einai
Mar26-04, 09:07 PM
Hi, I came across a problem which seems to be pretty simple, but I'm stuck :confused: .
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
-----------------------------------
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.
Thanks in advance!
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
-----------------------------------
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.
Thanks in advance!