2

I've just started working through Supersymmetric Gauge Field Theory and String Theory by D. Balin and A. Love and I got a question to something right at the beginning.

They are revisiting the Poincaré group and give the following construction of the generators:

For an infinitesimal translation it is

$$x'^\mu = x^\mu + a^\mu = x^\mu-ia_\lambda(P^\lambda)^\mu$$ and from this they conclude that the generator for translations is given by $$(P^\lambda)_\mu = i \delta^\lambda_\mu.\tag 1 \label{1}$$ I'm a little confused by this. Aren't the finite translations given by $$x'^\mu = \text{exp}(-i a_\lambda (P^\lambda)^\mu_\nu) x^\nu$$ and shouldn't the generator therefore read $$(P^\lambda)^\mu_\nu = i \delta^\mu_\nu \partial^\lambda \tag{2}$$ In other words, why is the generator not an operator valued matrix (as it would clearly be in my case.)

For Lorentz transformations they give the derivation $$x'^\mu = x^\rho + \omega^\rho_\sigma x^\sigma =x^\rho-\frac{1}{2} \omega_{\mu\nu}(M^{\mu\nu})^\rho_\sigma x^\sigma$$ and conclude $$(M^{\mu \nu})^{\rho \sigma} = i (\eta^{\mu \rho}\eta^{\nu \sigma} - \eta^{\mu \sigma}\eta^{\nu \rho}),\tag 3$$ which I understand a little better, as it at least makes sense with my understanding of generators. But with those two definitions, there is no way to make the commutator relation $$[M^{\mu \nu}, P^{\lambda}]=i (\eta^{\nu \lambda} P^\mu -\eta^{\mu \lambda}P^\nu) \tag 4$$work out. How would you even define this commutator, given that the $P^\mu$ in (1) has only one internal index?

On the other hand, to fulfill the commutation relation with the operator in (2) I think you also need the spacetime representation of the Lorentz generator, which is the typical $$L^{\mu\nu} = i(x^\mu \partial^\nu - x^\nu \partial^\mu).\tag 5$$

So I guess, my question boils down to: if (2) and (5) actually fulfill (4), then what is the translation generator corresponding to (3)? Which one of those representation is actually the action of the Poincare group on spacetime? I kind of want to say both but I'm just not sure.

Mauricio
  • 5,273
Herr_Mitesch
  • 866

1 Answers1

1

The notation of Bailin & Love is a bit confusing even though it `works'. A clearer derivation would be $$ x'^\mu = x^\mu + a^\mu = x^\mu + ia^\lambda (P_\lambda)^\mu_{\;\nu}x^\nu $$ from which it follows that $$ ia^\lambda (P_\lambda)^\mu_{\;\nu}x^\nu = a^\mu = a^\lambda \delta^\mu_{\;\lambda} $$ Thus, $$ (P_\lambda)^\mu_{\;\nu}x^\nu = -i\delta^\mu_{\;\lambda} $$

So, in the defining representation, the translation generator is not $$ (P_\lambda)^\mu_{\;\nu} = -i\delta^\mu_{\;\nu} \partial_\lambda $$ because $x^u$ is a point in the Minkowski manifold (and it being flat, can be regarded as a vector), and not a field.

If you introduce the partial derivative you are inadvertently regarding the point $x^\mu$ as a `coordinate vector field' $$ V^\mu(x^\nu) := x^\mu $$ which is a different thing, and in this case you are not in the defining representation anymore but the field representation. So you can write the translation generator as a differential operator, but then you should also include $L^{\mu\nu} = i(x^\mu\partial^\nu - x^\nu\partial^\mu)$ as well in the generators, and then you'll recover the correct commutation relations.