# Commutation Relations and Unitary Operators

1. Aug 25, 2007

### abode_x

I have a problem with deriving another result. Sorry I am new to this field. Please see the attached PDF - everything is there.

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2. Aug 26, 2007

### dextercioby

What you're trying to prove is wrong. Srednicki is wrong. If you make the unitary transformation of the 4-momentum operator the way it should be done, namely

$$\bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)U^{\dagger} (\Lambda)$$

then you get the correct RHS

$$\bar{P}^{\mu}(x')=U(\Lambda)P^{\mu}(x)PU^{\dagger} (\Lambda)= \Lambda^{\mu}{}_{\nu} P^{\nu}(x)$$

which leads to the commutation relation

$$\left[P^{\mu}(x),M^{\nu\lambda}(x)\right]_{-}= i\hbar \ g^{\mu[\nu}P^{\lambda]}(x)$$

Last edited: Aug 26, 2007
3. Sep 1, 2007

### Avodyne

abode_x: you got
$${i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}] =\delta\omega^\mu{}_\nu P^\nu$$
Next,
$$\delta\omega^\mu{}_\nu P^\nu =\delta\omega_{\rho\nu}g^{\rho\mu} P^\nu =\delta\omega_{\rho\sigma}g^{\rho\mu} P^\sigma$$
Next, use that $$\delta\omega_{\rho\sigma}$$ is antisymmetric, and so we can replace $$g^{\rho\mu} P^\sigma$$ with its antisymmetric part,
$${1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)$$
This gives
$${i\over2\hbar}\delta\omega_{\rho\sigma}[P^\mu,M^{\rho\sigma}]={1\over2}\delta\omega_{\rho\sigma}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)$$
For this to be true for any $$\delta\omega_{\rho\sigma}$$, it must be true of each coefficient. So,
$${i\over2\hbar}[P^\mu,M^{\rho\sigma}] ={1\over2}(g^{\rho\mu} P^\sigma-g^{\sigma\mu}P^\rho)$$
Multiply by $$-2i\hbar$$ and switch the two terms on the right to get
$$[P^\mu,M^{\rho\sigma}] =i\hbar(g^{\sigma\mu}P^\rho-g^{\rho\mu} P^\sigma)$$
Using $$g^{\sigma\mu}=g^{\mu\sigma}$$, this matches Srednicki's eq.(2.18).