3
$\begingroup$

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):

enter image description here

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?

$\endgroup$
2
$\begingroup$

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

$$\delta=\delta_1+\delta_2=\arctan \left(\frac{\frac{\epsilon}{2}\tan(2k'a)-\tan(2ka)}{1+\frac{\epsilon}{2}\tan(2k'a)\tan(2ka)}\right)$$

$\endgroup$
  • $\begingroup$ By 'scattering phase shift' do we mean the phase shift of the transmitted wave relative to the incoming wave? Is there any phase shift of the reflected wave? Is that considered as the 'scattering phase shift'? $\endgroup$ – omehoque Nov 17 '18 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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