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 question is two long and I'll be done if I can show that
    [a+,[a+, a]=0 and similarly
    [a,[a+, a]=0
    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
     
  2. jcsd
  3. Sep 30, 2009 #2
    The Jacobi Identity is [A,[B,C]] = [[A,B],C] + [B,[A,C]]
     
  4. Sep 30, 2009 #3
    why can't I see a button "delete topic" because this is the wrong topic, the original one is the other one
     
    Last edited: Sep 30, 2009
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)

Loading...