0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.)
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.