Thinking again…

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

One of the comments on my other blog made me think I should, perhaps, write something on waves again. The animation below shows the elementary wavefunction ψ = a·e^−iθ= ψ = a·e^−i·θ = a·e⁻ⁱ⁽^ω·t−k·x⁾ = a·e⁻⁽ⁱ^/ħ)·^(E·^t−p·x⁾ .We know this elementary wavefunction cannot represent a real-life particle. Indeed, the a·e^−i·θfunction implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |ψ(x, t)|² = |a·e⁻⁽ⁱ^/ħ)·^(E·^t−p·x⁾|² = |a|²·|e⁻⁽ⁱ^/ħ)·^(E·^t−p·x⁾|² = |a|²·1²= a² everywhere. Hence, the particle would be everywhere – and, therefore, nowhere really. We need to localize the wave – or build a wave packet. We can do so by introducing uncertainty: we then add a potentially infinite number of these elementary wavefunctions with slightly different values for E and p, and various amplitudes a. Each of these amplitudes will then reflect the contribution to the composite wave, which – in three-dimensional space – we can write as:

ψ(r, t) = e^{−i·(E/ħ)·t}·f(r)

As I explained in previous posts (see, for example, my recent post on reality and perception), the f(r) function basically provides some envelope for the two-dimensional e^−i·θ = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation, with r = (x, y, z), θ = (E/ħ)·t = ω·t and ω = E/ħ.

Note that it looks like the wave propagates from left to right – in the positive direction of an axis which we may refer to as the x-axis. Also note this perception results from the fact that, naturally, we’d associate time with the rotation of that arrow at the center – i.e. with the motion in the illustration, while the spatial dimensions are just what they are: linear spatial dimensions. [This point is, perhaps, somewhat less self-evident than you may think at first.]

Now, the axis which points upwards is usually referred to as the z-axis, and the third and final axis – which points towards us – would then be the y-axis, obviously. Unfortunately, this definition would violate the so-called right-hand rule for defining a proper reference frame: the figures below shows the two possibilities – a left-handed and a right-handed reference frame – and it’s the right-handed reference (i.e. the illustration on the right) which we have to use in order to correctly define all directions, including the direction of rotation of the argument of the wavefunction. 400px-Cartesian_coordinate_system_handedness Hence, if we don’t change the direction of the y– and z-axes – so we keep defining the z-axis as the axis pointing upwards, and the y-axis as the axis pointing towards us – then the positive direction of the x-axis would actually be the direction from right to left, and we should say that the elementary wavefunction in the animation above seems to propagate in the negative x-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of one axis of a left-handed frame makes it right-handed, and vice versa.]

Note my language when I talk about the direction of propagation of our wave. I wrote: it looks like, or it seems to go in this or that direction. And I mean that: there is no real traveling here. At this point, you may want to review a post I wrote for my son, which explains the basic math behind waves, and in which I also explained the animation below.

wave_opposite-group-phase-velocity

Note how the peaks and troughs of this pulse seem to move leftwards, but the wave packet (or the group or the envelope of the wave—whatever you want to call it) moves to the right. The point is: the pulse itself doesn’t travel left or right. Think of the horizontal axis in the illustration above as an oscillating guitar string: each point on the string just moves up and down. Likewise, if our repeated pulse would represent a physical wave in water, for example, then the water just stays where it is: it just moves up and down. Likewise, if we shake up some rope, the rope is not going anywhere: we just started some motion that is traveling down the rope. In other words, the phase velocity is just a mathematical concept. The peaks and troughs that seem to be traveling are just mathematical points that are ‘traveling’ left or right. That’s why there’s no limit on the phase velocity: it can – and, according to quantum mechanics, actually will – exceed the speed of light. In contrast, the group velocity – which is the actual speed of the particle that is being represented by the wavefunction – may approach – or, in the case of a massless photon, will actually equal – the speed of light, but will never exceed it, and its direction will, obviously, have a physical significance as it is, effectively, the direction of travel of our particle – be it an electron, a photon (electromagnetic radiation), or whatever.

Hence, you should not think the spin of a particle – integer or half-integer – is somehow related to the direction of rotation of the argument of the elementary wavefunction. It isn’t: Nature doesn’t give a damn about our mathematical conventions, and that’s what the direction of rotation of the argument of that wavefunction is: just some mathematical convention. That’s why we write a·e⁻ⁱ⁽^ω·t−k·x⁾ rather than a·eⁱ⁽^ω·t+k·x⁾ or a·eⁱ⁽^ω·t−k·x⁾: it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the counter-clockwise direction as the positive direction of an angle. There’s nothing more to it.

OK. That’s obvious. Let me now return to my interpretation of Einstein’s E = m·c² formula (see my previous posts on this). I noted that, in the reference frame of the particle itself (see my basics page), the elementary wavefunction a·e⁻⁽ⁱ^/ħ)·^(E·^t−p·x⁾ reduces to a·e⁻⁽ⁱ^/ħ)·^(E’·^t’⁾: the origin of the reference frame then coincides with (the center of) our particle itself, and the wavefunction only varies with the time in the inertial reference frame (i.e. the proper time t’), with the rest energy of the object (E’) as the time scale factor. How should we interpret this?

Well… Energy is force times distance, and force is defined as that what causes some mass to accelerate. To be precise, the newton – as the unit of force – is defined as the magnitude of a force which would cause a mass of one kg to accelerate with one meter per second per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s per second, i.e. 1 kg·m/s². So… Because energy is force times distance, the unit of energy may be expressed in units of kg·m/s²·m, or kg·m²/s², i.e. the unit of mass times the unit of velocity squared. To sum it all up:

1 J = 1 N·m = 1 kg·(m/s)²

This reflects the physical dimensions on both sides of the E = m·c² formula again but… Well… How should we interpret this? Look at the animation below once more, and imagine the green dot is some tiny mass moving around the origin, in an equally tiny circle. We’ve got two oscillations here: each packing half of the total energy of… Well… Whatever it is that our elementary wavefunction might represent in reality – which we don’t know, of course.

circle_cos_sin

Now, the blue and the red dot – i.e. the horizontal and vertical projection of the green dot – accelerate up and down. If we look carefully, we see these dots accelerate towards the zero point and, once they’ve crossed it, they decelerate, so as to allow for a reversal of direction: the blue dot goes up, and then down. Likewise, the red dot does the same. The interplay between the two oscillations, because of the 90° phase difference, is interesting: if the blue dot is at maximum speed (near or at the origin), the red dot reverses speed (its speed is, therefore, (almost) nil), and vice versa. The metaphor of our frictionless V-2 engine, our perpetuum mobile, comes to mind once more.

The question is: what’s going on, really?

My answer is: I don’t know. I do think that, somehow, energy should be thought of as some two-dimensional oscillation of something – something which we refer to as mass, but we didn’t define mass very clearly either. It also, somehow, combines linear and rotational motion. Each of the two dimensions packs half of the energy of the particle that is being represented by our wavefunction. It is, therefore, only logical that the physical unit of both is to be expressed as a force over some distance – which is, effectively, the physical dimension of energy – or the rotational equivalent of them: torque over some angle. Indeed, the analogy between linear and angular movement is obvious: the kinetic energy of a rotating object is equal to K.E. = (1/2)·I·ω². In this formula, I is the rotational inertia – i.e. the rotational equivalent of mass – and ω is the angular velocity – i.e. the rotational equivalent of linear velocity. Noting that the (average) kinetic energy in any system must be equal to the (average) potential energy in the system, we can add both, so we get a formula which is structurally similar to the E = m·c² formula. But is it the same? Is the effective mass of some object the sum of an almost infinite number of quanta that incorporate some kind of rotational motion? And – if we use the right units – is the angular velocity of these infinitesimally small rotations effectively equal to the speed of light?

I am not sure. Not at all, really. But, so far, I can’t think of any explanation of the wavefunction that would make more sense than this one. I just need to keep trying to find better ways to articulate or imagine what might be going on. 🙂 In this regard, I’d like to add a point – which may or may not be relevant. When I talked about that guitar string, or the water wave, and wrote that each point on the string – or each water drop – just moves up and down, we should think of the physicality of the situation: when the string oscillates, its length increases. So it’s only because our string is flexible that it can vibrate between the fixed points at its ends. For a rope that’s not flexible, the end points would need to move in and out with the oscillation. Look at the illustration below, for example: the two kids who are holding rope must come closer to each other, so as to provide the necessary space inside of the oscillation for the other kid. 🙂 kid in a rope The next illustration – of how water waves actually propagate – is, perhaps, more relevant. Just think of a two-dimensional equivalent – and of the two oscillations as being transverse waves, as opposed to longitudinal. See how string theory starts making sense? 🙂

rayleighwave The most fundamental question remains the same: what is it, exactly, that is oscillating here? What is the field? It’s always some force on some charge – but what charge, exactly? Mass? What is it? Well… I don’t have the answer to that. It’s the same as asking: what is electric charge, really? So the question is: what’s the reality of mass, of electric charge, or whatever other charge that causes a force to act on it?

If you know, please let me know. 🙂

Post scriptum: The fact that we’re talking some two-dimensional oscillation here – think of a surface now – explains the probability formula: we need to square the absolute value of the amplitude to get it. And normalize, of course. Also note that, when normalizing, we’d expect to get some factor involving π somewhere, because we’re talking some circular surface – as opposed to a rectangular one. But I’ll let you figure that out. 🙂

Reality and perception

It’s quite easy to get lost in all of the math when talking quantum mechanics. In this post, I’d like to freewheel a bit. I’ll basically try to relate the wavefunction we’ve derived for the electron orbitals to the more speculative posts I wrote on how to interpret the wavefunction. So… Well… Let’s go. 🙂

If there is one thing you should remember from all of the stuff I wrote in my previous posts, then it’s that the wavefunction for an electron orbital – ψ(x, t), so that’s a complex-valued function in two variables (position and time) – can be written as the product of two functions in one variable:

ψ(x, t) = e^{−i·(E/ħ)·t}·f(x)

In fact, we wrote f(x) as ψ(x), but I told you how confusing that is: the ψ(x) and ψ(x, t) functions are, obviously, very different. To be precise, the f(x) = ψ(x) function basically provides some envelope for the two-dimensional e^iθ = e^{−i·(E/ħ)·t} = cosθ + i·sinθ oscillation – as depicted below (θ = −(E/ħ)·t = ω·t with ω = −E/ħ). Circle_cos_sin When analyzing this animation – look at the movement of the green, red and blue dots respectively – one cannot miss the equivalence between this oscillation and the movement of a mass on a spring – as depicted below. spiral_s The e^{−i·(E/ħ)·t} function just gives us two springs for the price of one. 🙂 Now, you may want to imagine some kind of elastic medium – Feynman’s famous drum-head, perhaps 🙂 – and you may also want to think of all of this in terms of superimposed waves but… Well… I’d need to review if that’s really relevant to what we’re discussing here, so I’d rather not make things too complicated and stick to basics.

First note that the amplitude of the two linear oscillations above is normalized: the maximum displacement of the object from equilibrium, in the positive or negative direction, which we may denote by x = ±A, is equal to one. Hence, the energy formula is just the sum of the potential and kinetic energy: T + U = (1/2)·A²·m·ω² = (1/2)·m·ω². But so we have two springs and, therefore, the energy in this two-dimensional oscillation is equal to E = 2·(1/2)·m·ω² = m·ω².

This formula is structurally similar to Einstein’s E = m·c² formula. Hence, one may want to assume that the energy of some particle (an electron, in our case, because we’re discussing electron orbitals here) is just the two-dimensional motion of its mass. To put it differently, we might also want to think that the oscillating real and imaginary component of our wavefunction each store one half of the total energy of our particle.

However, the interpretation of this rather bold statement is not so straightforward. First, you should note that the ω in the E = m·ω² formula is an angular velocity, as opposed to the c in the E = m·c² formula, which is a linear velocity. Angular velocities are expressed in radians per second, while linear velocities are expressed in meter per second. However, while the radian measures an angle, we know it does so by measuring a length. Hence, if our distance unit is 1 m, an angle of 2π rad will correspond to a length of 2π meter, i.e. the circumference of the unit circle. So… Well… The two velocities may not be so different after all.

There are other questions here. In fact, the other questions are probably more relevant. First, we should note that the ω in the E = m·ω² can take on any value. For a mechanical spring, ω will be a function of (1) the stiffness of the spring (which we usually denote by k, and which is typically measured in newton (N) per meter) and (2) the mass (m) on the spring. To be precise, we write: ω² = k/m – or, what amounts to the same, ω = √(k/m). Both k and m are variables and, therefore, ω can really be anything. In contrast, we know that c is a constant: c equals 299,792,458 meter per second, to be precise. So we have this rather remarkable expression: c = √(E/m), and it is valid for any particle – our electron, or the proton at the center, or our hydrogen atom as a whole. It is also valid for more complicated atoms, of course. In fact, it is valid for any system.

Hence, we need to take another look at the energy concept that is used in our ψ(x, t) = e^{−i·(E/ħ)·t}·f(x) wavefunction. You’ll remember (if not, you should) that the E here is equal to E_n= −13.6 eV, −3.4 eV, −1.5 eV and so on, for n = 1, 2, 3, etc. Hence, this energy concept is rather particular. As Feynman puts it: “The energies are negative because we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

Now, this is the one and only issue I have with the standard physics story. I mentioned it in one of my previous posts and, just for clarity, let me copy what I wrote at the time:

Feynman gives us a rather casual explanation [on choosing a zero point for measuring energy] in one of his very first Lectures on quantum mechanics, where he writes the following: “If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·e^−iωt, with ħ·ω = E = m·c². Hence, we can write the amplitude for the two states, for example as:

e^{−i(E₁/ħ)·t} and e^{−i(E₂/ħ)·t}

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be:

e^{−i(E₁+A)·t/ħ} and e^{−i(E₂+A)·t/ħ}

All of his amplitudes would be multiplied by the same factor e^{−i(A/ħ)·t}, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy M_s·c², where M_s is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount M_g·c², where M_g is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount M_g·c², where M_g is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

So… Well… There you go. If we’d want to try to interpret our ψ(x, t) = e^{−i·(E_n/ħ)·t}·f(x) function as a two-dimensional oscillation of the mass of our electron, the energy concept in it – so that’s the E_nin it – should include all pieces. Most notably, it should also include the electron’s rest energy, i.e. its energy when it is not in a bound state. This rest energy is equal to 0.511 MeV. […] Read this again: 0.511 mega-electronvolt (10⁶ eV), so that’s huge as compared to the tiny energy values we mentioned so far (−13.6 eV, −3.4 eV, −1.5 eV,…).

Of course, this gives us a rather phenomenal order of magnitude for the oscillation that we’re looking at. Let’s quickly calculate it. We need to convert to SI units, of course: 0.511 MeV is about 8.2×10⁻¹⁴ joule (J), and so the associated frequency is equal to ν = E/h = (8.2×10⁻¹⁴ J)/(6.626×10⁻³⁴ J·s) ≈ 1.23559×10²⁰ cycles per second. Now, I know such number doesn’t say all that much: just note it’s the same order of magnitude as the frequency of gamma rays and… Well… No. I won’t say more. You should try to think about this for yourself. [If you do, think – for starters – about the difference between bosons and fermions: matter-particles are fermions, and photons are bosons. Their nature is very different.]

The corresponding angular frequency is just the same number but multiplied by 2π (one cycle corresponds to 2π radians and, hence, ω = 2π·ν = 7.76344×10²⁰ rad per second. Now, if our green dot would be moving around the origin, along the circumference of our unit circle, then its horizontal and/or vertical velocity would approach the same value. Think of it. We have this e^iθ = e^{−i·(E/ħ)·t} = e^i·ω·t = cos(ω·t) + i·sin(ω·t) function, with ω = E/ħ. So the cos(ω·t) captures the motion along the horizontal axis, while the sin(ω·t) function captures the motion along the vertical axis. Now, the velocity along the horizontal axis as a function of time is given by the following formula:

v(t) = d[x(t)]/dt = d[cos(ω·t)]/dt = −ω·sin(ω·t)

Likewise, the velocity along the vertical axis is given by v(t) = d[sin(ω·t)]/dt = ω·cos(ω·t). These are interesting formulas: they show the velocity (v) along one of the two axes is always less than the angular velocity (ω). To be precise, the velocity v approaches – or, in the limit, is equal to – the angular velocity ω when ω·t is equal to ω·t = 0, π/2, π or 3π/2. So… Well… 7.76344×10²⁰ meter per second!? That’s like 2.6 trillion times the speed of light. So that’s not possible, of course!

That’s where the amplitude of our wavefunction comes in – our envelope function f(x): the green dot does not move along the unit circle. The circle is much tinier and, hence, the oscillation should not exceed the speed of light. In fact, I should probably try to prove it oscillates at the speed of light, thereby respecting Einstein’s universal formula:

c = √(E/m)

Written like this – rather than as you know it: E = m·c² – this formula shows the speed of light is just a property of spacetime, just like the ω = √(k/m) formula (or the ω = √(1/LC) formula for a resonant AC circuit) shows that ω, the natural frequency of our oscillator, is a characteristic of the system.

Am I absolutely certain of what I am writing here? No. My level of understanding of physics is still that of an undergrad. But… Well… It all makes a lot of sense, doesn’t it? 🙂

Now, I said there were a few obvious questions, and so far I answered only one. The other obvious question is why energy would appear to us as mass in motion in two dimensions only. Why is it an oscillation in a plane? We might imagine a third spring, so to speak, moving in and out from us, right? Also, energy densities are measured per unit volume, right?

Now that‘s a clever question, and I must admit I can’t answer it right now. However, I do suspect it’s got to do with the fact that the wavefunction depends on the orientation of our reference frame. If we rotate it, it changes. So it’s like we’ve lost one degree of freedom already, so only two are left. Or think of the third direction as the direction of propagation of the wave. 🙂 Also, we should re-read what we wrote about the Poynting vector for the matter wave, or what Feynman wrote about probability currents. Let me give you some appetite for that by noting that we can re-write joule per cubic meter (J/m³) as newton per square meter: J/m³ = N·m/m³ = N/m². [Remember: the unit of energy is force times distance. In fact, looking at Einstein’s formula, I’d say it’s kg·m²/s² (mass times a squared velocity), but that simplifies to the same: kg·m²/s² = [N/(m/s²)]·m²/s².]

I should probably also remind you that there is no three-dimensional equivalent of Euler’s formula, and the way the kinetic and potential energy of those two oscillations works together is rather unique. Remember I illustrated it with the image of a V-2 engine in previous posts. There is no such thing as a V-3 engine. [Well… There actually is – but not with the third cylinder being positioned sideways.] two-timer-576-px-photo-369911-s-original

But… Then… Well… Perhaps we should think of some weird combination of two V-2 engines. The illustration below shows the superposition of two one-dimensional waves – I think – one traveling east-west and back, and the other one traveling north-south and back. So, yes, we may to think of Feynman’s drum-head again – but combining two-dimensional waves – two waves that both have an imaginary as well as a real dimension

dippArticle-14

Hmm… Not sure. If we go down this path, we’d need to add a third dimension – so w’d have a super-weird V-6 engine! As mentioned above, the wavefunction does depend on our reference frame: we’re looking at stuff from a certain direction and, therefore, we can only see what goes up and down, and what goes left or right. We can’t see what comes near and what goes away from us. Also think of the particularities involved in measuring angular momentum – or the magnetic moment of some particle. We’re measuring that along one direction only! Hence, it’s probably no use to imagine we’re looking at three waves simultaneously!

In any case… I’ll let you think about all of this. I do feel I am on to something. I am convinced that my interpretation of the wavefunction as an energy propagation mechanism, or as energy itself – as a two-dimensional oscillation of mass – makes sense. 🙂

Of course, I haven’t answered one key question here: what is mass? What is that green dot – in reality, that is? At this point, we can only waffle – probably best to just give its standard definition: mass is a measure of inertia. A resistance to acceleration or deceleration, or to changing direction. But that doesn’t say much. I hate to say that – in many ways – all that I’ve learned so far has deepened the mystery, rather than solve it. The more we understand, the less we understand? But… Well… That’s all for today, folks ! Have fun working through it for yourself. 🙂

Post scriptum: I’ve simplified the wavefunction a bit. As I noted in my post on it, the complex exponential is actually equal to e^{−i·[(E/ħ)·t −}^m·φ], so we’ve got a phase shift because of m, the quantum number which denotes the z-component of the angular momentum. But that’s a minor detail that shouldn’t trouble or worry you here.

An interpretation of the wavefunction

This is my umpteenth post on the same topic. 😦 It is obvious that this search for a sensible interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:

“The teaching of quantum mechanics these days usually follows the same dogma: firstly, the student is told about the failure of classical physics at the beginning of the last century; secondly, the heroic confusions of the founding fathers are described and the student is given to understand that no humble undergraduate student could hope to actually understand quantum mechanics for himself; thirdly, a deus ex machina arrives in the form of a set of postulates (the Schrödinger equation, the collapse of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given, so that the student cannot doubt that QM is correct; fifthly, the student learns how to solve the problems that will appear on the exam paper, hopefully with as little thought as possible.”

That’s obviously not the way we want to understand quantum mechanics. [With we, I mean, me, of course, and you, if you’re reading this blog.] Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’d like to understand it. Such statements should not prevent us from trying harder. So let’s look for better metaphors. The animation below shows the two components of the archetypal wavefunction – a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (π/2).

circle_cos_sin

It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.

two-timer-576-px-photo-369911-s-original

We will also think of the center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston is not at the center when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:

Piston 1	Piston 2	Motion of Piston 1	Motion Piston 2
Top	Center	Compressed air will push piston down	Piston moves down against external pressure
Center	Bottom	Piston moves down against external pressure	External air pressure will push piston up
Bottom	Center	External air pressure will push piston up	Piston moves further up and compresses the air
Center	Top	Piston moves further up and compresses the air	Compressed air will push piston down

When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealized limit situation. If not, check Feynman’s description of the harmonic oscillator.]

The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energy being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total energy, potential and kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives kinetic energy from the rotating mass of the shaft.

Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got two oscillators, our picture changes, and so we may have to adjust our thinking here.

Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillator—but the formulas are basically the same for any harmonic oscillator.

energy harmonic oscillator

The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a) is the maximum amplitude. Of course, you will also recognize ω₀ as the natural frequency of the oscillator, and Δ as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to π/2 and, hence, Δ would be 0 for one oscillator, and –π/2 for the other. [The formula to apply here is sinθ = cos(θ – π/2).] Also note that we can equate our θ argument to ω₀·t. Now, if a = 1 (which is the case here), then these formulas simplify to:

K.E. = T = m·v²/2 = m·ω₀²·sin²(θ + Δ) = m·ω₀²·sin²(ω₀·t + Δ)
P.E. = U = k·x²/2 = k·cos²(θ + Δ)

The coefficient k in the potential energy formula characterizes the force: F = −k·x. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to m·ω₀²., so that gives us the famous T + U = m·ω₀²/2 formula or, including a once again, T + U = m·a²·ω₀²/2.

Now, if we normalize our functions by equating k to one (k = 1), then the motion of our first oscillator is given by the cosθ function, 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)/dt = 2∙sinθ∙cosθ

Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by the sinθ function which, as mentioned above, 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 rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!

Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!

How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = m·c² equation, and it may not have any direction (when everything is said and done, it’s a scalar quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able to square this circle. 🙂

Schrödinger’s equation

Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of Schrödinger’s equation, which we write as:

i·ħ∙∂ψ/∂t = –(ħ²/2m)∙∇²ψ + V·ψ

We can do a quick dimensional analysis of both sides:

[i·ħ∙∂ψ/∂t] = N∙m∙s/s = N∙m
[–(ħ²/2m)∙∇²ψ] = N∙m³/m² = N∙m
[V·ψ] = N∙m

Note the dimension of the ‘diffusion’ constant ħ²/2m: [ħ²/2m] = N²∙m²∙s²/kg = N²∙m²∙s²/(N·s²/m) = N∙m³. Also note that, in order for the dimensions to come out alright, the dimension of V – the potential – must be that of energy. Hence, Feynman’s description of it as the potential energy – rather than the potential tout court – is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.

Now, Schrödinger’s equation – without the V·ψ term – can be written as the following pair of equations:

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

This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… ψ is a complex function, which we can write as a + i∙b. Likewise, ∂ψ/∂t is a complex function, which we can write as c + i∙d, and ∇²ψ can then be written as e + i∙f. If we temporarily forget about the coefficients (ħ, ħ²/m and V), then Schrödinger’s equation – including V·ψ term – amounts to writing something like this:

i∙(c + i∙d) = –(e + i∙f) + (a + i∙b) ⇔ a + i∙b = i∙c − d + e+ i∙f ⇔ a = −d + e and b = c + f

Hence, we can now write:

V∙Re(ψ) = −ħ∙Im(∂ψ/∂t) + (1/2)∙( ħ²/m)∙Re(∇²ψ)
V∙Im(ψ) = ħ∙Re(∂ψ/∂t) + (1/2)∙( ħ²/m)∙Im(∇²ψ)

This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:

V∙Re(ψ) = −ħ∙∂Im (ψ)/∂t + (1/2)∙( ħ²/m)∙∇²Re(ψ)
V∙Im(ψ) = ħ∙∂Re(ψ)/∂t + (1/2)∙( ħ²/m)∙∇²Im(ψ)

This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = –(ħ²/2m)∙∇² + V· (note the multiplication sign after the V, which we do not have – for obvious reasons – for the –(ħ²/2m)∙∇² expression):

H[Re (ψ)] = −ħ∙∂Im(ψ)/∂t
H[Im(ψ)] = ħ∙∂Re(ψ)/∂t

A dimensional analysis shows us both sides are, once again, expressed in N∙m. It’s a beautiful expression because – if we write the real and imaginary part of ψ as r∙cosθ and r∙sinθ, we get:

H[cosθ] = −ħ∙∂sinθ/∂t = E∙cosθ
H[sinθ] = ħ∙∂cosθ/∂t = E∙sinθ

Indeed, θ = (E∙t − p∙x)/ħ and, hence, −ħ∙∂sinθ/∂t = ħ∙cosθ∙E/ħ = E∙cosθ and ħ∙∂cosθ/∂t = ħ∙sinθ∙E/ħ = E∙sinθ. Now we can combine the two equations in one equation again and write:

H[r∙(cosθ + i∙sinθ)] = r∙(E∙cosθ + i∙sinθ) ⇔ H[ψ] = E∙ψ

The operator H – applied to the wavefunction – gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?

Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… 🙂

Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.

We may, therefore, say that three-dimensional space is actually being implied by the geometry of our V-2 engine. Now that is interesting, isn’t it? 🙂

Re-visiting the matter wave (II)

Pre-scriptum (dated 26 June 2020): This post did not suffer too much from the DMCA take-down of some material: only one or two illustrations from Feynman’s Lectures were removed. It is, therefore, still quite readable—even if my views on these matters have evolved quite a bit as part of my realist interpretation of QM. I have actually re-written Feynman’s first lectures of quantum mechanics to replace de Broglie’s concept of the matter-wave with what I think is a much better description of ‘wavicles’: one that fully captures their duality. That paper has got the same title (Quantum Behavior) as Feynman’s first lecture but you will see it is a totally different animal. 🙂

So you should probably not read the post below but my lecture(s) instead. 🙂 This is the link to the first one, and here you can look at the second one. Both taken together are an alternative treatment of the subject-matter which Feynman’s discusses in Lecture 1 to 9 of his Lectures (I use a big L for his lectures to show the required reverence for all of the Mystery Wallahs—Feynman included). Let me know what you think of them (I mean the lectures here, not the mystery wallahs).

Original post:

My previous post was, once again, littered with formulas – even if I had not intended it to be that way: I want to convey some kind of understanding of what an electron – or any particle at the atomic scale – actually is – with the minimum number of formulas necessary.

We know particles display wave behavior: when an electron beam encounters an obstacle or a slit that is somewhat comparable in size to its wavelength, we’ll observe diffraction, or interference. [I have to insert a quick note on terminology here: the terms diffraction and interference are often used interchangeably, but there is a tendency to use interference when we have more than one wave source and diffraction when there is only one wave source. However, I’ll immediately add that distinction is somewhat artificial. Do we have one or two wave sources in a double-slit experiment? There is one beam but the two slits break it up in two and, hence, we would call it interference. If it’s only one slit, there is also an interference pattern, but the phenomenon will be referred to as diffraction.]

We also know that the wavelength we are talking about it here is not the wavelength of some electromagnetic wave, like light. It’s the wavelength of a de Broglie wave, i.e. a matter wave: such wave is represented by an (oscillating) complex number – so we need to keep track of a real and an imaginary part – representing a so-called probability amplitude Ψ(x, t) whose modulus squared (│Ψ(x, t)│²) is the probability of actually detecting the electron at point x at time t. [The purists will say that complex numbers can’t oscillate – but I am sure you get the idea.]

You should read the phrase above twice: we cannot know where the electron actually is. We can only calculate probabilities (and, of course, compare them with the probabilities we get from experiments). Hence, when the wave function tells us the probability is greatest at point x at time t, then we may be lucky when we actually probe point x at time t and find it there, but it may also not be there. In fact, the probability of finding it exactly at some point x at some definite time t is zero. That’s just a characteristic of such probability density functions: we need to probe some region Δx in some time interval Δt.

If you think that is not very satisfactory, there’s actually a very common-sense explanation that has nothing to do with quantum mechanics: our scientific instruments do not allow us to go beyond a certain scale anyway. Indeed, the resolution of the best electron microscopes, for example, is some 50 picometer (1 pm = 1×10^–12 m): that’s small (and resolutions get higher by the year), but so it implies that we are not looking at points – as defined in math that is: so that’s something with zero dimension – but at pixels of size Δx = 50×10^–12 m.

The same goes for time. Time is measured by atomic clocks nowadays but even these clocks do ‘tick’, and these ‘ticks’ are discrete. Atomic clocks take advantage of the property of atoms to resonate at extremely consistent frequencies. I’ll say something more about resonance soon – because it’s very relevant for what I am writing about in this post – but, for the moment, just note that, for example, Caesium-133 (which was used to build the first atomic clock) oscillates at 9,192,631,770 cycles per second. In fact, the International Bureau of Standards and Weights re-defined the (time) second in 1967 to correspond to “the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the Caesium-133 atom at rest at a temperature of 0 K.”

Don’t worry about it: the point to note is that when it comes to measuring time, we also have an uncertainty. Now, when using this Caesium-133 atomic clock, this uncertainty would be in the range of ±9.2×10^–9 seconds (so that’s nanoseconds: 1 ns = 1×10^–9 s), because that’s the rate at which this clock ‘ticks’. However, there are other (much more plausible) ways of measuring time: some of the unstable baryons have lifetimes in the range of a few picoseconds only (1 ps = 1×10^–12 s) and the really unstable ones – known as baryon resonances – have lifetimes in the 1×10^–22 to 1×10^–24 s range. This we can only measure because they leave some trace after these particle collisions in particle accelerators and, because we have some idea about their speed, we can calculate their lifetime from the (limited) distance they travel before disintegrating. The thing to remember is that for time also, we have to make do with time pixels instead of time points, so there is a Δt as well. [In case you wonder what baryons are: they are particles consisting of three quarks, and the proton and the neutron are the most prominent (and most stable) representatives of this family of particles.]

So what’s the size of an electron? Well… It depends. We need to distinguish two very different things: (1) the size of the area where we are likely to find the electron, and (2) the size of the electron itself. Let’s start with the latter, because that’s the easiest question to answer: there is a so-called classical electron radius r_e, which is also known as the Thompson scattering length, which has been calculated as:

$r_\mathrm{e} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{m_{\mathrm{e}} c^2} = 2.817 940 3267(27) \times 10^{-15} \mathrm{m}$ As for the constants in this formula, you know these by now: the speed of light c, the electron charge e, its mass m_e, and the permittivity of free space ε_e. For whatever it’s worth (because you should note that, in quantum mechanics, electrons do not have a size: they are treated as point-like particles, so they have a point charge and zero dimension), that’s small. It’s in the femtometer range (1 fm = 1×10^–15 m). You may or may not remember that the size of a proton is in the femtometer range as well – 1.7 fm to be precise – and we had a femtometer size estimate for quarks as well: 0.7 m. So we have the rather remarkable result that the much heavier proton (its rest mass is 938 MeV/c²sas opposed to only 0.511 MeV MeV/c², so the proton is 1835 times heavier) is 1.65 times smaller than the electron. That’s something to be explored later: for the moment, we’ll just assume the electron wiggles around a bit more – exactly because it’s lighter. Here you just have to note that this ‘classical’ electron radius does measure something: it’s something ‘hard’ and ‘real’ because it scatters, absorbs or deflects photons (and/or other particles). In one of my previous posts, I explained how particle accelerators probe things at the femtometer scale, so I’ll refer you to that post (End of the Road to Reality?) and move on to the next question.

The question concerning the area where we are likely to detect the electron is more interesting in light of the topic of this post (the nature of these matter waves). It is given by that wave function and, from my previous post, you’ll remember that we’re talking the nanometer scale here (1 nm = 1×10^–9 m), so that’s a million times larger than the femtometer scale. Indeed, we’ve calculated a de Broglie wavelength of 0.33 nanometer for relatively slow-moving electrons (electrons in orbit), and the slits used in single- or double-slit experiments with electrons are also nanotechnology. In fact, now that we are here, it’s probably good to look at those experiments in detail.

The illustration below relates the actual experimental set-up of a double-slit experiment performed in 2012 to Feynman’s 1965 thought experiment. Indeed, in 1965, the nanotechnology you need for this kind of experiment was not yet available, although the phenomenon of electron diffraction had been confirmed experimentally already in 1925 in the famous Davisson-Germer experiment. [It’s famous not only because electron diffraction was a weird thing to contemplate at the time but also because it confirmed the de Broglie hypothesis only two years after Louis de Broglie had advanced it!]. But so here is the experiment which Feynman thought would never be possible because of technology constraints:

The insert in the upper-left corner shows the two slits: they are each 50 nanometer wide (50×10^–9 m) and 4 micrometer tall (4×10^–6 m). [The thing in the middle of the slits is just a little support. Please do take a few seconds to contemplate the technology behind this feat: 50 nm is 50 millionths of a millimeter. Try to imagine dividing one millimeter in ten, and then one of these tenths in ten again, and again, and once again, again, and again. You just can’t imagine that, because our mind is used to addition/subtraction and – to some extent – with multiplication/division: our mind can’t deal with with exponentiation really – because it’s not a everyday phenomenon.] The second inset (in the upper-right corner) shows the mask that can be moved to close one or both slits partially or completely.

Now, 50 nanometer is 150 times larger than the 0.33 nanometer range we got for ‘our’ electron, but it’s small enough to show diffraction and/or interference. [In fact, in this experiment (done by Bach, Pope, Liou and Batelaan from the University of Nebraska-Lincoln less than two years ago indeed), the beam consisted of electrons with an (average) energy of 600 eV and a de Broglie wavelength of 50 picometer. So that’s like the electrons used in electron microscopes. 50 pm is 6.6 times smaller than the 0.33 nm wavelength we calculated for our low-energy (70 eV) electron – but then the energy and the fact these electrons are guided in electromagnetic fields explain the difference. Let’s go to the results.

The illustration below shows the predicted pattern next to the observed pattern for the two scenarios:

We first close slit 2, let a lot of electrons go through it, and so we get a pattern described by the probability density function P₁ = │Φ₁│². Here we see no interference but a typical diffraction pattern: the intensity follows a more or less normal (i.e. Gaussian) distribution. We then close slit 1 (and open slit 2 again), again let a lot of electrons through, and get a pattern described by the probability density function P₂ = │Φ₂│². So that’s how we get P₁ and P₂.
We then open both slits, let a whole electrons through, and get according to the pattern described by probability density function P₁₂ = │Φ₁+Φ₂│², which we get not from adding the probabilities P₁ and P₂(hence, P₁₂ ≠ P₁ + P₂) – as one would expect if electrons would behave like particles – but from adding the probability amplitudes. We have interference, rather than diffraction.

But so what exactly is interfering? Well… The electrons. But that can’t be, can it?

The electrons are obviously particles, as evidenced from the impact they make – one by one – as they hit the screen as shown below. [If you want to know what screen, let me quote the researchers: “The resulting patterns were magnified by an electrostatic quadrupole lens and imaged on a two-dimensional microchannel plate and phosphorus screen, then recorded with a charge-coupled device camera. […] To study the build-up of the diffraction pattern, each electron was localized using a “blob” detection scheme: each detection was replaced by a blob, whose size represents the error in the localization of the detection scheme. The blobs were compiled together to form the electron diffraction patterns.” So there you go.]

Look carefully at how this interference pattern becomes ‘reality’ as the electrons hit the screen one by one. And then say it: WAW !

Indeed, as predicted by Feynman (and any other physics professor at the time), even if the electrons go through the slits one by one, they will interfere – with themselves so to speak. [In case you wonder if these electrons really went through one by one, let me quote the researchers once again: “The electron source’s intensity was reduced so that the electron detection rate in the pattern was about 1 Hz. At this rate and kinetic energy, the average distance between consecutive electrons was 2.3 × 106 meters. This ensures that only one electron is present in the 1 meter long system at any one time, thus eliminating electron-electron interactions.” You don’t need to be a scientist or engineer to understand that, isn’t it?]

While this is very spooky, I have not seen any better way to describe the reality of the de Broglie wave: the particle is not some point-like thing but a matter wave, as evidenced from the fact that it does interfere with itself when forced to move through two slits – or through one slit, as evidenced by the diffraction patterns built up in this experiment when closing one of the two slits: the electrons went through one by one as well!

But so how does it relate to the characteristics of that wave packet which I described in my previous post? Let me sum up the salient conclusions from that discussion:

The wavelength λ of a wave packet is calculated directly from the momentum by using de Broglie‘s second relation: λ = h/p. In this case, the wavelength of the electrons averaged 50 picometer. That’s relatively small as compared to the width of the slit (50 nm) – a thousand times smaller actually! – but, as evidenced by the experiment, it’s small enough to show the ‘reality’ of the de Broglie wave.
From a math point (but, of course, Nature does not care about our math), we can decompose the wave packet in a finite or infinite number of component waves. Such decomposition is referred to, in the first case (finite number of composite waves or discrete calculus) as a Fourier analysis, or, in the second case, as a Fourier transform. A Fourier transform maps our (continuous) wave function, Ψ(x), to a (continuous) wave function in the momentum space, which we noted as φ(p). [In fact, we noted it as Φ(p) but I don’t want to create confusion with the Φ symbol used in the experiment, which is actually the wave function in space, so Ψ(x) is Φ(x) in the experiment – if you know what I mean.] The point to note is that uncertainty about momentum is related to uncertainty about position. In this case, we’ll have pretty standard electrons (so not much variation in momentum), and so the location of the wave packet in space should be fairly precise as well.
The group velocity of the wave packet (v_g) – i.e. the envelope in which our Ψ wave oscillates – equals the speed of our electron (v), but the phase velocity (i.e. the speed of our Ψ wave itself) is superluminal: we showed it’s equal to (v_p) = E/p = c²/v = c/β, with β = v/c, so that’s the ratio of the speed of our electron and the speed of light. Hence, the phase velocity will always be superluminal but will approach c as the speed of our particle approaches c. For slow-moving particles, we get astonishing values for the phase velocity, like more than a hundred times the speed of light for the electron we looked at in our previous post. That’s weird but it does not contradict relativity: if it helps, one can think of the wave packet as a modulation of an incredibly fast-moving ‘carrier wave’.

Is any of this relevant? Does it help you to imagine what the electron actually is? Or what that matter wave actually is? Probably not. You will still wonder: How does it look like? What is it in reality?

That’s hard to say. If the experiment above does not convey any ‘reality’ according to you, then perhaps the illustration below will help. It’s one I have used in another post too (An Easy Piece: Introducing Quantum Mechanics and the Wave Function). I took it from Wikipedia, and it represents “the (likely) space in which a single electron on the 5d atomic orbital of an atom would be found.” The solid body shows the places where the electron’s probability density (so that’s the squared modulus of the probability amplitude) is above a certain value – so it’s basically the area where the likelihood of finding the electron is higher than elsewhere. The hue on the colored surface shows the complex phase of the wave function.

So… Does this help?

You will wonder why the shape is so complicated (but it’s beautiful, isn’t it?) but that has to do with quantum-mechanical calculations involving quantum-mechanical quantities such as spin and other machinery which I don’t master (yet). I think there’s always a bit of a gap between ‘first principles’ in physics and the ‘model’ of a real-life situation (like a real-life electron in this case), but it’s surely the case in quantum mechanics! That being said, when looking at the illustration above, you should be aware of the fact that you are actually looking at a 3D representation of the wave function of an electron in orbit.

Indeed, wave functions of electrons in orbit are somewhat less random than – let’s say – the wave function of one of those baryon resonances I mentioned above. As mentioned in my Not So Easy Piece, in which I introduced the Schrödinger equation (i.e. one of my previous posts), they are solutions of a second-order partial differential equation – known as the Schrödinger wave equation indeed – which basically incorporates one key condition: these solutions – which are (atomic or molecular) ‘orbitals’ indeed – have to correspond to so-called stationary states or standing waves. Now what’s the ‘reality’ of that?

The illustration below comes from Wikipedia once again (Wikipedia is an incredible resource for autodidacts like me indeed) and so you can check the article (on stationary states) for more details if needed. Let me just summarize the basics:

A stationary state is called stationary because the system remains in the same ‘state’ independent of time. That does not mean the wave function is stationary. On the contrary, the wave function changes as function of both time and space – Ψ = Ψ(x, t) remember? – but it represents a so-called standing wave.
Each of these possible states corresponds to an energy state, which is given through the de Broglie relation: E = hf. So the energy of the state is proportional to the oscillation frequency of the (standing) wave, and Planck’s constant is the factor of proportionality. From a formal point of view, that’s actually the one and only condition we impose on the ‘system’, and so it immediately yields the so-called time-independent Schrödinger equation, which I briefly explained in the above-mentioned Not So Easy Piece (but I will not write it down here because it would only confuse you even more). Just look at these so-called harmonic oscillators below:

A and B represent a harmonic oscillator in classical mechanics: a ball with some mass m (mass is a measure for inertia, remember?) on a spring oscillating back and forth. In case you’d wonder what the difference is between the two: both the amplitude as well as the frequency of the movement are different. 🙂 A spring and a ball?

It represents a simple system. A harmonic oscillation is basically a resonance phenomenon: springs, electric circuits,… anything that swings, moves or oscillates (including large-scale things such as bridges and what have you – in his 1965 Lectures (Vol. I-23), Feynman even discusses resonance phenomena in the atmosphere in his Lectures) has some natural frequency ω₀, also referred to as the resonance frequency, at which it oscillates naturally indeed: that means it requires (relatively) little energy to keep it going. How much energy it takes exactly to keep them going depends on the frictional forces involved: because the springs in A and B keep going, there’s obviously no friction involved at all. [In physics, we say there is no damping.] However, both springs do have a different k (that’s the key characteristic of a spring in Hooke’s Law, which describes how springs work), and the mass m of the ball might be different as well. Now, one can show that the period of this ‘natural’ movement will be equal to t₀ = 2π/ω₀= 2π(m/k)^1/2or that ω₀= (m/k)^–1/2. So we’ve got a A and a B situation which differ in k and m. Let’s go to the so-called quantum oscillator, illustrations C to H.

C to H in the illustration are six possible solutions to the Schrödinger Equation for this situation. The horizontal axis is position (and so time is the variable) – but we could switch the two independent variables easily: as I said a number of times already, time and space are interchangeable in the argument representing the phase (θ) of a wave provided we use the right units (e.g. light-seconds for distance and seconds for time): θ = ωt – kx. Apart from the nice animation, the other great thing about these illustrations – and the main difference with resonance frequencies in the classical world – is that they show both the real part (blue) as well as the imaginary part (red) of the wave function as a function of space (fixed in the x axis) and time (the animation).

Is this ‘real’ enough? If it isn’t, I know of no way to make it any more ‘real’. Indeed, that’s key to understanding the nature of matter waves: we have to come to terms with the idea that these strange fluctuating mathematical quantities actually represent something. What? Well… The spooky thing that leads to the above-mentioned experimental results: electron diffraction and interference.

Let’s explore this quantum oscillator some more. Another key difference between natural frequencies in atomic physics (so the atomic scale) and resonance phenomena in ‘the big world’ is that there is more than one possibility: each of the six possible states above corresponds to a solution and an energy state indeed, which is given through the de Broglie relation: E = hf. However, in order to be fully complete, I have to mention that, while G and H are also solutions to the wave equation, they are actually not stationary states. The illustration below – which I took from the same Wikipedia article on stationary states – shows why. For stationary states, all observable properties of the state (such as the probability that the particle is at location x) are constant. For non-stationary states, the probabilities themselves fluctuate as a function of time (and space of obviously), so the observable properties of the system are not constant. These solutions are solutions to the time-dependent Schrödinger equation and, hence, they are, obviously, time-dependent solutions.

We can find these time-dependent solutions by superimposing two stationary states, so we have a new wave function Ψ_N which is the sum of two others: Ψ_N = Ψ₁ + Ψ₂. [If you include the normalization factor (as you should to make sure all probabilities add up to 1), it’s actually Ψ_N = (2^–1/2)(Ψ₁ + Ψ₂).] So G and H above still represent a state of a quantum harmonic oscillator (with a specific energy level proportional to h), but so they are not standing waves.

Let’s go back to our electron traveling in a more or less straight path. What’s the shape of the solution for that one? It could be anything. Well… Almost anything. As said, the only condition we can impose is that the envelope of the wave packet – its ‘general’ shape so to say – should not change. That because we should not have dispersion – as illustrated below. [Note that this illustration only represent the real or the imaginary part – not both – but you get the idea.]

That being said, if we exclude dispersion (because a real-life electron traveling in a straight line doesn’t just disappear – as do dispersive wave packets), then, inside of that envelope, the weirdest things are possible – in theory that is. Indeed, Nature does not care much about our Fourier transforms. So the example below, which shows a theoretical wave packet (again, the real or imaginary part only) based on some theoretical distribution of the wave numbers of the (infinite number) of component waves that make up the wave packet, may or may not represent our real-life electron. However, if our electron has any resemblance to real-life, then I would expect it to not be as well-behaved as the theoretical one that’s shown below.

The shape above is usually referred to as a Gaussian wave packet, because of the nice normal (Gaussian) probability density functions that are associated with it. But we can also imagine a ‘square’ wave packet: a somewhat weird shape but – in terms of the math involved – as consistent as the smooth Gaussian wave packet, in the sense that we can demonstrate that the wave packet is made up of an infinite number of waves with an angular frequency ω that is linearly related to their wave number k, so the dispersion relation is ω = ak + b. [Remember we need to impose that condition to ensure that our wave packet will not dissipate (or disperse or disappear – whatever term you prefer.] That’s shown below: a Fourier analysis of a square wave.

While we can construct many theoretical shapes of wave packets that respect the ‘no dispersion!’ condition, we cannot know which one will actually represent that electron we’re trying to visualize. Worse, if push comes to shove, we don’t know if these matter waves (so these wave packets) actually consist of component waves (or time-independent stationary states or whatever).

[…] OK. Let me finally admit it: while I am trying to explain you the ‘reality’ of these matter waves, we actually don’t know how real these matter waves actually are. We cannot ‘see’ or ‘touch’ them indeed. All that we know is that (i) assuming their existence, and (ii) assuming these matter waves are more or less well-behaved (e.g. that actual particles will be represented by a composite wave characterized by a linear dispersion relation between the angular frequencies and the wave numbers of its (theoretical) component waves) allows us to do all that arithmetic with these (complex-valued) probability amplitudes. More importantly, all that arithmetic with these complex numbers actually yields (real-valued) probabilities that are consistent with the probabilities we obtain through repeated experiments. So that’s what’s real and ‘not so real’ I’d say.

Indeed, the bottom-line is that we do not know what goes on inside that envelope. Worse, according to the commonly accepted Copenhagen interpretation of the Uncertainty Principle (and tons of experiments have been done to try to overthrow that interpretation – all to no avail), we never will.

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