# [qft] Srednicki 2.3 Lorentz group generator commutator

• wasia
In summary, the derived equation (2.16) can be obtained by taking into account the relationship between the components of the antisymmetric matrix \delta\omega_{\rho\sigma}. By considering the 12 and 21 components separately and applying the constraint \delta\omega_{12}=-\delta\omega_{21}, the equation can be rewritten in the desired form.
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.

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.

weejee said:
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?

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.

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?

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 $$\delta\omega_{12} = -\delta\omega_{21}$$?

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

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 :)

## 1. What is the Lorentz group generator commutator in QFT?

The Lorentz group generator commutator is a mathematical expression that describes how the generators of the Lorentz group, which is a set of transformations that preserve the form of physical laws in special relativity, commute or do not commute with each other. In QFT, this commutator is essential for understanding the behavior of fields and particles under Lorentz transformations.

## 2. Why is the Lorentz group generator commutator important in QFT?

The Lorentz group generator commutator is important in QFT because it helps us understand how fields and particles transform under Lorentz transformations, which is crucial for making predictions about their behavior. It also plays a key role in the construction of gauge theories, which are fundamental in modern particle physics.

## 3. How is the Lorentz group generator commutator derived in QFT?

The Lorentz group generator commutator is derived using the commutation relations between the generators of the Lorentz group, which are derived from the Lorentz group algebra. These commutation relations are then used to construct the commutator between any two Lorentz group generators, which gives us the Lorentz group generator commutator in QFT.

## 4. What is the physical significance of the Lorentz group generator commutator?

The Lorentz group generator commutator has several physical significances in QFT. It is related to the symmetries of the theory, such as rotational and boost invariance, and it helps us understand how fields and particles transform under these symmetries. It also plays a crucial role in the quantization of fields and in the construction of gauge theories.

## 5. How does the Lorentz group generator commutator relate to other commutators in QFT?

The Lorentz group generator commutator is related to other commutators in QFT, such as the commutator between the Hamiltonian and momentum operators, or the commutator between different fields. It is also related to the commutator of the Poincaré group, which is a larger symmetry group that includes the Lorentz group. All these commutators are important for understanding the behavior of fields and particles in QFT.

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