Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

reading about the irreducible tensors and its commutation relations with the angular momentum one can find relations for $J_{z}$, $J_{+}$, $J_{-}$, but I was wondering, what about $J^2$ ?

from sakurai: (revised edition p.236) (sakurai napolitano edition p.232)

$$ [\bar{J}.\hat{n}, T_{q}^{k} ] = \sum_{q'} T_{q'}^{k} \langle {kq'} | \bar{J}.\hat{n} | {kq} \rangle $$

and for evaluating with $\hat{n}=\hat{z}$, $\hat{x}$, $\hat{y}$

$$ [J_{z}, T_{q}^{k} ] = \hbar q T_{q}^{k} $$

$$ [J_{\pm}, T_{q}^{k} ] = \hbar \sqrt{(k \mp q)(k \pm q + 1)} T_{q \pm 1}^{k} $$

But is there an expression for $ [ \bar{J}^{2}, T_{q}^{k} ] $ ? $$ [ \bar{J}^2, T_{q}^{k} ] = [ J_{x}^2+J_{y}^2+J_{z}^2, T_{q}^{k} ] = [ \frac{J_{+}J_{-} + J_{-}J_{+}}{2} + J_{z}^2 , T_{q}^{k} ] $$ $$ = \frac{1}{2} J_{+}[J_{-},T_{q}^{k}] + \frac{1}{2} [J_{+},T_{q}^{k}]J_{-} + \frac{1}{2} J_{-}[J_{+},T_{q}^{k}] + \frac{1}{2} [J_{-},T_{q}^{k}]J_{+} + J_{z}[ J_{z} , T_{q}^{k} ] + [ J_{z} , T_{q}^{k} ]J_{z} $$

$$ = \frac{1}{2} J_{+} \hbar \sqrt{(k + q)(k - q + 1)} T_{q - 1}^{k} + \frac{1}{2} \hbar \sqrt{(k - q)(k + q + 1)} T_{q + 1}^{k} J_{-} $$ $$ + \frac{1}{2} J_{-}\hbar \sqrt{(k - q)(k + q + 1)} T_{q + 1}^{k} + \frac{1}{2} \hbar \sqrt{(k + q)(k - q + 1)} T_{q - 1}^{k} J_{+} $$ $$ + \hbar q J_{z} T_{q}^{k} + \hbar q T_{q}^{k} J_{z} $$

I cant figure out what should i "sandwich" this with in order to obtain a useful expresion, if something.

Any other idea?

thanks in advance

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.