# What do free equations describe and what is the interaction supposed to do?

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3) Of course, a quadratic Poincare-invariant ''interaction'' changes a free theory into another free theory, so the renormalization can only have the effect of a change of mass. It is not really an interaction.

Let us assume that we found (invented, advanced) an interaction that does not lead to renormalizations. What we expect from such an interaction to describe with respect to the solutions of our free equations? In my opinion, an interaction changes the occupation numbers of "free states", which otherwise would be constant in time. What do you think of it?

What is this question? Of course interactions lead to changes in occupation numbers, and if you want an interaction with no renormalization effects, use a nonrelativistic field theory. This seems to be a rhetorical question.

@RonMaimon @dimension10: This is not a rhetorical question but a quest for clarification of the role of interaction terms in QFT.

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Any quantum system is governed by a Hamiltonian $H$ that generates the dynamics via the Schroedinger equation. If the quantum system can be manipulated by an experimenter, the possible choices by the experimenter can be described by parameters in the Hamiltonian. Thus it is more appropriate to write $H=H(b)$, where $b$ is a vector of external parameters that can be varied. (Usually just a few parameters are considered, but in case of a fully controllable external magnetic field, say, $b$ would be this external field - considered as an infinite-dimensional parameter vector.)

This gives a complete account of physical reality - reality knows not of interactions, only of dynamics and its generator $H$, and how the latter depends on experimental settings. Thus $H(b)$ tells, in principle, the full story.

The interpretation in terms of interactions is introduced by the physicist in order to be able to compute the dynamics. Typically, it is assumed that $H$ is close to a Hamiltonian $H_0$ for which the dynamics can be solved explicitly. In this case one defines the interaction to be $V:=(H-H_0)/g$, where $g$ is of the order of the deviation. Then $H=H_0+gV$, which is the standard starting point for perturbation theory. In the context of perturbation theory, $H_0$ is called the unperturbed system and $H$ the perturbed system. Typically, $H_0$ has an interpretation as a simplified, idealized physical model, and the terms in $V$ can be given an interpretation in terms of this idealized model. Thus one talks about interaction terms, and names the interactions according to the intuition coming from the idealized model.

Simple systems can be prepared in a way that $g$ is actually a controllable parameter; in this case, the unperturbed system is physically real, and $g$ can be expressed as a function of the control parameter $b$ in $H(b)$.

In many other cases - actually in the majority of real applications -, $g$ is treated as if it were a controllable parameter, though it cannot be varied in practice. (For example, this is the case for anharmonic oscillators arising in quantum chemistry of simple molecules.) In all these cases, the unperturbed, idealized system is fictitious - physically nonexistent. But being mathematically realizable and simple to solve, perturbation theory can be carried out successfully. Hence the same perturbation terminology is used.

However, in all cases where the idealized system is fictitious, the system itself doesn't determine the interactions - it is the choice of the fictitious idealization defining $H_0$ that determines it. Even in the case where the idealized system is physically realizable, and realized for a particular value $k_0$ of $k$, the system doesn't determine the interactions in case that several values of $k_0$ lead to an exactly solvable system and hence are eligible to define $H_0$.

For example, we may consider a harmonic oscillator with Hamiltonian $H=1/2(p^2/m+k q^2)$ with $m,k>0$. (A system of corresponding relativistic oscillators was the context where the OP's quote was taken from.) We may think of $b=(m,k)$ to be the controllable parameters. This system is exactly solvable, so we could take $H_0:=H$, and the interaction is zero. On the other hand, we might think of it as a perturbation of the free particle with Hamiltonian $H_0:=p^2/2m$ by an interaction $V=kq^2/2$; the system itself is the same, though what is considered an interaction is different.

Now perturbation theory works well only if the spectra are close, which is not the case for a harmonic oscillator and a free particle. In the latter case, perturbation theory is completely meaningless (just as naive perturbation theory in QFT).

But we could also take $H_0:=1/2(p^2/m+k_0 q^2)$ with $k_0\approx k$, which have a similar spectrum, and the interaction term would be $V=gq^2/2$ with small $g=k-k_0$. Perturbation theory now works (at least for the small eigenvalues), so this is a meaningful use of the term interaction. In this particular case it is possible to sum the perturbation series to infinite order, and one gets the correct results. In particular, the ground state frequency $\omega_0$ of the unperturbed system changes into the ground state frequency $\omega$ of the true system. One says that the frequency has been renormalized by resumming the series.  This is sort of an academic exercise for the exactly solvable harmonic oscillator; this is why in my remark that you had quoted I referred to this situation as ''It is not really an interaction''.

But the same freedom of choosing what deserves to be called the interaction becomes numerically very relevant already for the anharmonic oscillator with Hamiltonian$H=1/2(p^2/m+k q^2)+cq^4$. Here the textbook choice $H_0:=1/2(p^2/m+k q^2)$ that leads to the interaction $V=cq^4$ is much inferior compared to an adaptive choice of a harmonic oscillator Hamiltonian $H_0$ with an improved, renormalized frequency, typically determined by variational perturbation theory. (You can find plenty of references by entering "renormalization anharmonic oscillator" - without the quotation marks - into scholar.google.com.)

If the same is done in quantum field theory, mass takes the place of frequency, and one speaks of mass renormalization (and for other constants, charge renormalization and field renormalization). Note that in QED we are always in the situation that only the total Hamiltonian is realized in Nature, so $H_0$ has to be chosen by the physicist doing the perturbative analysis. Depending on what you declare to be $H_0$ you get different interaction terms. If you take an arbitrary free Hamiltonian with the same symmetries as in $H$ you get as interaction terms the same kinds of terms as if you simply choose the quadratic part of $H$ as $H_0$, but with subtracted coefficients. You can see that the origin of the subtraction terms in the QFT renormalization procedure lies in the fact that $H_0$ and hence the interactions are not determined by Nature but by how to make perturbation theory successful.

In short: If you choose the right $H_0$ (i.e., in QM, the one with the true ground state frequency, and in QFT, the one with the no-cutoff limit taken after having determined the optimal coefficients by low order loop calculations at fixed cutoff) then no further renormalization is needed. But the conventional $H_0$ that simply truncates the normally ordered Hamiltonian at second order is poorly adapted to the real physics and needs renormalization.

If a quantum system is described not directly by a Hamiltonian but instead in terms of a  Lagrangian $L$, the problem and its handling is essentially the same.

answered May 8, 2015 by (15,787 points)
edited May 10, 2015

Arnold, surely there's no guarantee that the Hamiltonian (the component of the generator of translations in a time-like direction) is an element of a given algebra of observables. We can define it in terms of active translations, something like $$P_\mu\hat\phi(x_1)\cdots\hat\phi(x_n)|0\rangle=\left.\frac{1}{\mathrm{i}}\frac{\partial}{\partial z^\mu}\hat\phi(x_1+z)\cdots\hat\phi(x_n+z)|0\rangle\right|_{z=0},$$ but in any case it's nonlocal. Specifically in the interacting QFT case, Haag's theorem has it that the interaction picture does not exist, there's no unitary transformation from initial to final, if we work with the free field.

@PeterMorgan: I do not understand what your Hamiltonian describes. Is it $H$ or $H_0$? Also, $H_0$ is not to be literally "free" Hamiltonian, but that whose solutions can be easily found. It may include some part of interaction.

@VladimirKalitvianski, $P_\mu$ as I've defined it here doesn't "describe" a dynamics as much as it is defined by the infinitesimal action of active translations. I would take the dynamics to be defined in some different way. $P_\mu$ as I've defined it here would apply equally well for an interacting field $\hat\xi(x)$ (for any commutation relations and vacuum state, assuming we can somehow construct a $\hat\xi(x)$) as it does for a free field $\hat\phi(x)$. For the free field, $P_\mu$ has the same expectation value as $\int k_\mu a^\dagger(k)a(k)\mathrm{d}^4k$ (up to a scale, depending on the definition of the field relative to the creation and annihilation operators), but for an interacting field $P_\mu$ might not be expressible as a function of the field and/or creation and annihilation operators (indeed, Haag's theorem says it will not be expressible within the free field algebra).

@PeterMorgan: The observable algebra contains many nonlocal operators - e.g., all exponentials of smeared fields, their products, and their linear combinations. It also has to contain the exponentials $e^{itH/\hbar}$ of the Hamiltonian $H$ (and the momentum operators), since translations must be unitarily implemented. Your formula is correct if $\hbar=1$, the fields are the interacting fields, and the $x_i$ are distinct.

Since nowhere a free field enters (unless you want to do perturbation theory), this does not contradict Haag's theorem. It is also independent of the S-matrix, as the latter is traditionally defined by a limit whose existence is difficult to establish, and exists at best if the right choice of $H_0$ is made. (Haag-Ruelle theory which guarantees the existence of an S-matrix without reference to an $H_0$ applies only in the case of massive theories where the IR problem is absent.)

Haag's theorem says not only that $H$ will not be expressible within the corresponding free field algebra, but that the interacting field operators won't be either. It says nothing about how $H$ is composed of interacting field operators.

@ArnoldNeumaier, thanks. I suppose I distinguish between observables and energy-momentum in the sense that Haag's §II.1.2 for the Wightman axioms and §III.1 for the Haag-Kastler axioms distinguish between operators $\hat\phi_{\!f}$, smeared by test functions taken from something like the function space of Laurent Schwartz, or algebras of local observables $\mathcal{A}\hspace{-0.07em}(\!\mathcal{O}\!)$, in contrast to the "representors $U\!(\!a,\alpha\!)$ of the symmetry transformations". Energy-momentum is different from local measurement insofar as we can make a change to the state at arbitrary space-like separation with the same effect on the energy-momentum as making the same change right next to an observer, whereas for local observables we require cluster separation.

@PeterMorgan: The algebra of observables in the Wightman sense is bigger than the union of the $A({\cal O})$, which contains only observables with compactly localized support. Indeed, the observables of primary interest in physics (that gave name to the concept of the observable algebra) are the various currents (which are local) and the associated conserved quantities (the spatial integral of the time component of  currents).

+1. All these(and recent discussions)  make me more favor the constructive approach, which in some sense is the way physicists have chosen, except that physicists do it non-rigorously, while the term "constructive field theory" seems to be reserved for those mathematically rigorous endeavors.

@JiaYiyang: "make me more favor the constructive approach" - over what? What do you mean by ''the constructive approach" in a nonrigorous setting?

@ArnoldNeumaier, it's too late in my time zone, I'll write a reply tomorrow.

By "make me more favor the constructive approach" I meant to compare axiomatic approach, algebraic approach and constructive approach, and I'm feeling constructive approach is the most natural and probably the most practical approach.

I reach this disposition by thinking about the meaning of cutoffs. I've wondered why cutoffs are so not welcomed, because for example we surely shouldn't expect the Minkowskian QFT to literally describe the realistic physics in infinite volume since we know well the large distance spacetime structure is not Minkowskian, so why are we not satisfied with a QFT in a finite volume? To me the reasonable answer is that, though there's nothing physically wrong with a finite volume QFT, there is something physically wrong if your QFT is not well controlled as a function of volume, for example if the cross section of your theory depends on the volume at the order of $O(V)$, which means the physics sensitively depends on large distance, which clearly is not the case for our physical reality. This is why it would be nice if the theory is constructed so that the physics depends on, say, $O(V^{-1})$. But once you have this you basically can remove the cutoff. So in this sense what we should expect from a QFT is a theory with cutoffs but well controlled, rather than insisting from the start that there shall be no cutoffs. The two perspectives might be mathematically equivalent at the end of the day, but I find the former much more natural.

It seems to me the constructive approach in its spirit is representing the first  perspective, while axiomatic and algebraic approach the latter. And with hindsight the former really is what physicists have been doing, for example the regularization and renormalization programs, but is most of the time only at the perturbative level, and this is what I meant by "non-rigorous constructive approach".

I still appreciate the clarification power that axiomatic and algebraic approach sometimes possess, but it seems hard to take them as starting point to make useful progress in physics.

@YiaYiyang as I understand it, cutoffs should naturally occure where the effective theory valid for a certain range of scales stops being valid, and new physics kicks in (at the GUT, QG scale etc for example). Aslo, it is not true that renormalization can only be done perturbatively, the exact (or functional) renrmalization group (ERG) method works non-perturbatively too for example.

@Dilaton, I think my earlier argument applies to effective field theories too, that's why I didn't bother to mention it. Whether your QFT can be UV/IR completed or not, you would need some good control. If your theory has a completion, it means it's perfectly controlled;if your theory is a EFT valid only in a certain range, it means the control is limited, and it still may describe physics at the regime of good control but not at other regimes where the control is bad.

Aslo, it is not true that renormalization can only be done perturbatively

I never claimed that, I said "most of the time", and "non-perturbative" does not equal "rigorous" in the context I'm speaking of, for example almost all work in lattice gauge theory can be considered non-perturbative, but few are rigorous in the constructive sense.

@JiaYiyang all quantum field theories are only effective theories with a limited domain of validity, as the physics effects needed to describe the system change when going to larger or smaller scales. For example the SM is expected to break down at some point when going to higher energies. This is why the LHC is looking for new physics. The new physics does not need to kick in at LHC reachable scales though.

Axiomatic, algebrai, constructive, perturbative, and lattice QFT should not be thought of as being competing choices but as being complementary techniques that reveal different insights into QFT.

In finite volume, there is neither Lorentz symmetry nor equilibrium thermodynamics. But both are essential for realistic physics. Thus one must be able to take the limit.

Whether one actually takes the limit or assumes that the cutoff is astronomically large doesn't make a significant difference once one assumes that a good large distance behavior. But the problem (of constructive QFT) in removing the finite volume constraint is precisely the difficulty in proving (for 4D YM, say) estimates of the sort you and Ron assume without qualms.