# Prove some relations but going round in circles

1. May 18, 2012

### Gregg

1. The problem statement, all variables and given/known data

I need to prove some relations but going round in circles.

$[\hat{J}_z, \hat{J}_+] = \hbar J_+$

I've got this:

$\left(a_+^{\dagger }a_+-a_-^{\dagger }a_-\right)\left(a_+^{\dagger }a_-\right)-\left(a_+^{\dagger }a_-\right)\left(a_+^{\dagger }a_+-a_-^{\dagger }a_-\right)$

2. Relevant equations

$J_{\pm} = \hbar a_{\pm}^{\dagger} a_{\mp}$

$J_z = \frac{\hbar}{2} \left(a_+^{\dagger }a_+-a_-^{\dagger }a_-\right)$

3. The attempt at a solution

I have these sort of obvious relations to work from

$[a_{\pm}^{\dagger}, a_{\pm}] = 1$
$[a_{\pm}^{\dagger}, a_{\mp}] = 0$
$[a_{\pm}^{\dagger}, a_{\mp}^{\dagger}] = 0$
$[a_{\pm}^{\dagger}, a_{\mp}] = 0$
$[a_{\pm}, a_{\mp}] = 0$

I start expanding and it gets too tough, what am I missing?

2. May 18, 2012

### Clever-Name

I'd suggest working in commutator notation instead of expanding it out completely. I.e. start from:
$$[\hat{J_z},\hat{J_+}] = [\frac{\hbar}{2} \left(a_+^{\dagger }a_+-a_-^{\dagger }a_-\right),\hbar a_{+}^{\dagger}a_{-}]$$

Then manipulate that using the general rules of algebra for commutators.

3. May 18, 2012

### Gregg

Using

$[A+B,C] = [A,C] + [B,C]$
$[AB,C] = [A,C]B + A[B,C]$
$[A,BC] = [A,B]C + B[A,C]$

It quickly boils down to

$\frac{\hbar^2}{2} \hat{a}_{+}^{\dagger} \hat{a}_{-}$

I am out by a factor of half but I'm sure it works!

4. May 18, 2012

### Gregg

Yes it works fine, I factored a half out of the commutator wrongly. Thanks!

5. May 18, 2012

### Clever-Name

Sweet! I'm glad it worked out. These types of problems can and will get messy sometimes. Good thing I didn't have to puzzle through it myself. :P