# Why do the two amplitudes need to match together through the region between the boxes?

This is an excerpt from Feynman's lectures 3;

http://www.feynmanlectures.caltech.edu/img/FLP_III/f07-03/f07-03_tc_big.svgz

Suppose we think of the situation in Fig. 7–3, which has two boxes held at the constant potentials $ϕ_1$ and $ϕ_2$ and a region in between where we will assume that the potential varies smoothly from one to the other. We imagine that some particle has an amplitude to be found in any one of the regions. We also assume that the momentum is large enough so that in any small region in which there are many wavelengths, the potential is nearly constant. We would then think that in any part of the space the amplitude ought to look like (7.18) with the appropriate $V$ for that part of the space. Let’s think of a special case in which $ϕ_1 = 0$, so that the potential energy there is zero, but in which $qϕ_2$ is negative, so that classically the particle would have more energy in the second box. Classically, it would be going faster in the second box—it would have more energy and, therefore, more momentum. Let’s see how that might come out of quantum mechanics. With our assumption, the amplitude in the first box would be proportional to $$e^{-(i/ℏ)[(W_\text{int}+p_1 ^2/2M+V_1)t−p_1⋅x]}$$ and the amplitude in the second box would be proportional to $$e^{−(i/ℏ)[(W_\text{int}+p_2 ^2/2M+V_2)t−p_2⋅x]}$$ (Let’s say that the internal energy is not being changed, but remains the same in both regions.) The question is: How do these two amplitudes match together through the region between the boxes? We are going to suppose that the potentials are all constant in time—so that nothing in the conditions varies. We will then suppose that the variations of the amplitude (that is, its phase) have the same frequency everywhere—because, so to speak, there is nothing in the “medium” that depends on time. If nothing in the space is changing, we can consider that the wave in one region “generates” subsidiary waves all over space which will all oscillate at the same frequency—just as light waves going through materials at rest do not change their frequency. If the frequencies in (7.21) and (7.22) are the same, we must have that $$W_\text{int} +p_1 ^2/2M+V_1=W_\text{int}+p_2 ^2/2M +V_2$$. Both sides are just the classical total energies, so Eq. (7.23) is a statement of the conservation of energy. In other words, the classical statement of the conservation of energy is equivalent to the quantum mechanical statement that the frequencies for a particle are everywhere the same if the conditions are not changing with time. It all fits with the idea that $ℏω=E$.

I am having a bit problem in understanding this. My questions are:

1. Why do the two amplitudes must match together through the region between the boxes? What is the reason? Feynman didn't mention.

2. What did Feynman want to mean by subsidiary waves?

3. How having the frequency constant, energy gets conserved?

• Unfortunately the image link is broken. I get a "Forbidden" error. Jul 30, 2015 at 18:32
• @Greenonline: Don't know how it happened:(
– user36790
Jul 31, 2015 at 2:04
• Could you repost the image to SE, i.e. imgur? Jul 31, 2015 at 2:30
• @Greenonline: Ok, I'm trying....
– user36790
Jul 31, 2015 at 2:31

The two amplitudes need to match because the wave function has to be continuous. If it were not continuous then you would have discontinuities in the probability function $\lvert \psi (x) \lvert ^2$ which leads to an absurd situation in which a point in space can accumulate or vanish particles.

I think that Feynman used the term of subsidiary waves just to point out that there is a stationary situation, which classically can only occur in a confined space if there are several waves reflecting and interfering in such a way that the wave form is static, or does not change in time in any point of space.

Finally, exactly from the amplitudes you get the frequencies since they have the form $e ^ {-{i (\omega t - k x)}} \rightarrow e ^ {-{i/ \hbar (E t - p x)}}$ which is typical form of waves. And having the frequencies equal means of course equating the corresponding terms from both equations, which happen to be the energies.

• @user36790 the wave form I put was intentionally $e^{i(\omega t - k x )}$ to emphasize the similitude between the wave function and the general form for waves (be it mechanical, electromagnetic, etc.) Jul 30, 2015 at 14:45
• Yes, sir, I know that but in order to have the same form as Feynman writes, I've altered that but if any problem looms, then sorry sir; I apologise; you can undo that:)
– user36790
Jul 30, 2015 at 15:04
• I understand your point. I made a modification to include also the form I intended. Thanks! Jul 30, 2015 at 15:18
• Oh, that's okay, sir; actually I went there to include a '-' sign you mistakenly forgot to write:)
– user36790
Jul 30, 2015 at 15:20
• I googled a bit & saw the wavefunction must be continuous as the divergence of probability current is always zero i.e. it must be conserved. Now, does this always happen? As I know from electric current, the case is steady-state. But what happens in non-steady state? Then the probability current can't be conserved, right? Is this possible for then the wavefunction can't be continuous, isn't it? Can you tell me more about the non-steady state a bit & what happens then to the wavefunction??
– user36790
Jul 31, 2015 at 2:52