# Lie algebra of the Poincare group

1. Oct 10, 2011

Given [M$^{\mu \nu}$,M$^{\rho\sigma}$] = -i($\eta^{\mu\rho}M^{\nu\sigma}+\eta^{\nu\sigma}M^{\mu\rho}-\eta^{\mu\sigma}M^{\nu\rho}-\eta^{\nu\rho}M^{\mu\sigma}$)

and [$P^{\mu},P^{\nu}$]=0

I need to show that

[M$^{\mu\nu},P^{\mu}$] = i$\eta^{\mu\rho}P^{\nu} - i\eta^{\nu\rho}P^{\mu}$

We've been given the following as a 'hint':

Lorentz transformation $x'^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}$,
$\Lambda$ is a 4x4 matrix such that xy is Lorentz invariant.

xy = $\eta_{\mu\nu}x^{\mu}y^{\nu} = x'y' = \eta_{\mu\nu}\Lambda^{\mu}_{\rho}x^{\rho}\Lambda^{\nu}_{\sigma}y^{\sigma}$

$\eta_{\mu\nu} = \eta_{\rho\sigma}\lambda^{\rho}_{\mu}\Lambda^{\sigma}_{\nu}$ (every Lambda solving this is a Lorentz transformation),

$\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu}+\eta^{\mu\rho}\omega_{\rho\nu}$

$\delta x^{\mu} = \eta^{\mu\rho}\omega_{\rho\nu} x^{\nu}$

--> $\omega_{\mu\nu} = - \omega_{\nu\mu}$

Then $\delta x^{\mu} = i \omega_{\rho\sigma}(M^{\rho\sigma})^{\mu}_{\nu} x^{\nu}$

Then have two expressions for $\delta x$.

My problem is basically that I do not see the relation between the question and the 'hint'. Presumably we can substitute something from the hint into the commutator we want to calculate, right? But what?

Then we are also asked to find differential operators M such that

$\delta x^{\mu} = \eta^{\mu\rho}\omega_{\rho\nu} x^{\nu}.$

We are given the answer:

$(M^{\rho\sigma})^{\mu}_{\nu} = L^{\rho\sigma}\delta^{\mu}_{\nu}$

where

$L^{\rho\sigma} = i (x^{\rho}∂^{\sigma} - x^{\sigma}∂^{\rho})$ and $P^{\mu} = -i∂^{\mu}$

but I do not know how to get there.

Last edited by a moderator: Oct 10, 2011
2. Oct 10, 2011

### dextercioby

What textbook are you using ? The first question is addressed in Weinberg Vol 1, the second in Pierre Ramond's <Field Theory> book.

3. Oct 11, 2011

Weinberg shows how to derive the commutators directly, not how to derive one from the other. I know I should still be able to figure this problem out from what he shows in the book, but I can't seem to make it work.

4. Oct 11, 2011

### dextercioby

1.You can't derive the commutators one from the other.

2. P's are generators of infinitesimal translations. The form is thus the partial derivative.

3. M's are generators for infinitesimal Lorentz transformations. A solution to the commutation relations is

$$M^{\mu\nu} = x^{[\mu}P^{\nu]}$$

as one can show by direct computation, using the differential form of P.

P.S. I retract what I said about Ramond. He's too sketchy.

5. Oct 12, 2011

### samalkhaiat

Write
$$\delta x^{\mu}=(1/2)\omega^{\rho \sigma}\left(\eta_{\nu\sigma}\delta^{\mu}_{\rho}-\eta_{\nu\rho}\delta^{\mu}_{\sigma}\right)x^{\nu}$$

then use
$$\delta^{\mu}_{\rho}=\partial_{\rho}x^{\mu}$$

to rewrite the above as

$$\delta x^{\mu}= \frac{i}{2} \omega^{\rho \sigma}L_{\rho \sigma}\delta^{\mu}_{\nu} x^{\nu}$$