Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it possible for a particle trapped in a 1D finite potential well to evolve from a even state to an odd state and vice-versa? Why?

share|cite|improve this question
up vote 5 down vote accepted

I assume OP means an even potential $V(x)~=~V(-x)$, e.g., a finite square well potential $V(x) ~\propto~ \theta(|x|-a) $.

Then the answer to the question(v1) is No.

Sketched proof: Under the assumption that $V$ is even, the Hamiltonian

$$H= \frac{p^2}{2m}+V(x)$$ then commutes with the parity operator $P$. So the operators $H$ and $P$ can be diagonalized simultaneously. So there exists a complete set of energy eigenstates that are either even or odd. Let's call them $e_i(x)=e_i(-x)$ and $o_j(x)=-o_j(-x)$, respectively. An initially even state


is hence a linear combination of even energy eigenstate only

$$\psi(x,t\!=\!0)~=~\sum_i c_i e_i(x). $$

The wave function

$$ \psi(x,t)~=~\sum_i c_i e_i(x) \exp\left[-\frac{\mathrm{i}t E_i}{\hbar}\right]~=~\psi(-x,t) $$

remains an even function also at a future time $t$, and can hence not become odd.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.