1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Harmonic Oscillator

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Given a quantum harmonic oscillator, calculate the following values:
    [itex]\left \langle n \right | a \left | n \right \rangle, \left \langle n \right | a^\dagger \left | n \right \rangle, \left \langle n \right | X \left | n \right \rangle, \left \langle n \right | P \left | n \right \rangle[/itex]

    2. Relevant equations
    Hamiltonian: [itex]H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2[/itex]
    Ladder operators:
    [itex]a=\sqrt{\frac{m\omega}{2\hbar}}\left ( X + \frac{i}{m\omega}P \right )[/itex]
    [itex]a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(X-\frac{i}{m\omega}P \right )[/itex]
    [itex]\left [ a,a^\dagger \right ] = 1[/itex]
    [itex]\left [ a^\dagger a,a \right ] = -a[/itex]
    N operator:
    [itex]N=a^\dagger a[/itex]
    [itex]N\left | n \right \rangle = n\left | n \right \rangle[/itex]
    [itex]a^\dagger \left | n \right \rangle = \sqrt{n+1} \left | n+1 \right \rangle[/itex]
    [itex]a \left | n \right \rangle = \sqrt{n} \left | n-1 \right \rangle[/itex]
    [itex]\left | n \right \rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle[/itex]

    3. The attempt at a solution
    [itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
    \sqrt{n} \left \langle n | n-1 \right \rangle =
    \sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle =
    \sqrt{n} \left ( \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle \right ) ^ \dagger \frac{(a^\dagger)^{n-1}}{\sqrt{(n-1)!}} \left | 0 \right \rangle =
    \sqrt{\frac{n}{n!(n-1)!}} \left \langle 0 \right | a^n (a^\dagger)^{n-1} \left | 0 \right \rangle[/itex]

    Not sure how to continue from here...
  2. jcsd
  3. Dec 5, 2013 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    When you get to [itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
    \sqrt{n} \left \langle n | n-1 \right \rangle [/itex] you should be able to see the answer.
  4. Dec 5, 2013 #3
    Is the answer zero because eigenvectors are always perpendicular to one another?
    Doesn't that mean that all the four values in the question are zero?
  5. Dec 5, 2013 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Yes, eigenvectors of a Hermitian operator that correspond to different eigenvalues are orthogonal.

    And, yes, all 4 are zero :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted