Should we reinvent wavefunction math?

Preliminary note: This post may cause brain damage. 🙂 If you haven’t worked yourself through a good introduction to physics – including the math – you will probably not understand what this is about. So… Well… Sorry. 😦 But if you have… Then this should be very interesting. Let’s go. 🙂

If you know one or two things about quantum math – Schrödinger’s equation and all that – then you’ll agree the math is anything but straightforward. Personally, I find the most annoying thing about wavefunction math are those transformation matrices: every time we look at the same thing from a different direction, we need to transform the wavefunction using one or more rotation matrices – and that gets quite complicated !

Now, if you have read any of my posts on this or my other blog, then you know I firmly believe the wavefunction represents something real or… Well… Perhaps it’s just the next best thing to reality: we cannot know das Ding an sich, but the wavefunction gives us everything we would want to know about it (linear or angular momentum, energy, and whatever else we have an operator for). So what am I thinking of? Let me first quote Feynman’s summary interpretation of Schrödinger’s equation (Lectures, III-16-1):

“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.”

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”. His analysis there is centered on the local conservation of energy, which makes me think Schrödinger’s equation might be an energy diffusion equation. I’ve written about this ad nauseam in the past, and so I’ll just refer you to one of my papers here for the details, and limit this post to the basics, which are as follows.

The wave equation (so that’s Schrödinger’s equation in its non-relativistic form, which is an approximation that is good enough) is written as:formula 1The resemblance with the standard diffusion equation (shown below) is, effectively, very obvious:formula 2As Feynman notes, it’s just that imaginary coefficient that makes the behavior quite different. How exactly? Well… You know we get all of those complicated electron orbitals (i.e. the various wave functions that satisfy the equation) out of Schrödinger’s differential equation. We can think of these solutions as (complex) standing waves. They basically represent some equilibrium situation, and the main characteristic of each is their energy level. I won’t dwell on this because – as mentioned above – I assume you master the math. Now, you know that – if we would want to interpret these wavefunctions as something real (which is surely what want to do!) – the real and imaginary component of a wavefunction will be perpendicular to each other. Let me copy the animation for the elementary wavefunction ψ(θ) = a·ei∙θ = a·ei∙(E/ħ)·t = a·cos[(E/ħ)∙t]  i·a·sin[(E/ħ)∙t] once more:

Circle_cos_sin

So… Well… That 90° angle makes me think of the similarity with the mathematical description of an electromagnetic wave. Let me quickly show you why. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Vψ term – which is just the equivalent of the the sink or source term S in the diffusion equation – disappears. Therefore, Schrödinger’s equation reduces to:

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

Now, the key difference with the diffusion equation – let me write it for you once again: ∂φ(x, t)/∂t = D·∇2φ(x, t) – is that Schrödinger’s 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:

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

Huh? Yes. These equations 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. [Now that we’re getting a bit technical, let me note that 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.] 🙂 OK. Onwards ! 🙂

The equations above make me 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

Now, these equations – and, I must therefore assume, the other equations above as well – effectively describe a propagation mechanism in spacetime, as illustrated below:

propagation

You know how it works for the electromagnetic field: it’s the interplay between circulation and flux. Indeed, circulation around some axis of rotation creates a flux in a direction perpendicular to it, and that flux causes this, and then that, and it all goes round and round and round. 🙂 Something like that. 🙂 I will let you look up how it goes, exactly. The principle is clear enough. 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.

Now, 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? I firmly believe they do. The obvious question then is the following: why wouldn’t we represent them as vectors, just like E and B? I mean… Representing them as vectors (I mean real vectors here – something with a magnitude and a direction in a real space – as opposed to state vectors from the Hilbert space) would show they are real, and there would be no need for cumbersome transformations when going from one representational base to another. In fact, that’s why vector notation was invented (sort of): we don’t need to worry about the coordinate frame. It’s much easier to write physical laws in vector notation because… Well… They’re the real thing, aren’t they? 🙂

What about dimensions? Well… I am not sure. However, because we are – arguably – talking about some pointlike charge moving around in those oscillating fields, I would suspect the dimension of the real and imaginary component of the wavefunction will be the same as that of the electric and magnetic field vectors E and B. We may want to recall these:

  1. E is measured in newton per coulomb (N/C).
  2. B is measured in newton per coulomb divided by m/s, so that’s (N/C)/(m/s).

The weird dimension of B is because of the weird force law for the magnetic force. It involves a vector cross product, as shown by Lorentz’ formula:

F = qE + q(v×B)

Of course, it is only one force (one and the same physical reality), as evidenced by the fact that we can 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). [Check it, because you may not have seen this expression before. Just take a piece of paper and think about the geometry of the situation.] 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, if we can agree on a suitable convention for the direction of rotation here, we may boldly write:

B = (1/c)∙ex×E = (1/c)∙iE

This is, in fact, what triggered my geometric interpretation of Schrödinger’s equation about a year ago now. I have had little time to work on it, but think I am on the right track. Of course, you should note that, for an electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneously (as shown below). So their phase is the same.

E and B

In contrast, the phase of the real and imaginary component of the wavefunction is not the same, as shown below.wavefunction

In fact, because of the Stern-Gerlach experiment, I am actually more thinking of a motion like this:

Wavefunction 2But that shouldn’t distract you. 🙂 The question here is the following: could we possibly think of a new formulation of Schrödinger’s equation – using vectors (again, real vectors – not these weird state vectors) rather than complex algebra?

I think we can, but then I wonder why the inventors of the wavefunction – Heisenberg, Born, Dirac, and Schrödinger himself, of course – never thought of that. 🙂

Hmm… I need to do some research here. 🙂

Post scriptum: You will, of course, wonder how and why the matter-wave would be different from the electromagnetic wave if my suggestion that the dimension of the wavefunction component is the same is correct. The answer is: the difference lies in the phase difference and then, most probably, the different orientation of the angular momentum. Do we have any other possibilities? 🙂

P.S. 2: I also published this post on my new blog: https://readingeinstein.blog/. However, I thought the followers of this blog should get it first. 🙂

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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).

Quantum-mechanical operators

We climbed a mountain—step by step, post by post. 🙂 We have reached the top now, and the view is gorgeous. We understand Schrödinger’s equation, which describes how amplitudes propagate through space-time. It’s the quintessential quantum-mechanical expression. Let’s enjoy now, and deepen our understanding by introducing the concept of (quantum-mechanical) operators.

The operator concept

We’ll introduce the operator concept using Schrödinger’s equation itself and, in the process, deepen our understanding of Schrödinger’s equation a bit. You’ll remember we wrote it as:

schrodinger 5

However, you’ve probably seen it like it’s written on his bust, or on his grave, or wherever, which is as follows:

simple

Grave

It’s the same thing, of course. The ‘over-dot’ is Newton’s notation for the time derivative. In fact, if you click on the picture above (and zoom in a bit), then you’ll see that the craftsman who made the stone grave marker, mistakenly, also carved a dot above the psi (ψ) on the right-hand side of the equation—but then someone pointed out his mistake and so the dot on the right-hand side isn’t painted. 🙂 The thing I want to talk about here, however, is the H in that expression above, which is, obviously, the following operator:

H

That’s a pretty monstrous operator, isn’t it? It is what it is, however: an algebraic operator (it operates on a number—albeit a complex number—unlike a matrix operator, which operates on a vector or another matrix). As you can see, it actually consists of two other (algebraic) operators:

  1. The ∇operator, which you know: it’s a differential operator. To be specific, it’s the Laplace operator, which is the divergence (·) of the gradient () of a function: ∇= · = (∂/∂x, ∂/∂y , ∂/∂z)·(∂/∂x, ∂/∂y , ∂/∂z) = ∂2/∂x2  + ∂2/∂y+ ∂2/∂z2. This too operates on our complex-valued function wavefunction ψ, and yields some other complex-valued function, which we then multiply by −ħ2/2m to get the first term.
  2. The V(x, y, z) ‘operator’, which—in this particular context—just means: “multiply with V”. Needless to say, V is the potential here, and so it captures the presence of external force fields. Also note that V is a real number, just like −ħ2/2m.

Let me say something about the dimensions here. On the left-hand side of Schrödinger’s equation, we have the product of ħ and a time derivative (is just the imaginary unit, so that’s just a (complex) number). Hence, the dimension there is [J·s]/[s] (the dimension of a time derivative is something expressed per second). So the dimension of the left-hand side is joule. On the right-hand side, we’ve got two terms. The dimension of that second-order derivative (∇2ψ) is something expressed per square meter, but then we multiply it with −ħ2/2m, whose dimension is [J2·s2]/[J/(m2/s2)]. [Remember: m = E/c2.] So that reduces to [J·m2]. Hence, the dimension of (−ħ2/2m)∇2ψ is joule. And the dimension of V is joule too, of course. So it all works out. In fact, now that we’re here, it may or may not be useful to remind you of that heat diffusion equation we discussed when introducing the basic concepts involved in vector analysis:

diffusion equation

That equation illustrated the physical significance of the Laplacian. We were talking about the flow of heat in, say, a block of metal, as illustrated below. The in the equation above is the heat per unit volume, and the h in the illustration below was the heat flow vector (so it’s got nothing to do with Planck’s constant), which depended on the material, and which we wrote as = –κT, with T the temperature, and κ (kappa) the thermal conductivity. In any case, the point is the following: the equation below illustrates the physical significance of the Laplacian. We let it operate on the temperature (i.e. a scalar function) and its product with some constant (just think of replacing κ by −ħ2/2m gives us the time derivative of q, i.e. the heat per unit volume.

heat flow

In fact, we know that is proportional to T, so if we’d choose an appropriate temperature scale – i.e. choose the zero point such that T (your physics teacher in high school would refer to as the (volume) specific heat capacity) – then we could simple write:

∂T/∂t = (κ/k)∇2T

From a mathematical point of view, that equation is just the same as ∂ψ/∂t = –(i·ħ/2m)·∇2ψ, which is Schrödinger’s equation for V = 0. In other words, you can – and actually should – also think of Schrödinger’s equation as describing the flow of… Well… What?

Well… Not sure. I am tempted to think of something like a probability density in space, but ψ represents a (complex-valued) amplitude. Having said that, you get the idea—I hope! 🙂 If not, let me paraphrase Feynman on this:

“We can think of Schrödinger’s equation as describing the diffusion of a probability amplitude from one point to another. In fact, the equation looks something like the diffusion equation we introduced when discussing heat flow, or the spreading of a gas. But there is one main difference: the imaginary coefficient in front of the time derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”

That says it all, right? 🙂 In fact, Schrödinger’s equation – as discussed here – was actually being derived when describing the motion of an electron along a line of atoms, i.e. for motion in one direction only, but you can visualize what it represents in three-dimensional space. The real exponential functions Feynman refer to exponential decay function: as the energy is spread over an ever-increasing volume, the amplitude of the wave becomes smaller and smaller. That may be the case for complex-valued exponentials as well. The key difference between a real- and complex-valued exponential decay function is that a complex exponential is a cyclical function. Now, I quickly googled to see how we could visualize that, and I like the following illustration:

decay

The dimensional analysis of Schrödinger’s equation is also quite interesting because… Well… Think of it: that heat diffusion equation incorporates the same dimensions: temperature is a measure of the average energy of the molecules. That’s really something to think about. These differential equations are not only structurally similar but, in addition, they all seem to describe some flow of energy. That’s pretty deep stuff: it relates amplitudes to energies, so we should think in terms of Poynting vectors and all that. But… Well… I need to move on, and so I will move on—so you can re-visit this later. 🙂

Now that we’ve introduced the concept of an operator, let me say something about notations, because that’s quite confusing.

Some remarks on notation

Because it’s an operator, we should actually use the hat symbol—in line with what we did when we were discussing matrix operators: we’d distinguish the matrix (e.g. A) from its use as an operator (Â). You may or may not remember we do the same in statistics: the hat symbol is supposed to distinguish the estimator (â) – i.e. some function we use to estimate a parameter (which we usually denoted by some Greek symbol, like α) – from a specific estimate of the parameter, i.e. the value (a) we get when applying â to a specific sample or observation. However, if you remember the difference, you’ll also remember that hat symbol was quickly forgotten, because the context made it clear what was what, and so we’d just write a(x) instead of â(x). So… Well… I’ll be sloppy as well here, if only because the WordPress editor only offers very few symbols with a hat! 🙂

In any case, this discussion on the use (or not) of that hat is irrelevant. In contrast, what is relevant is to realize this algebraic operator H here is very different from that other quantum-mechanical Hamiltonian operator we discussed when dealing with a finite set of base states: that H was the Hamiltonian matrix, but used in an ‘operation’ on some state. So we have the matrix operator H, and the algebraic operator H.

Confusing?

Yes and no. First, we’ve got the context again, and so you always know whether you’re looking at continuous or discrete stuff:

  1. If your ‘space’ is continuous (i.e. if states are to defined with reference to an infinite set of base states), then it’s the algebraic operator.
  2. If, on the other hand, your states are defined by some finite set of discrete base states, then it’s the Hamiltonian matrix.

There’s another, more fundamental, reason why there should be no confusion. In fact, it’s the reason why physicists use the same symbol H in the first place: despite the fact that they look so different, these two operators (i.e. H the algebraic operator and H the matrix operator) are actually equivalent. Their interpretation is similar, as evidenced from the fact that both are being referred to as the energy operator in quantum physics. The only difference is that one operates on a (state) vector, while the other operates on a continuous function. It’s just the difference between matrix mechanics as opposed to wave mechanics really.

But… Well… I am sure I’ve confused you by now—and probably very much so—and so let’s start from the start. 🙂

Matrix mechanics

Let’s start with the easy thing indeed: matrix mechanics. The matrix-mechanical approach is summarized in that set of Hamiltonian equations which, by now, you know so well:

new

If we have base states, then we have equations like this: one for each = 1, 2,… n. As for the introduction of the Hamiltonian, and the other subscript (j), just think of the description of a state:

essential

So… Well… Because we had used already, we had to introduce j. 🙂

Let’s think about |ψ〉. It is the state of a system, like the ground state of a hydrogen atom, or one of its many excited states. But… Well… It’s a bit of a weird term, really. It all depends on what you want to measure: when we’re thinking of the ground state, or an excited state, we’re thinking energy. That’s something else than thinking its position in space, for example. Always remember: a state is defined by a set of base states, and so those base states come with a certain perspective: when talking states, we’re only looking at some aspect of reality, really. Let’s continue with our example of energy states, however.

You know that the lifetime of a system in an excited state is usually short: some spontaneous or induced emission of a quantum of energy (i.e. a photon) will ensure that the system quickly returns to a less excited state, or to the ground state itself. However, you shouldn’t think of that here: we’re looking at stable systems here. To be clear: we’re looking at systems that have some definite energy—or so we think: it’s just because of the quantum-mechanical uncertainty that we’ll always measure some other different value. Does that make sense?

If it doesn’t… Well… Stop reading, because it’s only going to get even more confusing. Not my fault, however!

Psi-chology

The ubiquity of that ψ symbol (i.e. the Greek letter psi) is really something psi-chological 🙂 and, hence, very confusing, really. In matrix mechanics, our ψ would just denote a state of a system, like the energy of an electron (or, when there’s only one electron, our hydrogen atom). If it’s an electron, then we’d describe it by its orbital. In this regard, I found the following illustration from Wikipedia particularly helpful: the green orbitals show excitations of copper (Cu) orbitals on a CuOplane. [The two big arrows just illustrate the principle of X-ray spectroscopy, so it’s an X-ray probing the structure of the material.]

800px-CuO2-plane_in_high_Tc_superconductor

So… Well… We’d write ψ as |ψ〉 just to remind ourselves we’re talking of some state of the system indeed. However, quantum physicists always want to confuse you, and so they will also use the psi symbol to denote something else: they’ll use it to denote a very particular Ci amplitude (or coefficient) in that |ψ〉 = ∑|iCi formula above. To be specific, they’d replace the base states |i〉 by the continuous position variable x, and they would write the following:

Ci = ψ(i = x) = ψ(x) = Cψ(x) = C(x) = 〈x|ψ〉

In fact, that’s just like writing:

φ(p) = 〈 mom p | ψ 〉 = 〈p|ψ〉 = Cφ(p) = C(p)

What they’re doing here, is (1) reduce the ‘system‘ to a ‘particle‘ once more (which is OK, as long as you know what you’re doing) and (2) they basically state the following:

If a particle is in some state |ψ〉, then we can associate some wavefunction ψ(x) or φ(p)—with it, and that wavefunction will represent the amplitude for the system (i.e. our particle) to be at x, or to have a momentum that’s equal to p.

So what’s wrong with that? Well… Nothing. It’s just that… Well… Why don’t they use χ(x) instead of ψ(x)? That would avoid a lot of confusion, I feel: one should not use the same symbol (psi) for the |ψ〉 state and the ψ(x) wavefunction.

Huh? Yes. Think about it. The point is: the position or the momentum, or even the energy, are properties of the system, so to speak and, therefore, it’s really confusing to use the same symbol psi (ψ) to describe (1) the state of the system, in general, versus (2) the position wavefunction, which describes… Well… Some very particular aspect (or ‘state’, if you want) of the same system (in this case: its position). There’s no such problem with φ(p), so… Well… Why don’t they use χ(x) instead of ψ(x) indeed? I have only one answer: psi-chology. 🙂

In any case, there’s nothing we can do about it and… Well… In fact, that’s what this post is about: it’s about how to describe certain properties of the system. Of course, we’re talking quantum mechanics here and, hence, uncertainty, and, therefore, we’re going to talk about the average position, energy, momentum, etcetera that’s associated with a particular state of a system, or—as we’ll keep things very simple—the properties of a ‘particle’, really. Think of an electron in some orbital, indeed! 🙂

So let’s now look at that set of Hamiltonian equations once again:

new

Looking at it carefully – so just look at it once again! 🙂 – and thinking about what we did when going from the discrete to the continuous setting, we can now understand we should write the following for the continuous case:

equivalence

Of course, combining Schrödinger’s equation with the expression above implies the following:

equality

Now how can we relate that integral to the expression on the right-hand side? I’ll have to disappoint you here, as it requires a lot of math to transform that integral. It requires writing H(x, x’) in terms of rather complicated functions, including – you guessed it, didn’t you? – Dirac’s delta function. Hence, I assume you’ll believe me if I say that the matrix- and wave-mechanical approaches are actually equivalent. In any case, if you’d want to check it, you can always read Feynman yourself. 🙂

Now, I wrote this post to talk about quantum-mechanical operators, so let me do that now.

Quantum-mechanical operators

You know the concept of an operator. As mentioned above, we should put a little hat (^) on top of our Hamiltonian operator, so as to distinguish it from the matrix itself. However, as mentioned above, the difference is usually quite clear from the context. Our operators were all matrices so far, and we’d write the matrix elements of, say, some operator A, as:

Aij ≡ 〈 i | A | j 〉

The whole matrix itself, however, would usually not act on a base state but… Well… Just on some more general state ψ, to produce some new state φ, and so we’d write:

| φ 〉 = A | ψ 〉

Of course, we’d have to describe | φ 〉 in terms of the (same) set of base states and, therefore, we’d expand this expression into something like this:

operator 2

You get the idea. I should just add one more thing. You know this important property of amplitudes: the 〈 ψ | φ 〉 amplitude is the complex conjugate of the 〈 φ | ψ 〉 amplitude. It’s got to do with time reversibility, because the complex conjugate of eiθ = ei(ω·t−k·x) is equal to eiθ = ei(ω·t−k·x), so we’re just reversing the x- and tdirection. We write:

 〈 ψ | φ 〉 = 〈 φ | ψ 〉*

Now what happens if we want to take the complex conjugate when we insert a matrix, so when writing 〈 φ | A | ψ 〉 instead of 〈 φ | ψ 〉, this rules becomes:

〈 φ | A | ψ 〉* = 〈 ψ | A† | φ 〉

The dagger symbol denotes the conjugate transpose, so A† is an operator whose matrix elements are equal to Aij† = Aji*. Now, it may or may not happen that the A† matrix is actually equal to the original A matrix. In that case – and only in that case – we can write:

〈 ψ | A | φ 〉 = 〈 φ | A | ψ 〉*

We then say that A is a ‘self-adjoint’ or ‘Hermitian’ operator. That’s just a definition of a property, which the operator may or may not have—but many quantum-mechanical operators are actually Hermitian. In any case, we’re well armed now to discuss some actual operators, and we’ll start with that energy operator.

The energy operator (H)

We know the state of a system is described in terms of a set of base states. Now, our analysis of N-state systems showed we can always describe it in terms of a special set of base states, which are referred to as the states of definite energy because… Well… Because they’re associated with some definite energy. In that post, we referred to these energy levels as En (n = I, II,… N). We used boldface for the subscript n (so we wrote n instead of n) because of these Roman numerals. With each energy level, we could associate a base state, of definite energy indeed, that we wrote as |n〉. To make a long story short, we summarized our results as follows:

  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.

We’ll be working with some more subscripts in what follows, and these Roman numerals and the boldface notation are somewhat confusing (if only because I don’t want you to think of these subscripts as vectors), we’ll just denote EI, EII,…, En,…, EN as E1, E2,…, Ei,…, EN, and we’ll number the states of definite energy accordingly, also using some Greek letter so as to clearly distinguish them from all our Latin letter symbols: we’ll write these states as: |η1〉, |η1〉,… |ηN〉. [If I say, ‘we’, I mean Feynman of course. You may wonder why he doesn’t write |Ei〉, or |εi〉. The answer is: writing |En〉 would cause confusion, because this state will appear in expressions like: |Ei〉Ei, so that’s the ‘product’ of a state (|Ei〉) and the associated scalar (Ei). Too confusing. As for using η (eta) instead of ε (epsilon) to denote something that’s got to do with energy… Well… I guess he wanted to keep the resemblance with the n, and then the Ancient Greek apparently did use this η letter  for a sound like ‘e‘ so… Well… Why not? Let’s get back to the lesson.]

Using these base states of definite energy, we can write the state of the system as:

|ψ〉 = ∑ |ηi〉 C = ∑ |ηi〉〈ηi|ψ〉    over all (i = 1, 2,… , N)

Now, we didn’t talk all that much about what these base states actually mean in terms of measuring something but you’ll believe if I say that, when measuring the energy of the system, we’ll always measure one or the other E1, E2,…, Ei,…, EN value. We’ll never measure something in-between: it’s eitheror. Now, as you know, measuring something in quantum physics is supposed to be destructive but… Well… Let us imagine we could make a thousand measurements to try to determine the average energy of the system. We’d do so by counting the number of times we measure E1 (and of course we’d denote that number as N1), E2E3, etcetera. You’ll agree that we’d measure the average energy as:

E average

However, measurement is destructive, and we actually know what the expected value of this ‘average’ energy will be, because we know the probabilities of finding the system in a particular base state. That probability is equal to the absolute square of that Ccoefficient above, so we can use the P= |Ci|2 formula to write:

Eav〉 = ∑ Pi Ei over all (i = 1, 2,… , N)

Note that this is a rather general formula. It’s got nothing to do with quantum mechanics: if Ai represents the possible values of some quantity A, and Pi is the probability of getting that value, then (the expected value of) the average A will also be equal to 〈Aav〉 = ∑ Pi Ai. No rocket science here! 🙂 But let’s now apply our quantum-mechanical formulas to that 〈Eav〉 = ∑ Pi Ei formula. [Oh—and I apologize for using the same angle brackets 〈 and 〉 to denote an expected value here—sorry for that! But it’s what Feynman does—and other physicists! You see: they don’t really want you to understand stuff, and so they often use very confusing symbols.] Remembering that the absolute square of a complex number equals the product of that number and its complex conjugate, we can re-write the 〈Eav〉 = ∑ Pi Ei formula as:

Eav〉 = ∑ Pi Ei = ∑ |Ci|Ei = ∑ Ci*CEi = ∑ C*CEi = ∑ 〈ψ|ηi〉〈ηi|ψ〉E= ∑ 〈ψ|ηiEi〈ηi|ψ〉 over all i

Now, you know that Dirac’s bra-ket notation allows numerous manipulations. For example, what we could do is take out that ‘common factor’ 〈ψ|, and so we may re-write that monster above as:

Eav〉 = 〈ψ| ∑ ηiEi〈ηi|ψ〉 = 〈ψ|φ〉, with |φ〉 = ∑ |ηiEi〈ηi|ψ〉 over all i

Huh? Yes. Note the difference between |ψ〉 = ∑ |ηi〉 C = ∑ |ηi〉〈ηi|ψ〉 and |φ〉 = ∑ |ηiEi〈ηi|ψ〉. As Feynman puts it: φ is just some ‘cooked-up‘ state which you get by taking each of the base states |ηi〉 in the amount Ei〈ηi|ψ〉 (as opposed to the 〈ηi|ψ〉 amounts we took for ψ).

I know: you’re getting tired and you wonder why we need all this stuff. Just hang in there. We’re almost done. I just need to do a few more unpleasant things, one of which is to remind you that this business of the energy states being eigenstates (and the energy levels being eigenvalues) of our Hamiltonian matrix (see my post on N-state systems) comes with a number of interesting properties, including this one:

H |ηi〉 = Eii〉 = |ηiEi

Just think about what’s written here: on the left-hand side, we’re multiplying a matrix with a (base) state vector, and on the left-hand side we’re multiplying it with a scalar. So our |φ〉 = ∑ |ηiEi〈ηi|ψ〉 sum now becomes:

|φ〉 = ∑ H |ηi〉〈ηi|ψ〉 over all (i = 1, 2,… , N)

Now we can manipulate that expression some more so as to get the following:

|φ〉 = H ∑|ηi〉〈ηi|ψ〉 = H|ψ〉

Finally, we can re-combine this now with the 〈Eav〉 = 〈ψ|φ〉 equation above, and so we get the fantastic result we wanted:

Eav〉 = 〈 ψ | φ 〉 = 〈 ψ | H ψ 〉

Huh? Yes! To get the average energy, you operate on |ψ with H, and then you multiply the result with ψ|. It’s a beautiful formula. On top of that, the new formula for the average energy is not only pretty but also useful, because now we don’t need to say anything about any particular set of base states. We don’t even have to know all of the possible energy levels. When we have to calculate the average energy of some system, we only need to be able to describe the state of that system in terms of some set of base states, and we also need to know the Hamiltonian matrix for that set, of course. But if we know that, we can calculate its average energy.

You’ll say that’s not a big deal because… Well… If you know the Hamiltonian, you know everything, so… Well… Yes. You’re right: it’s less of a big deal than it seems. Having said that, the whole development above is very interesting because of something else: we can easily generalize it for other physical measurements. I call it the ‘average value’ operator idea, but you won’t find that term in any textbook. 🙂 Let me explain the idea.

The average value operator (A)

The development above illustrates how we can relate a physical observable, like the (average) energy (E), to a quantum-mechanical operator (H). Now, the development above can easily be generalized to any observable that would be proportional to the energy. It’s perfectly reasonable, for example, to assume the angular momentum – as measured in some direction, of course, which we usually refer to as the z-direction – would be proportional to the energy, and so then it would be easy to define a new operator Lz, which we’d define as the operator of the z-component of the angular momentum L. [I know… That’s a bit of a long name but… Well… You get the idea.] So we can write:

Lzav = 〈 ψ | Lψ 〉

In fact, further generalization yields the following grand result:

If a physical observable A is related to a suitable quantum-mechanical operator Â, then the average value of A for the state | ψ 〉 is given by:

Aav = 〈 ψ |  ψ 〉 = 〈 ψ | φ 〉 with | φ 〉 =  ψ 〉

At this point, you may have second thoughts, and wonder: what state | ψ 〉? The answer is: it doesn’t matter. It can be any state, as long as we’re able to describe in terms of a chosen set of base states. 🙂

OK. So far, so good. The next step is to look at how this works for the continuity case.

The energy operator for wavefunctions (H)

We can start thinking about the continuous equivalent of the 〈Eav〉 = 〈ψ|H|ψ〉 expression by first expanding it. We write:

e average continuous function

You know the continuous equivalent of a sum like this is an integral, i.e. an infinite sum. Now, because we’ve got two subscripts here (i and j), we get the following double integral:

double integral

Now, I did take my time to walk you through Feynman’s derivation of the energy operator for the discrete case, i.e. the operator when we’re dealing with matrix mechanics, but I think I can simplify my life here by just copying Feynman’s succinct development:

Feynman

Done! Given a wavefunction ψ(x), we get the average energy by doing that integral above. Now, the quantity in the braces of that integral can be written as that operator we introduced when we started this post:

H

So now we can write that integral much more elegantly. It becomes:

Eav = ∫ ψ*(xH ψ(x) dx

You’ll say that doesn’t look like 〈Eav〉 = 〈 ψ | H ψ 〉! It does. Remember that 〈 ψ | = ψ 〉*. 🙂 Done!

I should add one qualifier though: the formula above assumes our wavefunction has been normalized, so all probabilities add up to one. But that’s a minor thing. The only thing left to do now is to generalize to three dimensions. That’s easy enough. Our expression becomes a volume integral:

Eav = ∫ ψ*(rH ψ(r) dV

Of course, dV stands for dVolume here, not for any potential energy, and, of course, once again we assume all probabilities over the volume add up to 1, so all is normalized. Done! 🙂

We’re almost done with this post. What’s left is the position and momentum operator. You may think this is going to another lengthy development but… Well… It turns out the analysis is remarkably simple. Just stay with me a few more minutes and you’ll have earned your degree. 🙂

The position operator (x)

The thing we need to solve here is really easy. Look at the illustration below as representing the probability density of some particle being at x. Think about it: what’s the average position?

average position

Well? What? The (expected value of the) average position is just this simple integral: 〈xav = ∫ P(x) dx, over all the whole range of possible values for x. 🙂 That’s all. Of course, because P(x) = |ψ(x)|2 =ψ*(x)·ψ(x), this integral now becomes:

xav = ∫ ψ*(x) x ψ(x) dx

That looks exactly the same as 〈Eav = ∫ ψ*(xH ψ(x) dx, and so we can look at as an operator too!

Huh? Yes. It’s an extremely simple operator: it just means “multiply by x“. 🙂

I know you’re shaking your head now: is it that easy? It is. Moreover, the ‘matrix-mechanical equivalent’ is equally simple but, as it’s getting late here, I’ll refer you to Feynman for that. 🙂

The momentum operator (px)

Now we want to calculate the average momentum of, say, some electron. What integral would you use for that? […] Well… What? […] It’s easy: it’s the same thing as for x. We can just substitute replace for in that 〈xav = ∫ P(x) dformula, so we get:

pav = ∫ P(p) dp, over all the whole range of possible values for p

Now, you might think the rest is equally simple, and… Well… It actually is simple but there’s one additional thing in regard to the need to normalize stuff here. You’ll remember we defined a momentum wavefunction (see my post on the Uncertainty Principle), which we wrote as:

φ(p) = 〈 mom p | ψ 〉

Now, in the mentioned post, we related this momentum wavefunction to the particle’s ψ(x) = 〈x|ψ〉 wavefunction—which we should actually refer to as the position wavefunction, but everyone just calls it the particle’s wavefunction, which is a bit of a misnomer, as you can see now: a wavefunction describes some property of the system, and so we can associate several wavefunctions with the same system, really! In any case, we noted the following there:

  • The two probability density functions, φ(p) and ψ(x), look pretty much the same, but the half-width (or standard deviation) of one was inversely proportional to the half-width of the other. To be precise, we found that the constant of proportionality was equal to ħ/2, and wrote that relation as follows: σp = (ħ/2)/σx.
  • We also found that, when using a regular normal distribution function for ψ(x), we’d have to normalize the probability density function by inserting a (2πσx2)−1/2 in front of the exponential.

Now, it’s a bit of a complicated argument, but the upshot is that we cannot just write what we usually write, i.e. Pi = |Ci|2 or P(x) = |ψ(x)|2. No. We need to put a normalization factor in front, which combines the two factors I mentioned above. To be precise, we have to write:

P(p) = |〈p|ψ〉|2/(2πħ)

So… Well… Our 〈pav = ∫ P(p) dp integral can now be written as:

pav = ∫ 〈ψ|ppp|ψ〉 dp/(2πħ)

So that integral is totally like what we found for 〈xav and so… We could just leave it at that, and say we’ve solved the problem. In that sense, it is easy. However, having said that, it’s obvious we’d want some solution that’s written in terms of ψ(x), rather than in terms of φ(p), and that requires some more manipulation. I’ll refer you, once more, to Feynman for that, and I’ll just give you the result:

momentum operator

So… Well… I turns out that the momentum operator – which I tentatively denoted as px above – is not so simple as our position operator (x). Still… It’s not hugely complicated either, as we can write it as:

px ≡ (ħ/i)·(∂/∂x)

Of course, the purists amongst you will, once again, say that I should be more careful and put a hat wherever I’d need to put one so… Well… You’re right. I’ll wrap this all up by copying Feynman’s overview of the operators we just explained, and so he does use the fancy symbols. 🙂

overview

Well, folks—that’s it! Off we go! You know all about quantum physics now! We just need to work ourselves through the exercises that come with Feynman’s Lectures, and then you’re ready to go and bag a degree in physics somewhere. So… Yes… That’s what I want to do now, so I’ll be silent for quite a while now. Have fun! 🙂