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

+ 2 like - 3 dislike
650 views

From another thread:

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?

asked May 7, 2015

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.

## 1 Answer

+ 4 like - 0 dislike

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 (13,959 points)
edited May 10, 2015
Most voted comments show all comments

@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.

@Dilaton: Maybe I am wrong, but UV incompleteness does not necessary mean the cut-off is unavoidable. An incomplete theory may "work" even out of its region of validity, if properly formulated (I mean, no catastrophes).

@ArnoldNeumaier: Thank you for your answer. +1! (I still cannot vote up directly.)

@ArnoldNeumaier,

I'm not objecting usefulness of the study of taking infinite volume limit, but rather discussing which perspective is more natural: Is infinite volume a fundamental setting and finite volume an convenient artifact, or the other way around? I think the latter. The earlier discussion on perturbation theory and Haag's theorem can be naturally resolved by adopting this perspective and Ron's answer there: at any stage of the construction we are doing scattering theory in finite volume, and we throw out the vacuum bubbles at this finite volume, which is a well defined procedure. Then we take the infinite volume limit, and through this procedure we get a good control of the S-matrix, all this just means our constructed S-matrix won't depend sensitively on large distance stuff, which is all we really wanted. This is a legitimate way to get a limitedly controlled theory, and this control is good enough to study much interesting physics, though not all of it. However from axiomatic/algebraic philosophy the success of perturbation theory looks still mysterious, though not in direct contradiction.

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.

I'm not on Ron's side regarding this issue, and my whole comment was promoting the importance of the work in constructive field theory direction!

@JiaYiyang: Of course, there is nothing uncommon about two  wavepackets coming near to each other, collide and then go away. But this happens in finite time. If you look at what happens for $t\to\pm\infty$ (which is intrinsic to any discussion of the S-matrix), you'll see that the wavepackets spread and ultimately (after sufficiently many revolutions if necessary) recombine in a more and more delocalized manner, and asymptotically one cannot say anything. One is and remains in a superposition of bound states. Nothing like an S-matrix appears.

On the other hand, chemist compute S-matrices (and certain information derived from them -  chemical reaction rates) from computations in finite volumes. But this requires a lot of attention to details. In particular, you cannot use a manifold without boundary but need to impose artificial absorbing boundary conditions, to avoid that what I describes happens. This changes the Hamiltonian to a nonhermitian operator; hence the dynamics becomes nonunitary. How to get nevertheless a unitary S-matrix (or rather a unitary approximation to the intended one) is a nontrivial story that involves the true S-matrix (on the unbounded flat space) and a significant amount of theory.

@VladimirKalitvianski: But what I said shows that the problems appear far earlier than you locate them. In particular, IR problems already appear when a single charged particle scatters in the presence of an external electromagnetic field. Scattering is no longer simply described by an S-matrix anymore but they can still be solved using coherent states. In fact, coherent states are sufficient to solve all IR problems of QED.

Most recent comments show all comments

@ArnoldNeumaier,

Thus, effectively, you must treat scattering as if it happens in the tangent space of the space M at the point where the scattering event happens, rather than in a compact space, even though the true universe might be compact. Nothing should depend on more than the region where your experiments take place, and this is effectively flat space.

Exactly! This is why I think the constructive philosophy looks more natural at least on the surface. The stipulation of a sensible infinite volume limit just means our physics should be insensitive to large distance: when you do collision experiment in your lab, the shape of the universe shoudn't matter. To me this stipulation is not so much connected to the requirement of exact symmetry,  that's all I'm saying. This philosophy naturally encompasses both UV/IR completed QFT and effective field theory.

@JiaYiyang: But still, if you want to describe your local scattering by an S-matrix you must make the idealization of a spacetime with a timelike Killing vector and noncompact space slices. Thus talking about compact space in the context of scattering amounts to choosing the wrong model for the analysis.

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.