Prove some relations but going round in circles

  • Thread starter Gregg
  • Start date
  • #1
459
0

Homework Statement




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)##

Homework Equations



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

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

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?
 

Answers and Replies

  • #2
380
1
I'd suggest working in commutator notation instead of expanding it out completely. I.e. start from:
[tex]
[\hat{J_z},\hat{J_+}] = [\frac{\hbar}{2} \left(a_+^{\dagger }a_+-a_-^{\dagger }a_-\right),\hbar a_{+}^{\dagger}a_{-}]
[/tex]

Then manipulate that using the general rules of algebra for commutators.
 
  • #3
459
0
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
459
0
Yes it works fine, I factored a half out of the commutator wrongly. Thanks!
 
  • #5
380
1
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
 

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