[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: $$[M^{\mu\nu},M^{\rho\sigma}]=i\hbar (g^{\mu\rho}M^{\nu\sigma}- g^{\nu\rho}M^{\mu\sigma}+g^{\nu\sigma}M^{\mu\rho}-g^{\mu\sigma}M^{\nu\rho})$$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:

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

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

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

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

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

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