Physics Asked by Captain HD on July 1, 2021
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 🙁
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.
Correct answer by J. Murray on July 1, 2021
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