# 1d potential V (-x)=-V (x) eigenfunctions.

1. Oct 5, 2014

### sudipmaity

1. The problem statement, all variables and given/known data
Show that for a 1d potential V (-x)=-V (x), the eigen functions of the Schrödinger equation are either symmetric/ anti-symmetric functions of x.

2. Relevant equations

3. The attempt at a solution
I really don't know how to do it for odd potential.
Let me show you how I am doing it for even potential.
V (x)=V (-x)
- (h2 / 2μ) [d2ψ(x)/dx2] + V (x) ψ(x)=E ψ (x).... (1)
Making x to -x transformation we get
- (h2 / 2μ) [d2ψ(-x)/dx2] + V (x) ψ(-x)=E ψ (-x)..(2)
where i use V (-x)=V (x)
Comparing (1) and (2) we see that ψ (x) and ψ(-x) eigenfunctions belongs to the same
energy E.
For a non degenerate state, ψ (-x ) must be a multiple of ψ (x):
Ψ (-x) = λψ (x).
Clearly ψ (x)= λψ (-x)=λ2ψ (x).
λ2=1 or λ=+/- 1
So ψ(-x)=+/- ψ(x)
Now if i use odd potential in (2) eigenfunctions no longer belong to same energy E.
The hamiltonian becomes weird for negative potential.

2. Oct 5, 2014

### BvU

Hello Sudip,
I like your solution to the opposite problem (even potential), but I don't see how you have proven that there is no degeneracy.
My book (Merzbacher, quantum mechanics, about 50 years old :-) ) only says that if $\psi(x)$ is a solution, then ${1\over 2} [\psi(x)+\psi(-x)]$ and ${1\over 2} [\psi(x)-\psi(-x)]$ are also solutions.

If I imagine an odd-parity potential, e.g. a well from -a to 0 and a barrier from 0 to +a, I find it hard to believe the either/or statement can be proven. An even simpler V is a step from -V0 to + V0 at x = 0.

I think any reasonable odd-parity V has to go to 0 when away from x=0, so there eigenfunctions are close to plane waves.

Sorry I can't help you any better. Hope someone else can; if no: let us know if you find a sensible solution from teacher or by yourself.

3. Oct 6, 2014

### vela

Staff Emeritus
I agree that the problem is flawed.

4. Oct 6, 2014

### sudipmaity

How did I prove non degeneracy??
Well , 1d bound states are non degenerate aren't they?
I would like to quote somethings from a book by Dr.S.N.Ghosal(student of Emilio Segre) here :
"The above classification of wavefunctions [λ=+/-1 so ψ(-x)=+/- ψ(x)] according to parity is possible only if Hamiltonian H^ remains invariant under parity operation i.e., if the interaction potential V (x) remain unchanged by parity operation, V (-x)=V(x)"
Doesn't this statement raises question on the feasibility of the problem?
I want to know in a simple yes or no is it possible to find symmetric as well antisymmetric functions for a potential well
V(x)=-V0 for -a <x <0
V (x)=0 elsewhere.
Can't seem to find the above problem anywhere.A purely odd potential.

5. Oct 6, 2014

### vela

Staff Emeritus
No, it's not possible. Your V(x) isn't odd either.

6. Oct 6, 2014

### sudipmaity

Sorry.
Will you say :
V (x)= -V (a) for -a <x <0
V (x)=0 elsewhere
is odd??

7. Oct 6, 2014

### vela

Staff Emeritus
No. An odd function satisfies V(-x) = -V(x) for all x. For your potential, V(a/2) = 0 while V(-a/2) = -V(a) ≠ 0 (I assume).

BvU gave you an example of a simple odd potential: V(x) = V0 for x>0 and -V0 for x<0. Try solving the Schrodinger equation with this potential. You'll find the solutions aren't odd or even.

8. Oct 6, 2014

### BvU

I have to correct myself: after all the step function is a counter-example. V can't go to $\infty$, but it doesn't have to go to 0 to be (mathematically - and thereby also physically) acceptable.

I bring in another interesting potential function: a periodic square well . When x=0 is chosen such that V is even, eigenfunctions are even, but a shift of 1/4 period makes the potential odd, but not the eigenfunctions.