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

1. Sep 30, 2009

### meanyack

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$$\psi$$n=$$\sqrt{n}$$$$\psi$$n-1
a+$$\psi$$n=$$\sqrt{n+1}$$$$\psi$$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 $$\psi$$n for this operator but it gives something like
[a,[a,a+]]$$\psi$$n=(n3/2+1)$$\psi$$n-(n3/2+n1/2)$$\psi$$n-1
which does not seem to be zero, trivially.

2. Sep 30, 2009

### Dick

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