- #1

- 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]

-----------------------------------

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. I don't think I'm on the right track actually.

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. I don't think I'm on the right track actually.

Thanks in advance!

Last edited: