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I Raising the ladder operators to a power

  1. Dec 8, 2018 #1
    Hi! I am working on homework and came across this problem:


    I know X = ((ħ/(2mω))1/2 (a + a+))

    And if I raise X to the 5th, its becomes X5 = ((ħ/(2mω))5/2 (a + a+)5)

    What I'm wondering is, is there anyway to be able to solve this without going through all of the iterations the raising and lowering operators will create being raised to the 5th power? Or will I just have to tackle every strain this will produce?
  2. jcsd
  3. Dec 8, 2018 #2

    Paul Colby

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    One thing you might notice is that only terms with equal numbers of ##a^+## and ##a## will be non zero.
  4. Dec 8, 2018 #3
    XD Oh man, thanks, that actually makes this problem a cakewalk
  5. Dec 8, 2018 #4

    king vitamin

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    Even more generally, there is something called Wick's theorem which is invaluable to computing these sorts of expectation values. It essentially uses the trick pointed out by Paul Colby, but more explicitly reduces computing harmonic oscillator expectation values to a combinatorics problem (rather than laboriously doing all the commutations).
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