
#1
Feb2213, 07:51 PM

P: 146

I am confused with SU(2). How do you prove that the generators are pauli matrices (1/2 sigma)? I would appreciate any link or reference with details of all algebraic steps, taking one from the conditions of unitarity and det=1 to the well know exponential form.
In general, how do you find generators of arbitrary special unitary group? Thanks. 



#2
Feb2213, 09:23 PM

Sci Advisor
P: 1,732





#3
Feb2313, 02:11 AM

P: 987

https://docs.google.com/viewer?a=v&q...am6R4szBZXBeA 



#4
Feb2413, 10:16 PM

P: 146

generators of SU(2)
Thanks for replies, I also found it in Arfken.
But I still wonder, what is the SYSTEMATIC procedure of finding out the generators of arbitrary SU(N)? Do any algorithms exist? For example I don't quite understand how to systematically obtain GellMann matrices. 



#5
Feb2413, 10:46 PM

P: 796

The GellMann matrices are a basis for the space of traceless hermitian 3x3 matrices. For SU(N) in general, there are N^21 generators. If you look over the GellMann matrices for SU(3) and the Pauli matrices for SU(2) you can probably generalize to come up with a systematic way to enumerate N^21 linearly independent traceless hermitian NxN matrixes for any N.




#6
Feb2413, 11:21 PM

P: 146

How to do that?




#7
Feb2413, 11:29 PM

P: 233

It's easy just from the definition of the lie algebra of su(n). You need x*=x and tr(x)=0. Think of the lie algebra as a vector space and show that the pauli spin matrices span it for su(2). So they are the generators. That is to say any 2x2 matrix such that x*=1 and tr(x)=0 can be spanned by a linear combination of Pauli matrices.




#8
Feb2413, 11:39 PM

P: 233

I think you're also getting confused between SU(2) and su(2). The Pauli matrices span the lie algebra su(2). The unitary condition on SU(2) imposes the antihermitian on su(2) x*=x and det U= 1 imposes the tr (x)=0 condition.




#9
Feb2413, 11:43 PM

P: 146

So there is no rigorous algorithm for general su(n) case?




#10
Feb2513, 12:10 AM

P: 233

http://en.wikipedia.org/wiki/General...28Hermitian.29 It can be shown that these span the lie algebra su(n). There are rigorous algorithms, but that is a subject unto itself. For su(n+1) it's pretty simple since the dimension of the vector space can be shown to be n(n+2) so you need that many linearly independent basis vectors to span it. So take for example su(3) you need 2(2+2)=8 linearly independent antihermitian traceless matrices to span it. So the algorithm is one that can create that many of those kind of matrices for every n.




#11
Feb2513, 12:20 AM

P: 146

Awesome entry on wiki. Thanks! That sounds like what I was really looking for.




#12
Feb2513, 11:06 AM

P: 146

Just one more question. How do you prove the commutation relations for generators?
I understand that one can pick up the generators given by your link and explicitly prove the commutations and then, because it should be independent of representation, state that it's the general form. But nevertheless how to prove them without reference to explicit form of the generators? 


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