxfunction. Taking the complex conjugate of this function amounts to reversing the direction of t and x which, once again, evokes that idea of going back in spacetime.

I feel there’s some more fundamental principle here at work, on which I’ll try to reflect a bit more. Perhaps we can also do something with that relationship between the multiplicative inverse of a complex number and its complex conjugate, i.e. z−1 = z*/|z|2. I’ll check it out. As for now, however, I’ll leave you to do that, and please let me know if you’ve got any inspirational ideas on this. 🙂

So… Well… Goodbye as for now. I’ll probably talk about the Hamiltonian in my next post. I think we really did a good job in laying the groundwork for the really hardcore stuff, so let’s go for that now. 🙂

Post Scriptum: On the Uncertainty Principle and other rules

After writing all of the above, I realized I should add some remarks to make this post somewhat more readable. First thing: not all of the rules are there—obviously! Most notably, I didn’t say anything about the rules for adding or multiplying amplitudes, but that’s because I wrote extensively about that already, and so I assume you’re familiar with that. [If not, see my page on the essentials.]

Second, I didn’t talk about the Uncertainty Principle. That’s because I didn’t have to. In fact, we don’t need it here. In general, all popular accounts of quantum mechanics have an excessive focus on the position and momentum of a particle, while the approach in this and my previous post is quite different. Of course, it’s Feynman’s approach to QM really. Not ‘mine’. 🙂 All of the examples and all of the theory he presents in his introductory chapters in the Third Volume of Lectures, i.e. the volume on QM, are related to things like:

  • What is the amplitude for a particle to go from spin state +S to spin state −T?
  • What is the amplitude for a particle to be scattered, by a crystal, or from some collision with another particle, in the θ direction?
  • What is the amplitude for two identical particles to be scattered in the same direction?
  • What is the amplitude for an atom to absorb or emit a photon? [See, for example, Feynman’s approach to the blackbody radiation problem.]
  • What is the amplitude to go from one place to another?

In short, you read Feynman, and it’s only at the very end of his exposé, that he starts talking about the things popular books start with, such as the amplitude of a particle to be at point (x, t) in spacetime, or the Schrödinger equation, which describes the orbital of an electron in an atom. That’s where the Uncertainty Principle comes in and, hence, one can really avoid it for quite a while. In fact, one should avoid it for quite a while, because it’s now become clear to me that simply presenting the Uncertainty Principle doesn’t help all that much to truly understand quantum mechanics.

Truly understanding quantum mechanics involves understanding all of these weird rules above. To some extent, that involves dissociating the idea of the wavefunction with our conventional ideas of time and position. From the questions above, it should be obvious that ‘the’ wavefunction does actually not exist: we’ve got a wavefunction for anything we can and possibly want to measure. That brings us to the question of the base states: what are they?

Feynman addresses this question in a rather verbose section of his Lectures titled: What are the base states of the world? I won’t copy it here, but I strongly recommend you have a look at it. 🙂

I’ll end here with a final equation that we’ll need frequently: the amplitude for a particle to go from one place (r1) to another (r2). It’s referred to as a propagator function, for obvious reasons—one of them being that physicists like fancy terminology!—and it looks like this:

propagator

The shape of the e(i/ħ)·(pr12function is now familiar to you. Note the r12 in the argument, i.e. the vector pointing from r1 to r2. The pr12 dot product equals |p|∙|r12|·cosθ = p∙r12·cosθ, with θ the angle between p and r12. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r12. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r12 is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r12|·cosθ = p∙r12. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r12/ħ. The function then reduces to (ħ/p)·eix/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while it’s going to infinity when very near. [Note that the absolute square, using cos(x)/x and sin(x)/x yields the same graph as squaring 1/x—obviously!]

graph