Commutator of position operator and field in scalar QFT

Given an energy momentum tensor $$T^{\mu \nu} = \partial^{\mu} \phi \partial^{\nu} \phi - \eta^{\mu \nu}\mathcal{L}$$, the standard definition of the momentum operator is $$P^{\mu} \equiv \int d^3x T^{\mu 0} = \int d^3 x (\partial^0 \phi \partial^{\nu} \phi - \eta^{0\nu} \mathcal{L})$$. I am not able to verify the claim that $$i[P^{\mu},\phi(x)]=\partial^{\mu}\phi$$.

We usually assume that $$[\phi(x), \pi(y)] = i \delta(x-y);$$ I am not quite sure why we identify $$\partial^0 \phi$$ as $$\pi(y)$$ in this case. Further, $$[P^{\mu},\phi(x)] = \int d^3 x \;(\pi (\partial^{\mu}\phi)\phi - \eta^{0\nu}\mathcal{L}\phi) - \phi \int d^3 x (\partial^0 \phi \partial^{\nu} \phi - \eta^{0\nu} \mathcal{L})$$, and I can't seem to simplify this expression. Perhaps the commutation is to be carried out in the integrand, in which case, by commutation of $$\phi$$ with $$\mathcal{L}$$ and $$\partial^{\mu}\phi$$, one would instead get $$[P^{\mu},\phi(x)]= \int d^3 x \; \partial^{\mu} \phi[\pi,\phi] = -i \partial^{\mu} \phi$$, which is the desired relation. However, I do not see why, formally speaking, the commutation should be carried out in the integrand.

I would also appreciate a line or two providing the intuition behind $$P^{\mu}$$'s definition and how it agrees with to our notion of momentum in elementary QM.

• $$\partial^0\phi$$ is the canonical momentum because that's defined as $$\pi\sim\frac{\delta S}{\delta \dot\phi}\,,$$ as in classical mechanics.
• The definition of $$P^\mu$$ basically relies on Noether's theorem: The energy-momentum tensor is constructed to be the conserved current associated with spacetime translations (i.e. $$\partial_\mu T^{\mu\nu}=0$$ because translation leave the action invariant), and the conserved charge is obtained by a spatial integral of the $$0$$-component. (Commonly, you have a vector current and thus a scalar charge, but in this case, it's a two-tensore current and a vector charge.)
• You have an expression $$\left(\int \text{d}^3 x T^{\mu0}(x)\right)\phi(y) - \phi(y) \left(\int \text{d}^3 x T^{\mu0}(x)\right)$$. (Note that it helps to be more explicit in your arguments -- this expression is not a function of $$x$$!) Now where elso would you "perform the commutator" other than $$\int \text{d}^3 x\, \left[T^{\mu0}(x),\phi(y)\right]$$ ? (If you worry about well-definedness, given that there are delta functions involved, you would need to integrate against test functions. The result hold in the sense of operator-valued distributions.)
• I don't quite get what you mean by your last point. Explicitly, I have $[\int d^3 x \; \pi(y)\partial^{\mu}\phi(x), \phi(x)]$, right? The second term in the commutator will have $\phi(x)$ land to the left of the integral! – SystematicDisintegration Nov 15 '18 at 12:06
• @SystematicDisintegration: Ah, maybe that's your problem: The momentum identification is not $\pi=\dot\phi(y)$. Rather, it is $\pi(x)=\dot\phi(x)$ (or equivalently $\pi(y)=\dot\phi(y)$), i.e. $\pi=\dot\phi$, and the arguments (spacetime locations) match on both sides). – Toffomat Nov 15 '18 at 12:16