# Group Theory

1. Feb 16, 2004

### Norman

Problem:
a) To determine the number of generators needed for the group O(n) we write a rotation matrix as:
$$R=e^{-i\theta J}$$
where $J$ is an n x n matrix, Hermitian and imaginary, and therefore anti-symmetric. The number of indepedent parameters $\theta$ (and hence the number of generators) is the number of independent matrices. This number can be found by counting the number of parameters required to make up any n x n antisymmetric matrix. This is n(n-1)/2- WHY?
b)Show for any n:
$$[J_{ij},J_{kl}]=\plusminus (\delta_{ij}J_{il}-\delta_{ik}J_{jl}-\delta_{jl}J_{ik}+\delta_{il}J_{ik})$$

where $J_{ij}$ are two index objects with matrix elements:

$$(J_{ij})_{kl} = -i(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk})$$

and
$$[J_{ij},J_{kl}]$$
is the commutator

Ok...
So part a):
I am a little confused. I know that the matrix must be imaginary and hermitian, but I don't think that is enough to prove that only n(n-1)/2 parameters are required to make a n x n antisymmetric matrix. In fact I am not even sure what determines whether the parameters are independent. Is a complex number and its conjugate independent? If not, then I think I understand. But if not I am lost.

part b) No clue.

I have never taken a group theory class and this was thrown into a Quantum Mechanics homework set so I am pretty lost. Any help would really be appreciated.

Last edited: Feb 16, 2004
2. Feb 17, 2004

### HallsofIvy

Staff Emeritus
No, a complex number, a+ bi, and its conjugate, a- bi, are definitely NOT independent! Especially if we are given that the numbers are all imaginary so it is really bi and -bi. Clearly an imaginary, Hermitian matrix is anti-symmetric. Now, calculate how many "choices" you could make for the values in an anti-symmetric matrix: aij= -aji.

In particular, all the entries on the main diagonal (i= j) must be 0: aii= -aii means aii= 0 so we cannot make any choices for them. There are, of course, exactly n diagonal elements in an n by n matrix, leaving n2-n. If we "choose" any one of those, say aij then its "opposite", aji is fixed. That is, we can "choose" exactly half of the numbers off the main diagonal (choose all those above the main diagonal for example and all those below are automatically fixed as their negatives). We can "choose" (n2-n)/2= n(n-1)/2 values.

b) The "commutator"is , by definition, given by
$$[J_{ij},J_{kl}]= J_{ij}J_{kl}-J{kl}J{ij}$$
Since you are told that $$(J_{ij})_{kl} = -i(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk})$$, go ahead a put those into that formula and see what you get!

3. Feb 17, 2004

### Norman

HallsofIvy,

First of all, thankyou so much for responding. I am really not comfortable with Group Theory yet and it is a great relief that my intuition about part a) was correct.
For part b) I am a little confused still. I only know the $kl$ components of the matrix. How do I write $J_{ij}$ and $J_{kl}$ in a form in which I can just plug them into the commutator?
Thanks again for the help.
Norm

4. Feb 18, 2004

### Norman

Help... still stuck.

5. Feb 18, 2004

### Norman

Does anyone think that this should actually be:
$$[J_{ij},J_{kl}]= -i (\delta_{jk}J_{il}-\delta_{ik}J_{jl}-\delta_{jl}J_{ik}+\delta_{il}J_{ik})$$
???????
Any help would really be appreciated.
Thanks.

Last edited: Feb 18, 2004
6. Feb 18, 2004

### NateTG

Re: Re: Group Theory

That's certainly looks better since it's symetric. I'm not quite following the notation though, so I can't give you a stronger answer.

7. Feb 18, 2004

### Norman

Re: Re: Re: Group Theory

The way I was told to think about it is that:
$$(J_{ij})_{kl} = -i(\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk})$$
is the $kl^{th}$ component of the matrix $J_{ij}$ so all you do is sum over k and l for matrix multiplication. But I have no clue if that is correct or not and if I am understanding this at all. It is very frustrating.
Cheers,
Norman

8. Feb 18, 2004

### NateTG

OK, that makes a little bit more sense.

From group theory we have that
$$[ab]=b^{-1}a^{-1}ba$$

You may be able to grind it out from there by figuring out what the inverse of $$J_{il}$$ looks like.

9. Feb 19, 2004

### Norman

is:
$$[ab]=b^{-1}a^{-1}ba$$
the commutator or just multiplication?
Thanks,
Norman

10. Feb 19, 2004

### NateTG

$$[ab]$$ is shorthand for the commutator of $$a$$ and $$b$$. The RHS of that equation is a general expression for the commutator. If you multiply $$ab$$ by it, you get $$ba$$ so it commutes them.