Quantum Well & Barrier w/ Phase Shift

[darkfall13]

Homework Statement

V(x) = [tex]\inf[/tex] if x [tex]\leq[/tex] 0
= -V if 0 [tex]<[/tex] x [tex]<[/tex] a
= 0 if x [tex]>[/tex] a

NOTE: E>V

Find the wave functions in each region and the stated phase shift.

Homework Equations

Schrodinger Eq.

Instructions note that wave functions are:
[tex]\Psi_I[/tex] = [tex]A\sin{kx}[/tex]
[tex]\Psi_{II}[/tex] = [tex]B\sin{k'x+\phi}[/tex]

where,

k = [tex]\sqrt{2m(V+E)}/\hbar[/tex]
k' = [tex]\sqrt{2mE}/\hbar[/tex]

Show [tex]2\phi = 2\left[\cot^{-1}\left(\frac{k}{k'}\cot\left({ka}\right)\right)-k'a\right][/tex]

The Attempt at a Solution

I can get Region I just fine (0:a well)

[tex]\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2} - V \Psi = E \Psi [/tex]
[tex]-\frac{d^2 \Psi}{dx^2} = \left(E+V\right)\Psi\left(\frac{2m}{\hbar^2}[/tex]
let [tex]k = \frac{\sqrt{2m\left(V+E\right)}}{\hbar}[/tex]
[tex] \Psi_x\left(x\right) = A\sin{kx} [/tex]

Region II

[tex]\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2} = E \Psi [/tex]
[tex]-\frac{d^2\Psi}{dx^2} = \frac{2mE}{\hbar^2}\Psi [/tex]
[tex] \Psi_{II}\left(x\right) = B\sin{k'x} [/tex]
where [tex]k' = sqrt{2mE}{\hbar}[/tex]

But my problem lies in the phase shift. How is it related to the wave functions and how can it be calculated?

 

[Jhero]

The wave functions in each region can be calculated using the Schrodinger equation, as you have correctly done. However, to calculate the phase shift, we need to look at the boundary conditions at x = 0 and x = a. At these points, the wave function must be continuous and the derivative of the wave function must be continuous as well. This means that the wave function and its derivative must have the same value at these points in both regions.

Using this condition, we can set up the following equations:

\Psi_I\left(0\right) = \Psi_{II}\left(0\right)
\Psi_I'\left(0\right) = \Psi_{II}'\left(0\right)

Substituting in the wave functions we have already found, we get:

A\sin{0} = B\sin{\phi}
Ak = Bk'\cos{\phi}

Solving for \phi, we get:

\phi = \cot^{-1}\left(\frac{k}{k'}\cot\left(ka\right)\right) - k'a

This is the phase shift between the wave functions in Region I and Region II. I hope this helps you with your calculations. Let me know if you have any further questions. Good luck with your work!