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Lorentz generators

by alphaone
Tags: generators, lorentz
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alphaone
#1
Feb13-07, 12:16 PM
P: 46
Could somebody please show me the calculation which shows that the M_mu_nu representation of the Lorentz generators gives rise to a (1,0)+(0,0) representation? Thanks in advance
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dextercioby
#2
Feb14-07, 12:21 AM
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P: 11,894
I don't think this is possible. [itex] M_{\mu\nu} [/itex] is different for every representation and the calculations are actually the other way around. The generators are computed by knowing how the spinors behave under restricted LT's.
alphaone
#3
Feb14-07, 02:41 AM
P: 46
Thanks for the reply. I am sorry probably my notation is uncommon. When I said M_{\mu\nu} I meant the representation of the Lorentz generators when acting on a Lorentz 4-vector(so antisymmetric matrices, when all indices are raised). Also the way I learned it we started at differrent reps of the Lorentz generators and then afterwards defined the fields the transformation could act on and deduced its properties - seems to me to be some sort of chicken and egg problem. However I thought that it should be possible to compute that the vector representation is (1,0)+(0,0) as this is basically the spin of the object.

dextercioby
#4
Feb14-07, 03:51 AM
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HW Helper
P: 11,894
Lorentz generators

Actually the vector representation is (1/2,1/2). (1,0)+(0,0) is a reducibile representation and is made up of a self-dual 2-form and a scalar.


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