# Field Renormalization vs. Interaction Picture

• I

## Summary:

Renormalized fields give the wrong values for the free Hamiltonian.

## Main Question or Discussion Point

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.

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A. Neumaier
2019 Award
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.

In the free case, Z=1.
Of course. Does this address my question?

A. Neumaier
2019 Award
Of course. Does this address my question?
Yes, since 1 doesn't equal infinity. Thus the free theory needs no renormalization beyond normal ordering.

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
2019 Award
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
Gold Member
2019 Award
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).

• odietrich
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.

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
2019 Award
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 $H_0$.

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
2019 Award
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.

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
2019 Award
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.

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
2019 Award
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.

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vanhees71
Gold Member
2019 Award
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
$$\lim_{x^0 \rightarrow \pm \infty} \hat{\phi}(x) = \sqrt{Z} \hat{\phi}_{\text{in}/\text{out}}(x).$$

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.

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 $10$.

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vanhees71
Gold Member
2019 Award
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
2019 Award
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.

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• vanhees71
vanhees71