- #1
einai
- 27
- 0
Hi, I came across a problem which seems to be pretty simple, but I'm stuck .
Given a Hamiltonian:
[tex]H=\frac{\vec{p}^2}{2m}+V(\vec{x})[/tex]
If |E> is a bound state of the Hamiltonian with energy eigenvalue E, show that: [tex]<E| \vec{p} |E>=0[/tex]
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So I've been trying something like this:
[tex]\frac{1}{2m}<E|\vec{p} \cdot \vec{p}|E> + <E|V(\vec{x})|E> = E<E|E> = E[/tex]
but I have no idea how to proceed from here.
Thanks in advance!
Given a Hamiltonian:
[tex]H=\frac{\vec{p}^2}{2m}+V(\vec{x})[/tex]
If |E> is a bound state of the Hamiltonian with energy eigenvalue E, show that: [tex]<E| \vec{p} |E>=0[/tex]
-----------------------------------
So I've been trying something like this:
[tex]\frac{1}{2m}<E|\vec{p} \cdot \vec{p}|E> + <E|V(\vec{x})|E> = E<E|E> = E[/tex]
but I have no idea how to proceed from here.
Thanks in advance!