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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?

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1 Answer 1

up vote 3 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

$$\psi(x,t\!=\!0)~=~\psi(-x,t\!=\!0)$$

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.

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