How Do S^{\mu\nu} Matrices Form a Lorentz Algebra Representation?

[vputz]

Homework Statement

As part of a problem set, I'm trying to show that the $$S^{\mu\nu}$$ matrices defined by $$S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$$ constitute a representation of the Lorentz algebra.

Well, luckily P&S assure me that by repeated use of
{$$\gamma^\mu,\gamma^\nu$$}$$=2g^{\mu\nu}\times 1_n$$

"...it is easy to verify that these matrices satisfy the commutation relations"
[$$J^{\mu\nu},J^\rho\sigma] = i(g^{\nu\rho}J^{\mu\sigma} - g^{\mu\rho}J^{\nu\sigma} - g^{\nu\sigma}J^{\mu\rho} + g^{\mu\sigma}J^{\nu\rho})$$

Homework Equations

The Attempt at a Solution

Well, it is just ducky that it is easy to verify (grin). But it is not turning out to be easy for me! If I approach it by expanding [$$S^{\mu\nu},S^{\rho\sigma}$$], I wind up with a lot of gamma matrices; writing $$\gamma^\mu$$ just as $$\mu$$, it looks something like

$$\mu\nu\rho\sigma - \mu\nu\sigma\rho - \nu\mu\rho\sigma + \nu\mu\sigma\rho - \rho\sigma\mu\nu + \rho\sigma\nu\mu + \sigma\rho\mu\nu - \sigma\rho\nu\mu$$

Now, if I'm moving toward the Lorentz algebra, I need to get terms of the form $$g^{\nu\rho}S^{\mu\sigma}$$ after making use of the anticommutator relation--but I only see those pairings of indices in the middle of the groups of four. I am certain that I'm missing some obvious commutation or anticommutation trick to break those groups up so that I can make use of the anticommutation identity to introduce the metric and thus get toward a solution, but I've been staring at it for hours and no luck. I tried approaching it the other way (expanding the right hand side of the Lorentz algebra equation above) and wound up with sixteen groups of four gammas with different indices and still no obvious way to reduce them and show equality.

Stated a little differently, in order to get a $$g^{\mu\rho}S^{\mu\sigma}$$ term, I need a {$$\nu,\rho$$}[$$\mu,\sigma$$] term, which if I expand it comes out to be something like

$$\nu\rho\mu\sigma - \nu\rho\sigma\mu + \rho\nu\mu\sigma - \rho\nu\sigma\mu$$

My original expansion does contain some terms with the right first and last indices, but the middle terms reversed from what I want. It also does NOT have some terms I need (for example, terms beginning and ending with $$\mu$$ and $$\nu$$).

I know it should be straightforward and I'm just not seeing the link. Any suggestions?

 

[nrqed]

Homework Statement

As part of a problem set, I'm trying to show that the $$S^{\mu\nu}$$ matrices defined by $$S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$$ constitute a representation of the Lorentz algebra.

Well, luckily P&S assure me that by repeated use of
{$$\gamma^\mu,\gamma^\nu$$}$$=2g^{\mu\nu}\times 1_n$$

"...it is easy to verify that these matrices satisfy the commutation relations"
[$$J^{\mu\nu},J^\rho\sigma] = i(g^{\nu\rho}J^{\mu\sigma} - g^{\mu\rho}J^{\nu\sigma} - g^{\nu\sigma}J^{\mu\rho} + g^{\mu\sigma}J^{\nu\rho})$$

Homework Equations

The Attempt at a Solution

Well, it is just ducky that it is easy to verify (grin). But it is not turning out to be easy for me! If I approach it by expanding [$$S^{\mu\nu},S^{\rho\sigma}$$], I wind up with a lot of gamma matrices; writing $$\gamma^\mu$$ just as $$\mu$$, it looks something like

$$\mu\nu\rho\sigma - \mu\nu\sigma\rho - \nu\mu\rho\sigma + \nu\mu\sigma\rho - \rho\sigma\mu\nu + \rho\sigma\nu\mu + \sigma\rho\mu\nu - \sigma\rho\nu\mu$$

Now, if I'm moving toward the Lorentz algebra, I need to get terms of the form $$g^{\nu\rho}S^{\mu\sigma}$$ after making use of the anticommutator relation--but I only see those pairings of indices in the middle of the groups of four. I am certain that I'm missing some obvious commutation or anticommutation trick to break those groups up so that I can make use of the anticommutation identity to introduce the metric and thus get toward a solution, but I've been staring at it for hours and no luck. I tried approaching it the other way (expanding the right hand side of the Lorentz algebra equation above) and wound up with sixteen groups of four gammas with different indices and still no obvious way to reduce them and show equality.

Stated a little differently, in order to get a $$g^{\mu\rho}S^{\mu\sigma}$$ term, I need a {$$\nu,\rho$$}[$$\mu,\sigma$$] term, which if I expand it comes out to be something like

$$\nu\rho\mu\sigma - \nu\rho\sigma\mu + \rho\nu\mu\sigma - \rho\nu\sigma\mu$$

My original expansion does contain some terms with the right first and last indices, but the middle terms reversed from what I want. It also does NOT have some terms I need (for example, terms beginning and ending with $$\mu$$ and $$\nu$$).

I know it should be straightforward and I'm just not seeing the link. Any suggestions?

Isn't the difference between the two expression trivially zero? (I did not chekc it in details so I may be wrong but this is what my gut feeling is). I mean, if you regroup all the extra 8 terms you got from the second way (starting from the answer) do they add up to zero? My guess is that you would simply have to add zero in different ways to your first result ( I mean 0 in the form $\alpha \beta \gamma \delta - \alpha \beta \gamma \delta$ where alpha, etc are some combinations of the mu,nu, rho sigma).

 

[vputz]

Well, I would certainly like it to be so! But when I expand the right hand side of the second way, I get terms looking like $$\eta^{\nu\rho}S^{\mu\sigma} = \frac{i}{4}(\gamma^{\nu\rho}+\gamma^\rho\nu)[\gamma^\mu,\gamma^\sigma]$$.

Writing those out, I seem to get the terms (again, writing $$\gamma^\mu$$ just as $$\mu$$)...

$$\nu\rho\mu\sigma - \nu\rho\sigma\mu + \rho\nu\mu\sigma - \rho\nu\sigma\mu + \mu\sigma\nu\rho - \mu\sigma\rho\nu + \sigma\mu\nu\rho - \sigma\mu\rho\nu - \mu\rho\nu\sigma + \mu\rho\sigma\nu - \rho\mu\nu\sigma + \mu\rho\sigma\nu - \nu\sigma\mu\rho + \nu\sigma\rho\mu - \sigma\nu\mu\rho + \sigma\nu\rho\mu$$

(whew! I find myself wishing I understood Maxima well enough to do this sort of thing).

Anyway, I don't see a lot of terms that easily cancel, which leads me to believe I'm missing a minor (but trivial) manipulation--or that I've made a mistake in the expansion.

 

[nrqed]

Well, I would certainly like it to be so! But when I expand the right hand side of the second way, I get terms looking like $$\eta^{\nu\rho}S^{\mu\sigma} = \frac{i}{4}(\gamma^{\nu\rho}+\gamma^\rho\nu)[\gamma^\mu,\gamma^\sigma]$$.

Writing those out, I seem to get the terms (again, writing $$\gamma^\mu$$ just as $$\mu$$)...

$$\nu\rho\mu\sigma - \nu\rho\sigma\mu + \rho\nu\mu\sigma - \rho\nu\sigma\mu + \mu\sigma\nu\rho - \mu\sigma\rho\nu + \sigma\mu\nu\rho - \sigma\mu\rho\nu - \mu\rho\nu\sigma + \mu\rho\sigma\nu - \rho\mu\nu\sigma + \mu\rho\sigma\nu - \nu\sigma\mu\rho + \nu\sigma\rho\mu - \sigma\nu\mu\rho + \sigma\nu\rho\mu$$

(whew! I find myself wishing I understood Maxima well enough to do this sort of thing).

Anyway, I don't see a lot of terms that easily cancel, which leads me to believe I'm missing a minor (but trivial) manipulation--or that I've made a mistake in the expansion.

Sorry, just a few seconds of thought made me realize I was wrong.

I think the trick is this: you start from your expression for $[ S^{\mu \nu}, S^{\rho \sigma} ]$ which gives you 4 terms. Now, you don't have all the indices in the right place. What you do is that you move the indices where you want them using repeatedly, say, $\mu \sigma = - \sigma \mu + g^{\sigma \mu}$ This will bring the indices where you want them while doubling the number of terms and, hopefully, giving you the correct final result. I think this will work.

 

[vputz]

Boggle boggle boggle... hmm, I'll play with it if I have time. Thanks!

(I "solved" that problem about 4/5 of the way by brute force--actually calculating matrices and their commutators; hooray for Maxima. But I'd like to do it the right way. Sure wish I knew how to make a CAS like Maxima play nice with "order matters" things like gamma matrices)

 

[Avodyne]

Somewhat easier is to first compute $[S^{\mu\nu},\gamma^\rho]$, then use $[A,BC]=[A,B]C+B[A,C]$ to compute $[S^{\mu\nu},\gamma^\rho\gamma^\sigma]$, then antisymmetrize the result on $\rho\sigma$.