# Need help with simple quantum problem

1. Mar 26, 2004

### einai

Hi, I came across a problem which seems to be pretty simple, but I'm stuck .

Given a Hamiltonian:
$$H=\frac{\vec{p}^2}{2m}+V(\vec{x})$$

If |E> is a bound state of the Hamiltonian with energy eigenvalue E, show that: $$<E| \vec{p} |E>=0$$

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So I've been trying something like this:

$$\frac{1}{2m}<E|\vec{p} \cdot \vec{p}|E> + <E|V(\vec{x})|E> = E<E|E> = E$$

but I have no idea how to proceed from here. I don't think I'm on the right track actually.