Lorentz Transformation of Scalar Fields

[waht]

Homework Statement

Working on an exercise from Srednicki's QFT and something is not clear.

Show that

$$[\varphi(x), M^{uv}] = \mathcal{L}^{uv} \varphi(x)$$

where

$$\mathcal{L}^{uv} = \frac{\hbar}{i} (x^u \partial^v - x^v \partial^u )$$

Homework Equations

(1) $$U(\Lambda)^{-1} \varphi(x) U(\Lambda) = \varphi(\Lambda^{-1}x)$$

(2) $$\Lambda = 1 + \delta\omega$$

where [tex]\delta\omega [/itex] is an infinitesimal, and<br /> <br /> (3) $$U(\Lambda) = I + \frac{i}{2\hbar} \delta\omega_{uv} M^{uv}$$<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Got the left side of (1) equal to<br /> <br /> $$\varphi(x) + \frac{i}{2\hbar}\delta\omega_{uv}[\varphi(x), M^{uv}]$$<br /> <br /> but not sure what to do with the right side and how to get the desired derivatives. I suspect<br /> it has something to do with the transformation (1) of its derivative, but so far no luck.<br /> <br /> $$U(\Lambda)^{-1} \partial^u \varphi(x) U(\Lambda) = \Lambda^{u}_{ p} \bar{\partial}^p \varphi(\Lambda^{-1}x)$$[/tex]

 

[borgwal]

1) What does Eq. (2) give you for the inverse of \Lambda?

2) use that expression to Taylor expand the right-hand side of your Eq. (1): that should also give you \phi(x) plus something...

 

[waht]

1) What does Eq. (2) give you for the inverse of \Lambda?

Would it be

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

and taking the second order $O(\delta\omega^2)$ to zero.

2) use that expression to Taylor expand the right-hand side of your Eq. (1): that should also give you \phi(x) plus something...

If I were to Taylor that then would get something like $\varphi(0) + \varphi^{,}(0)$

but that doesn't seem right

 

[borgwal]

Would it be

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

and taking the second order $O(\delta\omega^2)$ to zero.



If I were to Taylor that then would get something like $\varphi(0) + \varphi^{,}(0)$

but that doesn't seem right

The first part is correct, the second isn't (but getting close): check your Taylor expansion: what's \phi(x+\epsilon)?