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.


Just an outline:

  • $\partial^0\phi$ 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.)
  • $\begingroup$ 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! $\endgroup$ – SystematicDisintegration Nov 15 '18 at 12:06
  • $\begingroup$ @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). $\endgroup$ – Toffomat Nov 15 '18 at 12:16
  • $\begingroup$ Ah, I think I understand after looking at your edit! $\endgroup$ – SystematicDisintegration Nov 15 '18 at 12:42

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