Why do we drop the renormalization term in momentum Klein-Gordon Field Theory?

Question

I'm following Peskin & Schroeder's book on QFT. I managed to prove expression (2.33) which gives us the 3-momentum operator for the Klein-Gordon Theory: $$\mathbf{P}=\int \frac{d^3p}{(2\pi)^3}\mathbf{p}a_\mathbf{p}^\dagger a_\mathbf{p} + renorm.$$ However, the book drops the renormalization term that is proportional to $\delta^3(\mathbf{0})$. I understand that, in the case of energy we cannot measure changes in energy, so that term is irrelevant. But why is this so in the case for momentum? Is it because we can choose a frame that moves alongside that infinite momentum term? That solution doesn't really make sense, since it would imply infinite velocity (or mass)

Answer 1

The normal ordering procedure in flat spacetime is nothing but the subtraction of the (infinite) contribution of the Minkowski vacuum.

As a matter of fact, an added finite renormalization term could remain. $${\bf P}^\mu = \int {\bf p}^\mu a^\dagger _pa_p d^3p + t^\mu I$$ However, referring to the four momentum, it should have the form $t^\mu I$ for some Lorentz invariant four vector, since the theory has to be Lorentz covariant. $$U_\Lambda {\bf P}^\mu U^\dagger_\Lambda = {\Lambda^\mu}_\nu {\bf P}^\nu\:.$$ So that $$U_\Lambda \int {\bf p}^\mu a^\dagger _pa_p d^3p U^\dagger_\Lambda + t^\mu U_\Lambda I U_\Lambda^\dagger ={\Lambda^\mu}_\nu \int {\bf p}^\nu a^\dagger _pa_p d^3p + {\Lambda^\mu}_\nu t^\nu I\:.$$ Since $$U_\Lambda \int {\bf p}^\mu a^\dagger _pa_p d^3p U^\dagger_\Lambda = {\Lambda^\mu}_\nu \int {\bf p}^\nu a^\dagger _pa_p d^3p\:,$$ we conclude that $${\Lambda^\mu}_\nu t^\nu = t^\mu$$ As we are dealing with a tensor of order $1$, the only possibility is $t^\mu=0$.

Answer 2

1. During quantization, classical quantities are typically replaced by normal ordered operators to ensure correct action on the vacuum state, cf. e.g. my Phys.SE answer here.

2. Nevertheless, in OP's case of the 3-momentum operator $\hat{\bf P}$ a would-be normal ordering constant $$ \int_{\mathbb{R}^3}\frac{d^3p}{(2\pi)^3} {\bf p}~[\hat{a}_{\bf p},\hat{a}^{\dagger}_{\bf p}]~\stackrel{(2.29)}{=}~\int_{\mathbb{R}^3}d^3p~{\bf p}~\delta^3({\bf 0})~\hat{\bf 1}~=~0, $$ vanishes if we use a parity invariant regularization, cf. above comment by joseph h.

Answer 3

That term has nothing to do with renormalization, it’s just a result of ordering ambiguity in transitioning from classical physics to quantum physics.