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Confused by simple quantum problem

  1. Mar 26, 2004 #1
    Hi, I came across a problem which seems to be pretty simple, but I'm stuck :confused: .

    Given a Hamiltonian:

    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!
  2. jcsd
  3. Mar 27, 2004 #2


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    Theres 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.
  4. Mar 27, 2004 #3
    for a bound state, the wavefunction drops to zero at infinity, which allows you to use integration by parts to show that

    [tex]\langle p\rangle=m\frac{d\langle x\rangle}{dt}[/tex]

    and in a stationary state (i.e. energy eigenvalue), all expectation values are time independent, so the derivative vanishes.
    Last edited: Mar 27, 2004
  5. Mar 27, 2004 #4
    Thanks, Haelfix and lethe :redface: .

    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,

    [tex]<E|-i\hbar \frac{d}{dx}|E> = <E|E'> =0[/tex]

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

    Hmm....I don't think I'm doing the right thing.
    Last edited: Mar 27, 2004
  6. Mar 27, 2004 #5
    I think I figured it out. I used the commutation relation p = - i m hbar*[H,x].
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