I am calculating the phase shift from a 1-dimensional potential well. This seems extremely simple, but I am just getting so confused by it.
Let there be a potential well of depth $V_0$ and spatial width $2a$. Consider a scattering solution with mass parameter $m$ and energy $E$, incident from the left. The wavefunctions to the left and right of the potential well are given by
$$\psi_{-\infty}(x)=Ae^{ikx}+Be^{-ikx}$$
$$\psi_{\infty}(x)=E^{ikx}$$
Where $k=\sqrt{2mE}/\hbar$, $A$ is in arbitrary incoming-wave complex amplitude, $B$ is the reflected wave complex amplitude, and $E$ is the transmitted wave (post-well incidence of course) complex amplitude. What I'm concerned with is the phase shift of the transmitted wave relative to the incoming wave, so what I want is $E$ in terms of $A$. This can be calculated as
$$\frac{E}{A}=\frac{e^{-2ika}}{\cos(2k'a)-i\frac{\epsilon}{2}\sin(2k'a)}$$
Where the (new) parameters $k'$ and $\epsilon$ are given by
$$k'=\frac{\sqrt{2m(E+V_0)}}{\hbar},\,\,\,\,\epsilon=\frac{k'}{k}+\frac{k}{k'}$$
By taking the square modulus, one arrives at the correct expression for the transmission coefficient (I won't write it here, but its a safety check).
Here's my problem. The phase shift would simply be given by the phase of $E/A$. When I calculate the phase (or just look at the expression and write it down), I get
$$\delta=\arctan \left(\frac{\epsilon}{2}\tan (2k'a)\right) – 2ka$$
Here is a sample plot of this (convenient units, random parameter values):
What this is saying is the that phase-shift increases without bound (or moving seamlessly through $90^o$ and $-90^o$. I would expect the phase shift to decrease with increasing energy, eventually settling at zero. I'm pretty sure I've messed up something. Could you help me out?
Best Answer
I figured it out. The problem was that I was simply adding the two phase shifts, ignoring the periodicity in the tangent and arctangent functions. In order to get the proper result/plot, one can resort to the addition formula for tangents
$$\delta=\delta_1+\delta_2=\arctan \left(\frac{\frac{\epsilon}{2}\tan(2k'a)-\tan(2ka)}{1+\frac{\epsilon}{2}\tan(2k'a)\tan(2ka)}\right)$$