Why Does the Expectation Value of Momentum Vanish in a Quantum Bound State?

[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.

Thanks in advance!

 

[Doc Al]

FYI: This question has been answered in the quantum physics forum.

 

[M98Ranger]

Hi there! It looks like you're on the right track, but there are a few things you can do to simplify your approach. First, remember that the momentum operator \vec{p} is a vector, so it can be written as \vec{p}=\vec{p}_x+\vec{p}_y+\vec{p}_z. This means that the dot product \vec{p}\cdot\vec{p} can be expanded as:

\vec{p}\cdot\vec{p} = (\vec{p}_x+\vec{p}_y+\vec{p}_z)\cdot(\vec{p}_x+\vec{p}_y+\vec{p}_z) = \vec{p}_x^2 + \vec{p}_y^2 + \vec{p}_z^2

Since we're interested in the expectation value of \vec{p}, we can write this as:

<E|\vec{p}\cdot\vec{p}|E> = <E|\vec{p}_x^2 + \vec{p}_y^2 + \vec{p}_z^2|E>

Now, remember that in a bound state, the energy eigenvalue E is fixed, so we can pull it out of the expectation value. This gives us:

<E|\vec{p}\cdot\vec{p}|E> = E<E|\vec{p}_x^2 + \vec{p}_y^2 + \vec{p}_z^2|E>

Using the definition of the momentum operator, we can write \vec{p}_i^2 = \frac{\vec{p}_i\cdot\vec{p}_i}{2m} for each component i=x,y,z. This gives us:

<E|\vec{p}\cdot\vec{p}|E> = E<E|\frac{\vec{p}_x\cdot\vec{p}_x+\vec{p}_y\cdot\vec{p}_y+\vec{p}_z\cdot\vec{p}_z}{2m}|E>

Now, using the commutation relation [\vec{p}_i,\vec{p}_j]=0 for all i,j=x,y,z, we can rearrange this expression as:

<E|\vec{p}\cdot\vec{p}|E> = E<E|\frac{\vec{p}_x