# Wave function for a potential barrier

1. Feb 7, 2015

### Joy Prakash Das

Suppose I have a region from 0 to L. I have a barrier potential V from x1 to x2, such that 0<x1<x2<L. The potential is 0 everywhere. I have obtained the wave-function by considering the propagation from left to right. Now if I consider the propagation from right to left should I get a different wave function?

2. Feb 7, 2015

### vanhees71

Yes! I guess you look for the stationary solution (time-independent Schrödinger equation). Then implicitly you assume a time dependence $\Psi(t,x)=\exp(-\mathrm{i} E t) \psi_E(x)$. For wave packets starting at the left, you have (by assumption!) only an outgoing wave packet at the right and waves going in both directions elsewhere, because at your potential steps the wave gets reflected. Now you can formulate the boundary conditions for the case of a wave packet starting at the right going left yourself!

You have a degeneracy in the 1D SchrEq, because of the symmetry under reflections of momentum: $x \rightarrow x$, $p \rightarrow -p$.

3. Feb 7, 2015

### Joy Prakash Das

Thanks vanhees for the reply. I am not considering time dependence. Suppose I am considering a periodic case where L is the period. In that case even the rightmost side will have both incoming and outgoing wave packet. Now the thing is that when I change the wave vector k to -k ,I get another wavefunction but if I solve it I get the same coefficients as in the case of propagation from left to right. This means I get the same wavefunction for both left to right and right to left propagation.

4. Feb 7, 2015

### vanhees71

Ok, then I need the specific problem. For scattering states (i.e., energy eigenstates to energy eigenvalues in the continuous part of the spectrum) you have this degeneracy, i.e., two different (!) eigenfunctions to the same eigenvalue, which is due to the left-right-moving symmetry.

A very thorough discussion of the 1d Schrödinger equation can be found in Messiah's textbook (there's a cheap Dover edition in 1 volume).

5. Feb 7, 2015

### Joy Prakash Das

That was exactly my question. For same eigenvalue E, you have two eigenfunctions, one for +k(moving right) and the other for -k(moving left). I just wanted to know whether these two wavefunctions will be the same or different. I tried solving it, but realised that the wavefunction comes out to be the same. I might have done some mistakes.

6. Feb 7, 2015

### vanhees71

Hm, in general they are different. To make progress here, I'd need the full problem, i.e., the potential.

7. Feb 7, 2015

### Joy Prakash Das

The full problem is this :
There is a one dimensional potential given by :-
V(x) = 0 for 0< x < x1 (Region i)
= V for x1 < x <x2 (Region ii)
= 0 for x2 < x < L (Region iii)

The potential is periodic, meaning V(x+L) = V(x).
Find the wavefunction of this potential everywhere in space.

I have done it by considering the propagation from left to right. What is I now consider it from right to left. In other words, the wave starts from region iii.