I Field Renormalization vs. Interaction Picture

[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.