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SHO ladder operators & some hamiltonian commutator relations

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    For the SHO, find these commutators to their simplest form:
    [itex] [a_{-}, a_{-}a_{+}] [/itex]
    [itex]
    [a_{+},a_{-}a_{+}]
    [/itex]
    [itex]
    [x,H]
    [/itex]
    [itex]
    [p,H]
    [/itex]


    2. Relevant equations



    3. The attempt at a solution
    I though this would be an easy problem but I am stuck on the first two parts. Here's what I did at first:
    [itex]
    [a_{-}, a_{-}a_{+}]\psi = a_{-}(n+1)\psi_{n} - a_{-}n\psi_{n} = a_{-}\psi_{n}
    [/itex]
    [itex]
    = \sqrt{n}\psi_{n-1}
    [/itex]

    [itex]
    [a_{+}, a_{-}a_{+}]\psi = a_{+}(n+1)\psi_{n} - a_{-}a_{+}\sqrt{n+1}\psi_{n+1}
    [/itex]
    [itex]
    = (n+1)^{3/2}\psi_{n+1} - (n+1)^{3/2}\psi_{n+1} = 0
    [/itex]


    Now what I am confused about is the [itex]\psi_{n-1}[/itex] term in the first commutator. Surely there is a general form of the commutator without the test wavefunction? And I can't just drop this term and have root of n as the result. So did I do something wrong?

    I tried the first part again using the explicit form of the ladder operators, in terms of H, p, x, with all of the constants. What I have gotten so far looks quite messy and involves [itex] [p,H] [/itex] and [itex] [x,H] [/itex], which I've yet to compute and are the last two parts of the problem... So it seems this route is not the easiest?

    Any assistance is appreciated!
     
    Last edited: Oct 15, 2013
  2. jcsd
  3. Oct 15, 2013 #2
    Duh! The first answer is just [itex] a_{-}[/itex]

    And I didnt realize you could just use the "product rule" for commutators to simplify the algebra a bit!

    All set now
     
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