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

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)

• But aren't you integrating over all possible momentum $\bf p$ and therefore you'd expect a final result of zero? That is, for each $\bf p$ you'll have a $\bf -p$? Feb 26 at 2:32
• In other words, the terms in the integral $p_i a_{\bf p}a_{-{\bf p}}$ and $p_i a^{\dagger}_{{\bf p}}a^{\dagger}_{-{\bf p}}$ are antisymmetric (summing terms with $\bf p$ and $\bf -p$ will give zero). Feb 26 at 3:24

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 $$t^\mu=0$$.
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.