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Most of the time you discuss the step potential case when $E>V_0$, you consider an electron (or a beam of electrons) travelling from a region of space $x<0$ in which $V=0$, to a region $x>0$ in which $V=V_0$, and I think I have understood that case:

the electron has enough energy to cross the step, but due to uncertainity there's a chance of it being pushed back, so the wavefunction modulus will be smaller after the step, and also the wavelength decreases following the rule $λ=\frac{2π\hbar}{\sqrt{2mE}}$ for $x<0$ and $λ'=\frac{2π\hbar}{\sqrt{2m(E-V_0})}$ for $x>0$

I can't understand what would happen if we considered the case of an electron travelling from a region with $V=V_0$ to a region with $V=0$: how could it be pushed back at the step even if the potential energy is decreasing, so the passage should be eased?

I ask this because in an exercise the solution says that the chances of reflection $R_1$ and $R_2 $ are equal in both case of electron "stepping up" and electron "stepping down"

in the bottom drawing the dotted lines represent the reflected waves.

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  • $\begingroup$ This is one of the many quirks of quantum mechanics. Why can't the particle bounce back from a potential drop? Our classical intuition often fails us. $\endgroup$ – Hanting Zhang Nov 19 '18 at 18:03
  • $\begingroup$ @HantingZhang I think you're right in a technical sense, but it's still important to build an intuition for the workings of quantum mechanics. To stop at the point of lacking intuition because of inherent quantum 'weirdness' is I think unsatisfactory. $\endgroup$ – CuriousHegemon Nov 19 '18 at 18:25
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Edit. To begin with, I don't understand the drawings showing incident and reflected waves. They are complex. At the step only imaginary parts vanish, for all wave components.


As to your question, perhaps an analogy could help. Consider a string, or better two strings of different linear densities tied together. It you send a wave on the strings, at the junction point there will be a partial reflection, both if the wave is coming from the lighter string towards the heavier and the other way around.

Using the electronics language, it is the impedance mismatch which causes reflection. Here the place of impedance is taken by the ratio $\psi'/\psi$. Schrödinger equation is $$\psi'' + k^2 \psi = 0.$$ If $\psi=e^{ikx}$ then impedance $Z$ is $$Z = \psi'/\psi = ik$$ and changes at the junction.

The same happens for your problem with a potential step.

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  • $\begingroup$ Thanks for the analogy, really helpful, I don't know much about waves in strings but I've studied EM waves in transmission lines and dielectric interfaces, and I knew also of the same analogy for the potential barrier, in which there are peak of transmission when the wavelength "fits" spacially in the barrier, but I never thought of it for the step. $\endgroup$ – tiamat Nov 22 '18 at 16:04
  • $\begingroup$ The drawings are not mine, but from the professor's solution of an exam, so I hoped of them being correct, I don't know why he didn't include the transmitted wave. Still I don't understand why you say the reflected wave can't be in phase or antiphase: phase(R1) =0 and phase(R2)=180 $\endgroup$ – tiamat Nov 22 '18 at 16:08
  • $\begingroup$ Still I don't understand why you say the reflected wave can't be in phase or antiphase: phase(R1) =0 and phase(R2)=180 You're right. I had a faulty reminder, maybe referring to the barrier, not to the step. I'm going to edit my reply. But still don't agree with the drawing. Incoming, reflected and transmitted wavefunctions are complex. What do those sinusoids mean? If you take the origin at the step, only the imaginary parts vanish there. As to reflection coefficients, they are OK. But can you compute transmission coefficients? $\endgroup$ – Elio Fabri Nov 24 '18 at 13:29
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To get the physical intuition here, consider a physical apparatus which creates the potential step in the case of an electron. Imagine a series of plane wire grids or meshes, connected by a battery:

4 grids, two at one potential, 2 at another

The battery gives positive electric potential to region B compared to region A, so electrons (which are negatively charged) will feel a drop down in potential energy when they cross from A to B. That is to say, they experience a large force to the right in the narrow gap between regions A and B where the voltage changes. This is where the step occurs in the potential energy diagram. The electron waves, arriving at this step, experience this strong 'kick', which abruptly forces them to change shape into waves of shorter wavelength. This picture helps one's physical intuition, I think, to see why it is that the electron waves might respond to this situation by being partially reflected.

Note that it is the abruptness of the kick that causes the reflection. Let $L$ be the width of the narrow region between the two central meshes, where the potential is changing (and so the wavelength is also changing). If the wavelength is small enough, i.e. when $\lambda \ll L$, then the waves can adjust their wavelength continuously as they pass through this region and in this case the reflection tends to zero. When $L \ll \lambda$, on the other hand, the reflection happens (and is calculated in the usual way by stating the solution to Schrodinger's equation and making sure the wavefunction and its gradient are continuous). The "step" potential is an idealisation which takes $L=0$. That is not physically possible; it is really an approximation which is suited to the case $L \ll \lambda$.

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the electron has enough energy to cross the step, but due to uncertainty there's a chance of it being pushed back

$ $

how could it be pushed back at the step even if the potential energy is decreasing, so the passage should be eased?

There is an issue with your interpretation of the wavefunction here. The wavefunction is not the electron. Before we measure the electron to be somewhere, it is not located anywhere. Therefore, we cannot talk about the electron reaching the step and bouncing off or being pushed back.

What we can talk about is how the wavefunction evolves over time according to the Schrodinger equation. Just like how Newton's laws tell us how the classical trajectory of a particle evolves over time, the Schrodinger equation can tell us how the wavefunction evolves.

So what really is being reflected or transmitted across the step is just the wavefunction, which encodes the probability of measuring the electron to be at some position. The step influences these probabilities.

In the case of an initial Gaussian packet approaching the step, there are solutions allowed that have a wave packet traveling in the opposite direction that correspond to measuring the electron to have a momentum in the opposite direction as the initial direction. But we cannot apply classical reasoning, as stated above. The wavefunction reflecting off of the step is not the same thing as the electron reflecting off the step. This is how the wavefunction evolves according to the Schrodinger equation, and so this ends up being a possiblity of our measurements.

I must also note that in the case of the step up, the wavefunction reflection is not due to uncertainty. For both step scenarios, the cause of wavefunction reflection is due to the same reason as discussed above.

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One way to think about whole this is by realizing two things:

  • Schrödinger equation gives the time evolution of a wave function.
  • For unitary time independent Hamiltonian (which is the case in your example), taking complex conjugation of Schröginger equation and inverting the direction of the time gives you back the original equation.

So, naively, the complex conjugation of wave function evolves backward in time just the same way the wave function evolves forward in time. If there are multiple wavefunction that mix, then the complex conjugation may result in relative phase differences. However, for one wavefunction, complex conjugation will not change the observables.

Therefore, knowing that wavefunction gets reflected when hitting a higher potential, it is only natural to expect a similar reflection when we go to a lower potential as going from lower to higher potential in forward time is equivalent to going from higher to lower potential in backward time. In fact, realizing this also reveals that the reflection has nothing to do with potential increase. It simply follows from the fact that the potential changes there.

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  • $\begingroup$ This doesn't actually work. The time-reversed configuration of reflection on the uphill step isn't a reflection on a downhill step. Instead, it is two incoming waves from either side, whose amplitude and phase are carefully tuned that the transmitted wave going uphill exactly cancels out the outgoing reflection of the incoming uphill wave. OP's configuration has both incoming and outgoing waves on the uphill side, and you can't get that with a simple time reversal. $\endgroup$ – Emilio Pisanty Nov 23 '18 at 11:39
  • $\begingroup$ The difference is boundary conditions: As you correctly say, just reflecting the equation itself is insufficient, you need to change the boundary condition in time; in other words, one needs to use $t=\infty$ final condition instead of $t=0$ initial condition. I intentionally avoided that and instead said "naively" to avoid clutter. If we wanna be strictly rigorous, I guess we can think of QM as 1d QFT and use S-matrix arguments to relate incoming and outgoing states. $\endgroup$ – Soner Nov 23 '18 at 11:47
  • $\begingroup$ That's a lot of jargon but none of that fixes the problem. OP was working in the stationary regime and if you time-reverse a stationary situation it is still stationary - there is absolutely no role for a "$t=\infty" final condition" or a "$t=0$ initial condition". The method just doesn't work - the two configurations are not time-reversed copies of each other - and no amount of burying the argument in jargon is going to fix it. $\endgroup$ – Emilio Pisanty Nov 23 '18 at 11:50
  • $\begingroup$ I think you misunderstood the question.OP is asking a scattering question:There is a wave function propagating and scattering off a potential change.This wave function is not necessarily a plane wave (which is the static solution),it may be a gaussian wave packet as well (or actually any L2 function).If we restrict ourselves to a static case (which OP is not requesting),then we can forget about the scattering theory and focus on the static solutions, for which the boundary conditions for time becomes the boundary conditions in space, and time reflection becomes spatial reflection.(PART 1) $\endgroup$ – Soner Nov 23 '18 at 12:07
  • $\begingroup$ However, if we remain general, the situation is not amenable to static picture and one needs to use scattering theory.As I wrote earlier, details are not important,but one can naively view this process both forward in time and backward in time to get the correct intuition: Change in the wave function is not because of increase in potential but because of the change of the potential.This is simply the S-matrix arguments (relating in and out states) combined with LSZ formalism (relating S matrix to correlation functions) and CPT theorem (time invariance in correlation functions). (PART 2) $\endgroup$ – Soner Nov 23 '18 at 12:09

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