Consider the scattering of a quantum particle in one dimension, caused by a step in the potential (this appears in many undergrad level QM books):

$$ V(x) = \begin{cases} V_1 & x<0 \\ V_2 & x>0\end{cases}. $$

The particle is incident from the left, so it's wavefunction is:

$$ \psi(x) = \begin{cases} e^{i k_1 x} + r e^{-i k_1 x} & x<0 \\ t e^{i k_2 x} & x>0\end{cases}, $$

where $k_i =\sqrt{2m(E-V_i)}/\hbar$.
Matching the wavefunction and its derivative at $x=0$ gives:

$$ r = \frac{k_1-k_2}{k_1+k_2} ~~~;~~~ t = \frac{2 k_1}{k_1+k_2}.$$

Now we put another step in the potential at some distance $L$, which makes it a box potential:

$$ V(x) = \begin{cases} V_1 & x<0 \\ V_2 & 0<x<L \\ V_1 & L<x\end{cases}. $$

We solve this in a similar manner as before, with the wavefunction:

$$ \psi(x) = \begin{cases} e^{i k_1 x} + r e^{-i k_1 x} & x<0 \\ a e^{i k_2 x} + b e^{-i k_2 x} & 0<x<L \\ t e^{i k_1 x} & L<x \end{cases}. $$

Matching the wavefunction and its derivative at $x=0,L$ gives:

$$ r = \frac{k_1^2-k_2^2}{k_1^2+k_2^2+2 i k_1 k_2 \cot{(k_2 L)} } ~~~;~~~ t = \text{(something)}.$$

How come the second scattering problem doesn't reproduce the first scattering problem in the limit $L \rightarrow \infty$?
I'm looking only at the value of $r$. I send a particle in, it scatters, and I get something back with an amplitude $r$. It seems unphysical that if the potential changed at $x=L$, it changes the scattering at $x=0$, no matter how far $L$ is.


1 Answer 1


In 1D particles propagate without decay. That is, the propagator of a free particle (or of a particle under constant potential lower than its energy, see comment below) does not decay with distance. Therefore, the particle will "sense" any change in the potential at any distance.

Think of the ripples created by a stone thrown into a lake. If the lake's surface is the usual 2D, the waves decay with distance and therefore you'll see a different pattern for different sizes of lakes.

If, on the other hand, the lake is 1D (a wave guide), then the waves do not decay with distance and you'll feel the boundary of your lake no matter how far it is (the only thing that'll change is the phase with which the reflected waves will return - exactly like in your calculation).


BTW - If $E<V_2$ then $k_2$ is imaginary, and therefore $\cot(k_2L)\to -i$ for $L\to\infty$ and the two expressions coincide. This is because in this case the propagator does decay with distance.

  • $\begingroup$ Small correction: $\cot(k_2 L) \rightarrow i ~ \text{sgn}[\text{Im}(k_2)]$. About the rest I need to think a little... $\endgroup$
    – Joe
    Jan 27, 2013 at 14:44
  • $\begingroup$ @Joe Well, unless your function grows exponentially for large $x$, you have no freedom in choosing the sign of $k$. $\endgroup$
    – yohBS
    Jan 27, 2013 at 21:24
  • $\begingroup$ That's true, but in fact in order to avoid divergence we need $\text{Im}(k_2)>0$, which means $\cot(k_2 L) \rightarrow i$, and then the limit reproduces the step potential case only up to a sign change in $k_2$. But actually I think I can live with that. Other than that I accept your answer :) $\endgroup$
    – Joe
    Jan 28, 2013 at 7:12

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.