Canonical commutation relation in QFT

Question

The canonical commutation relation in QFT with say one (non-free) scalar real field $\phi$ is $$[\phi(\vec x,t),\dot \phi(\vec y,t)]=i\hbar\delta^{(3)}(\vec x-\vec y).$$

Is this equation satisfied by the bare field or the renormalized one?

Answer 1

The canonical commutation relation you wrote is valid for the bare fields. For renormalized fields it changes as follows. Consider a renormalized real scalar field $\phi(x)$. By the spectral decomposition, we know that $$\langle 0 | \phi(x)\phi(y)|0\rangle = \int\frac{d^4p}{(2\pi)^4}e^{-ip(x-y)}\int_0^{\infty}d\mu^2\rho(\mu^2)\frac{i}{p^2-\mu^2+i\varepsilon} $$ where $$ \rho(\mu^2) = \sum_n |\langle n | \phi(0)|0\rangle |^2\delta^{(4)}(p^2_n-\mu^2)$$ is the spectral density and the $|n\rangle$ is a complete set of states with momenta $p_n^{\mu}$. It follows that $$\langle 0 | [\dot{\phi}(x),\phi(y)]|0\rangle = \int\frac{d^4p}{(2\pi)^4}\int_0^{\infty}d\mu^2\rho(\mu^2)\frac{p_0}{p^2-\mu^2+i\varepsilon}\left(e^{-ip(x-y)}+e^{+ip(x-y)}\right) $$ We can use the distributional identity $$\frac{1}{p^2-\mu^2+i\varepsilon}=\mathcal{P}\frac{1}{p^2-\mu^2}-i\pi\delta^{(4)}(p^2-\mu^2) $$ where $\mathcal{P}$ is the Cauch principal value. When $x^0 = y^0$ we can substite this distributional identity into the spectral decomposition for the commutator and integrate over $p_0$. The term with the Cauchy principal value vanishes, while the delta-term yields $$\langle 0 | [\dot{\phi}(x),\phi(y)]|0\rangle\rvert_{x^0=y^0}=-i\delta^{(3)}(\textbf{x}-\textbf{y})\int_0^{\infty}d\mu^2\rho(\mu^2) $$ So the identity contribution to equal-time commutator for a renormalized field and its renormalized canonical momentum differs from its bare counterpart by the infinite factor $\int_0^{\infty}d\mu^2\rho(\mu^2)$.

Answer 2

Ok, here's my best attempt at an answer. I say the following without full confidence as I've never seen this particular detail discussed in any textbook (maybe it's buried in some unreadable way in Weinberg?), and it's also an issue I've worried about over the years. My conclusion is that the answer depends on whether you wish to perform bare perturbation theory or renormalized perturbation theory.

If you wish to perform bare perturbation theory, then your Lagrangian looks like $$\mathcal L_{bare} = \mathcal L^0_{free} + \mathcal L^0_{int},$$ where $\mathcal L^0$ indicates that the densities are made up of bare fields, masses, and couplings. You compute the canonically conjugate bare momenta of the bare fields in the bare free Lagrangian, and then you impose the canonical quantization conditions on those fields. Computing physical observables is particularly difficult in this approach because there are infinite shifts between the bare quantities and measured quantities.

Alternatively, if you wish to perform renormalized perturbation theory, then your Lagrangian looks like $$\mathcal L_{renormalized} = \mathcal L^r_{free} + \mathcal L^r_{int},$$ where $\mathcal L^r$ indicates that the densities are made up of renormalized fields, masses, and couplings. You compute the canonically conjugate renormalized momenta of the renormalized fields in the renormalized free Lagrangian, and then you impose the canonical quantization conditions. The advantage of this approach is that you're always working with finite quantities, as the counterterms absorb the UV divergences in a natural way. If you're working in the on-shell (OS) renormalization scheme, the masses are the physical masses, and your propagators are simple. In the MSbar renormalization scheme, your propagators now have an implicit coupling dependence which connects the renormalized masses $m_r$ to the physical masses.

It's probably worth providing some explicit examples. Suppose we consider $\phi^4$ theory, in which case we start off with the bare Lagrangian $$\mathcal L_{bare} \equiv \frac12\partial_\mu\phi_0\partial^\mu\phi_0 - \frac12m_0^2\phi_0^2 - \frac{\lambda_0}{4!}\phi_0^4.$$ Should we wish to perform bare perturbation theory, we'd take $$\mathcal L_{free}^0 \equiv \frac12\partial_\mu\phi_0\partial^\mu\phi_0 - \frac12m_0^2\phi_0^2$$ and impose $$[\phi_0(t_0,\vec x),\Pi_0(t_0,\vec y)]=i\delta^{(3)}(\vec x-\vec y),$$ where $\Pi_0(x)\equiv\partial\mathcal L/\partial\dot\phi_0 = \dot\phi_0(x)$.

Suppose, rather, that we wish to perform renormalized perturbation theory with dimensional regularization. Then we'd define $$\phi_0\equiv\sqrt{Z_\phi}\phi_r,\quad Z_\phi\equiv 1 + \delta_\phi \\ m_0\equiv\frac{\sqrt{Z_m}}{\sqrt{Z_\phi}}m_r, \quad Z_m \equiv 1 + \frac{1}{m_r^2}\delta_m \\ \lambda_0 \equiv \frac{Z_\lambda}{Z_\phi^2}\mu^\epsilon\lambda_r,\quad Z_\lambda\equiv1+\frac{1}{\lambda_r}\delta_\lambda.$$ Then $\mathcal L_{renormalized} = \mathcal L^r_{free}+\mathcal L^r_{int}$ where $$\mathcal L^r_{free} \equiv \frac12\partial_\mu\phi_r\partial^\mu\phi_r - \frac12m_r^2\phi_r^2 \\ \mathcal L^r_{int}\equiv \frac12\delta_\phi\partial_\mu\phi_r\partial^\mu\phi_r - \frac12\delta_m\phi_r^2-\frac{\delta_\lambda\mu^\epsilon}{4!}\phi_r^4,$$ and we'd impose the canonical quantization conditions $$[\phi_r(t_0,\vec x),\Pi_r(t_0,\vec y)] = i\delta^{(3)}(\vec x-\vec y),$$ where $\Pi_r(x) \equiv\partial\mathcal L/\partial\dot\phi_r = \dot\phi_r(x)$.