# Equation 2.14 in Srednicki

• nrqed
In summary, the individual discusses how he defines U(1 + \delta \omega) and then considers U(\Lambda^{-1} \Lambda' \Lambda) with \Lambda' = 1 + \delta \omega'. He then argues that U(\Lambda^{-1} \Lambda' \Lambda) \approx \delta \omega_{\mu \nu} \Lambda^{\mu}_{\, \, \rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma}, but the other individual does not understand how this is true and questions whether the \omega is meant to be \omega' instead. The expert summarizer explains the steps and calculations that lead to this conclusion.

#### nrqed

Homework Helper
Gold Member
He defines

$$U(1 + \delta \omega) \approx 1 + \frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu}$$

Then he considers

$$U(\Lambda^{-1} \Lambda' \Lambda)$$

with $\Lambda' = 1 + \delta \omega'$
He then says that

$$U(\Lambda^{-1} \Lambda' \Lambda) \approx \delta \omega_{\mu \nu} \Lambda^{\mu}_{\, \, \rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma}$$

I don't see why this is true. (by the way, I assume the $\omega$ is actually meant to be $\omega'$ ). I don't see how the $\Lambda^{-1} \Lambda$ turned into the expression on the right.

thanks

We have:

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

Now, from 2.5 we know ${(\Lambda^{-1})^\mu}_\nu = {\Lambda_\nu}^{\mu}$, so this becomes (expanding $\Lambda'$):

$$= {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma \left( {\delta^\nu}_\rho + {\delta \omega^\nu}_\rho \right)$$

So that the linear term is (with both indices down):

$${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$

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StatusX said:
We have:

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

Now, from 2.5 we know ${(\Lambda^{-1})^\mu}_\nu = {\Lambda_\nu}^{\mu}$, so this becomes (expanding $\Lambda'$):

$$= {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma \left( {\delta^\nu}_\rho + {\delta \omega^\nu}_\rho \right)$$

So that the linear term is (with both indices down):

$${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$

Thank you very much StatusX for all your help, it is very much appreciated.

Makes perfect sense !

how is

$${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ equal to $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?

The index structure inside the U should be $${}^a{}_b$$ and outside U nothing (full contracted), so what we can write $${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ equal to as is:
$$\Lambda^{\nu \mu}{\Lambda^\rho}_\sigma {\delta \omega _\nu}_\rho$$ and then what? how can one conclude that the linear term with "both indices down" is $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?

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I mean, is it an "argument by analogy" or is there a more profound way to find these things out?

malawi_glenn said:
how is

$${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ equal to $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?
It isn't. Let's write $\Lambda^{-1}\Lambda'\Lambda=\Lambda''$. You're trying to find the first order term of $U(\Lambda'')=U(1+\delta\omega'')=I+\frac i 2 \delta\omega''_{\mu\nu}M^{\mu\nu}$.

You already know that $(\delta\omega'')^\mu{}_\sigma=\Lambda_\nu{}^\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho$. That implies that $(\delta\omega'')_\mu{}_\sigma=\Lambda_\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho=\Lambda^\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega_\nu{}_\rho$.

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Everything is OK.

$$U \left( \Lambda^{-1} \Lambda ' \Lambda \right) = U \left( 1 + \delta \Omega \right) = 1 + \frac{i}{2 \hbar} \delta \Omega_{\mu \nu} M^{\mu \nu},$$

where

$$\delta \Omega = \Lambda^{-1} \delta \omega \Lambda.$$

Consequently,

$${\delta \Omega^\mu}_\sigma = {\Lambda_\nu}^\mu \delta {\omega^\nu}_\rho {\Lambda^\rho}_\sigma$$

and

$$\delta \Omega_{\mu \sigma} = {\Lambda^\nu}_\mu \delta \omega_{\nu \rho} {\Lambda^\rho}_\sigma .$$

I hope I haven't screwed up the indicies too much.

Thanx Fredrik and George, I had those two things in mind, you confirmed them :-)

## 1. What is Equation 2.14 in Srednicki?

Equation 2.14 in Srednicki refers to a specific equation presented in the book "Quantum Field Theory" by Mark Srednicki. It is a mathematical representation of a physical phenomenon or relationship between variables in the field of quantum mechanics.

## 2. What does Equation 2.14 in Srednicki represent?

Equation 2.14 in Srednicki represents a mathematical description of a specific physical system or process. It may represent the behavior of particles, interactions between particles, or other fundamental principles in quantum mechanics.

## 3. How is Equation 2.14 in Srednicki derived?

The derivation of Equation 2.14 in Srednicki may vary depending on the context and topic being discussed in the book. However, it is typically derived using mathematical principles and equations from quantum mechanics and related fields.

## 4. Can Equation 2.14 in Srednicki be applied to real-world situations?

Yes, Equation 2.14 in Srednicki can be applied to real-world situations in quantum mechanics, such as the behavior of particles in a particular physical system. However, its applicability may depend on the specific context and assumptions made in its derivation.

## 5. Why is Equation 2.14 in Srednicki important?

Equation 2.14 in Srednicki is important because it provides a mathematical framework for understanding and describing physical phenomena in quantum mechanics. It allows scientists to make predictions and test theories about the behavior of particles and their interactions, leading to a deeper understanding of the universe.