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Creation and Annihilation Operators

  1. Dec 9, 2012 #1
    We know that

    a|n> = √n | √(n-1)>

    and

    a' |n> = √(n+1) | n + 1 >

    so, If we use this to find

    <n|a'a|n>

    it would be equal to n

    <n|a'a|n> = n

    Am I correct?

    I'm not really sure about my calculations.

    I operate with a first so.

    <n|a'a|n>
    <n|a' √n | √(n-1)>

    = n

    ????

    Can someone please help me with the complete solution?
     
  2. jcsd
  3. Dec 9, 2012 #2

    mfb

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    Staff: Mentor

    <n|a'a|n> = <n|a' √n | (n-1)> (no square root for the state)
    = √n <n|a' | (n-1)> (√n is a scalar, you can pull it out)
    = √n <n|√n | n>
    = √n √n <n|n>
    =n
     
  4. Dec 9, 2012 #3
    Thanks :)) yah no √ for the state. Sorry.
     
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