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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

    TSny

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    Homework Helper
    Gold Member

    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

    TSny

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    Homework Helper
    Gold Member

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

    And, yes, all 4 are zero :smile:
     
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