Proving momentum expectation value vanishes for bound states

[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.

Thanks in advance!

 

[Haelfix]

there's several ways to do this, one elegant way, one brute force way, one abstract mathematical way (probably not suitable if this is a first course).

I'll give you a hint on the brute force way. You are going to want to think of what the operator P is. Strictly speaking, in three dimensions it looks like

P = -i hbar * del. In one dimension its p = -i hbar d/dx

Use the Schroedinger formalism and plow away =)

The abstract method hint is to think of what P does to your state space. Hmm, it looks like a translational operator. Maybe what you are looking for is a statement of translational symmetry.

 

[lethe]

for a bound state, the wavefunction drops to zero at infinity, which allows you to use integration by parts to show that

$$\langle p\rangle=m\frac{d\langle x\rangle}{dt}$$

and in a stationary state (i.e. energy eigenvalue), all expectation values are time independent, so the derivative vanishes.

 

[einai]

Thanks, Haelfix and lethe .

P = -i hbar * del. In one dimension its p = -i hbar d/dx

Use the Schroedinger formalism and plow away =)

I thought about using the operator form of p, but I wasn't sure how it acts on the energy eigenstate |E>. Can I just say that after it takes the x derivative of |E>, the state becomes orthorgonal to the original |E>, ie,

$$<E|-i\hbar \frac{d}{dx}|E> = <E|E'> =0$$

because |E'> is now orthorgonal to |E>?

Hmm...I don't think I'm doing the right thing.

 

[einai]

I think I figured it out. I used the commutation relation p = - i m hbar*[H,x].
Thanks.