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I have looked at the questions on this stack exchange and did not find a single convincing answer. Please absolutely remember the mathematical definition of only 4 things as you read this. Probability density, probability current, and the transmission and reflection coefficients. In this case the potential energy will jump straight to a constant value of $V$ after the point $x=0$ (assume continuity of wavefunctions).

In a potential step function where $V=0$ before and $V=V$ after and where $E<V$. why is it the case that a right-moving matter wave into region 2 (where V is non-zero) has a decaying exponential describing it whereas at the same time my lecturer says that the particle is reflected? He proved this by showing that R=1 in this case and that T=0. I understood that but I feel like the idea of a probability current is being confused together incorrectly with the idea of probability density. The decaying exponential in region 2 (where V was finite to the right of the point x=0) is associated with the probability density despite E<V. and this means that it can be transmitted through the potential step. He also proved that the transmission coefficient is 0 but to me that doesn't mean the particle DOESN'T get transmitted. It just means that the probability density describing the matter wave in region 2 is a stationary wave (meaning there is no rate of change in probability density). This is because current is defined as the rate of change in density with time (I'm emitting the overused words "probability").

Why is this reasoning not true? T = j_transmitted / j_incident and j_transmitted=0 is not the same thing as saying that the probability of the particle being found in region 2 is 0. Not. The. Same. Thing :(

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  • $\begingroup$ The probability density is not the same as the probability current. Current can be 0 and we could still have a finite density for either region of the potential step system. Here is the question, if the probability density is a non-zero decaying exponential in region 2 then why is it said that the particle is still completely reflected when T=0, R=1? I say that because T=0 just simply means that the transmitted current is 0, not the actual probability density itself. This would therefore mean that the particle can still be found (“transmitted”) in region 2. Does the reasoning not follow? $\endgroup$
    – Captain HD
    Oct 25, 2020 at 21:29
  • $\begingroup$ See tunneling but essentially solving Shrodingers equation for all regions requires the probability density to decay rather than disappear. $\endgroup$
    – user220348
    Oct 25, 2020 at 23:34
  • $\begingroup$ I don’t feel like this answers it for me though. I will try YouTube as a last resort $\endgroup$
    – Captain HD
    Oct 26, 2020 at 17:41

1 Answer 1

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He also proved that the transmission coefficient is 0 but to me that doesn't mean the particle DOESN'T get transmitted. It just means that the probability density describing the matter wave in region 2 is a stationary wave (meaning there is no rate of change in probability density). This is because current is defined as the rate of change in density with time

This is true. However, in a scattering event you start with a particle which is far to the left of the barrier, which means that the wavefunction vanishes for $x>0$. As the particle approaches the barrier the wavefunction changes (obviously), but as demonstrated by your instructor, the wavefunction inside the barrier remains zero, and the particle is completely reflected.

The problem is that you are looking at non-normalizable energy eigenstates as though real particles have them as wavefunctions. This is not the case. Real particles exist in superpositions of these energy eigenstates, and the different eigenstates interfere with each other to keep the probability density inside the step equal to zero.

To put it a different way, if at $t=0$ the wavefunction vanishes for $x>0$ - as it would for a particle which is incident on the step from far away to the left - then we know that the wavefunction vanishes for $x>0$ for all $t$ because the probability current (and therefore the rate of change of the probability density) is zero for $x>0$. If the probability density starts at zero and cannot change with time, then it's zero forever.

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  • $\begingroup$ I didn't understand most of what was said as highly technical words that I am still getting used to were used but I think I know how to explain it in my own words. The particle is coming in from the far left and as such the probability density at t=0 is zero inside the barrier. When it arrives it will have a transmittion coefficient of 0 which translates in my understanding as meaning that the probability density inside the barrier does NOT change because current is zero where current is the rate of change of density. The density was 0 as it was coming towards the barrier remember? $\endgroup$
    – Captain HD
    Oct 29, 2020 at 3:46
  • $\begingroup$ This means that if the density was initially 0 and that the current inside the barrier will be 0 that it will therefore remain as 0 and hence what I was taught about a non-zero probability was only true for the case that the transmission coefficient is >0 (non-zero). If all of that is true, then why did nobody put it like that! $\endgroup$
    – Captain HD
    Oct 29, 2020 at 3:48
  • $\begingroup$ @CaptainHD Your first comment is correct, your second is not. Consider a particle which is incident on the step from the left which exists as a superposition of energy eigenfunctions with $E<V$. Each eigenfunction is non-zero within the barrier, but they add up to zero when you put them together. If they cancel like this at $t=0$, then they cancel forever because (as you have noted) the probability current is zero (and so the probability density is constant) for $x>0$. $\endgroup$
    – J. Murray
    Oct 29, 2020 at 4:00
  • $\begingroup$ @CaptainHD For $x<0$, the probability current is not constant, and so the probability density does change. In particular, the wave packet approaches the barrier and is reflected, like a pulse on a rope being reflected from a fixed end. $\endgroup$
    – J. Murray
    Oct 29, 2020 at 4:06
  • $\begingroup$ Thank you for your help and time. You should have given a reply on here so that I could upvote it although you probably don't care about the voting system. Can't have made physics sound more political than that. (never mind you did post a reply I just haven't been on here in a while). Also sorry I can't upvote it. My reputation is below 15 $\endgroup$
    – Captain HD
    Oct 31, 2020 at 6:33

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