# Divergence in the total momentum opertator in QFT

The classical expression for the total momentum operator is $$P^{i} = -\int d^3x \, \pi(x) \, \partial_{i} \phi(x),$$

which, after second quantisation, using $$\hat{\phi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \frac{1}{ \sqrt{2 E_{k}}} \left( \hat{a}_{k} + \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x}$$ and $$\hat{\pi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \sqrt{\frac{E_{k}}{2}} \left( \hat{a}_{k} - \hat{a}^{\dagger}_{-k} \right) e^{i k \cdot x},$$

it becomes: $$\int \frac{d^3p}{(2\pi)^3} \, p^i \left( \hat{a}_{p}^{\dagger} \hat{a}_{p} + \frac12 (2\pi)^3 \delta^{(3)}(0) \right).$$

What is the significance of the diverging second term?

I know we get a similar expression for the ground state energy, which we just ignore by arguing that absolute energy is not an observable since we can only measure energy differences. But momentum is surely an observable? We can measure absolute momentum yes?

• That second term is not a divergence. You have an odd integral so it vanishes. – FenderLesPaul Dec 30 '14 at 18:43
• You are integrating the momentum $\vec{p}$ over all directions so it adds up to zero. More explicitly you can send $\vec{p} \rightarrow -\vec{p}$ which gives $\int d^3p \vec{p}\delta^3(0) = -\int d^3p \vec{p}\delta^3(0) \Rightarrow \int d^3p \vec{p} \delta^3(0) = 0$. – FenderLesPaul Dec 30 '14 at 19:03

Well, quantization of a classical system may not be a unique procedure. Classically, all variables commute, and, say, $\pi ~\partial_i \phi = \partial_i\phi~\pi.$
When we quantize, why should we choose $\hat{\pi}\hat{\phi}$ ordering, as OP does in his example? Why not instead the opposite $\hat{\phi}\hat{\pi}$ ordering? Or perhaps symmetrize? Or use normal ordering? Or something else?
• OK I see that it vanishes because $p^i$ is odd, I'm happy – SuperCiocia Dec 30 '14 at 18:59