I Field Renormalization vs. Interaction Picture

[maline]

TL;DR Summary

Renormalized fields give the wrong values for the free Hamiltonian.

When introducing renormalization of fields, we define the "free Lagrangian" to be the kinetic and mass terms, using the renormalized fields. The remaining kinetic term is treated as an "interaction" counterterm. If we write down the Hamiltonian, the split between "free" and "interaction" terms, used in defining the interaction picture, will presumably be the same. So apparently field renormalization involves multiplying the free Hamiltonian by a factor $Z^{-1}$.

How does this make sense? The free Hamiltonian has a well-defined scale - it is just the sum of the particle number operators for given momenta, with the appropriate energies (plus an infinite constant). And this only works with a particular normalization of the field operators: the derivation uses the canonical commutation/anticommutation relations, implying canonical normalization.

 

[A. Neumaier]

Summary:: Renormalized fields give the wrong values for the free Hamiltonian.

When introducing renormalization of fields, we define the "free Lagrangian" to be the kinetic and mass terms, using the renormalized fields. The remaining kinetic term is treated as an "interaction" counterterm. If we write down the Hamiltonian, the split between "free" and "interaction" terms, used in defining the interaction picture, will presumably be the same. So apparently field renormalization involves multiplying the free Hamiltonian by a factor $Z^{-1}$.

How does this make sense? The free Hamiltonian has a well-defined scale - it is just the sum of the particle number operators for given momenta, with the appropriate energies (plus an infinite constant). And this only works with a particular normalization of the field operators: the derivation uses the canonical commutation/anticommutation relations, implying canonical normalization.

In the free case, Z=1.

 

[maline]

In the free case, Z=1.

Of course. Does this address my question?

 

[A. Neumaier]

Of course. Does this address my question?

Yes, since 1 doesn't equal infinity. Thus the free theory needs no renormalization beyond normal ordering.

 

[maline]

Yes, since 1 doesn't equal infinity. Thus the free theory needs no renormalization beyond normal ordering.

I did not ask about the free theory. I asked about the "free Hamiltonian" $H_0=H-V$ used in defining the interaction picture of the theory, which is needed to derive the Feynman diagram formalism from canonical QFT.

 

[A. Neumaier]

I did not ask about the free theory. I asked about the "free Hamiltonian" $H_0=H-V$ used in defining the interaction picture of the theory, which is needed to derive the Feynman diagram formalism from canonical QFT.

But this is the Hamiltonian of the free theory, no renormalization there. The factors appear in the interaction, not in the free term.
The point is that given a Hamiltonian for a system one is in principle free to do perturbation theory around any free Hamiltonian. In the renormalization context, one needs to take the physical one, whereas the parameters in the original Hamiltonian are bare parameters to be determined by the renormalization conditions.

 

[vanhees71]

To answer the original question. You start from ficitious entities, described by the free Hamiltonian, called "bare particles", e.g., in QED represented by a Dirac field describing electrons and positrons. Since bare particles don't interact, they don't carry an electromagnetic field. Now you add the interaction term with the electromagnetic field, and this implies that the electron interacts with the electromagnetic field and as a charge carrier (including a magnetic moment automatically too) also has an electric field around it. This electric field carries energy, momentum, and angular momentum. Since the energy is equivalent to a mass in the famous sense of $E_0=m c^2$ (note that there are only invariant masses in modern theory), the corresponding mass of the electromagnetic field adds to the bare mass of the non-interacting electron. Unfortunately with our physicists' sloppy math we add an infinite mass in each order of the socalled self-energy diagram, but of course the bare mass is not observable, because electrons carry their electromagnetic field around. Thus we can lump another infinity to the bare mass such as to cancel the infinite mass contribution of the electromagnetic field such that the observed mass of the electron is left at any order of perturbation theory. The same holds for the normalization of the electron-positron Dirac field and the the electromagnetic field, i.e., you have to lump the infinities to the bare normalization constants, which don't occur in any observable quantities at the end. Last but not least the same holds for the electric charge, which is renormalized by the same mathematical reshuffling of infinities. What's also renormalized is the magnetic moment of the electron, but fortunately that's a finite contribution of the perturbative series (in the radiative corrections to the proper three-point vertex function), and is among the most precise predictions of any physical theory ever (in comparison with the as amazing accuracy of the experimental determination of this quantity).

QED (and the entire standard model) is a very special kind of quantum field theory: It's Dyson-renormalizable, i.e., there are only a finite number of constants (wave-function normalization, masses, and coupling constants) needed to lump all the infinities from the sloppy treatment of the math into the corresponding unobservable "bare quantities".

What also enters necessarily into the game of this renormalization procedure is an energy-momentum scale, the socalled renormalization scale, at which you determine the said constants through measurement of observable quantities concerning appropriate types scattering reactions (e.g., the electromagnetic interaction strength is taken at low energies, leading to a fine-structure constant of about 1/137 by looking, e.g., at the Cross section for elastic Coulomb scattering; if you look at larger scattering energies (as at the $Z$-boson-mass scale) you find a larger value of about 1/128).

This hints at another important interpretation of the renormalization procedure, which so far looks a bit artificial in just eliminating mathematical blunder by subtracting infinities such as to get finite results for the observable quantities: That's the Wilsonian view on the renormalization procedure, which interprets QFTs like the standard model as effective theories, dependent on the resolution at which I look at the interacting particles in scattering experiments. The higher the scattering energy, the finer spatial resolution I get in investigating the inner structures of the scattering particles.

At the extreme in the standard model with this respect are the strongly interacting particles. At the fundamental level, writing down the bare Lagrangian of quantum chromodynamics you deal with quarks and gluons of Quantum Chromodynamics (forgetting for a moment about the electromagnetic and weak interactions also described by the standard model). Then you switch on the interaction and realize very fascinating properties: Contrary to the similar looking case of QED the strong coupling constant decreases with larger renormalization scales. This is called asymptotic freedom, and it applies that we can't use perturbation theory for the strong interactions among quarks and gluons for low-energy scattering energies, but it works well at large scattering energies. Indeed at low energies we cannot find any quarks and gluons. All we find looking at strongly interacting objects are bound states of quarks and gluons, the hadrons (among the protons and neutrons, which are the building blocks of the atomic nuclei of all the matter around us), which carry no net-color-charge. Now you can use electrons to scatter on the hadrons and investigate their inner structure, and indeed it turns out that a proton consists of three socalled valence quarks and an entire cloud of quarks, antiquarks, and gluons, and the more energetic the interactions get, the more details are revealed (encoded in socalled parton-distribution functions or even generalized parton-distribution functions).

 

[maline]

But this is the Hamiltonian of the free theory, no renormalization there. The factors appear in the interaction, not in the free term.

But the factors do appear in the free term! When we write $\partial^\mu\phi\partial_\mu\phi$, this $\phi$ is the renormalized field. It does not satisfy the canonical commutation relations. So this "free Hamiltonian" after renormalization has been changed by a factor $Z^{-1}$ compared to the Hamiltonian of the free theory.

 

[Demystifier]

So this "free Hamiltonian" after renormalization has been changed by a factor $Z^{-1}$ compared to the Hamiltonian of the free theory.

A Hamiltonian multiplied by a constant is physically equivalent to the original Hamiltonian. If you consider the equations of motion, you can see that the two Hamiltonians describe the same physics expressed in different units of time.

 

[maline]

A Hamiltonian multiplied by a constant is physically equivalent to the original Hamiltonian. If you consider the equations of motion, you can see that the two Hamiltonians describe the same physics expressed in different units of time.

Multiplying the Hamiltonian by a constant is a physical change. It changes the actual rate at which the dynamics occur, not just the arbitrary units. For instance, multiplying the Hamiltonian by a large number will cause objects to move faster than $c$.

To see that this change is not a matter of units, note that the Schrodinger equation $$\partial_t |\psi>=-i H |\psi>$$ is dimensionally correct as it stands (using $\hbar=1$). There is no room to insert an arbitrary dimension-full constant such as a unit of time.

 

[A. Neumaier]

But the factors do appear in the free term! When we write $\partial^\mu\phi\partial_\mu\phi$, this $\phi$ is the renormalized field. It does not satisfy the canonical commutation relations. So this "free Hamiltonian" after renormalization has been changed by a factor $Z^{-1}$ compared to the Hamiltonian of the free theory.

When we do renormalized perturbation theory we write the Hamiltonian $H$ first in terms of the (already scaled) renormalized field. Then we get the interaction $V=H-H_0$ by subtracting a quadratic expression $H_0$ in this renormalized field, defined by the physical mass. This $H_0$ is the Hamiltonian of a free theory, which we use to determine time-dependent Hamiltonian of the interaction picture. The multipliers only appear in the interaction terms, not in $H_0$.

 

[maline]

When we do renormalized perturbation theory we write the Hamiltonian $H$ first in terms of the (already scaled) renormalized field. Then we get the interaction $V=H-H_0$ by subtracting a quadratic expression $H_0$ in this renormalized field, defined by the physical mass. This $H_0$ is the Hamiltonian of a free theory, which we use to determine time-dependent Hamiltonian of the interaction picture. The multipliers only appear in the interaction terms, not in $H_0$.

Once again: the fields in a free theory satisfy canonical commutation relations, while renormalized fields do not. Therefore it seems clear that a quadratic expression in renormalized fields, without further multipliers, cannot be the Hamiltonian of a free theory.

 

[A. Neumaier]

Once again: the fields in a free theory satisfy canonical commutation relations, while renormalized fields do not.

You confuse two different notions of canonical commutation relations.

Both $H_0$ and $H$ are independent of time because of translation invariance, and their definition only involves the fields at a fixed time. The renormalized field operators are causal and hence satisfy canonical equal time commutation relations. From the Schrödinger equation, the quadratic Hamiltonian $H_0$ produces a free renormalized field with 4-dimensional CCR, while the nonquadratic Hamiltonian $H$ produces an interacting renormalized field where the CCR is valid only for space-like arguments, consistent with causality.

 

[maline]

The renormalized field operators are causal and hence satisfy canonical equal time commutation relations

The fields at spacelike separated points do of course commute/anticommute. But the canonical relations also determine a scale: the RHS of the CCR is a simple delta function, without numerical factors. For renormalized fields this is not the case.

 

[A. Neumaier]

The fields at spacelike separated points do of course commute/anticommute. But the canonical relations also determine a scale: the RHS of the CCR is a simple delta function, without numerical factors. For renormalized fields this is not the case.

This actually proves that your assumption that the fields in the Hamiltonian represent the renormalized field is incorrect. (The fact that the renormalization factor is infinite means that the mathematical formulas at the singularity - the light cone, in particular at equal spacetime position - and all arguments based on them become meaningless.) Indeed, the field operators figuring in the Hamiltonian acts on a Fock space, but the renormalized field operators (obtained after taking the renormalization limit) don't. The whole Fock space structure (used to set up the renormalization schemes in the conventional way with infinite constants) disappears in the renormalization limit. What survives under renormalization are just the n-point correlation functions.

Thus the correct picture is the following: One finitely rescales the field and the couplings to define a new Hamiltonian with counterterms, interprets the field operator in this formal expression in the canonical way, calculates the regularized n-point correlation functions perturbatively, sets the counterterms to meet the renormalization conditions, then removes the regularization by taking the appropriate limit (which makes the counterterms approach zero or infinity), and gets the (both retarded and time-ordered) renormalized n-point correlation functions. From these, assuming the series defining these at each order converge, the renormalized field operators and the Hilbert space they act on can be constructed using Wightman's reconstruction theorem. This is consistent with the description given in the textbooks.

 

[maline]

One finitely rescales the field and the couplings to define a new Hamiltonian with counterterms, interprets the field operator in this formal expression in the canonical way,

I don't understand. You are saying that as long as we treat $Z$ as finite, the renormalized fields do satisfy the CCR without numerical factors? How does this make sense? We "rescale" the field without changing the commutator?

 

[A. Neumaier]

I don't understand. You are saying that as long as we treat $Z$ as finite, the renormalized fields do satisfy the CCR without numerical factors?

No. There are no renormalized fields on the Fock space level, i.e., before the final limit in the n-point functions has been performed.

We "rescale" the field without changing the commutator?

One changes the action by inserting multiplicative factors wherever we can without spoiling symmetry and renormalizability (and informally interprets them as scalings). But the field is rescaled only in the Hamiltonian, not in the equal time CCR. You can see this by looking at the actual computations made.

 

[Demystifier]

Multiplying the Hamiltonian by a constant is a physical change. It changes the actual rate at which the dynamics occur, not just the arbitrary units. For instance, multiplying the Hamiltonian by a large number will cause objects to move faster than $c$.

To see that this change is not a matter of units, note that the Schrodinger equation $$\partial_t |\psi>=-i H |\psi>$$ is dimensionally correct as it stands (using $\hbar=1$). There is no room to insert an arbitrary dimension-full constant such as a unit of time.

When you change the units of time, then $\hbar$ also changes its value. You have to account for that too.

 

[vanhees71]

The interacting fields indeed satisfy other commutation relations and have another normalization. If this were not so, the fields were not interacting, i.e., for interacting fields $Z \neq 1$. For a pretty good and careful discussion, see the famous book by Bjorken and Drell (the 2nd volume about "Quantum Field Theory"; stay away from the 1st volume "Relativistic Quantum Mechanics", which is utmost confusing from a modern point of view since Dirac's hole theory is just a very complicated version of the modern version of QED in terms of a relativistic QFT), i.e., the asymptotic free field obeys the weak (!) limit
$$\lim_{x^0 \rightarrow \pm \infty} \hat{\phi}(x) = \sqrt{Z} \hat{\phi}_{\text{in}/\text{out}}(x).$$

 

[maline]

When you change the units of time, then $\hbar$ also changes its value. You have to account for that too.

So let's just say that the standard convention $\hbar=1$ eliminates the freedom to change the unit of time without changing the unit of energy, and that if you change both, you don't get new factors in the Hamiltonian.

At any rate, in renormalization theory we certainly are keeping $\hbar$ fixed, so the change of units you suggest does not solve the issue I raised.

 

[maline]

One changes the action by inserting multiplicative factors wherever we can without spoiling symmetry and renormalizability (and informally interprets them as scalings). But the field is rescaled only in the Hamiltonian, not in the equal time CCR. You can see this by looking at the actual computations made.

I'm sorry, I still don't understand. Would you mind spelling this out in detail?

The interacting fields indeed satisfy other commutation relations and have another normalization. If this were not so, the fields were not interacting, i.e., for interacting fields $Z \neq 1$. For a pretty good and careful discussion, see the famous book by Bjorken and Drell (the 2nd volume about "Quantum Field Theory"; stay away from the 1st volume "Relativistic Quantum Mechanics", which is utmost confusing from a modern point of view since Dirac's hole theory is just a very complicated version of the modern version of QED in terms of a relativistic QFT), i.e., the asymptotic free field obeys the weak (!) limit
$$\lim_{x^0 \rightarrow \pm \infty} \hat{\phi}(x) = \sqrt{Z} \hat{\phi}_{\text{in}/\text{out}}(x).$$

Yes, this agrees with my current understanding. But it brings us back to my original question: the free part of the Hamiltonian equals the sum of frequencies of particles, if and only if it is written using fields that satisfy the CCR. So how can we do perturbation theory using fields with a different normalization?

Also, are the two of you (Vanhees71 and A. Neumaier) in agreement here?

 

[Demystifier]

So let's just say that the standard convention $\hbar=1$ eliminates the freedom to change the unit of time without changing the unit of energy, and that if you change both, you don't get new factors in the Hamiltonian.

OK, fair enough, now I see that I was wrong. The solution of your problem is, in fact, entirely different.

The idea is the following. The physical quantities one is really interested about are matrix elements of the S-matrix. Via the LSZ formalism, they are computed from the $n$-point functions. Hence the $n$-point functions are the only abstract theoretical entities one is really interested about. All the other abstract theoretical entities (Hamiltonians, field operators, ...) are just auxiliary mathematical quantities that do not make sense on its own, except as an intermediate tool towards computing the $n$-point functions. In other words, a strange new factor in the Hamiltonian does not matter, as long as it leads to the physically right behavior of the $n$-point functions. That's because, in this formalism, one does not study physical effects by solving the Schrodinger equation with the modified Hamiltonian. Instead, one studies the physical effects by computing the S-matrix from the modified $n$-point functions.

The above is the "philosophy" lying behind the so called Gell-Mann and Low approach to renormalization, common in particle physics. It is quite unintuitive from the point of view of more traditional formulation of quantum theory based on the Schrodinger equation.

Fortunately, there is an alternative Wilson approach to renormalization, which is much more intuitive. It is much more popular in statistical physics and condensed matter. It's used in particle physics too, but less often. In the Wilson approach the Hamiltonian plays a central role. The Hamiltonian gets an additional factor too, but the intuitive picture of that is entirely different. In the Wilson approach you replace a more fundamental Hamiltonian $H$ (with a larger number of degrees of freedom) with an effective Hamiltonian $H'$ (with a smaller number of degrees of freedom). In the so called "renormalizable" theories, $H'$ takes a similar form as $H$, but there is the extra factor that accounts for the effect of the fundamental degrees of freedom in $H$ that are ignored in $H'$. In this sense $H$ and $H'$ are different but their large-distance effects are the same. $H'$ contains a smaller number of degrees of freedom, but it's compensated by an additional factor (usually larger than 1) in front of it.

Here is a classical picture one may have in mind in the Wilson approach. Suppose that 10 men push a car. The Hamiltonian $H$ contains 10 degrees of freedom corresponding to 10 men. But the effect of pushing can be described by $H'$ describing only 1 super-man, who is as strong as 10 real men. Hence $H'$ must contain the extra factor of $10$.

 

[vanhees71]

I'm sorry, I still don't understand. Would you mind spelling this out in detail?Yes, this agrees with my current understanding. But it brings us back to my original question: the free part of the Hamiltonian equals the sum of frequencies of particles, if and only if it is written using fields that satisfy the CCR. So how can we do perturbation theory using fields with a different normalization?

Also, are the two of you (Vanhees71 and A. Neumaier) in agreement here?

Usually @A. Neumaier agree about standard notions of QFT, but this he should judge.

The free part of the Hamiltonian does not equal the sum of frequencies of particles. Written in the interaction picture in terms of particle-number operators, it rather reads (in normal-ordered form)
$$\hat{H}_0=\sum_{\sigma} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} E(\vec{p}) [\hat{N}(\vec{p,\sigma}) + \hat{\bar{N}}(\vec{p},\sigma)],$$
where $E(\vec{p})=\sqrt{\vec{p}^2+m^2}>0$, and $\hat{N}$ and $\hat{\bar{N}}$ are the occupation-number operators of field modes with definite momentum $\vec{p}$ and spin-magnetic quantum number $\sigma \in \{-s,-s+1,\ldots,s \}$ where $s \in \{0,1/2,1,\ldots \}$ is the spin quantum number of the particular sort of particles under investigation.

It's of course clear that when you take into account interactions all physical quantities must be renormalized, and you have corresponding renormalization conditions ensuring that you get the correct normalization of energies, momenta and any other physical quantities as well as S-matrix elements. A change of the renormalization prescription must not change the outcome for physical quantities or the observable cross sections defined via the S-matrix elements. This idea leads to the renormalization-group equations, which describe how the renormalized quantities (wave-function normalizations, masses, couplings) change when changing the renormalization prescription. Depending on which renormalization scheme you use these quantities need not necessarily be the observed quantities. E.g., a renormalized mass is only the physical mass of the particle if you can use a socalled on-shell renormalization scheme, which is usually not the case as soon as massless particles are involved in the fields. The observed mass is rather defined as the poles in the two-point (one-particle) Green's function as a function of energy, and this observable mass is independent of the renormalization scheme and a physical quantity.

 

[A. Neumaier]

Yes, this agrees with my current understanding. But it brings us back to my original question: the free part of the Hamiltonian equals the sum of frequencies of particles, if and only if it is written using fields that satisfy the CCR. So how can we do perturbation theory using fields with a different normalization?

Also, are the two of you (Vanhees71 and A. Neumaier) in agreement here?

Yes, we agree.

As I mentioned, one does not scale the CCR when preparing the perturbation scheme, hence scaling is strictly speaking a misnomer, part of the inaccurate laisse faire manner in which renormalization is typically introduced. More rigorously, what is called field renormalization is just the introduction of a new constant into the Hamiltonian, needed to make the renormalization work. The effect is that the new Hamiltonian depends on enough constants to be fixed by renormalization conditions.

The field in this parameterized Hamiltonian remains a free field (of asymptotic particles) with standard CCR, since the Fock machinery is defined only for such a field.

 

[vanhees71]

Of course all this is quite subtle, as everything having to do with the definition of what "asymptotic free fields/states" are. These are needed to define the prime observable quantities aimed at to be calculated in perturbative "vacuum QFT", namely cross sections with the corresponding transition-rate matrix elements (or S-matrix elements). All this can be found in the literature under "LSZ reduction formalism" or the like. As I said, I find the discussion in the old textbook by Bjorken and Drell pretty illuminating. Also the discussion in the book by Itzykson and Zuber is nice. An alternative elegant formulation is using the path-integral formalism defining various generating functionals. This can be found in the book by Bailin and Love. My own attempt can be found in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

All this is of course far from being mathematically rigorous, and I'm not sure, whether there are mathematically fully rigorous formulations. I doubt it ;-).

 

[A. Neumaier]

All this is of course far from being mathematically rigorous, and I'm not sure, whether there are mathematically fully rigorous formulations. I doubt it ;-).

Causal perturbation theory is mathematically rigorous, but gives like all other techniques only a perturbative series. It works rigorously on the free field level and produces the interacting fields very late in the development - after the full S-matrix has been constructed. (For details see the second edition of the QED book by Scharf.)

 

[vanhees71]

Well, if this counts as rigorous, then perturbative QFT is rigorous :-), because what Scharff writes is equivalent to the usual less rigorous methods. It's just using one possible (in fact physically very intuitive) regularization of the ill-defined products of operator valued distributions.

 

[A. Neumaier]

Well, if this counts as rigorous, then perturbative QFT is rigorous :-), because what Scharff writes is equivalent to the usual less rigorous methods. It's just using one possible (in fact physically very intuitive) regularization of the ill-defined products of operator valued distributions.

Causal perturbation theory is nonrigorously equivalent to traditionally nonrigorous renormalization, but rigorously unique.

Unlike the traditional methods it is manifestly covariant and neither introduces infinite terms, nor a cutoff, nor physically meaningless nonintegral dimensions, nor undefined infinite-dimensional measures. This is the reason why it counts as rigorous, while the traditional methods don't.

 

[Demystifier]

Well, if this counts as rigorous, then perturbative QFT is rigorous :-), because what Scharff writes is equivalent to the usual less rigorous methods. It's just using one possible (in fact physically very intuitive) regularization of the ill-defined products of operator valued distributions.

Something related I noted in 2013:
https://www.physicsforums.com/threads/haags-theorem-in-qft.728744/post-4610861

 

[maline]

If there is no disagreement here, how are these two statements consistent?

The interacting fields indeed satisfy other commutation relations and have another normalization.

The field in this parameterized Hamiltonian remains a free (bare) field with standard CCR

Is it that you are referring to different formalisms?

 

[A. Neumaier]

If there is no disagreement here, how are these two statements consistent?Is it that you are referring to different formalisms?

We refer to different objects. The field in the parmeterized Hamiltonian with the $Z$#s is a free field (of asymptotic particles) acting on a Fock space and satisfying free field equations, not the renormalized field acting on a non-Fock space and satisfying interacting field equations. (The underlying math is that of the ''two Hilbert space approach'' to perturbative scattering theory.)

 

[maline]

I'm afraid I didn't really understand any of your answers, so here is my own attempt. Please let me know whether I'm making sense.

It seems to me that we don't need to talk about rescaling the fields at all. We can produce the counterterms using only the canonically normalized fields throughout, as follows:

First, the $\delta m$ counterterm is an actual modification of the Lagrangian, on physical grounds: the self-interactions of a single particle will naturally add to its mass, but we need $m$ (used in $H_0$) to equal the physical mass, so the only option is to explicitly subtract the added mass by assuming a new interaction with the form of a mass term (which happens to be infinite but whatever). We find the correct value for the coupling constant $\delta m$ by demanding that the pole in the propagator be at $p^2=-m^2$.

Next, the LSZ theorem tell us that when using Feynman diagrams to calculate S-matrix elements,, without any field rescaling, we must multiply the result by a factor $\sqrt{Z_i}$ for each incoming and outgoing particle of species $i$, where each $\sqrt{Z_i}$ is the inner product between a state created from the vacuum by the field operator - an eigenstate of the free Hamiltonian - and the corresponding one-particle state, an eigenstate of the full Hamiltonian. But in order to treat these factors perturbatively, we manipulate them as follows:

Instead of applying the factors to S-matrix elements or diagrams, we multiply each interaction vertex (including the $\delta m$ terms) by $\sqrt{Z_i}$ for every field of species $i$ appearing in the interaction, and divide every internal line of species $i$ by $Z_i$. Each internal line meets two field operators at vertices, so all these factors will be cancelled, except for those for field operators that attach to the external lines. So the total effect is just to insert the factors for external lines, as needed.

Finally, we expand the $Z^{-1}_i$ factors in the propagators. Using a scalar field for simplicity:
$$\frac 1 Z = \frac 1 {1-(1-Z)} = \Sigma_{n=0}^\infty (1-Z)^n$$. Therefore,$$\frac 1 {Z(p^2+m^2-\epsilon i)}=$$ $$=\frac 1 {(p^2+m^2-\epsilon i)}+\frac 1 {(p^2+m^2-\epsilon i)}(1-Z)(p^2+m^2-\epsilon i)\frac 1 {(p^2+m^2-\epsilon i)}+...$$

So the final effect is the same as creating a new interaction $(1-Z)(p^2+m^2-\epsilon i)$ that can be inserted any number of times in any internal line.

One last detail is that in QED, it is customary to absorb the factor $\sqrt{Z_3}$ for the photon field in the interaction into a redefinition of the coupling constant $e$, the "renormalized charge".

In summary, all the actual calculations can be justified without ever defining a "renormalized field". In particular, there is no reason to change $H_0$ at all, so my problem disappears.

I still don't see how to make sense of the standard presentations of this material, but if I have one version that works I am happy.

 

[A. Neumaier]

In summary, all the actual calculations can be justified without ever defining a "renormalized field". In particular, there is no reason to change $H_0$ at all, so my problem disappears.

Yes. One just introduce the maximal number of parameters in the Lagrangian (whether motivated in fuzzy or even incorrect terms doesn't matter) , and writes them as differences or sums of the free term used for the perturbation and a counterterm. This includes the factor before the derivative term, which (by convention) is unity in the free case. Whatever field appears in the Lagrangian is still a free field - since only then it is known what the free Hamiltonain and the interaction means in terms of creation and annihilation operators.

 

[maline]

One just introduce the maximal number of parameters in the Lagrangian (whether motivated in fuzzy or even incorrect terms doesn't matter) , and writes them as differences or sums of the free term used for the perturbation and a counterterm. This includes the factor before the derivative term, which (by convention) is unity in the free case.

If we have that much freedom, couldn't we just add an interaction term proportional to $p^2+m^2$, without also multiplying the vertices by $\sqrt Z$ factors, and without the restriction $0\le Z\le 1$? I think we cannot change the factor before the derivative term. In a theory with only one field, this is clear: multiplying the entire Lagrangian by any numerical factor has no effect, so the kinetic term is the one chosen to have a fixed scale and remove that freedom.

Instead, I explained how we can get the $1-Z$ counterterms without having any such interaction in the Lagrangian, because these counterterms are equivalent to the factors needed for the overlap between the true particle states and the "bare particles" produced by the field, as required by LSZ.

 

[A. Neumaier]

If we have that much freedom, couldn't we just add an interaction term proportional to $p^2+m^2$, without also multiplying the vertices by $\sqrt Z$ factors, and without the restriction $0\le Z\le 1$? I think we cannot change the factor before the derivative term.

A family of field theories is just given by providing a family of Lagrangians - how you construct the latter is irrelevant. What counts is that the form given has enough freedom to make the renormalization process work. This requires that every term has a counterterm. In particular, there must be a factor before the derivative term.

The restriction $0\le Z\le 1$ plays no role in the renormalization procedure and is never checked since $Z$ is a bare quantity without physical meaning.

$p$ is not a variable in the Lagrangian, so I don't understand your proposal about adding an interaction term proportional to $p^2+m^2$.

In a theory with only one field, this is clear: multiplying the entire Lagrangian by any numerical factor has no effect, so the kinetic term is the one chosen to have a fixed scale and remove that freedom.

This holds for a classical theory but not for a quantum theory when, as usual, $\hbar$ is fixed at 1, since scaling the Lagrangian changes the expression in the functional integral.

 

[maline]

In particular, there must be a factor before the derivative term.

Well, I think I just provided a derivation where this is not necessary...

The restriction 0≤Z≤10≤Z≤10\le Z\le 1 plays no role in the renormalization procedure and is never checked since ZZZ is a bare quantity without physical meaning.

This is indeed an issue that also bothers me quite a bit. The $Z$ factors are defined in terms of a real inner product between normalized states, so $0\le Z\le 1$ is a physical restriction, but in fact $(1-Z)$ is found to diverge to each order in perturbation theory. I have been told that this is the same issue as the existence of the Landau pole, that it indeed shows that theories such as QED and $\phi^4$ are incomplete at high energies, and that for asymptotically free theories such as QCD there is a geometric series over the orders in perturbation theory that shows that the factor goes to zero. I don't know whether any of this is correct.

You seem to be saying that the values of the counterterms in fact have nothing to do with the inner-product-based definition of $Z$. But than leaves the question: why do we require that the vertices get a counterterm with the same value (to the appropriate power) as the propagator kinetic counterterm?

ppp is not a variable in the Lagrangian, so I don't understand your proposal about adding an interaction term proportional to p2+m2p2+m2p^2+m^2.

Of course I mean $\partial^\mu \phi \partial_\mu \phi +m^2$. I wrote $p^2+m^2$, being the corresponding vertex contribution.

This holds for a classical theory but not for a quantum theory when, as usual, ℏℏ\hbar is fixed at 1, since scaling the Lagrangian changes the expression in the functional integral.

Good point, my mistake. The CCR have $[\phi(x),\pi(y)]=i\delta(x-y)$, where $\pi(x)=\frac{\delta L}{\delta(\partial_t \phi)}$. So changing the scale of the Lagrangian changes the CCR, just as rescaling the fields would.

But this also settles the issue: The terms in the Lagrangian that include time derivatives of the field cannot be arbitrarily rescaled, because their scale encodes the commutation structure of the field operators. So the only way to get a $p^2 +m^2$ counterterm is to use the the $Z$ factors from the LSZ formula, either the way I described or with some other derivation involving a "renormalized field" (which I still don't understand).

 

[A. Neumaier]

Of course I mean $\partial^\mu \phi \partial_\mu \phi +m^2$. I wrote $p^2+m^2$, being the corresponding vertex contribution.

This has nothing to do with $p^2+m^2$, and doesn't even have meaningful units. I think you meant
$\partial^\mu \phi \partial_\mu \phi +m^2\phi ^2$
but this is just a special case of a weighted sum of $\partial^\mu \phi \partial_\mu \phi$ and $\phi ^2$,
which are already present in the Lagrangian and provide two of the counterterms. Nothing new if you add your combination to the Lagrangian.

But this also settles the issue: The terms in the Lagrangian that include time derivatives of the field cannot be arbitrarily rescaled, because their scale encodes the commutation structure of the field operators. So the only way to get a $p^2 +m^2$ counterterm is to use the the $Z$ factors from the LSZ formula, either the way I described or with some other derivation involving a "renormalized field" (which I still don't understand).

There are no renormalized fields on the level of perturbation theory - all this necessarily happens in a Fock space with the standard free meaning of the operators. The commutation can be chosen completely independent of the Lagrangian and still make mathematical sense. The bare coefficients have no meaning in the relativistic setting; the constraint on $Z$ is a remnant of the nonrelativistic theory where all renormalizations are finite and the interacting fields are series in products of the bare fields.

The textbook derivations are largely historical baggage motivated by the nonrelativistic theory (which breaks down in the relativistic case due to UV divergences). If you want to have a logically sound setting for the renormalization stuff you need to make a fresh start and work with [URL='https://www.physicsforums.com/insights/causal-perturbation-theory/']causal perturbation theory[/URL] as in the books by Scharf; I recommend the second edition of his QED book. Then everything is (perturbatively) well-defined and well-motivated in modern, logically sound terms.

 

[maline]

I think you meant
$\partial^\mu \phi \partial_\mu \phi +m^2\phi ^2$

Yes, of course. That was a simple mis-type.

Nothing new if you add your combination to the Lagrangian.

I am not trying to add anything new. I'm sorry if my notation made this unclear. I am just asking whether we can get a counterterm that depends on $p^2$ by simply multiplying the $\partial^\mu \phi \partial_\mu \phi$ term in the Lagrangian by an arbitrary constant, as you seem to be suggesting. It seems to me that we cannot do this, because in the canonical formalism, it would imply a change in the CCR of the fields. Instead, we need a specific justification for where the $(1-Z)(p^2+m^2)$ counterterm "comes from".

The commutation can be chosen completely independent of the Lagrangian and still make mathematical sense

Sure, it can make sense, but it doesn't follow the canonical formalism. And that formalism is the only thing that given the Lagrangian a meaning.

By the way, I need to retract a wrong argument that I made above:

But than leaves the question: why do we require that the vertices get a counterterm with the same value (to the appropriate power) as the propagator kinetic counterterm?

In gauge theory, this has a straightforward answer: the relationship between the fermion propagator and the interaction vertex is fixed by gauge invariance, specifically the Ward identity. For non-gauge vertices, including the $Z$ factors in the interaction terms has no real effect, because we will anyway include another counterterm proportional to each such interaction.

 

[vanhees71]

Well, I think I just provided a derivation where this is not necessary...

This is indeed an issue that also bothers me quite a bit. The $Z$ factors are defined in terms of a real inner product between normalized states, so $0\le Z\le 1$ is a physical restriction, but in fact $(1-Z)$ is found to diverge to each order in perturbation theory. I have been told that this is the same issue as the existence of the Landau pole, that it indeed shows that theories such as QED and $\phi^4$ are incomplete at high energies, and that for asymptotically free theories such as QCD there is a geometric series over the orders in perturbation theory that shows that the factor goes to zero. I don't know whether any of this is correct.

Isn't there a theorem (or is it rather a conjection?) that if $Z=1$, the theory is non-interacting? Also in perturbation theory the formal $Z$ factors or the counterterms $\delta Z$ are infinite, but that's due to the improper handling of the distributions and products among them. The important thing is that (in some cases properly resummed) perturbative S-matrix elements don't violate unitarity constraints in the realm, where perturbation theory is applicable (i.e., for small renormalized couplings). For gauge theories, indeed the Ward identities are the key issue. If they are not fulfilled you annot guarantee unitarity, and that's why then the theory becomes inconsistent, because it means that gauge symmetry has been broken.

 

[A. Neumaier]

Sure, it can make sense, but it doesn't follow the canonical formalism. And that formalism is the only thing that given the Lagrangian a meaning.

In the canonical formalism you can specify an arbitrary Lagrangian with arbitrary parameters and the standard commutation rule.

If you want more than I explained then you simply try to put more meaning into an ill-defined recipe than is sensible.