9
$\begingroup$

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.

$\endgroup$
3
+200
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I cannot see why the renormalized fields do not obey canonical commutation relations (CCR), since if we make a transformation on a system which breaks CCRs, then the “physics” of the system will change. So, I think we should rescale the renormalized fields so that they obey CCRs again. $\endgroup$ – AlQuemist Jul 31 '17 at 7:25
  • $\begingroup$ @PhilosophiaeNaturalis: The physics of the system does change. Renormalized fields and bare fields don't obey the same equations. They differ by factors of $Z$. $\endgroup$ – JeffDror Jul 31 '17 at 12:07
  • $\begingroup$ I have no problem with accepting that CCR need not hold if we don't explicitly enforce them by field rescaling. I am a little puzzled by the fact that this is never mentioned in textbooks. However, I am not quite sure wheter it's bare fields that satisfy CCR. Correlations functions of renormalized fields are finite, hence so is spectral density function. Therefore CCR hold up to a finite rescaling constant. Bare fields differ from renormalized fields by infinite rescalings needed to cancel UV divergences. $\endgroup$ – Blazej Jul 31 '17 at 12:15
  • $\begingroup$ Hence I would conclude that commutation relations of bare fields are in fact meaningless when regularization is taken away. This agrees with the point of view that bare fields (and renormalization constants) cease to exist for $\Lambda \to \infty$. We only expect the renormalized fields to be an approximation of some genuine, existing QFT. $\endgroup$ – Blazej Jul 31 '17 at 12:16
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Note also: the Weinberg sum rule quoted in the OP is typically derived in an on-shell normalisation scheme (and, in essence, can be taken to define such a scheme). In other schemes, the sum rule clearly remains valid, after introducing the necessary factors of $Z$. $\endgroup$ – AccidentalFourierTransform Aug 1 '17 at 16:21
  • $\begingroup$ As explained in the comment to the previous answer, I am not sure whether this answer is fully correct. You claim that bare fields satisfy CCR (without extra Z factors) and then that renormalized fields in some specific scheme satisfy Weinberg sum rule. I believe these statements are contradictory, because these fields are related by divergent rescaling. Possibly I am missing some fundamental point here. $\endgroup$ – Blazej Aug 2 '17 at 8:57
  • $\begingroup$ If $\phi$ (renormalised or not) satisfies $\langle\phi\phi\rangle\sim \rho$, then $\bar\phi\equiv c\phi$ satisfies $\langle\bar\phi\bar\phi\rangle\sim\bar\rho$, where $\bar\rho\equiv c^2\rho$. There is nothing contradictory about this: the spectral density of an arbitrary operator depends on the normalisation of the operator itself. If $c$ is divergent, then either $\rho$ or $\bar\rho$ will be divergent (or perhaps both of them). The usual postulate (that only works for renormalisable theories) is that renormalised fields have finite correlation functions (and thus, spectral densities). $\endgroup$ – AccidentalFourierTransform Aug 2 '17 at 15:08

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.