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![[parent]](https://images.physicslibrary.org/images/uparrow.png) Viewing Message
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| ``Re: Why we can say a spinor be a representation of SU(2)?''
by hobo_physicist on 2007-03-01 19:26:46 |
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| You are actually correct. The spinor is a vector and the representation should be a matrix in this case. It's due to bad agreed terminology amongst physicists and mathematicians. I think it's because that once you've figured out the eigenstates or eigenvectors which are spinors, then you can work out the whole representation.
eg for spin 1 have e.vectors
(1,0,0), (0,1,0) and (0,0,1) with J_z eigenvalues
1,0,-1 respectively in units of h-bar = 1
You can also find J^2 this way. Then you can work out what the ladders J_+ and J_- are. You do this by looking at how
J_+ (0,1,0) = k (1,0,0)
where k is some proportionality constant which you can look up in books. It's the same for all representations and depends only on j and m. This gives you one entry. Then do the same for J_- and then from the work out what J_x and J_y are. Then after that you have the generators. To get the group representation itself you need to exponentiate these matrices eg. rotation by phi about z is
Exp{-i*phi*J_z}
Read pg. 23 for more details
http://paul.metcalfe.googlepages.com/fqm.pdf
Hope it helps.
hobo |
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