[qft] Srednicki 2.3 Lorentz group generator commutator

[wasia]

Homework Statement

Verify that (2.16) follows from (2.14). Here \Lambda is a Lorentz transformation matrix, U is a unitary operator, M is a generator of the Lorentz group.

Homework Equations

2.8: \delta\omega_{\rho\sigma}=-\delta\omega_{\sigma\rho}

M^{\mu\nu}=-M^{\nu\mu}

2.14: U(\Lambda}^{-1})M^{\mu\nu}U(\Lambda})=\Lambda^\mu_\rho\Lambda^\nu_\sigma M^{\rho\sigma}

2.16: [M^{\mu\nu},M^{\rho\sigma}]=i\hbar (g^{\mu\rho}M^{\nu\sigma}-<br /> g^{\nu\rho}M^{\mu\sigma}+g^{\nu\sigma}M^{\mu\rho}-g^{\mu\sigma}M^{\nu\rho})<br />2.12: U(1+\delta\omega)=1+{i \over 2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}

The Attempt at a Solution

I assume that \Lambda is a small transformation, as hinted by Srednicki: \Lambda=1+\delta\omega and rewrite the 2.14:

(1-{i \over 2\hbar}\delta\omega_{\rho\sigma}M^{\rho\sigma})M^{\mu\nu}(1+{i \over 2\hbar}\delta\omega_{\rho\sigma}M^{\rho\sigma})=<br /> (\delta^\mu_\rho+\delta\omega^\mu_\rho)(\delta^\nu_\sigma+\delta\omega^\nu_\sigma)M^{\rho\sigma}.

Then I cross out the M^{\mu\nu} that come on the both sides, throw out the double-omega pieces and rewrite the omegas in the following manner

\delta\omega^\nu_\sigma=g^{\rho\nu}\delta\omega_{\rho\sigma}

I come to

<br /> [M^{\mu\nu},M^{\rho\sigma}]=2i\hbar (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})<br />

which seems to be incorrect. Where do I make the mistakes, if I do? How to derive the (2.16) without involvement of certain expression of the Lorentz generator?

Thanks.

 

[weejee]

you shouldn't just throw away the omega. Take into accout the fact that omega_ab and omega_ba are not independent but related by the relation omega_ab=-omega_ba. i.e. when you consider omega_ab, you shoud also consider omega_ba.

 

[wasia]

you shouldn't just throw away the omega. Take into accout the fact that omega_ab and omega_ba are not independent but related by the relation omega_ab=-omega_ba. i.e. when you consider omega_ab, you shoud also consider omega_ba.

I throw away the double omega's, that is \delta\omega^\mu_\rho\delta\omega^\nu_\sigma. I think this is reasonable, because this kind of term is second-order (very small).

I have also come to having \delta\omega_{\rho\sigma} on both sides, before cancelling them out (that is, we are only dealing with asymmetric pieces).

Or have I misunderstood your comment?

 

[weejee]

I guess the expressioni "throw away" in my comment was misleading.
I'll explain it again.

What you actually got in the calculation must have been something like this.
\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})

Now, you want to take away \delta\omega_{\rho\sigma} from the above expression, as the above equation holds for arbitrary \delta\omega

If each component \delta\omega_{\rho\sigma} is truly arbitrary, you can do that. Yet, in your case, \delta\omega is antisymmetric so that some of its components are related. You should take this fact into account.

 

[wasia]

Yes, I have got the equation

\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})

but M^{\rho\sigma} is antisymmetric, too, and I thought this gives us the right to cancel the \delta\omega_{\rho\sigma} 's.

Thank you for pointing out the mistake, however, I still do not know how I take into account that some of the components are related. Should I try to write a new equation with interchanged indices \rho\leftrightarrow\sigma and try to combine the two? Is there some resource in the net that would give me this kind of elementary understanding about equations with antisymmetric coefficients?

 

[weejee]

In the above equation, \rho and \sigma are summed over. Let's expand this summation and look into the 12 and 21 components, as an example.

\delta\omega_{\rho\sigma}[M^{\mu\nu},M^{\rho\sigma}]=2i\hbar \delta\omega_{\rho\sigma} (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma})

->

... + \delta\omega_{12}[M^{\mu\nu},M^{12}] + \delta\omega_{21}[M^{\mu\nu},M^{21}] +...= ... + 2i\hbar \delta\omega_{12} (g^{\mu 2}M^{1\nu}-g^{1\nu}M^{\mu 2}) + 2i\hbar \delta\omega_{21} (g^{\mu 1}M^{2\nu}-g^{2\nu}M^{\mu 1})+...

First, let's assume every component of \delta\omega is independent. Then, for the above equation to hold for an arbitrary \delta\omega, the following equations should be satisfied.

...
[M^{\mu\nu},M^{12}] = 2i\hbar (g^{\mu 2}M^{1\nu}-g^{1\nu}M^{\mu 2})
[M^{\mu\nu},M^{21}] = 2i\hbar (g^{\mu 1}M^{2\nu}-g^{2\nu}M^{\mu 1})
...

What if there is a constraint that \delta\omega_{12} = -\delta\omega_{21}?

 

[wasia]

\delta\omega_{12}[M^{\mu\nu}, M^{12}]-\delta\omega_{12}[M^{\mu\nu}, M^{21}]=2i\hbar\delta\omega_{12}(g^{\mu2}M^{1\nu}-g^{1\nu}M^{\mu2}-g^{\mu 1}M^{2\nu}+g^{2\nu}M^{\mu 1}

that, by crossing out 2\delta\omega_{12} and writing rho instead of 1 and sigma instead of 2 is

<br /> [M^{\mu\nu},M^{\rho\sigma}]=i\hbar (g^{\mu\sigma}M^{\rho\nu}-g^{\rho\nu}M^{\mu\sigma}-g^{\mu\rho}M^{\sigma\nu}+g^{\sigma\nu}M^{\mu\rho})<br />

and this is the same as (2.16). Thank you very much, weejee, that was a really great help :)