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Proof of how [a+,[a+,a]]=0

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Actually the original question is too long but I need help for this tiny part that confuses me. I'll be done if I can show that
    [a+,[a+,a]]=0 and similarly
    [a,[a+,a]]=0
    [a,[a,a+]]=0 etc
    where a+ is the raising and a is the lowering ladder operator in quantum oscillator.

    2. Relevant equations
    I tried the formulas
    [A,[B,C]]= -[C,[A,B]] -[B,[C,A]] and
    a[tex]\psi[/tex]n=[tex]\sqrt{n}[/tex][tex]\psi[/tex]n-1
    a+[tex]\psi[/tex]n=[tex]\sqrt{n+1}[/tex][tex]\psi[/tex]n+1

    3. The attempt at a solution

    When I use the formula I found
    [a+,[a+, a]]= -[a,[a+,a+]]- [a+,[a,a+]]
    On the rhs, first term is zero but I have no idea about the second

    Also I tried to use [tex]\psi[/tex]n for this operator but it gives something like
    [a,[a,a+]][tex]\psi[/tex]n=(n3/2+1)[tex]\psi[/tex]n-(n3/2+n1/2)[tex]\psi[/tex]n-1
    which does not seem to be zero, trivially.
     
  2. jcsd
  3. Sep 30, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are making this way too complicated. Isn't [a,a^(+)]=1? Isn't that what a raising/lowering operator means?
     
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