# Commutators of Lorentz generators

• nrqed
So it would be considered at a graduate level. In summary, equation 2.14 of Srednicki involves the use of U(\Lambda) and M^{\mu \nu} to derive the commutation relation of M_{\mu \nu}. By substituting \Lambda = 1 + \delta \omega and using the anti-symmetric property of M, the commutator can be extracted. This concept is typically studied at a graduate level, after two books on quantum mechanics have been studied.

#### nrqed

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Equation 2.14 of Srednicki is:

$$U(\Lambda)^{-1} M^{\mu \nu} U(\Lambda) = \Lambda^{\mu}_{\,\,\rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma}$$

He says that writing $\Lambda = 1 + \delta \omega$, one obtains the usual commutation relation of the $M_{\mu \nu}$:

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

I get, starting from the first equation

$$[ M^{\mu \nu},M^{\rho \sigma} ] \delta \omega_{\rho \sigma} = i (\eta^\mu_\rho \delta \omega^\nu_\sigma M^{\rho \sigma} + \eta^\nu_\sigma \delta \omega^\mu_\rho M^{\rho \sigma} )$$

I don't see how to extract the commutator because the omega on the right side contain one of the indices that is not summed over. Its' not clear to me how to proceed. Any suggestion?

I think it should work if you stick in a kronecker delta.

Check the right hand side of the equation, you can write

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

Edit:

Also M is anti-symmetric, which means you can write it as

$$M^{uv} = \frac{1}{2} (M^{uv} - M^{vu})$$

Last edited:
what level of quantum physics is this? grad?

waht said:
Check the right hand side of the equation, you can write

$$\delta\omega^\nu_\sigma = \delta\omega_{\rho\sigma} \eta^{\rho \nu}$$
Of course! (Slapping myself right now...I should have seen it). Thanks!

thoughtgaze said:
what level of quantum physics is this? grad?
This is from a book on quantum field theory. You would normally study two books on quantum mechanics before this.

## 1. What are commutators of Lorentz generators?

Commutators of Lorentz generators are mathematical operators used to describe the transformations between inertial reference frames in special relativity. They represent how the generators of boosts and rotations commute with each other.

## 2. How are commutators of Lorentz generators calculated?

The commutator of two Lorentz generators is calculated by taking the difference between the two generators, applying them to a vector, and then taking the difference between the resulting vectors. This is represented by the mathematical notation [A,B] = AB - BA.

## 3. What is the physical significance of commutators of Lorentz generators?

Commutators of Lorentz generators describe the non-commutative nature of boosts and rotations in special relativity. They play a crucial role in understanding the structure of spacetime and the laws of physics, and are used in the development of quantum field theories.

## 4. How do commutators of Lorentz generators relate to the Lorentz transformations?

Lorentz transformations are the mathematical representation of how physical quantities, such as space and time, appear differently in different inertial reference frames. Commutators of Lorentz generators are used to calculate the transformations between these frames and how they interact with each other.

## 5. What is the importance of studying commutators of Lorentz generators?

Studying commutators of Lorentz generators is important for understanding the fundamental principles of special relativity and the structure of spacetime. It also has practical applications in fields such as particle physics and cosmology, where the effects of relativity are significant.