# Quantum mechanics and forces: the classical limit

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:

This post continues along the same line as the two previous ones: the objective is to get a better ‘feel’ for those weird amplitudes by showing how they are being affected by a potential − such as an electric potential, or a gravitational field, or a magnetic field − and how they relate to a classical analysis of a situation. It’s what Feynman tries to do too. The best example of such ‘familiarization exercises’, by far, is the comparison between a classical analysis of a force field, and the quantum-mechanical one, so let’s quickly go through that.

The classical situation is depicted below. We’ve got a particle moving in the x-direction and entering a region where the potential − again, think of an electrostatic field, for example, as it probably inspires the V (for voltage, I assume) – varies in the y-direction. As you know, the potential (energy) is measured by doing work against the force, and the work done by the force is the W = ∫F•ds integral along the path. Conversely, one can differentiate both sides of that W = ∫F•ds equation to get the force, so we can write: F = dW/ds. That explains the F = −∂V/∂y expression in the illustration below. Again, think of a positively charged particle that’s being attracted from a high or positive voltage (high V) to a lower or negative voltage.

As a result of the force, the particle is being accelerated transversely, i.e. in the y-direction, and so it adds some momentum in the y-direction (py) to the initial momentum p, which is all in the x-direction (p = px). Of course, the force only acts as the particle is passing through the area, so that’s during a time interval that’s equal to w/v, with w being the ‘width’ of the area. Now, the vertical momentum builds from 0 to py, and we can calculate py using Newton’s Law: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt. It goes as follows:

We can then use a simple trigonometric formula to find the angle of deflection δθ:

Done! Now we need to show we get the same result with our quantum-mechanical approach or, at the very least, show how our quantum-mechanical approach actually works.

Of course, the assumption is that everything is on a very large scale as compared with the wavelength of our probability amplitudes. We then know that the presence of a potential V will just add to the energy E that we are to use in our wavefunction. Hence, in any small region, the amplitude will vary as:

a·e(i/ħ)[(Eint + p2/(2m) + V)·t − px]

Now, in the previous post, we explained that the energy conservation principle does not have any impact on the temporal variation of the wavefunction. Hence, Eint + p2/(2m) + V effectively remains constant, and the effect of a changing V is on p, and works out through the −px term in the exponent of our complex-valued wave function. To be specific, in a region where V is larger, p will be smaller and, therefore, we’ll have a larger wavelength λ = h/p. [Sorry if you can’t follow here: please do check out what I wrote in my two previous posts. It’s not that difficult.]

The illustration below shows how it works. The wavelength is the distance between successive wave nodes, which you should think of as surfaces where the phase of the amplitude is zero, or as wavefronts. So we’ve got a change in the angle of the wave nodes as well. How does that come about? Look at the two paths a and b: there’s a difference in potential between them, which we denote by ΔV, and which can be approximated as ΔV = (∂V/∂y)·D, with D the separation between the two paths.

Now, if Eint + p2/(2m) + V is a constant, then ΔV must be equal to −Δ[p2/(2m)]. Now what can we do with that? We’re talking a differential really, so we should apply the Δf(x) = [df(x)/dx]·Δx formula, which gives us:

Δ[p2/(2m)] = (p/m)·Δp = −ΔV ⇔ Δp = −(m/p)·ΔV

[Note that the d[f(x)] = [df(x)/dx]·dx formula is an interesting one: you’ll surely have come across something like Δ[1/λ], and wondered what you could do with that. Now you know: Δ[1/λ] = Δλ/λ2.]

Now, we’re interested in Δx, so how can we get some equation for that? Here we need to use the wavenumber k, and the derivation is not so easy—unfortunately! The (p/m)·Δp = −ΔV equation tells us that the wave number will be different along the two paths, and we can write the difference as Δk = Δ(p/ħ) = Δp/ħ. Now, we also now that the wavenumber is expressed in radians per unit distance. In addition, we also know that the wavenumber is the derivative of the phase with respect to distance, so we can approximately calculate the amount by which the phase on path b is ‘ahead’ of the phase on path a as the wave leaves the area, as:

Δ(phase) = Δk·w

Now, Δk = Δp/ħ, so Δ(phase) = Δp·w/ħ = −[m/(p·ħ)]·ΔV·w. Now, along path b, we will have some wavenumber k which, as per the definition of k, is equal to the derivative of the phase with respect to distance, so the ratio of the change in phase along path and the incremental distance Δx should equal the wave number k along path b, so we can write:

Δ(phase)/Δx = k ⇔ Δx = k·Δ(phase) = (λ/2π)·Δ(phase) = (ħ/p)·Δ(phase)

Combining that with Δ(phase) = −[m/(p·ħ)]·ΔV·w identity, we get:

Δx = −(ħ/p)·[m/(p·ħ)]·ΔV·w = −[m/p2]·ΔV·w

Phew! Just hang in there. We’re not done yet. I need to refer to the illustration now to show that Δx can also be approximated by D·δθ, so we write: Δx = D·δθ. [Again: it’s just plain triangle geometry, and the small angle approximation.] Combining both formulas for Δx yields:

δθ = −[m/p2]·ΔV·w/D

Now we can replace p/m by v and use our ΔV = (∂V/∂y)·D approximation to replace ΔV/D by (∂V/∂y), so we get:

δθ = −[1/(p·v)]·(∂V/∂y)·w = −[w/(p·v)]·(∂V/∂y)

So we get the very same equation for δθ as the one we got when applying classical mechanics. That’s a great result—also I have to admit that the differential analysis here was very painful and, hence, difficult to reproduce. In any case, what you should be interested in, is the result: as Feynman summarizes it,

We get the same particle motions we get from Newton’s F= m·a law, provided we assume that a potential contributes a phase to the probability amplitude equal to V·t/ħ. Hence, in the classical limit, quantum mechanics agrees with Newtonian mechanics.”

However, he also cautions that “the result we have just got is correct only if the potential variations are slow and smooth”, which actually defines what is referred to as the “classical limit”.

Let me, to conclude this post, present another example out of Feynman’s ‘illustrations’ of what a field or a potential does with amplitudes. Trust me: we’ll need it later. As Feynman is really succinct here, I will just copy it. The quote describes what happens in a magnetic field, as opposed to the electrostatic field which I mentioned above. Just read it: it’s easy. [If it’s too small, just click on it, and the text size will be OK.] However, while it’s easy to understand, the consequences, on which I’ll write a separate post, are quite deep.

So… Well… That’s really it for today. 🙂

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