# Scattering from a box potential of width $L$ doesn't reproduce a step potential in the limit $L \rightarrow \infty$

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.

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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.
 Small correction: $\cot(k_2 L) \rightarrow i ~ \text{sgn}[\text{Im}(k_2)]$. About the rest I need to think a little... – Joe Jan 27 at 14:44 @Joe Well, unless your function grows exponentially for large $x$, you have no freedom in choosing the sign of $k$. – yohBS Jan 27 at 21:24 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 :) – Joe Jan 28 at 7:12