V(x) is an arbitrary repulsive potential localized at a position along the x-axis as shown below.

The solution of the Schroedinger equation must be of the
form

ψ(x) = A e^{ikx} + B e^{-ikx} for x <
-a,

ψ(x) = C e^{ikx} + D e^{-ikx} for x > a.

Write

,

and assume A and B are arbitrary complex numbers. Use
conservation of flux to show that

|S_{11}|^{2}
+ |S_{21}|^{2} = |S_{12}|^{2} + |S_{22}|^{2}
= 1,

and that

S_{11}S_{12}^{*}
+ S_{21}S_{22}^{*} = 0.

Show that __S__ is a unitary matrix.

Solution:

- Concepts:

Probability flux - Reasoning:

Flux conservation requires that the flux into a region is equal to the flux out of that region. - Details of the calculation:

Flux: j(x,t) =[ħ/(2mi)][ψ*(x,t) ∂ψ(x,t) ∂x - ψ*(x,t) ∂ψ(x,t) ∂x]

Here j(x,t) = [ħk/m][|A|^{2}- |B|^{2}] for x < -a and j(x,t) = [ħk/m][|C|^{2}- |D|^{2}] for x > a.Flux conservation requires that |A|

^{2}- |B|^{2}= |C|^{2}- |D|^{2}, or |A|^{2}+ |D|^{2}= |C|^{2}+ |B|^{2}.

(Flow into the region where V(x) is not zero equals flow out of the region where V(x) is not zero.)

C = S_{11}A + S_{12}D. B = S_{21}A + S_{22}D.

|C|^{2}+ |B|^{2}= |S_{11}|^{2}|A|^{2}+ |S_{12}|^{2}|D|^{2}+ S_{11}S_{12}^{*}AD^{*}+ S_{11}^{*}S_{12}A^{*}D + |S_{21}|^{2}|A|^{2}+ |S_{22}|^{2}|D|^{2}+ S_{21}S_{22}^{*}AD^{*}+ S_{21}^{*}S_{22}A^{*}D

= (|S_{11}|^{2}+ |S_{21}|^{2})|A|^{2}+ (|S_{12}|^{2}+ |S_{22}|^{2})|D|^{2}+ (S_{11}S_{12}^{*}+ S_{21}S_{22}^{*})AD^{*}+ (S_{11}^{*}S_{12}+S_{21}^{*}S_{22})A^{*}D

= |A|^{2}+ |D|^{2}.

This implies that

|S_{11}|^{2}+ |S_{21}|^{2}= 1, |S_{22}|^{2}+ |S_{12}|^{2}= 1, S_{11}S_{12}^{*}+ S_{21}S_{22}^{*}= 0..

Similarly

,

S is unitary.