Playing with amplitudes

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:

Let’s play a bit with the stuff we found in our previous post. This is going to be unconventional, or experimental, if you want. The idea is to give you… Well… Some ideas. So you can play yourself. 🙂 Let’s go.

Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point a to point b. If we identify point a by the position vector r₁ and point b by the position vector r₂, and using Dirac’s fancy bra-ket notation, then it’s written as:

propagator

So we have a vector dot product here: p∙r₁₂ = |p|∙|r₁₂|· cosθ = p∙r₁₂·cosα. The angle here (α) is the angle between the p and r₁₂ vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the α angle doesn’t matter: |e^i·θ/r|² = 1/r². Hence, for the probability, we get: P = | 〈r₂|r₁〉 |² = 1/r₁₂². Always ! Now that’s strange. The θ = p∙r₁₂/ħ argument gives us a different phase depending on the angle (α) between p and r₁₂. But… Well… Think of it: cosα goes from 1 to 0 when α goes from 0 to ±90° and, of course, is negative when p and r₁₂ have opposite directions but… Well… According to this formula, the probabilities do not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconic Lectures, give us a meaningless formula?

Maybe. We may also note this function looks like the elementary wavefunction for any particle, which we wrote as:

ψ(x, t) = a·e⁻ⁱ^∙θ = a·e⁻ⁱ^∙^{(E∙t − p∙x)/ħ}= a·e⁻ⁱ^∙^(E∙t)/ħ·eⁱ^∙^(p∙x)/ħ

The only difference is that the 〈r₂|r₁〉 sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carrying some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625×1.6×10⁻¹⁹ J = 9×10⁻¹⁹ J. Hence, their momentum is equal to p = E/c = (9×10⁻¹⁹ N·m)/(3×10⁵ m/s) = 3×10⁻²⁴ N·s. That’s tiny but that’s only because newtons and seconds are enormous units at the (sub-)atomic scale. As for the distance, we may want to use the thickness of a playing card as a starter, as that’s what Young used when establishing the experimental fact of light interfering with itself. Now, playing cards in Young’s time were obviously rougher than those today, but let’s take the smaller distance: modern cards are as thin as 0.3 mm. Still, that distance is associated with a value of θ that is equal to 13.6 million. Hence, the density of our wavefunction is enormous at this scale, and it’s a bit of a miracle that Young could see any interference at all ! As shown in the table below, we only get meaningful values (remember: θ is a phase angle) when we go down to the nanometer scale (10⁻⁹ m) or, even better, the angstroms scale ((10⁻⁹ m).

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagator function but something that is more general (read: more meaningful) at our (limited) level of knowledge. As I’ve been reading Feynman for quite a while now – like three or four years 🙂 – I think… Well… Yes. That’s it. Feynman wants us to think about it. 🙂 Are you joking again, Mr. Feynman? 🙂 So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a to point b by the position vector along some path r. So, then, in line with what we wrote in our previous post, let’s say p·r (momentum over a distance) is the action (S) we’d associate with this particular path (r) and then see where we get. So let’s write the formula like this:

ψ = a·e^i·θ = (1/r)·e^i·S/ħ = e^i·p∙r/ħ/r

We’ll use an index to denote the various paths: r₀ is the straight-line path and r_i is any (other) path. Now, quantum mechanics tells us we should calculate this amplitude for every possible path. The illustration below shows the straight-line path and two nearby paths. So each of these paths is associated with some amount of action, which we measure in Planck units: θ = S/ħ. alternative paths

The time interval is given by t = t₀= r₀/c, for all paths. Why is the time interval the same for all paths? Because we think of a photon going from some specific point in space and in time to some other specific point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from point a to point b at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain how it does that. 🙂

Now, if we would think of the photon actually traveling along this or that path, then this implies its velocity along any of the nonlinear paths will be larger than c, which is OK. That’s just the weirdness of quantum mechanics, and you should actually not think of the photon actually traveling along one of these paths anyway although we’ll often put it that way. Think of something fuzzier, whatever that may be. 🙂

So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths i and j is given by:

δS = p·r_j − p·r_i = p·(r_j − r_i) = p·Δr

I’ll explain the δS < 2πħ/3 thing in a moment. Let’s first pause and think about the uncertainty and how we’re modeling it. We can effectively think of the variation in S as some uncertainty in the action: δS = ΔS = p·Δr. However, if S is also equal to energy times time (S = E·t), and we insist t is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write δS as ΔS = ΔE·t. But, of course, E = E = m·c² = p·c, so we will have an uncertainty in the momentum as well. Hence, the variation in S should be written as:

δS = ΔS = Δp·Δr

That’s just logical thinking: if we, somehow, entertain the idea of a photon going from some specific point in spacetime to some other specific point in spacetime along various paths, then the variation, or uncertainty, in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Δp as ΔE/c, so we get the following:

δS = ΔS = Δp·Δr = ΔE·Δr/c = ΔE·Δt with Δt = Δr/c

So we have the two expressions for the Uncertainty Principle here: ΔS = Δp·Δr = ΔE·Δt. Just be careful with the interpretation of Δt: it’s just the equivalent of Δr. We just express the uncertainty in distance in seconds using the (absolute) speed of light. We are not changing our spacetime interval: we’re still looking at a photon going from a to b in t seconds, exactly. Let’s now look at the δS < 2πħ/3 thing. If we’re adding two amplitudes (two arrows or vectors, so to speak) and we want the magnitude of the result to be larger than the magnitude of the two contributions, then the angle between them should be smaller than 120 degrees, so that’s 2π/3 rad. The illustration below shows how you can figure that out geometrically.Hence, if S₀ is the action for r₀, then S₁ = S₀ + ħ and S₂ = S₀ + 2·ħ are still good, but S₃ = S₀ + 3·ħ is not good. Why? Because the difference in the phase angles is Δθ = S₁/ħ − S₀/ħ = (S₀ + ħ)/ħ − S₀/ħ = 1 and Δθ = S₂/ħ − S₀/ħ = (S₀ + 2·ħ)/ħ − S₀/ħ = 2 respectively, so that’s 57.3° and 114.6° respectively and that’s, effectively, less than 120°. In contrast, for the next path, we find that Δθ = S₃/ħ − S₀/ħ = (S₀ + 3·ħ)/ħ − S₀/ħ = 3, so that’s 171.9°. So that amplitude gives us a negative contribution.

Let’s do some calculations using a spreadsheet. To simplify things, we will assume we measure everything (time, distance, force, mass, energy, action,…) in Planck units. Hence, we can simply write: S_n = S₀ + n. Of course, n = 1, 2,… etcetera, right? Well… Maybe not. We are measuring action in units of ħ, but do we actually think action comes in units of ħ? I am not sure. It would make sense, intuitively, but… Well… There’s uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So there’s some randomness everywhere. 😦 So let’s leave that question open as for now.

We will also assume that the phase angle for S₀ is equal to 0 (or some multiple of 2π, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: ΔS_n = S_n − S₀ = n, and the associated phase angle θ_n = Δθ_n is the same. In short, the amplitude for each path reduces to ψ_n = e^i·n/r₀. So we need to add these first and then calculate the magnitude, which we can then square to get a probability. Of course, there is also the issue of normalization (probabilities have to add up to one) but let’s tackle that later. For the calculations, we use Euler’s r·e^i·θ = r·(cosθ + i·sinθ) = r·cosθ + i·r·sinθ formula. Needless to say, |r·e^i·θ|² = |r|²·|e^i·θ|² = |r|²·(cos²θ + sin²θ) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the ψ₀ + ψ₁ +ψ₂ + … sum as Ψ.

Now, we also need to see how our ΔS = Δp·Δr works out. We may want to assume that the uncertainty in p and in r will both be proportional to the overall uncertainty in the action. For example, we could try writing the following: ΔS_n = Δp_n·Δr_n = n·Δp₁·Δr₁. It also makes sense that you may want Δp_n and Δr_n to be proportional to Δp₁ and Δr₁ respectively. Combining both, the assumption would be this:

Δp_n = √n·Δp₁and Δr_n = √n·Δr₁

So now we just need to decide how we will distribute ΔS₁ = ħ = 1 over Δp₁ and Δr₁ respectively. For example, if we’d assume Δp₁ = 1, then Δr₁ = ħ/Δp₁ = 1/1 = 1. These are the calculations. I will let you analyze them. 🙂 newnew Well… We get a weird result. It reminds me of Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?

partial reflection

Hmm… Maybe it does. 🙂 Look at the graph more carefully. The peaks sort of oscillate out so… Well… That might make sense… 🙂

Does it? Are we doing something wrong here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of that’s part of the Wikipedia article on Feynman’s path integral formulation of quantum mechanics). So what’s wrong, if anything? Well… Our paths differ by some fixed amount of action, which doesn’t quite reflect the geometric approach that’s used in those animations. The graph below shows how the distance r varies as a function of n. geometry

If we’d use a model in which the distance would increase linearly or, preferably, exponentially, then we’d get the result we want to get, right?

Well… Maybe. Let’s try it. Hmm… We need to think about the geometry here. Look at the triangle below. triangle side If b is the straight-line path (r₀), then ac could be one of the crooked paths (r_n). To simplify, we’ll assume isosceles triangles, so a equals c and, hence, r_n = 2·a = 2·c. We will also assume the successive paths are separated by the same vertical distance (h = h₁) right in the middle, so h_b = h_n = n·h₁. It is then easy to show the following:This gives the following graph for r_n = 10 and h₁= 0.01. r graph

Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. The photon will have to travel faster in order to cover the extra distance in the same time, so its momentum will be higher. Let’s think about the velocity. Let’s start with the first path (n = 1). In order to cover the extra distance Δr₁, the velocity c₁ must be equal to (r₀ + Δr₁)/t = r₀/t + Δr₁/t = c + Δr₁/t = c₀ + Δr₁/t. We can write c₁ as c₁ = c₀ + Δc₁, so Δc₁ = Δr₁/t. Now, the ratio of p₁ and p₀ will be equal to the ratio of c₁ and c₀ because p₁/p₀= (mc₁)/mc₀) = c₁/c₀. Hence, we have the following formula for p₁:

p₁ = p₀·c₁/c₀ = p₀·(c₀ + Δc₁)/c₀ = p₀·[1 + Δr₁/(c₀·t) = p₀·(1 + Δr₁/r₀)

For p_n, the logic is the same, so we write:

p_n = p₀·c_n/c₀ = p₀·(c₀ + Δc_n)/c₀ = p₀·[1 + Δr_n/(c₀·t) = p₀·(1 + Δr_n/r₀)

Let’s do the calculations, and let’s use meaningful values, so the nanometer scale and actual values for Planck’s constant and the photon momentum. The results are shown below. original

Pretty interesting. In fact, this looks really good. The probability first swings around wildly, because of these zones of constructive and destructive interference, but then stabilizes. [Of course, I would need to normalize the probabilities, but you get the idea, right?] So… Well… I think we get a very meaningful result with this model. Sweet ! 🙂 I’m lovin’ it ! 🙂 And, here you go, this is (part of) the calculation table, so you can see what I am doing. 🙂 newnew

The graphs below look even better: I just changed the h₁/r₀ ratio from 1/100 to 1/10. The probability stabilizes almost immediately. 🙂 So… Well… It’s not as fancy as the referenced animation, but I think the educational value of this thing here is at least as good ! 🙂 great

🙂 This is good stuff… 🙂

Post scriptum (19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r₁ = , r₂, r₂,etcetera, and then we calculate the action for it, and the amplitude, and then we add the amplitude to the sum. But, surely, we should count these paths twice, in two-dimensional space, that is. Think of the graph: we have positive and negative interference zones that are sort of layered around the straight-line path, as shown below. zones

In three-dimensional space, these lines become surfaces. Hence, rather than adding one arrow for every δ having one contribution only, we may want to add… Well… In three-dimensional space, the formula for the surface around the straight-line path would probably look like π·h_n·r₁, right? Hmm… Interesting idea. I changed my spreadsheet to incorporate that idea, and I got the graph below. It’s a nonsensical result, because the probability does swing around, but it gradually spins out of control: it never stabilizes. revised That’s because we increase the weight of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. 🙂 I’ll you look for the right formula, OK? Let me know when you found it. 🙂

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Some thoughts on the nature of reality

Original post:

Some other comment on an article on my other blog, inspired me to structure some thoughts that are spread over various blog posts. What follows below, is probably the first draft of an article or a paper I plan to write. Or, who knows, I might re-write my two introductory books on quantum physics and publish a new edition soon. 🙂

Physical dimensions and Uncertainty

The physical dimension of the quantum of action (h or ħ = h/2π) is force (expressed in newton) times distance (expressed in meter) times time (expressed in seconds): N·m·s. Now, you may think this N·m·s dimension is kinda hard to imagine. We can imagine its individual components, right? Force, distance and time. We know what they are. But the product of all three? What is it, really?

It shouldn’t be all that hard to imagine what it might be, right? The N·m·s unit is also the unit in which angular momentum is expressed – and you can sort of imagine what that is, right? Think of a spinning top, or a gyroscope. We may also think of the following:

[h] = N·m·s = (N·m)·s = [E]·[t]
[h] = N·m·s = (N·s)·m = [p]·[x]

Hence, the physical dimension of action is that of energy (E) multiplied by time (t) or, alternatively, that of momentum (p) times distance (x). To be precise, the second dimensional equation should be written as [h] = [p]·[x], because both the momentum and the distance traveled will be associated with some direction. It’s a moot point for the discussion at the moment, though. Let’s think about the first equation first: [h] = [E]·[t]. What does it mean?

Energy… Hmm… In real life, we are usually not interested in the energy of a system as such, but by the energy it can deliver, or absorb, per second. This is referred to as the power of a system, and it’s expressed in J/s, or watt. Power is also defined as the (time) rate at which work is done. Hmm… But so here we’re multiplying energy and time. So what’s that? After Hiroshima and Nagasaki, we can sort of imagine the energy of an atomic bomb. We can also sort of imagine the power that’s being released by the Sun in light and other forms of radiation, which is about 385×10²⁴ joule per second. But energy times time? What’s that?

I am not sure. If we think of the Sun as a huge reservoir of energy, then the physical dimension of action is just like having that reservoir of energy guaranteed for some time, regardless of how fast or how slow we use it. So, in short, it’s just like the Sun – or the Earth, or the Moon, or whatever object – just being there, for some definite amount of time. So, yes: some definite amount of mass or energy (E) for some definite amount of time (t).

Let’s bring the mass-energy equivalence formula in here: E = mc². Hence, the physical dimension of action can also be written as [h] = [E]·[t] = [mc]²·[t] = (kg·m²/s²)·s = kg·m²/s. What does that say? Not all that much – for the time being, at least. We can get this [h] = kg·m²/s through some other substitution as well. A force of one newton will give a mass of 1 kg an acceleration of 1 m/s per second. Therefore, 1 N = 1 kg·m/s² and, hence, the physical dimension of h, or the unit of angular momentum, may also be written as 1 N·m·s = 1 (kg·m/s²)·m·s = 1 kg·m²/s, i.e. the product of mass, velocity and distance.

Hmm… What can we do with that? Nothing much for the moment: our first reading of it is just that it reminds us of the definition of angular momentum – some mass with some velocity rotating around an axis. What about the distance? Oh… The distance here is just the distance from the axis, right? Right. But… Well… It’s like having some amount of linear momentum available over some distance – or in some space, right? That’s sufficiently significant as an interpretation for the moment, I’d think…

Fundamental units

This makes one think about what units would be fundamental – and what units we’d consider as being derived. Formally, the newton is a derived unit in the metric system, as opposed to the units of mass, length and time (kg, m, s). Nevertheless, I personally like to think of force as being fundamental: a force is what causes an object to deviate from its straight trajectory in spacetime. Hence, we may want to think of the quantum of action as representing three fundamental physical dimensions: (1) force, (2) time and (3) distance – or space. We may then look at energy and (linear) momentum as physical quantities combining (1) force and distance and (2) force and time respectively.

Let me write this out:

Force times length (think of a force that is acting on some object over some distance) is energy: 1 joule (J) = 1 newton·meter (N). Hence, we may think of the concept of energy as a projection of action in space only: we make abstraction of time. The physical dimension of the quantum of action should then be written as [h] = [E]·[t]. [Note the square brackets tell us we are looking at a dimensional equation only, so [t] is just the physical dimension of the time variable. It’s a bit confusing because I also use square brackets as parentheses.]
Conversely, the magnitude of linear momentum (p = m·v) is expressed in newton·seconds: 1 kg·m/s = 1 (kg·m/s²)·s = 1 N·s. Hence, we may think of (linear) momentum as a projection of action in time only: we make abstraction of its spatial dimension. Think of a force that is acting on some object during some time. The physical dimension of the quantum of action should then be written as [h] = [p]·[x]

Of course, a force that is acting on some object during some time, will usually also act on the same object over some distance but… Well… Just try, for once, to make abstraction of one of the two dimensions here: time or distance.

It is a difficult thing to do because, when everything is said and done, we don’t live in space or in time alone, but in spacetime and, hence, such abstractions are not easy. [Of course, now you’ll say that it’s easy to think of something that moves in time only: an object that is standing still does just that – but then we know movement is relative, so there is no such thing as an object that is standing still in space in an absolute sense: Hence, objects never stand still in spacetime.] In any case, we should try such abstractions, if only because of the principle of least action is so essential and deep in physics:

In classical physics, the path of some object in a force field will minimize the total action (which is usually written as S) along that path.
In quantum mechanics, the same action integral will give us various values S – each corresponding to a particular path – and each path (and, therefore, each value of S, really) will be associated with a probability amplitude that will be proportional to some constant times e^−i·θ = e^i·(S/ħ). Because ħ is so tiny, even a small change in S will give a completely different phase angle θ. Therefore, most amplitudes will cancel each other out as we take the sum of the amplitudes over all possible paths: only the paths that nearly give the same phase matter. In practice, these are the paths that are associated with a variation in S of an order of magnitude that is equal to ħ.

The paragraph above summarizes, in essence, Feynman’s path integral formulation of quantum mechanics. We may, therefore, think of the quantum of action expressing itself (1) in time only, (2) in space only, or – much more likely – (3) expressing itself in both dimensions at the same time. Hence, if the quantum of action gives us the order of magnitude of the uncertainty – think of writing something like S ± ħ, we may re-write our dimensional [ħ] = [E]·[t] and [ħ] = [p]·[x] equations as the uncertainty equations:

ΔE·Δt = ħ
Δp·Δx = ħ

You should note here that it is best to think of the uncertainty relations as a pair of equations, if only because you should also think of the concept of energy and momentum as representing different aspects of the same reality, as evidenced by the (relativistic) energy-momentum relation (E² = p²c² – m₀²c⁴). Also, as illustrated below, the actual path – or, to be more precise, what we might associate with the concept of the actual path – is likely to be some mix of Δx and Δt. If Δt is very small, then Δx will be very large. In order to move over such distance, our particle will require a larger energy, so ΔE will be large. Likewise, if Δt is very large, then Δx will be very small and, therefore, ΔE will be very small. You can also reason in terms of Δx, and talk about momentum rather than energy. You will arrive at the same conclusions: the ΔE·Δt = h and Δp·Δx = h relations represent two aspects of the same reality – or, at the very least, what we might think of as reality.

Uncertainty

Also think of the following: if ΔE·Δt = h and Δp·Δx = h, then ΔE·Δt = Δp·Δx and, therefore, ΔE/Δp must be equal to Δx/Δt. Hence, the ratio of the uncertainty about x (the distance) and the uncertainty about t (the time) equals the ratio of the uncertainty about E (the energy) and the uncertainty about p (the momentum).

Of course, you will note that the actual uncertainty relations have a factor 1/2 in them. This may be explained by thinking of both negative as well as positive variations in space and in time.

We will obviously want to do some more thinking about those physical dimensions. The idea of a force implies the idea of some object – of some mass on which the force is acting. Hence, let’s think about the concept of mass now. But… Well… Mass and energy are supposed to be equivalent, right? So let’s look at the concept of energy too.

Action, energy and mass

What is energy, really? In real life, we are usually not interested in the energy of a system as such, but by the energy it can deliver, or absorb, per second. This is referred to as the power of a system, and it’s expressed in J/s. However, in physics, we always talk energy – not power – so… Well… What is the energy of a system?

According to the de Broglie and Einstein – and so many other eminent physicists, of course – we should not only think of the kinetic energy of its parts, but also of their potential energy, and their rest energy, and – for an atomic system – we may add some internal energy, which may be binding energy, or excitation energy (think of a hydrogen atom in an excited state, for example). A lot of stuff. 🙂 But, obviously, Einstein’s mass-equivalence formula comes to mind here, and summarizes it all:

E = m·c²

The m in this formula refers to mass – not to meter, obviously. Stupid remark, of course… But… Well… What is energy, really? What is mass, really? What’s that equivalence between mass and energy, really?

I don’t have the definite answer to that question (otherwise I’d be famous), but… Well… I do think physicists and mathematicians should invest more in exploring some basic intuitions here. As I explained in several posts, it is very tempting to think of energy as some kind of two-dimensional oscillation of mass. A force over some distance will cause a mass to accelerate. This is reflected in the dimensional analysis:

[E] = [m]·[c²] = 1 kg·m²/s² = 1 kg·m/s²·m = 1 N·m

The kg and m/s²factors make this abundantly clear: m/s² is the physical dimension of acceleration: (the change in) velocity per time unit.

Other formulas now come to mind, such as the Planck-Einstein relation: E = h·f = ω·ħ. We could also write: E = h/T. Needless to say, T = 1/f is the period of the oscillation. So we could say, for example, that the energy of some particle times the period of the oscillation gives us Planck’s constant again. What does that mean? Perhaps it’s easier to think of it the other way around: E/f = h = 6.626070040(81)×10⁻³⁴ J·s. Now, f is the number of oscillations per second. Let’s write it as f = n/s, so we get:

E/f = E/(n/s) = E·s/n = 6.626070040(81)×10⁻³⁴ J·s ⇔ E/n = 6.626070040(81)×10⁻³⁴ J

What an amazing result! Our wavicle – be it a photon or a matter-particle – will always pack 6.626070040(81)×10⁻³⁴ joule in one oscillation, so that’s the numerical value of Planck’s constant which, of course, depends on our fundamental units (i.e. kg, meter, second, etcetera in the SI system).

Of course, the obvious question is: what’s one oscillation? If it’s a wave packet, the oscillations may not have the same amplitude, and we may also not be able to define an exact period. In fact, we should expect the amplitude and duration of each oscillation to be slightly different, shouldn’t we? And then…

Well… What’s an oscillation? We’re used to counting them: n oscillations per second, so that’s per time unit. How many do we have in total? We wrote about that in our posts on the shape and size of a photon. We know photons are emitted by atomic oscillators – or, to put it simply, just atoms going from one energy level to another. Feynman calculated the Q of these atomic oscillators: it’s of the order of 10⁸(see his Lectures, I-33-3: it’s a wonderfully simple exercise, and one that really shows his greatness as a physics teacher), so… Well… This wave train will last about 10^–8seconds (that’s the time it takes for the radiation to die out by a factor 1/e). To give a somewhat more precise example, for sodium light, which has a frequency of 500 THz (500×10¹²oscillations per second) and a wavelength of 600 nm (600×10^–9meter), the radiation will lasts about 3.2×10^–8seconds. [In fact, that’s the time it takes for the radiation’s energy to die out by a factor 1/e, so(i.e. the so-called decay time τ), so the wavetrain will actually last longer, but so the amplitude becomes quite small after that time.] So… Well… That’s a very short time but… Still, taking into account the rather spectacular frequency (500 THz) of sodium light, that makes for some 16 million oscillations and, taking into the account the rather spectacular speed of light (3×10⁸m/s), that makes for a wave train with a length of, roughly, 9.6 meter. Huh? 9.6 meter!? But a photon is supposed to be pointlike, isn’it it? It has no length, does it?

That’s where relativity helps us out: as I wrote in one of my posts, relativistic length contraction may explain the apparent paradox. Using the reference frame of the photon – so if we’d be traveling at speed c,’ riding’ with the photon, so to say, as it’s being emitted – then we’d ‘see’ the electromagnetic transient as it’s being radiated into space.

However, while we can associate some mass with the energy of the photon, none of what I wrote above explains what the (rest) mass of a matter-particle could possibly be. There is no real answer to that, I guess. You’ll think of the Higgs field now but… Then… Well. The Higgs field is a scalar field. Very simple: some number that’s associated with some position in spacetime. That doesn’t explain very much, does it? 😦 When everything is said and done, the scientists who, in 2013 only, got the Nobel Price for their theory on the Higgs mechanism, simply tell us mass is some number. That’s something we knew already, right? 🙂

The reality of the wavefunction

The wavefunction is, obviously, a mathematical construct: a description of reality using a very specific language. What language? Mathematics, of course! Math may not be universal (aliens might not be able to decipher our mathematical models) but it’s pretty good as a global tool of communication, at least.

The real question is: is the description accurate? Does it match reality and, if it does, how good is the match? For example, the wavefunction for an electron in a hydrogen atom looks as follows:

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. So it presumes the duration of each oscillation is some constant. Why? Well… Look at the formula: this thing has a constant frequency in time. It’s only the amplitude that is varying as a function of the r = (x, y, z) coordinates. 🙂 So… Well… If each oscillation is to always pack 6.626070040(81)×10⁻³⁴ joule, but the amplitude of the oscillation varies from point to point, then… Well… We’ve got a problem. The wavefunction above is likely to be an approximation of reality only. 🙂 The associated energy is the same, but… Well… Reality is probably not the nice geometrical shape we associate with those wavefunctions.

In addition, we should think of the Uncertainty Principle: there must be some uncertainty in the energy of the photons when our hydrogen atom makes a transition from one energy level to another. But then… Well… If our photon packs something like 16 million oscillations, and the order of magnitude of the uncertainty is only of the order of h (or ħ = h/2π) which, as mentioned above, is the (average) energy of one oscillation only, then we don’t have much of a problem here, do we? 🙂

Post scriptum: In previous posts, we offered some analogies – or metaphors – to a two-dimensional oscillation (remember the V-2 engine?). Perhaps it’s all relatively simple. If we have some tiny little ball of mass – and its center of mass has to stay where it is – then any rotation – around any axis – will be some combination of a rotation around our x- and z-axis – as shown below. Two axes only. So we may want to think of a two-dimensional oscillation as an oscillation of the polar and azimuthal angle. 🙂

oscillation of a ball

Thinking again…

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)

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. 🙂

Re-visiting electron orbitals (III)

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

In my previous post, I mentioned that it was not so obvious (both from a physical as well as from a mathematical point of view) to write the wavefunction for electron orbitals – which we denoted as ψ(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) – as the product of two other functions in one variable only.

[…] OK. The above sentence is difficult to read. Let me write in math. 🙂 It is not so obvious to write ψ(x, t) as:

ψ(x, t) = e^{−i·(E/ħ)·t}·ψ(x)

As I mentioned before, the physicists’ use of the same symbol (ψ, psi) for both the ψ(x, t) and ψ(x) function is quite confusing – because the two functions are very different:

ψ(x, t) is a complex-valued function of two (real) variables: x and t. Or four, I should say, because x = (x, y, z) – but it’s probably easier to think of x as one vector variable – a vector-valued argument, so to speak. And then t is, of course, just a scalar variable. So… Well… A function of two variables: the position in space (x), and time (t).
In contrast, ψ(x) is a real-valued function of one (vector) variable only: x, so that’s the position in space only.

Now you should cry foul, of course: ψ(x) is not necessarily real-valued. It may be complex-valued. You’re right. You know the formula: wavefunction Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from x = (x, y, z) to r = (r, θ, φ), and that the function is also a function of the two quantum numbers l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Y_l,m(θ, φ) and F_l,m(r) respectively. F_l,m(r) was a real-valued function alright, but the Y_l,m(θ, φ) had that e^i·m·φ factor in it. So… Yes. You’re right: the Y_l,m(θ, φ) function is real-valued if – and only if – m = 0, in which case e^i·m·φ = 1. Let me copy the table from Feynman’s treatment of the topic once again: spherical harmonics 2 The P_l^m(cosθ) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the P_l^m(cosθ) is a real-valued function. The point is the following:the ψ(x, t) is a complex-valued function because – and only because – we multiply a real-valued envelope function – which depends on position only – with e^{−i·(E/ħ)·t}·e^i·m·φ = e^{−i·[(E/ħ)·t −}^m·φ].

[…]

Please read the above once again and – more importantly – think about it for a while. 🙂 You’ll have to agree with the following:

As mentioned in my previous post, the e^i·m·φ factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak, and the whole e^{−i·[(E/ħ)·t −}^m·φ] factor just disappears when we’re calculating probabilities.
The envelope function gives us the basic amplitude – in the classical sense of the word: the maximum displacement from the zero value. And so it’s that e^{−i·[(E/ħ)·t −}^m·φ] that ensures the whole expression somehow captures the energy of the oscillation.

Let’s first look at the envelope function again. Let me copy the illustration for n = 5 and l = 2 from a Wikimedia Commons article. Note the symmetry planes:

Any plane containing the z-axis is a symmetry plane – like a mirror in which we can reflect one half of the shape to get the other half. [Note that I am talking the shape only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
Likewise, the plane containing both the x– and the y-axis is a symmetry plane as well.

n = 5

The first symmetry plane – or symmetry line, really (i.e. the z-axis) – should not surprise us, because the azimuthal angle φ is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the e^i·m·φ factor with the e^{−i·(E/ħ)·t}, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction. OK. Clear enough – I hope. 🙂 But why is the the xy-plane a symmetry plane too? We need to look at that monstrous formula for the P_l^m(cosθ) function here: just note the cosθ argument in it is being squared before it’s used in all of the other manipulation. Now, we know that cosθ = sin(π/2 − θ). So we can define some new angle – let’s just call it α – which is measured in the way we’re used to measuring angle, which is not from the z-axis but from the xy-plane. So we write: cosθ = sin(π/2 − θ) = sinα. The illustration below may or may not help you to see what we’re doing here. angle 2 So… To make a long story short, we can substitute the cosθ argument in the P_l^m(cosθ) function for sinα = sin(π/2 − θ). Now, if the xy-plane is a symmetry plane, then we must find the same value for P_l^m(sinα) and P_l^m[sin(−α)]. Now, that’s not obvious, because sin(−α) = −sinα ≠ sinα. However, because the argument in that P_l^m(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−sinα]² = [sinα]²= sin²α. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that d^l+m/dx^l+m operator, and so you should check what happens with the minus sign there. 🙂

[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:

A definite energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelope function in three dimensions, really – which has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
The e^{−i·[(E/ħ)·t −}^m·φ] factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow, captures the energy of the oscillation.

It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.

spherical

The e^i·m·φ factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the e^{−i·[(E/ħ)·t} factor, which gives us the two-dimensional oscillation that captures the energy of the state.

Circle_cos_sin

Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep Blue page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing force on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing force on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the mass of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.

So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit of mass, but then I am not sure how to define that unit right now (it’s probably not the kilogram!).

Now, there is another thing we should note here: we’re actually sort of de-constructing a rotation (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis. Hence, in essence, we’re actually talking about something that’s spinning. In other words, we’re actually talking some torque around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or −E, in other words, but… Well… I need to explore this further – as should you! 🙂

Let me just add one more note on the e^i·m·φ factor. It sort of defines the geometry of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals n = 4 and l = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for m = ±1, we’ve got one appearance of that color only. For m = ±1, the same color appears at two ends of the ‘tubes’ – or tori (plural of torus), I should say – just to sound more professional. 🙂 For m = ±2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetry is equal to 3. Check that Wikimedia Commons article for higher values of n and l: the shapes become very convoluted, but the observation holds. 🙂

l = 3

Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. 🙂

Post scriptum: You should do some thinking on whether or not these m = ±1, ±2,…, ±l orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s not a real difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.

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Feynman’s Seminar on Superconductivity (1)

Original post:

The ultimate challenge for students of Feynman’s iconic Lectures series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does not present the detail of each and every step in the development and, therefore, we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we would be able to go through each and every step. Well… Let’s see. Feynman throws a lot of stuff in here—including, I suspect, some stuff that may not be directly relevant, but that he sort of couldn’t insert into all of his other Lectures. So where do we start?

It took me one long maddening day to figure out the first formula:It says that the amplitude for a particle to go from a to b in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from a to b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant. I stared at this for quite a while, but then I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shift the phase of the amplitude of a particle with an amount equal to:

Hence, if we write 〈b|a〉 for A = 0 as 〈b|a〉_{A = 0} = C·eⁱ^θ, then 〈b|a〉 in A will, naturally, be equal to 〈b|a〉 _{in A}= C·eⁱ⁽^θ+φ) = C·eⁱ^θ·eⁱ^φ = 〈b|a〉_{A = 0}·eⁱ^φ, and so that explains it. 🙂 Alright… Next. Or… Well… Let us briefly re-examine the concept of the vector potential, because we’ll need it a lot. We introduced it in our post on magnetostatics. Let’s briefly re-cap the development there. In Maxwell’s set of equations, two out of the four equations give us the magnetic field: ∇•B = 0 and c²∇×B = j/ε₀. We noted the following in this regard:

The ∇•B = 0 equation is true, always, unlike the ∇×E = 0 expression, which is true for electrostatics only (no moving charges). So the ∇•B = 0 equation says the divergence of B is zero, always.
The divergence of the curl of a vector field is always zero. Hence, if A is some vector field, then div(curl A) = ∇•(∇×A) = 0, always.
We can now apply another theorem: if the divergence of a vector field, say D, is zero—so if ∇•D = 0—then $D will be the$ the curl of some other vector field $C, so we can write: D = \nabla \times C . Applying this to \nabla • B = 0, we can write:$

If ∇•B = 0, then there is an A such that B = ∇×A

So, in essence, we’re just re-defining the magnetic field (B) in terms of some other vector field. To be precise, we write it as the curl of some other vector field, which we refer to as the (magnetic) vector potential. The components of the magnetic field vector can then be re-written as:

We need to note an important point here: the equations above suggest that the components of B depend on position only. In other words, we assume static magnetic fields, so they do not change with time. That, in turn, assumes steady currents. We will want to extend the analysis to also include magnetodynamics. It complicates the analysis but… Well… Quantum mechanics is complicated. Let us remind ourselves here of Feynman’s re-formulation of Maxwell’s equations as a set of two equations (expressed in terms of the magnetic (vector) and the electric potential) only:

Wave equation for A

Wave equation for potential

These equations are wave equations, as you can see by writing out the second equation:

It is a wave equation in three dimensions. Note that, even in regions where we do no have any charges or currents, we have non-zero solutions for φ and A. These non-zero solutions are, effectively, representing the electric and magnetic fields as they travel through free space. As Feynman notes, the advantage of re-writing Maxwell’s equations as we do above, is that the two new equations make it immediately apparent that we’re talking electromagnetic waves, really. As he notes, for many practical purposes, it will still be convenient to use the original equations in terms of E and B, but… Well… Not in quantum mechanics, it turns out. As Feynman puts it: “E and B are on the other side of the mountain we have climbed. Now we are ready to cross over to the other side of the peak. Things will look different—we are ready for some new and beautiful views.”

Well… Maybe. Appreciating those views, as part of our study of quantum mechanics, does take time and effort, unfortunately. 😦

The Schrödinger equation in an electromagnetic field

Feynman then jots down Schrödinger’s equation for the same particle (with charge q) moving in an electromagnetic field that is characterized not only by the (scalar) potential Φ but also by a vector potential A:

schrodinger

Now where does that come from? We know the standard formula in an electric field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:

i·ħ·∂ψ/∂t = −(1/2)·(ħ²/m)∇²ψ + V·ψ

Of course, it is easy to see that we replaced V by q·Φ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ), because Φ is, obviously, the potential energy of the unit charge. It’s also easy to see we can re-write −ħ²·∇²ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i² = 1/(−1) = −1. 🙂 Alright. So it’s just that −q·A term in the (ħ/i)∇ − q·A expression that we need to explain now.

Unfortunately, that explanation is not so easy. Feynman basically re-derives Schrödinger’s equation using his trade-mark historical argument – which did not include any magnetic field – with a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that (ħ/i)·∇ operator – which is the momentum operator– ought to be replaced by a new momentum operator:

new-momentum-operator

So… Well… There we are… 🙂 So far, so good? Well… Maybe.

While, as mentioned, you won’t be interested in the mathematical argument, it is probably worthwhile to reproduce Feynman’s more intuitive explanation of why the operator above is what it is. In other words, let us try to understand that −qA term. Look at the following situation: we’ve got a solenoid here, and some current I is going through it so there’s a magnetic field B. Think of the dynamics while we turn on this flux. Maxwell’s second equation (∇×E = −∂B/∂t) tells us the line integral of E around a loop will be equal to the time rate of change of the magnetic flux through that loop. The ∇×E = −∂B/∂t equation is a differential equation, of course, so it doesn’t have the integral, but you get the idea—I hope. solenoid

Now, using the B = ∇×A equation we can re-write the ∇×E = −∂B/∂t as ∇×E = −∂(∇×A)/∂t. This allows us to write the following:

∇×E = −∂(∇×A)/∂t = −∇×(∂A/∂t) ⇔ E = −∂A/∂t

This is a remarkable expression. Note its derivation is based on the commutativity of the curl and time derivative operators, which is a property that can easily be explained: if we have a function in two variables—say x and t—then the order of the derivation doesn’t matter: we can first take the derivative with respect to x and then to t or, alternatively, we can first take the time derivative and then do the ∂/∂x operation. So… Well… The curl is, effectively, a derivative with regard to the spatial variables. OK. So what? What’s the point?

Well… If we’d have some charge q, as shown in the illustration above, that would happen to be there as the flux is being switched on, it will experience a force which is equal to F = qE. We can now integrate this over the time interval (t) during which the flux is being built up to get the following:

∫₀^t F = ∫₀^t m·a = ∫₀^t m·dv/dt = m·v_t= ∫₀^t q·E = −∫₀^t q·∂A/∂t = −q·A_t

Assuming v₀ and A₀are zero, we may drop the time subscript and simply write:

m·v = −q·A

The point is: during the build-up of the magnetic flux, our charge will pick up some (classical) momentum that is equal to p = m·v = −q·A. So… Well… That sort of explains the additional term in our new momentum operator.

Note: For some reason I don’t quite understand, Feynman introduces the weird concept of ‘dynamical momentum’, which he defines as the quantity m·v + q·A, so that quantity must be zero in the analysis above. I quickly googled to see why but didn’t invest too much time in the research here. It’s just… Well… A bit puzzling. I don’t really see the relevance of his point here: I am quite happy to go along with the new operator, as it’s rather obvious that introducing changing magnetic fields must, obviously, also have some impact on our wave equations—in classical as well as in quantum mechanics.

Local conservation of probability

The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is the equation of continuity for probabilities. I find it brilliant, because it confirms my interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:

“An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy. If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”

This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.

So what is the flow – or probability current as Feynman refers to it? Well… Here’s the formula:

probability-current-2

Huh? Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:

probability-current-1

So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on how he explains how pairs of electrons start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature can no longer break up electron pairs if temperature comes close to the zero point.

Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentation on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:

“We make another approximation by forgetting that the electron has spin. […] The non-relativistic Schrödinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron’s point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.”

To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stable that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particular Lecture), so I must assume the theory is well accepted now. 🙂

Of course, you’ll shout now: Hey! Hydrogen is not a metal! Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. 🙂 And I am really talking the latest breakthrough: Science just published the findings of this experiment last month! 🙂 🙂 In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.

So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last Lecture because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. 😦

Don’t give up ! I am struggling with the nitty-gritty too ! 🙂

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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? 🙂

All what you ever wanted to know about the photon wavefunction…

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

Original post:

This post is, essentially, a continuation of my previous post, in which I juxtaposed the following images:

Both are the same, and then they’re not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction. You’ve seen that one before. In this case, the x-axis represents time, so we’re looking at the wavefunction at some particular point in space. ]You know we can just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it’s not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it’s the same function. Is it also the same reality?

Yes and no. And I would say: more no than yes—in this case, at least. Note that the animation does not show the accompanying magnetic field vector (B). That vector is equally essential in the electromagnetic propagation mechanism according to Maxwell’s equations, which—let me remind you—are equal to:

∂B/∂t = –∇×E
∂E/∂t = ∇×B

In fact, I should write the second equation as ∂E/∂t = c²∇×B, but then I assume we measure time and distance in equivalent units, so c = 1.

You know that E and B are two aspects of one and the same thing: if we have one, then we have the other. To be precise, B is always orthogonal to E in the direction that’s given by the right-hand rule for the following vector cross-product: B = e_x×E, with e_x the unit vector pointing in the x-direction (i.e. the direction of propagation). The reality behind is illustrated below for a linearly polarized electromagnetic wave.

E and b

The B = e_x×E equation is equivalent to writing B= i·E, which is equivalent to:

B = i·E = eⁱ^(π/2)·eⁱ^{(kx − ωt)} = cos(kx − ωt + π/2) + i·sin(kx − ωt + π/2)

= −sin((kx − ωt) + i·cos(kx − ωt)

Now, E and B have only two components: E_yand E_z, and B_yand B_z. That’s only because we’re looking at some ideal or elementary electromagnetic wave here but… Well… Let’s just go along with it. 🙂 It is then easy to prove that the equation above amounts to writing:

B_y= cos(kx − ωt + π/2) = −sin(kx − ωt) = −E_z
B_z= sin(kx − ωt + π/2) = cos(kx − ωt) = E_y

We should now think of E_y and E_zas the real and imaginary part of some wavefunction, which we’ll denote as ψ_E = eⁱ^{(kx − ωt)}. So we write:

E = (E_y, E_z) = E_y + i·E_z= cos(kx − ωt) + i∙sin(kx − ωt) = Re(ψ_E) + i·Im(ψ_E) = ψ_E = eⁱ^{(kx − ωt)}

What about B? We just do the same, so we write:

B = (B_y, B_z) = B_y + i·B_z= ψ_B = i·E = i·ψ_E = −sin(kx − ωt) + i∙sin(kx − ωt) = − Im(ψ_E) + i·Re(ψ_E)

Now we need to prove that ψ_E and ψ_B are regular wavefunctions, which amounts to proving Schrödinger’s equation, i.e. ∂ψ/∂t = i·(ħ/m)·∇²ψ, for both ψ_E and ψ_B. [Note I use the Schrödinger’s equation for a zero-mass spin-zero particle here, which uses the ħ/m factor rather than the ħ/(2m) factor.] To prove that ψ_E and ψ_B are regular wavefunctions, we should prove that:

Re(∂ψ_E/∂t) = −(ħ/m)·Im(∇²ψ_E) and Im(∂ψ_E/∂t) = (ħ/m)·Re(∇²ψ_E), and
Re(∂ψ_B/∂t) = −(ħ/m)·Im(∇²ψ_B) and Im(∂ψ_B/∂t) = (ħ/m)·Re(∇²ψ_B).

Let’s do the calculations for the second pair of equations. The time derivative on the left-hand side is equal to:

∂ψ_B/∂t = −iω·ieⁱ^{(kx − ωt)}= ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

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

∇²ψ_B= ∂²ψ_B/∂x²= i·k²·eⁱ^{(kx − ωt)}= k²·cos(kx − ωt) + i·k²·sin(kx − ωt)

So the two equations for ψ_Bare equivalent to writing:

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

So we see that both conditions are fulfilled if, and only if, ω = k²·(ħ/m).

Now, we also demonstrated in that post of mine that Maxwell’s equations imply the following:

∂B_y/∂t = –(∇×E)_y = ∂E_z/∂x = ∂[sin(kx − ωt)]/∂x = k·cos(kx − ωt) = k·E_y
∂B_z/∂t = –(∇×E)_z = – ∂E_y/∂x = – ∂[cos(kx − ωt)]/∂x = k·sin(kx − ωt) = k·E_z

Hence, using those B_y= −E_zand B_z= E_yequations above, we can also calculate these derivatives as:

∂B_y/∂t = −∂E_z/∂t = −∂sin(kx − ωt)/∂t = ω·cos(kx − ωt) = ω·E_y
∂B_z/∂t = ∂E_y/∂t = ∂cos(kx − ωt)/∂t = −ω·[−sin(kx − ωt)] = ω·E_z

In other words, Maxwell’s equations imply that ω = k, which is consistent with us measuring time and distance in equivalent units, so the phase velocity is c = 1 = ω/k.

So far, so good. We basically established that the propagation mechanism for an electromagnetic wave, as described by Maxwell’s equations, is fully coherent with the propagation mechanism—if we can call it like that—as described by Schrödinger’s equation. We also established the following equalities:

ω = k
ω = k²·(ħ/m)

The second of the two de Broglie equations tells us that k = p/ħ, so we can combine these two equations and re-write these two conditions as:

ω/k = 1 = k·(ħ/m) = (p/ħ)·(ħ/m) = p/m ⇔ p = m

What does this imply? The p here is the momentum: p = m·v, so this condition implies v must be equal to 1 too, so the wave velocity is equal to the speed of light. Makes sense, because we actually are talking light here. 🙂 In addition, because it’s light, we also know E/p = c = 1, so we have – once again – the general E = p = m equation, which we’ll need!

OK. Next. Let’s write the Schrödinger wave equation for both wavefunctions:

∂ψ_E/∂t = i·(ħ/m_E)·∇²ψ_E, and
∂ψ_B/∂t = i·(ħ/m_B)·∇²ψ_B.

Huh? What’s m_E and m_E? We should only associate one mass concept with our electromagnetic wave, shouldn’t we? Perhaps. I just want to be on the safe side now. Of course, if we distinguish m_E and m_B, we should probably also distinguish p_E and p_B, and E_E and E_B as well, right? Well… Yes. If we accept this line of reasoning, then the mass factor in Schrödinger’s equations is pretty much like the 1/c² = μ₀ε₀ factor in Maxwell’s (1/c²)·∂E/∂t = ∇×B equation: the mass factor appears as a property of the medium, i.e. the vacuum here! [Just check my post on physical constants in case you wonder what I am trying to say here, in which I explain why and how c defines the (properties of the) vacuum.]

To be consistent, we should also distinguish p_E and p_B, and E_E and E_B, and so we should write ψ_Eand ψ_B as:

ψ_E = eⁱ^{(k_Ex − ω_Et)}, and
ψ_B = eⁱ^{(k_Bx − ω_Bt)}.

Huh? Yes. I know what you think: we’re talking one photon—or one electromagnetic wave—so there can be only one energy, one momentum and, hence, only one k, and one ω. Well… Yes and no. Of course, the following identities should hold: k_E = k_B and, likewise, ω_E= ω_B. So… Yes. They’re the same: one k and one ω. But then… Well… Conceptually, the two k’s and ω’s are different. So we write:

p_E = E_E = m_E, and
p_B = E_B = m_B.

The obvious question is: can we just add them up to find the total energy and momentum of our photon? The answer is obviously positive: E = E_E + E_B, p = p_E + p_B and m = m_E + m_B.

Let’s check a few things now. How does it work for the phase and group velocity of ψ_Eand ψ_B? Simple:

v_g = ∂ω_E/∂k_E = ∂[E_E/ħ]/∂[p_E/ħ] = ∂E_E/∂p_E = ∂p_E/∂p_E = 1
v_p = ω_E/k_E = (E_E/ħ)/(p_E/ħ) = E_E/p_E = p_E/p_E = 1

So we’re fine, and you can check the result for ψ_Bby substituting the subscript E for B. To sum it all up, what we’ve got here is the following:

We can think of a photon having some energy that’s equal to E = p = m (assuming c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψ_Eand ψ_B.
Schrödinger’s equation applies to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (p_E =E_E = m_E = p_B =E_B = m_B), but they’re conceptually different. They must be: if not, we’d get a phase and group velocity for the wave that doesn’t make sense.

Of course, the phase and group velocity for the sum of the ψ_Eand ψ_Bwaves must also be equal to c. This is obviously the case, because we’re adding waves with the same phase and group velocity c, so there’s no issue with the dispersion relation.

So let’s insert those p_E =E_E = m_E = p_B =E_B = m_B values in the two wavefunctions. For ψ_E, we get:

ψ_E= eⁱ^{[k_Ex − ω_Et)}= eⁱ^{[(p_E/ħ)·x − (E_E/ħ)·t]}

You can do the calculation for ψ_Byourself. Let’s simplify our life a little bit and assume we’re using Planck units, so ħ = 1, and so the wavefunction simplifies to ψ_E= eⁱ^{·(p_E·x − E_E·t)}. We can now add the components of E and B using the summation formulas for sines and cosines:

1. B_y+ E_y = cos(p_B·x − E_B·t + π/2) + cos(p_E·x − E_E·t) = 2·cos[(p·x − E·t + π/2)/2]·cos(π/4) = √2·cos(p·x/2 − E·t/2 + π/4)

2. B_z+ E_z = sin(p_B·x − E_B·t+π/2) + sin(p_E·x − E_E·t) = 2·sin[(p·x − E·t + π/2)/2]·cos(π/4) = √2·sin(p·x/2 − E·t/2 + π/4)

Interesting! We find a composite wavefunction for our photon which we can write as:

E + B = ψ_E+ ψ_B= E + i·E = √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}= √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}= √2·eⁱ^(π/4)·E

What a great result! It’s easy to double-check, because we can see the E + i·E = √2·eⁱ^(π/4)·E formula implies that 1 + i should equal √2·eⁱ^(π/4). Now that’s easy to prove, both geometrically (just do a drawing) or formally: √2·eⁱ^(π/4) = √2·cos(π/4) + i·sin(π/4eⁱ^(π/4) = (√2/√2) + i·(√2/√2) = 1 + i. We’re bang on! 🙂

We can double-check once more, because we should get the same from adding E and B = i·E, right? Let’s try:

E + B = E + i·E = cos(p_E·x − E_E·t) + i·sin(p_E·x − E_E·t) + i·cos(p_E·x − E_E·t) − sin(p_E·x − E_E·t)

= [cos(p_E·x − E_E·t) – sin(p_E·x − E_E·t)] + i·[sin(p_E·x − E_E·t) – cos(p_E·x − E_E·t)]

Indeed, we can see we’re going to obtain the same result, because the −sinθ in the real part of our composite wavefunction is equal to cos(θ+π/2), and the −cosθ in its imaginary part is equal to sin(θ+π/2). So the sum above is the same sum of cosines and sines that we did already.

So our electromagnetic wavefunction, i.e. the wavefunction for the photon, is equal to:

ψ = ψ_E+ ψ_B= √2·eⁱ^{(p·x/2 − E·t/2 + π/4)} = √2·eⁱ^(π/4)·eⁱ^{(p·x/2 − E·t/2)}

What about the √2 factor in front, and the π/4 term in the argument itself? No sure. It must have something to do with the way the magnetic force works, which is not like the electric force. Indeed, remember the Lorentz formula: the force on some unit charge (q = 1) will be equal to F = E + v×B. So… Well… We’ve got another cross-product here and so the geometry of the situation is quite complicated: it’s not like adding two forces F₁and F₂to get some combined force F = F₁and F₂.

In any case, we need the energy, and we know that its proportional to the square of the amplitude, so… Well… We’re spot on: the square of the √2 factor in the √2·cos product and √2·sin product is 2, so that’s twice… Well… What? Hold on a minute! We’re actually taking the absolute square of the E + B = ψ_E+ ψ_B= E + i·E = √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}wavefunction here. Is that legal? I must assume it is—although… Well… Yes. You’re right. We should do some more explaining here.

We know that we usually measure the energy as some definite integral, from t = 0 to some other point in time, or over the cycle of the oscillation. So what’s the cycle here? Our combined wavefunction can be written as √2·eⁱ^{(p·x/2 − E·t/2 + π/4)}= √2·eⁱ^{(θ/2 + π/4)}, so a full cycle would correspond to θ going from 0 to 4π here, rather than from 0 to 2π. So that explains the √2 factor in front of our wave equation.

Bingo! If you were looking for an interpretation of the Planck energy and momentum, here it is. And, while everything that’s written above is not easy to understand, it’s close to the ‘intuitive’ understanding to quantum mechanics that we were looking for, isn’t it? The quantum-mechanical propagation model explains everything now. 🙂 I only need to show one more thing, and that’s the different behavior of bosons and fermions:

The amplitudes of identitical bosonic particles interfere with a positive sign, so we have Bose-Einstein statistics here. As Feynman writes it: (amplitude direct) + (amplitude exchanged).
The amplitudes of identical fermionic particles interfere with a negative sign, so we have Fermi-Dirac statistics here: (amplitude direct) − (amplitude exchanged).

I’ll think about it. I am sure it’s got something to do with that B= i·E formula or, to put it simply, with the fact that, when bosons are involved, we get two wavefunctions (ψ_Eand ψ_B) for the price of one. The reasoning should be something like this:

I. For a massless particle (i.e. a zero-mass fermion), our wavefunction is just ψ = eⁱ^{(p·x − E·t)}. So we have no √2 or √2·eⁱ^(π/4)factor in front here. So we can just add any number of them – ψ₁+ ψ₂+ ψ₃+ … – and then take the absolute square of the amplitude to find a probability density, and we’re done.

II. For a photon (i.e. a zero-mass boson), our wavefunction is √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}, which – let’s introduce a new symbol – we’ll denote by φ, so φ = √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}. Now, if we add any number of these, we get a similar sum but with that √2·eⁱ^(π/4)factor in front, so we write: φ₁+ φ₂+ φ₃+ … = √2·eⁱ^(π/4)·(ψ₁+ ψ₂+ ψ₃+ …). If we take the absolute square now, we’ll see the probability density will be equal to twice the density for the ψ₁+ ψ₂+ ψ₃+ … sum, because

|√2·eⁱ^(π/4)·(ψ₁+ ψ₂+ ψ₃+ …)|² = |√2·eⁱ^(π/4)|²·|ψ₁+ ψ₂+ ψ₃+ …)|² = 2·|ψ₁+ ψ₂+ ψ₃+ …)|²

So… Well… I still need to connect this to Feynman’s (amplitude direct) ± (amplitude exchanged) formula, but I am sure it can be done.

Now, we haven’t tested the complete √2·eⁱ^(π/4)·eⁱ^{(p·x − E·t)/2}wavefunction. Does it respect Schrödinger’s ∂ψ/∂t = i·(1/m)·∇²ψ or, including the 1/2 factor, the ∂ψ/∂t = i·[1/2m)]·∇²ψ equation? [Note we assume, once again, that ħ = 1, so we use Planck units once more.] Let’s see. We can calculate the derivatives as:

∂ψ/∂t = −√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·E/2)
∇²ψ = ∂²[√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}]/∂x²= √2·eⁱ^(π/4)·∂[√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·p/2)]/∂x = −√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(p²/4)

So Schrödinger’s equation becomes:

−i·√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(i·E/2) = −i·(1/m)·√2·eⁱ^(π/4)·e^{−i∙[p·x − E·t]/2}·(p²/4) ⇔ 1/2 = 1/4!?

That’s funny ! It doesn’t work ! The E and m and p² are OK because we’ve got that E = m = p equation, but we’ve got problems with yet another factor 2. It only works when we use the 2/m coefficient in Schrödinger’s equation.

So… Well… There’s no choice. That’s what we’re going to do. The Schrödinger equation for the photon is ∂ψ/∂t = i·(2/m)·∇²ψ !

It’s a very subtle point. This is all great, and very fundamental stuff! Let’s now move on to Schrödinger’s actual equation, i.e. the ∂ψ/∂t = i·(ħ/2m)·∇²ψ equation.

Post scriptum on the Planck units:

If we measure time and distance in equivalent units, say seconds, we can re-write the quantum of action as:

1.0545718×10⁻³⁴N·m·s = (1.21×10⁴⁴N)·(1.6162×10⁻³⁵m)·(5.391×10⁻⁴⁴ s)

⇔ (1.0545718×10⁻³⁴/2.998×10⁸) N·s² = (1.21×10⁴⁴N)·(1.6162×10⁻³⁵/2.998×10⁸ s)(5.391×10⁻⁴⁴ s)

⇔ (1.21×10⁴⁴N) = [(1.0545718×10⁻³⁴/2.998×10⁸)]/[(1.6162×10⁻³⁵/2.998×10⁸ s)(5.391×10⁻⁴⁴ s)] N·s²/s²

You’ll say: what’s this? Well… Look at it. We’ve got a much easier formula for the Planck force—much easier than the standard formulas you’ll find on Wikipedia, for example. If we re-interpret the symbols ħ and c so they denote the numerical value of the quantum of action and the speed of light in standard SI units (i.e. newton, meter and second)—so ħ and c become dimensionless, or mathematical constants only, rather than physical constants—then the formula above can be written as:

F_P newton = (ħ/c)/[(l_P/c)·t_P] newton ⇔ F_P = ħ/(l_P·t_P)

Just double-check it: 1.0545718×10⁻³⁴/(1.6162×10⁻³⁵·5.391×10⁻⁴⁴) = 1.21×10⁴⁴. Bingo!

You’ll say: what’s the point? The point is: our model is complete. We don’t need the other physical constants – i.e. the Coulomb, Boltzmann and gravitational constant – to calculate the Planck units we need, i.e. the Planck force, distance and time units. It all comes out of our elementary wavefunction! All we need to explain the Universe – or, let’s be more modest, quantum mechanics – is two numerical constants (c and ħ) and Euler’s formula (which uses π and e, of course). That’s it.

If you don’t think that’s a great result, then… Well… Then you’re not reading this. 🙂