1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 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

    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
    which does not seem to be zero, trivially.
  2. jcsd
  3. Sep 30, 2009 #2


    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Proof of how [a+,[a+,a]]=0
  1. Rabcd = 0 proof (Replies: 1)