Normal ordering versus no normal ordering

[geoduck]

Within the path integral formalism, in a $\phi^4$ theory, the one-loop self-energy is divergent. Moreover the loop does not depend on external momenta.

However, if you normal order the interaction in the operator formalism, then the one-loop self-energy is zero, since it represents a contraction of operators that are in normal order, which is zero.

How can the two results be the same? Also is it possible to normal order an interaction in the path integral representation?

Also, both the operator formalism and the path integral formalism lead to the same Feynman rules, so shouldn't they lead to the same result? But how then in the operator formalism do we get zero?

 

[Chopin]

The operator formalism requires the operators to be time-ordered, not normal ordered. So normal ordering leads to a different equation than time ordering--it's only natural that they would give different results.

If you start with the time-ordered operator string, you can turn it into a sum of normal-ordered operator strings which is equal to it, by making use of Wick's Theorem. In addition to a normal-ordered string of operators, which does indeed have a vacuum expectation value of 0, you also get a term with no operators left in it, but multiplied by a C-number whose value is determined by contracting the various operators with each other. This term produces the non-zero VEV, and is equivalent to what you get if you use the path integral formalism.

 

[geoduck]

I think for the operator formalism to give the same result as the path integral formalism, you aren't allowed to normal order the Hamiltonian in the operator formalism.

To include normal ordering in the path integral formalism, I guess you have to add counter-terms to your Lagrangian $\mathcal L_I+ \Delta \mathcal L_I$ such that $\mathcal L_I+\Delta \mathcal L_I= :\mathcal L_I:$

 

[vanhees71]

In the path-integral formalism you (tacitly) to Weyl ordering, i.e., you sort local operator products such that all canonical field momenta to the left of all fields, while in the operator formalism you can define local operator products (in the formal interaction picture of perturbation theory!) to be normal ordered.

Note that the ordering of local operator products is always an indetermined expression to begin with, as the canonical equal-time commutation relations show, because these commutators are distributions like the $\delta$ distribution that become undetermined when also the spatial arguments become equal.

As your example of the tadopole diagram in $\phi^4$ theory shows the solution is renormalization theory. The UV infinities in perturbation theory need to be renormalized, and the tadpole diagram is simply a mass-renormalization part that is determined by the choice of the renormalization scheme. If you have chosen to normal order the local operator products, the corresponding tadpole loop is 0 by definition, but this doesn't affect the final result for the renormalized self-energy (polarization tensor) of the $\phi$ field, because this is completely determined by the choice of the renormalization scheme. Further, the result for physical observables (S-matrix elements) is independent of the choice of the renormalization scheme, and thus whether or not you do normal ordering doesn't affect any physical results.

 

[geoduck]

In the path-integral formalism you (tacitly) to Weyl ordering, i.e., you sort local operator products such that all canonical field momenta to the left of all fields, while in the operator formalism you can define local operator products (in the formal interaction picture of perturbation theory!) to be normal ordered.

I thought Weyl ordering was mixing p's and q's equally, and taking the average. So if you had pq, then Weyl ordered it would be 1/2(pq+qp). If you have p^2q, it would be: 1/3(ppq+pqp+qpp)?

If you have chosen to normal order the local operator products, the corresponding tadpole loop is 0 by definition, but this doesn't affect the final result for the renormalized self-energy (polarization tensor) of the $\phi$ field, because this is completely determined by the choice of the renormalization scheme. Further, the result for physical observables (S-matrix elements)

What's the point of calculating a tadpole then? Isn't it easier to have all diagrams with propagators that close on themselves be zero? Zero is nice!

 

[Avodyne]

You are correct about Weyl ordering.

There is no point in calculating tadpoles if what you want at the end is renormalized amplitudes in an on-shell scheme.

 

[Demystifier]

For most theories, taking the normal-ordered Hamiltonian : is equivalent to replacing H with H-const. Here const can be interpreted as the vacuum energy, which (if gravity is not taken into account) plays no physical role. Likewise, in path integral quantization you can add a constant to the Hamiltonian/Lagrangian/action, without changing physics.

 

[vanhees71]

I don't know, where I read the definition of Weyl ordering being the one you use in the path-integral formulation, but in any case there you have to bring all conjugate field momenta to the left of all field operators. See, e.g.,

http://fias.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf

In vacuum QFT you can usually forget about tadpole diagrams. However, in gauge theories it can be important to include them formally to have gauge invariance in the sense of fulfilling Ward-Takahashi or Slavnov-Taylor identities at all stages of the calculation. E.g., if you calculate the gluon self-energy in QCD you need the tadpole with the four-gluon vertex to keep the (dimensionally regularized) gluon polarization tensor transverse. The same is true in scalar QED for the "sea-gull tadpole" in the photon polarization.

 

[geoduck]

For most theories, taking the normal-ordered Hamiltonian : is equivalent to replacing H with H-const. Here const can be interpreted as the vacuum energy, which (if gravity is not taken into account) plays no physical role. Likewise, in path integral quantization you can add a constant to the Hamiltonian/Lagrangian/action, without changing physics.

Does putting a product of operators in normal order (creation operators on left, annihilation operators on right), always produce a finite quantity when the operators are taken at the same spacetime point?

For the free Hamiltonian it does, as the expectation value of the Hamiltonian between any states is finite. But the Hamiltonian is integrated over space though. Is the expectation value of the Hamiltonian density between two states (the energy density) finite?

Obviously any operator in normal order is finite between the vacuum states. But that does not imply that the operator is finite between say one-particle states. The identity operator for example is finite between vacuum states, but infinite between one-particle states.

 

[geoduck]

In vacuum QFT you can usually forget about tadpole diagrams. However, in gauge theories it can be important to include them formally to have gauge invariance in the sense of fulfilling Ward-Takahashi or Slavnov-Taylor identities at all stages of the calculation. E.g., if you calculate the gluon self-energy in QCD you need the tadpole with the four-gluon vertex to keep the (dimensionally regularized) gluon polarization tensor transverse. The same is true in scalar QED for the "sea-gull tadpole" in the photon polarization.

Is maintaining gauge invariance for something like tadpoles even important? It's my understanding you can regularize a theory in any way, breaking all sorts of symmetries, but the end result after renormalization will have all the symmetries. So you can break symmetries at the regularization step, but the renormalized result will respect all the symmetries? So you can choose any regulator, or order your operators however you want (neglecting tadpoles for instance via normal ordering), but after you renormalize, you're fine?

 

[Demystifier]

Does putting a product of operators in normal order (creation operators on left, annihilation operators on right), always produce a finite quantity when the operators are taken at the same spacetime point?

Yes.

The identity operator for example is finite between vacuum states, but infinite between one-particle states.

The identity operator between vacuum states is the same as identity operator between one-particle states.

 

[vanhees71]

Is maintaining gauge invariance for something like tadpoles even important? It's my understanding you can regularize a theory in any way, breaking all sorts of symmetries, but the end result after renormalization will have all the symmetries. So you can break symmetries at the regularization step, but the renormalized result will respect all the symmetries? So you can choose any regulator, or order your operators however you want (neglecting tadpoles for instance via normal ordering), but after you renormalize, you're fine?

This is not completely true. An important example are anomalies. An anomaly is the explicit breaking of a symmetry of the classical action when quantizing the field-theoretical model. Take, e.g., massless QCD or pure Yang-Mills theory: The classical action invariant under scale transformations, because these models do not contain any dimensionful parameters. The quantized theory, however, needs to be renormalized at some momentum scale. The reason is that you cannot renormalize the divergent diagrams at the point where all external momenta are vanishing, because there the diagrams like the gluon self-energy have a singularity. In this way you introduce a scale into the system, which breaks scale invariance explicitly. This is known as the "trace anomaly", because the conformal symmetry of the classical action implies that the trace of the energy-momentum tensor vanishes, which is not the case for the quantized model.

Of course, a local gauge symmetry must not be broken explicitly in any way explicitly, because this would spoil the inner consistency of the model immediately. The local gauge symmetry implies the socalled Ward-Takahashi or Slavnov-Taylor identities, which lead to constraints on the proper vertex functions. Thus you can also keep the normal-ordering description, leading to the vanishing of all tadpole diagrams (in vacuum QFT only of course, at finite temperature they are important contributions, leading to thermal masses). Then the Ward-Takahashi identities lead to constraints to the counter terms for the renormalized quantities that partially determine the renormalization conditions for the divergent proper vertex functions. This procedure leads to gauge-covariant quantities and gauge-invariant physical observables like S-matrix elements.

This is also important in connection with anomalies of local gauge theories. E.g., the classical action of massless QCD is invariant under the axial U(1) transformations as well as under the usual phase-factor invariance U(1), but in the quantized theory not both currents can be conserved simultaneously due to an anomaly (this is the modern version of the Adler-Bell-Jackiw anomaly). Gauge invariance, however dictates that the usual U(1) current must be conserved as a necessary condition. Thus you have to renormalize the theory such that only the $\mathrm{U}(1)_{\mathrm{A}}$ is broken and the axial-scalar current not conserved, and this anomaly is very welcome, because it leads to the correct prediction for the decay width of neutral pions to two photons.

 

[andrien]

Does putting a product of operators in normal order (creation operators on left, annihilation operators on right), always produce a finite quantity when the operators are taken at the same spacetime point?

Yes.

The answer is of course 'No'.If you are working in a free field theory,you can subtract out the singular vacuum expectation value from the simple product of field operators defining it as normal ordering.There is no direct switching from it to interacting field theory in which one has to resort to operator product expansion in which the singularity is contained in coefficient functions.
As far as the tadpole goes, one can avoid it by giving a linear counterterm in lagrangian to cancel the vacuum expectation value arising from it,however in the cases where spontaneous symmetry breaking occurs one has to be a bit careful in avoiding it.However one must consider them to preserve gauge invariance.

 

[geoduck]

This is not completely true. An important example are anomalies. An anomaly is the explicit breaking of a symmetry of the classical action when quantizing the field-theoretical model. Take, e.g., massless QCD or pure Yang-Mills theory: The classical action invariant under scale transformations, because these models do not contain any dimensionful parameters. The quantized theory, however, needs to be renormalized at some momentum scale. The reason is that you cannot renormalize the divergent diagrams at the point where all external momenta are vanishing, because there the diagrams like the gluon self-energy have a singularity. In this way you introduce a scale into the system, which breaks scale invariance explicitly. This is known as the "trace anomaly", because the conformal symmetry of the classical action implies that the trace of the energy-momentum tensor vanishes, which is not the case for the quantized model.

But can't you find the anomaly with any regulator? I vaguely recall for example in the chiral anomaly, if you use dimensional regularization, the anomaly is the result of problems defining γ₅ in d-dimensions. But if you use something else, say Pauli-Villars, the anomaly is due to linear shifts not allowed for divergent expressions.

I think the anomaly can appear at the level of regularization - you don't have to have renormalization for the anomaly? For the dilation anomaly, even if you could renormalize at zero momentum, you still have to regulate first, and the regulator itself breaks scale invariance? If you regulate using dimensional-regularization, then the coupling is no longer dimensionless in d-dimensions, and scale invariance is broken there, and not due to the introduction of scale $\mu$?

So I still think you might be allowed to use any regulator you want, even if you have anomalies.

 

[vanhees71]

The point is that in the case of an anomlay you cannot find any regularization that obeys all symmetries. You need additional constraints to decide which symmetries should "survive" the quantization of the model. As I tried to say in my previous posting, often the additional constraints are dictated by consistency conditions. E.g., you are not allowed to break a local gauge symmetry in any way explicitly. If a local gauge symmetry is anomalously broken, the whole model doesn't make sense anymore, because then you populate unphysical degrees of freedom leading to problems with causality and unitarity of the S matrix. At the end the additional constraints are used to define the renormalization conditions of the theory and at the end only the renormalized theory has a proper physical interpretation, and this should be independent of the regularization used.

Take dimensional regularization and the ABJ anomaly. Here, 't Hooft and Veltman invented an additional rule to solve the "$\gamma_5$ problem" by defining $\gamma_5$ to anticommute with $\gamma^0,\ldots,\gamma^3$ but commute with the $\gamma$-matrices with higher indices. This rule, however, does nothing else than shifting the anomaly solely to the $\mathrm{U}(1)_A$ part of the symmetry and keep to usual $U(1)$ phase symmetry intact.

If technically possible, I prefer to look at renormalization from the BPHZ point of view. You don't regularize the divergent (loop) integrals but read the Feynman rules as rules for the integrands and then do the subtractions, according to the renormalization conditions chosen such as to fullfil all constraints, to get directly the finite contributions to the corresponding proper vertex function. This point of view has the advantage that you don't need to worry about the properties of a specific regularization scheme.

Of course, often clever regularization schemes help to simplify a calculation significantly. E.g., one of the breakthroughs in renormalization theory was the idea of dimensional regularization (Bollini, Giambiagi and then of course 't Hooft, Veltman).