Renormalization and canonical commutation relations

Question

My question is whether canonical commutation relations hold for renormalized quantum fields. Below I show reasoning which caused by doubts.

Consider a relativistic scalar QFT. We have spectral decomposition of two-point function $$ \langle \Omega | \phi(x_1) \phi(x_2) | \Omega \rangle = \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) \Delta_+(x_1-x_2,m^2), $$ where $\rho \geq 0$ is called spectral density function and distribution $\Delta_+$ is defined as $$ \Delta_+ (x,m^2) = \int \frac{\mathrm d ^3 p}{(2 \pi)^3 2p^0} e^{-ipx}, $$ with integral evaluated over the positive frequency ($p^0 \geq 0$) mass-shell $p^2=m^2$. I assumed above that field $\phi$ has no vacuum expectation value. If we take the difference of the first formula with itself with $x_2$ and $x_1$ interchanged, set $x_2 = 0$, take derivative with respect to $x_1^0$ and set $x_1^0=0$ we get canonical commutator on the left hand side. By comparing with the right hand side one obtains the Weinberg sum rule for the spectral density: $$ \int \frac{\mathrm d m^2}{2 \pi} \rho(m^2) = 1. $$ What bothers me is that value of this integral depends on the values of finite parts of renormalization constants. Hence it is not renormalization scheme and scale independent. I checked some simple examples and it turned out to be possible to enforce this relation as renormalization condition and fix the value of wavefunction renormalization. However, I don't think this is what is usually done.

Answer 1

The commutation relations for renormalized fields are different then those of the bare fields by factors of the wavefunction renormalization. As an example, consider a complex scalar field, $\phi$. The bare fields obey, e.g., $$ \left[ \phi (x) , \phi (y) ^\dagger \right] = \int \frac{ d^3p }{ (2\pi)^3 } e ^{ i p \cdot x } $$ while the renormalized fields ($ \phi _r \equiv \phi /\sqrt{ Z} $) obey, $$ \left[ \phi _r (x) , \phi _r (y ) ^\dagger \right] = Z \int \frac{ d^3p }{ (2\pi)^3 } e ^{ i p \cdot x } $$

Should we be troubled by this? I don't think so. The important conclusion with regards to the commutation relations is they vanish for space-like points in order to be consistent with special relativity. Other than that they don't play a significant role here.

As a side point, the time-ordered product of fields is also different for the bare and renormalized fields. This leads to a modification of the propagator as I suspect you are already aware.

Answer 2

The relevant axiom.

Any (canonical) field, renormalised or not, satisfies, by postulate, $$ [\phi,\pi]=\delta $$ where $\pi$ is the field conjugate to $\phi$. In Lagrangian field theory, $$ \pi\overset{\mathrm{def}}=\frac{\partial \mathcal L}{\partial\dot\phi} $$

Case 1.

If $\phi$ is an unrenormalised field, $$ \mathcal L=\frac12\dot\phi^2_{\mathrm{un}}+\cdots $$ then $$ [\phi_{\mathrm{un}},\dot\phi_{\mathrm{un}}]=\delta $$

Case 2.

On the other hand, if $\phi$ is a renormalised field, $$ \mathcal L=\frac12Z\dot\phi^2_{\mathrm{re}}+\cdots $$ then $$ [\phi_{\mathrm{re}},Z\dot\phi_{\mathrm{re}}]=\delta $$

The Upshot.

In conclusion, the canonical commutators, when expressed in terms of (canonical) phase-space variables, are independent of the normalisation of the fields. When expressed in terms of, say, configuration-space variables, they depend on the normalisation of the fields.