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.

In the paper arXiv:1004.5489 The origin of the hidden supersymmetry, the author use {Qa,Qa}={Qb,Qb}=2H, {Qa,Qb}=0 for N=2 hidden SUSY, which is different from what I was taught: {Qa,Qa}={Qb,Qb}=0, {Qa,Qb}=2H. I think they are controversially different. Is any of them wrong? If not, could anyone help by explaining some more?

share|improve this question
3  
The indices don't match for $\{Q_a,Q_b\} = 2H.$ What happens if you put $a=b$, for example? On the RHS, you should have some object that carries the right indices (such as $\delta_{ab}$, if $a,b$ live in some Euclidean space). –  Vibert Mar 2 '13 at 12:12
    
Yes that was confusing. Now I fixed the notation. Thanks a lot. –  Simon Mar 2 '13 at 20:06
    
Could anybody give a hand on this? Isn't it supposed not to be a hard question...? –  Simon Mar 5 '13 at 18:02
1  
You're thinking about this the wrong way. $a$ and $b$ are NOT fixed labels, i.e. you should never write down $\{Q_a,Q_a\} = \ldots$ or something like that. If you're really confused, you can write $\{Q_1,Q_1\} = \ldots,$ $\{Q_1,Q_2\} = 0$ etc. They are truly indices that can transform; you could rotate the supercharges, or write down $T_{\pm} = Q_1 \pm Q_2$, for example. The authors' $\{Q_a,Q_b\} = 2\delta_{ab} H$ is definitely right (up to a possible choice of conventions, of course). –  Vibert Mar 5 '13 at 18:48
1  
It's actually the same as with normal QM commutators you learned years ago. At the beginning, you write $[x,p_x] = \ldots$ etc., but at some point you grow up and just put $[x_i,p_j] = i\hbar \delta_{ij}$. The latter is manifestly $SO(3)$ invariant (if $i = 1,2,3$). –  Vibert Mar 5 '13 at 18:52
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

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.