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Particle Number Operator (Hermitian?)

  1. Nov 17, 2013 #1
    Particle Number Operator (Hermitian??)

    Hey guys,

    I'm studying the quantic harmonic oscillator and I'm using "Cohen-Tannoudji Quantum Mechanics Volume 1".

    At some point he introduced the particle number operator, N, such that:

    N=a+.a , where a+ is the conjugate operator of a.

    The line of proof that they used to show that N is hermitian is the following:

    N+=a+.(a+)+=N , where N+ is the conjugate operator of N.

    Somehow, in ways I cannot understand, this proves it.

    But everytime I try and prove it myself I get to the conclusion that N is not hermitian.

    Can someone be kind enough and help me with this :)

    Thank you
     
  2. jcsd
  3. Nov 17, 2013 #2

    George Jones

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    Did you mean to write this? It should be N^+ = (a^+ a)+ = a+ (a+)+ = a+ a = N.
     
  4. Nov 17, 2013 #3
    Hi George,

    I'm not sure I get the transition:

    (a+. a)+=a+.(a+)+

    Back me up here, if I have the operator (AB) and I apply the conjugate it gives me:

    (AB)+=A+.B+

    Right?

    (I seem to be missing something really obvious here :s)
     
  5. Nov 17, 2013 #4

    George Jones

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    No, (AB)+ = B+ A+
     
  6. Nov 17, 2013 #5

    Oh...I seem to have misread that property (shame on me).

    Thank you :)
     
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