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Creation and annihilation operators

  1. Apr 17, 2015 #1
    - a|n>=C|n-1>
    - a+|n>=D|n+1>

    And because |n-1> is normalized, <n-1|n-1>=1: (<n|a+)(a|n>)=C2
    Thus, <n|a+a|n>=C2

    Where a is the annihilation operator and a+ is the creation operator

    I don't understand this as isn't <n|a+=<n+1|D , thus <n|a+a|n> =<n+1|DC|n-1> instead?

    This is from the quantum physics for dummies.
     
  2. jcsd
  3. Apr 17, 2015 #2

    Orodruin

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    No, the annihilation operator raises ket states, which it acts on from the left, but lowers bra states. The relation is simply the conjugation of a|n> = C|n-1>. Conjugation switches the order of the expressions, turn bras into kets, kets into bras, and conjugates operators.
     
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