Consider a potential well $V(x)$ given by \begin{align} V(x) = \begin{cases} 0 \qquad &x < 0 \\ -V_{0} \qquad & 0 < x < L \\ 0 \qquad &x > L \end{cases} \end{align} where $V_{0} > 0$. A free particle with energy $E > 0$ is incident from the left. I was asked to derive the transmission coefficient $T$ given by \begin{align} T = \Bigg[ 1 + \frac{V_{0}^{2}}{4E(E+V_{0})} \sin^{2} \Bigg( \frac{L}{\hbar} \sqrt{2m(E+V_{0})} \Bigg) \Bigg]^{-1} \end{align} and I can derive it.

My question is that when the potential well has a very large depth (i.e. $V_{0} \to \infty $), we have $T \to 0$ from the above expression. It seems very non-trivial because the potential well should attract the particle very strongly when $V_{0} \to \infty$. Is there any physical explanation of such behaviour of $T$?

  • 2
    $\begingroup$ Why do you say that $\lim_{V_0 \to \infty} T = 0$? $\endgroup$ – valerio Oct 20 '16 at 10:53
  • $\begingroup$ Possible duplicate of Transparence of an infinite square well? $\endgroup$ – sammy gerbil Oct 20 '16 at 11:10
  • $\begingroup$ Becasue at large $V_{0}$, the coefficient in front of $ \sin^2 $ is of order $V_{0}$ and $\sin^2$ is bounded. Hence the whole expreesion tends to zero. $\endgroup$ – K_inverse Oct 20 '16 at 11:14

Actually, it is easy to see that $T=1$ whenever

$$\frac{L}{\hbar} \sqrt{2m(E+V_0)}=n \pi \ \ \ n \in \mathbb N$$

You can see a representation of $T(V_0)$ below, with $E=1$ and $L \sqrt{2m}/\hbar=1$ (it's just to have a qualitative representation so we don't care about the units).

enter image description here

An interesting thing is that the energies corresponding to complete transmission are

$$E_n +V_0= \frac{(n \pi \hbar)^2}{2 m L^2} \ \ \ n \in \mathbb N $$

which are exactly equal to the energies corresponding to the bound states of the infinite square well.

Quoting R.W. Robinett, Quantum Mechanics, chapter 11:

This effect is easily understood in wave terms as due to the complete destructive interference between waves scattered at the first “step” (for which there is a phase change on reflection) and the second (for which there is no phase change)


What is true is that the average value of $T$ calculated in the interval $[0, V]$ decreases with the length of the interval $V$. This can be seen by numerically evaluating

$$\bar T (V) = \frac 1 V \int_0^V T(V_0) d V_0$$

It is evident that

$$\lim_{V \to \infty} \bar T (V)= \lim_{V \to \infty} \frac 1 V \int_0^V T(V_0) d V_0 = 0$$

since the area of the peaks is always more or less equal but the distance between them increases proportionally to $n$.

About the physical reason of this effect, I'm not entirely sure, but it all boils down to destructive interference between the waves scattered at the two "steps" of the potential.

PS: This app is great.

  • $\begingroup$ Oh thanks a lot!! Btw, it reminds me something in classical EM, the frustrated total internal reflection although it is not exactly the same since $\psi(x)$ in $ 0 < x < L$ is not an evanescent wave. $\endgroup$ – K_inverse Oct 20 '16 at 13:24
  • $\begingroup$ @QMM You're welcome. Yes, there is indeed a strong analogy between the two phenomena. $\endgroup$ – valerio Oct 20 '16 at 14:08

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