5
$\begingroup$

In standard 1D textbook step potential problems, for $E>V_0$ with $V_0$ the step potential for $x>0$, one finds that for $x<0$ that eigensolutions are of the form $$\psi(x) = A_1e^{i\sqrt{2ME/h^2}x} + B_1e^{-i\sqrt{2ME/h^2}x}$$ and for $x>0$ that $$\psi(x) = A_2e^{i\sqrt{2M(E-V_0)/h^2}x}$$ where we've enforced unstated boundary conditions far from the $x=0$ interface to assert that there is no left-travelling wave for $x>0$.

My question is, why do textbooks state that $|A_2/A_1|^2$ is the "probability of escape". Is this just a manner of speaking, or is there real physical basis to speaking like this? My confusion lies in that probabilities in coordinate representation should be associated with a volume integral of the modulus squared of the wavefunction over an appropriate region.

$\endgroup$
3
  • $\begingroup$ Related: Why can we treat quantum scattering problems as time-independent? $\endgroup$
    – Qmechanic
    Mar 28, 2023 at 16:31
  • $\begingroup$ @Qmechanic Thank you as always for your help in providing these links. This question is massively helpful. $\endgroup$
    – EE18
    Mar 28, 2023 at 19:37
  • $\begingroup$ It's physics. The first thing you should ask in physics is "what's the experiment here?". So, how would you measure your coefficients? $\endgroup$
    – John Doty
    Mar 29, 2023 at 13:52

3 Answers 3

1
$\begingroup$

One way to think about this is in terms of probability current. If your wavefunction describes a steady flux of particles incident on the potential step, the probability current at point $x$ can be thought of as the number of particles passing through point $x$ per unit time. The probability current (ignoring the constant multiplier) is $|A_1|^2 - |B_1|^2$ to the left of the step and $|A_2|^2$ to the right. On the other hand, the probability current you can identify with just an incident stream of particles described by $A_1 e^{ikx}$ is $|A_1|^2$. The flux of reflected particles is $|B_1|^2$ and that of the transmitted particles is $|A_2|^2$, so it is natural to identify $|B_1|^2/|A_1|^2$ as the probability that each particle gets reflected, and likewise for transmitted particles.

More explicitly, we can think about what happens to a normalized wave packet describing a single particle of energy expected value $E$, heading toward the potential step. At its initial position well to the left of the step, the wave packet will be a superposition of the stationary states that you have expressed, with a tight energy distribution around $E$. If you let the system evolve, the wave packet will reach the potential step and split into two: one wave packet reflected and the other transmitted. If you then integrate the probability density $|\psi|^2$ to the left and to the right of $x = 0$, you will find $|B_1|^2/|A_1|^2$ and $|A_2|^2/|A_1|^2$, respectively, corresponding to actual probabilities of reflection and transmission.

$\endgroup$
0
$\begingroup$

My question is, why do textbooks state that $|A2/A1|^2$ is the "probability of escape". Is this just a manner of speaking, or is there real physical basis to speaking like this?

This is likely terminology borrowed by your professor or textbook author from the field of nuclear physics, where one often, treats a similar "barrier" problem (image source): enter image description here

The difference is that in the nuclear escape problem the coordinate axis is the radius, i.e., the waves on the left of the barrier are really not freely propagating but represent (quasi-)bound states. On the other hand, the extent of the central potential well is often big enough to treat the particles as standing waves rather than by solving exactly the Schrödinger equation (aka Bohr-Sommerfeld quantization/quasi-classical approximation.)

When applied to one-dimensional scattering (or any scattering) "probability of escape" is a misnomer. However, it still can be viewed as probability - more precisely "transmission probability", if we normalize the incident wave to unity and speak of "probability flux/current" (However, one also frequently uses normalization to the number of particles or the particle flux, in which case probabilistic interpretation does not apply.)

Related: Conceptual question about reflection probability in the Dirac potential (discussion of 1D scattering matrix.)

$\endgroup$
0
$\begingroup$

Reflection coefficient can be stated as ratio of beam reflected flux to incident flux :

$$ \Gamma = \frac {\Phi_{out}}{\Phi_{in}} \tag 1$$

Flux is basically particle amount per unit time, so : $$ \Gamma = \frac {n_{out}\cdot t^{-1}}{n_{in}\cdot t^{-1}} = \frac {n_{out}}{n_{in}} \tag 2 ,$$

which in essence is part of particles being reflected, which in a classical sense is probability in the range $[0..1]$ that arbitrary particle will be reflected. (Like the probability that you will take a particular suit card from the standard 52 card deck is $\frac {13}{52}=\frac 14$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.