Why the even Eigen states have less energy than odd ones?

1. Dec 24, 2011

hokhani

In the symmetrical potentials the Eigenstates are either even or odd. why the even states have
less energy than odd states?

2. Dec 24, 2011

Staff Emeritus
What makes you think this is true?

Consider a 1D harmonic oscillator. The n=1 state has less energy than n = 2. The n = 3 state has less energy than n = 4. The n = 5 state has less energy than n = 6.

3. Dec 26, 2011

hokhani

Sorry;
I meant, why the ground state in symmetric potentials is always an even state?

4. Dec 26, 2011

aesir

For one dimensional problems the eigenstates are ordered by their number of nodes (the points in which the wavefunctions is 0).
So an odd wavefunction can not be the ground state.

5. Dec 26, 2011

hokhani

All right, thanks
But, what is the analytic justification here?

6. Dec 27, 2011

aesir

I found a general demonstration in Courant & Hilbert "Methods of Mathematical Physics" Vol.I p 452 (you can read it from google books).

But for one dimension it can be intuitively understood by elementary methods.

Consider a real function which solves the equation for a generic energy E, but obeys the border condition only on one side, say the left one, so E is not an eigenvalue.
For border conditions take as usual that the function vanishes at infinity.
When E is very low (with respect to the potential) you can prove that it has no nodes (because it has the same sign on the whole domain), and that it does not go to 0 on the right side (for example something like e^x).
Then let E grow until it obeys the border condition on the right side: this is the fundamental level and it still has no nodes.
Then for bigger E, looking at the signs of f(x) and f''(x) you can prove that the number of nodes continue increasing.

Does this make any sense to you?

7. Dec 28, 2011

hokhani

Yes; to some extent.
Thanks.
But i am trying to explain it by using the concept of curvature of the wave function.

Last edited: Dec 28, 2011
8. Dec 28, 2011