Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Discussion regarding Haag's theorem and the (lack of) connection between this and renormalisation of physical states

+ 0 like - 0 dislike
2902 views

This is a discussion moved from http://physicsoverflow.org/22400/understand-success-perturbation-theory-despite-haags-theorem , about the unitary inequivalence of free and interacting field theory states, and the lack of any connection between this issue and renormalization of physical particle states.

asked Aug 18, 2014 in Chat by SchrodingersCatVoter (-10 points) [ revision history ]
edited Aug 19, 2014 by dimension10

4 Answers

+ 0 like - 0 dislike

In my opinion, the problem is not in the interaction picture, but in interaction operator. Roughly speaking, it contains a "wrong" part and a "good" part. The wrong part contribution is removed with renormalization (subtraction), and the good part is treated perturbatively. I cannot say that the corresponding series become good and successful, no. The "good" part contains a "strong interaction" in the sense that it should be taken into account exactly to obtain physically and mathematically reasonable results. Otherwise the series suffer the IR divergence and cannot be used. In other words, the success is not that immediate and renormalization is not the only fix of the theory formulation, but after summation of the soft diagrams to all orders (which is equivalent to taking their contribution exactly via another initial approximation), the results are successful.

Why are the final results successful? Because many ingredients are taken from experiment and implemented in a way to reproduce the experimental data. The Maxwell theory, the Lorentz theory, etc., - all they have an experimental support and foundations. So, in the end, we obtain something that gives the radiation and other effects as necessary.

answered Aug 18, 2014 by Vladimir Kalitvianski (22 points) [ no revision ]

The above comment has no relevance to Haag's theorem, which is about infrared issues, and doesn't care about the type of interaction (it applies even in theories which are nonrelativistic and have a constant density condensate in the ground state, or in relativistic theories which are cut off on the lattice). It is therefore off topic, and I think it should be hidden.

Ron, it is, probably, I who cannot express myself properly. I just wanted to say how "success of perturbative series" was achieved. Namely, we finally consider scattering of interacting particles, not bare ones.

Out of courtesy, I think it is ok to wait a day or two before hiding, so that the off-topicness is clear to all. I don't want to make a controversial judgement (the top comment by VK really is off topic--- this theorem is not about renormalization at all)

Ron, I just say that renormalization is not enough to get a "success", don't you see it? Also, an inclusive picture we finally use contains infinte number of particles, so I do not understand why you object it.

This is not about "success", it's a mathematical theorem expressing the fact that there is a limitation of the formalism of Hilbert spaces, a state containing a constant density of particles is not included. That's all.

Sorry, the question is about "success", and this constant density of particles or pairs in vacuum is not clear to me at all.

Perhaps you should learn some old-fasioned perturbation theory (pre-Feynman, ordinary QM perturbation theory), and apply it to the Fock space of phonons in a solid. What is the state where you move the solid by 1 cm to the right? If you move a finite chunk of the solid 1cm to the right, and smoothly interpolate this to 0 motion at the boundaries, this state, so long as the displaced region has finite volume, can be represented as a coherent state of phonons. But now what if you move the entire (infinite) solid to the right? Is the translated state still in the Fock space? This is an example of a superselection rule.

If you introduce a single defect, say a single heavier atom, into the solid, you can describe the phonon polarization cloud around the heavy atom, and it's in the Fock space. Now consider introducing a density of defects into the solid, say every hundred atom in a regular pattern is slightly heavier. Is the new phonon state in the Fock space of the unperturbed phonons? This is an example of a new vacuum being unrelated unitarily to an old one.

When you have an interaction term, the Fock vacuum is no longer an eigenstate of the Hamiltonian, because it's an eigenstate of H. Now suppose you use time-independent perturbation theory to describe the new eigenstates of the Hamiltonian. You describe coefficients that mix you with 2 particle, 4 particle, 6 particle states. Everything is normal on a finite volume lattice, but in the limit of infinite volume, the interacting vacuum is no longer unitarily related to the old vacuum, because all the coefficients for mixing with finitely many particle states go to zero, the vacuum has a finite density of particles, so infinitely many particles overall, and this is not in the Fock space.

This is no different than the translated solid having infinitely many phonons as compared to the untranslated solid. It's not deep physically, it's a mathematical nuisance of infinite volume limits.

This is a question about the inner products of different vacuum states, when you can consider two states as part of the same Hilbert space. It's a question which is best formulated in a pre-modern formalism, so forget you know what a Feynman diagram is, and just naively apply quantum mechanics to field theory, the way the founders of quantum mechanics did. The comments here are too low level, since it is assumed that one already knows old-fasioned perturbation theory and Dyson-style derivation of perturbation theory.

+ 0 like - 0 dislike

(from below Ron Maimon's answer to this question http://physicsoverflow.org/22400/understand-success-perturbation-theory-despite-haags-theorem) Thanks to Ron, I can now formulate the explanation of "success" as follows: we, in fact, calculate scattering of real, dressed, physical particles in QFT. The scattering and the dressing "phenomena" are calculated perturbatively, so in the end the results are quite reasonable.

answered Aug 19, 2014 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Aug 19, 2014 by Ron Maimon
Most voted comments show all comments

No, this has nothing to do with renormalization! It has nothing to do with bare or dressed anything! It's about something else, that the Hilbert space of a field theory does not include states with infinitely many particles, those states are outside the Hilbert space description, so that an interacting vacuum is not unitarily linked to a free particle Hilbert space, because it has a finite density of particles, and therefore infinitely many particles overall.

Your comments are off topic here, because the issue has nothing to do with renormalizing the interaction, bare or dressed particles, or anything of the sort. The same issue occurs in a nonrelativistic field theory with a chemical potential and repulsion that has a minimum at nonzero density, and here there is no renormalization, no ambiguities, no subtraction, no nothing, just a finite density of particles, leading to infinitely many particles overall. There are no states in Fock space with $N=\infty$, they just aren't there.

It is annoying that you are repeating the same things in this question--- the Haag theorem has nothing to do with renormalization.

By saying "infinitely many particles", do you mean coherent states or something else? I do not understand how vacuum can contain particles in QFT.

The interaction picture has a bare vacuum, in the Dyson formalism, there is a turning-on function and interaction which you use to adiabatically link the bare vacuum to an interacting vacuum, and you can ask, in Dyson's formalism, "what is the inner product of the bare vacuum with the interacting vacuum".

The problem is that it is zero, that's Haag's theorem, because the state of the interacting vacuum contains a density of particle pairs, and infinitely many particles in the infinite volume limit. It's not deep, I said it three times already. It's nothing to do with renormalization, it's about the structure and content of Hilbert Fock spaces, and which kinds of states they are capable of describing. They can't describe states with a constant density of particles.

Which pairs, Ron? Real or virtual? Please, be clear.

This question can't be answered, as this is not a question in scattering. Particles are only "real" or "virtual" when you ask about infinite time limits, or external legs in a covariant formalism. This is a mathematical question about a non-covariant formalism, about Fock space. Here you have a Fock space bare vacuum, you expand one state (the interacting vacuum) in terms of other states (Fock states of definite particle number). The interacting vacuum is a state, there is a Fock space of bare particles, and you can ask how many bare particles the interacting vacuum has, when you turn on the interaction. The answer is "infinitely many" even when you have an ultraviolet regulator, and that makes it outside the Fock space, so the description fails. It's not deep.

These comments are not particularly proper level, because this question presumes familiarity with Fock space and Haag's theorem! I think they should also be hidden.

Most recent comments show all comments

It's not about power, we can move the comments to chat, so that they aren't deleted, but I really think they are useless to everyone, because they are about ancient topics from the 1930's-1950s, they aren't about modern issues. The issues you first brought up regarding the form of the interaction, bare or dressed or whatever, are completely irrelevant to the question of Haag's theorem, because no matter what, the interacting vacuum will not be in the same Hilbert space as the bare vacuum.

I know you didn't downvote, I'll move the comments to chat soon.

We just focus on different things - you explain the Haag's theorem and I explain the success of the perturbation theory where the "dressing" is an essential part of transition from the bare to the real stuff. Agree, scattering of the real particles is not concerned with the Haag's theorem.

+ 0 like - 1 dislike

As I said in another comment, I never agreed that my comments were off-topic, I agreed to have my comments hidden. We all have different backgrounds and our explanations may look weird to each other.

As to the Haag's theorem, it can indeed be explained in simple terms, much simpler than reasonings about bare and dressed particles in a volume and the corresponding vacua in QFT. Let us consider two (for instance, 1D) Hamiltonians \(H_1\) and \(H_2\) with their eigenfunctions \(\varphi_n\) and \(\psi_n\) correspondingly, the eigenvalues being \(E1_n\) and \(E2_n\). The projection of \(\varphi_0\) to \(\psi_0\) (vacua) is a number, say, \(p_0\), which can be less than 1. In QFT the Hamiltonians have products of \(N\) such vacuum states, one for each mode, so when one makes a projection of the \(H_1\) vacuum state to the \(H_2\) vacuum state, one obtains something like \(p_0^N\), where \(N\) is a number of independent modes. No wonder this projection vanishes when \(N\to\infty\). It is a banality indeed and renormalization has nothing to do with this simple observation.

I never invoked renormalization as an explanation of the Haag's theorem. I was trying to explain the success of the perturbation theory despite the Haag's theorem.

Now, let me explain my confusion with those particles or pairs (or, better, eigenstates \(\varphi_n e^{-iE1_n t}\)) that are "present" in the vacuum state \(\psi_0\). They are not present literally, as "observable" eigenstates of \(H_1\). The eigenfunctions \(\varphi_n\) in the spectral decomposition \(\psi_0=\sum_{n=0}^\infty p_n\varphi_n\) are just dumb numbers describing the difference between the ground states \(\psi_0\) and \(\varphi_0\). You will never find an excited state of \(H_1\), say \(\varphi_5 e^{-iE1_5 t}\), in the ground state \(\psi_0 e^{-iE2_0 t}\) of \(H_2\)

\(\psi_0 e^{-iE2_0 t}=e^{-iE2_0 t}\sum_{n=0}^\infty p_n\varphi_n \ne\sum_{n=0}^\infty p_n\varphi_n e^{-iE1_n t}\)!

The vacuum state \(\psi_0 e^{-iE2_0 t}\) has a certain energy and no other energies, approximate or exact, can be observed in it. It is not a superposition of physical eigenstates, so no "particles" (or excitations) of \(H_1\) are "present" in the vacuum state \(\psi_0 e^{-iE2_0 t}\).

Finally, let me explain the "dressing" mechanism, which always takes place in course of perturbative calculations. By "dressing" I mean a transition from the eigenstates \(\varphi_n e^{-iE1_n t}\) to \(\psi_n e^{-iE2_n t}\). (In your language, it is a transition from the "bare" states to the "interacting" ones.) Although it looks like we start our scattering calculations from the "bare" states, in fact we always deal with the interacting theory, we scatter the eigenstates of the interacting Hamiltonian. As the difference between \(H_2\) and \(H_1\) is treated perturbatively, it is taken into account step by step to obtain finally the eigenstates of \(H_2\). It is they who are scattered, but the scattering phenomenon and perturbatively "dressing" are mixed in our calculations which makes an impression of physical presence of "bare" states. No, they are absent by definition. The easiest way to see it is to perform a calculation without any scattering. It will be just perturbatively "dressing", i.e., building the eigenstates of \(H_2\) with help of dumb functions \(\varphi_n\). Now it is easy to understand the "success of the perturbation theory despite the Haag's theorem": we factually obtain the results of scattering of "dressed" particles, i.e., we only deal with one sole Hamiltonian \(H_2\) and its eigenstates, including its vacuum state. If you look at a physically meaningful diagram, you will see the "dressed" incoming and outgoing lines: the electron is permanently coupled to the EM field degrees of freedom and excites them, when scattered. It is "hairy", "dressed" external lines that provide the calculation success; otherwise it is zero indeed.

answered Aug 19, 2014 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Aug 19, 2014 by Vladimir Kalitvianski

That's philosophy. The "bare states" are "absent by definition" only if you define a formalism where they are absent. Dyson explicitly defines the formalism so they are present, and the question is about Dyson's vacuum, not Kalitvianski's.

In Stueckelberg style perturbation theory, the vacuum bubbles are clearly irrelevant for scattering, the same is reproduced in modern perturbation theory. The vacuum bubbles are only for calculating the cosmological constant.

So all modern practitioners do calculations relative to the true vacuum, with scattering states defined as excitations over the true vacuum, and scattering events ignoring vacuum bubbles, which factor out in modern coveriant formalisms (that's one of the advantages of modern formalisms). The relation between bare and interacting vacuum only shows up when you have a Dyson style formalism, and it is only well defined in a finite volume and with an ultraviolet regulator, so that you have finite dimensional quantum mechanics.

But the question is about the hypothetical unitary relation between bare and interacting states, which is explicitly used in Dyson's derivation, and explicitly criticized by Haag. The theorem is rather banal, I agree, but it does show that if you want to define the unitary transformation between bare and interacting vacuum, you had better have both an infrared and ultraviolet regulator, i.e. you better be in a finite box.

As far as we deal with $H_2$, the bare states are absent physically.

But Dyson doesn't deal with $H_2$, he is dealing with $H_0$ and $f(t) H_i$, and adiabatically links the vacuum of $H_0$ to the vacuum of $H_0 + H_i$. The question is why he gets the right answer with such an unphysical formalism.

In fact, nothing is unphysical in his formal construction. His switching function $f(t)$ is in fact always on in our calculations and it serves solely as a pretext of not taking into account anything from $H_{int}$ in the initial approximation. It is just a perturbation theory for $H_2$, nothing else. I see it this way. There is no adiabatic linking, as a matter of fact.

That's in "our calculations", your highness. The question is, surprisingly enough, not asking about our calculations, but rather about his calculations, where the function is not always on. That's why all your comments are totally off topic (and infuriating).

 In QFT the Hamiltonians have products of N such vacuum states, one for each mode, so when one makes a projection of the $H_1$ vacuum state to the $H_2$ vacuum state, one obtains something like $p_0^N$, where N is a number of independent modes. No wonder this projection vanishes when N→∞.

This seems to be a very neat way of seeing it! But according to the previous discussion with @RonMaimon, it should vanish as $\frac{1}{\sqrt{N}}$, I wonder where the discrepancy comes from.

If he calculates \(\mathrm{Texp}(-i\int_{-\infty}^t H_0 dt')\) for some reason, then indeed, my comments are irrelevant and infuriating, sorry. I just thought the function $f(t)$ was always equal to 1 while interaction, so it permitted to take into account everything.

Well, I got tired.

+ 0 like - 1 dislike

Ron Maimon wrote "In the Dyson-style interaction picture, there's a turning on-and-off function \(f(t)\) that describes the interaction strength. This turning on and off is supposed to be adiabatic, so that the bare vacuum slowly relaxes into the interacting vacuum, the scattering happens, and then you turn off the interaction again."

OK let us see it on a toy model. Let us consider a particle in a 1D box and with some regular time-independent potential \(V(z)\) in it. For simplicity we can choose a linear potential. Then the exact eigenfunctions \(\psi_n(z)\) will be combinations of Airy functions. It can physically be a neutron jumping in the gravitational field between two mirrors, so \(V(z)=g\cdot z\) and \(\psi_n(z)=N_n\left[Ai(z-z_n(L))-Bi(z-z_n(L))\frac{Ai(-z_n(L))}{Bi(-z_n(L)}\right]\), where \(z_n(L)\) are the roots of the equation \(\psi_n(L)=0\).

The "bare" eigenfunctions are just \(\varphi_n (z)\propto\sin\frac{\pi (n+1)z}{L};\;n\ge0\). Let us compare the ground (vacuum) states in case when \(L\) is sufficiently big to have only one Airy function in \(\psi_0\). Qualitatively the ground state functions are given in Fig. 1.

They are different.

We can represent \(\psi_0 (z)\) as a spectral sum \(\psi_0(z)=\sum p_n \varphi_n (z)\), so what? Does that mean the interacting vacuum  \(\psi_0\) "contains" bare states or bare particles? No. No, Ron, no!

There is no way to observe physically those \(\varphi_n\) in \(\psi_0\), nor exists a mathematical way to "extract" \(\varphi_n\) from \(\psi_0\). Only their sum makes sense, namely, it represents the ground state. This statement about dumb functions \(\varphi_n\) is valid for any function \(F(z)\) expanded in the Fourier series. Saying that interaction populates the vacuum state with bare particles ("condensat", according to Lubosh Motl or "constant density" particles, according to you) is a very sloppy and misleading way of expressing this simple Fourier decomposition. This point should be clear.

Now, Dyson interaction picture. Although some people use the adiabatic function \(f(t)\) in the time evolution calculations, in fact it serves just to start perturbative calculations from the zeroth-order approximation (or "bare" states in the above sense). If you analyze carefully what is going on with such a calculation, you will see nothing, but dressing. It means even in the initial state we have a dressed state rather than a bare one. Let us see it. For that, let us calculate the "time evolution" of a bare vacuum state \(\varphi_0(z) e^{-iE_0^{(0)} t}\) due to turning the potential \(V(z)\) on with help of \(f(t)\) (no time-dependent scattering potential is considered for instance). The Dyson's solution can be represented as a time ordered exponential with the interaction potential \(V(z)\) in it. If you make the first order calculation, you will see that the potential \(V(z)\) transforms \(\varphi_0(z) e^{-iE_0^{(0)} t}\) into \(\left[\varphi_0 (z) +\sum_{m>0}\frac{V_{m0}\varphi_m(z)}{E_0^{(0)}-E_m^{(0)}}\right]e^{-i(E_0^{(0)}+V_{00}) t}=\psi_0^{(1)} e^{-iE_0^{(1)} t}\). In the k-th order, the Dyson's solution gives \(\psi_0^{(k)} e^{-iE_0^{(k)} t}\) where \(\psi_0^{(k)}\) and \(E_0^{(k)}\) are just the usual stationary perturbation theory solutions of the k-th order. So the Dyson's solution is the dressed vacuum state \(\psi_0 e^{-iE_0 t}\) at any time \(t\), Ron. At any time means at the initial time too. There is no trace of \(f(t)\) here because it is always on.

If we add now a scattering potential, we will factually start from the dressed states, we will scatter them, and we will finish with dressed states, and no bare states will be involved, strictly speaking (Fig. 2). Representing the dressed states as spectral decompositions over bare states does not involve the latter physically. This is the true picture in QFT. The scattering phenomenon (or, better, an additional time-dependent external potential) makes the initial dressed state \(\psi_0 e^{-iE_0 t}\) evolve into a physical superposition \(\psi_0 (z)e^{-iE_0 t} \to\sum C_n\psi_n(z)e^{-iE_n t}\) with observable stated  \(\psi_n(z)\), \(C_n(t)\) being the occupation number amplitudes. The external potential turns on and off itself, without our help.

Unfortunately, our perturbative spectral representation of solutions \(\psi_n(z)\) in course of occupation amplitude calculations confuses two perturbative calculations - dressing calculation (=time-independent thing) and scattering calculation (=time evolution of occupation number amplitudes \(C_n(t)\)). That's the reason of your opinion about bare particles being present in the initial/final states and in the physical vacuum.

answered Aug 20, 2014 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Aug 20, 2014 by Vladimir Kalitvianski

The Airy function solution can be expanded in the sine-wave solutions, and this is the entire content of the statement that the Airy function contains "bare particles". It just means what you said, this expansion. Since you can do this expansion in the 1d box problem, or in finite dimensional Quantum Mechanics (including finite volume lattice field theory), it is mathematically correct to say "the interacting vacuum is made of bare particles" if you interpret it to mean exactly this expansion and nothing more.

This expansion is not very physical in the case of field theory, because you can't prepare a bare vacuum state, or measure relative  to bare states, rather you can only do scattering to probe the vacuum. But  so what! This expansion is mathematically well defined in finite volume, and that's what the Haag theorem (and the question) are about.

Your comments are well known, nobody disputes that you can define the perturbation theory and scattering to refer only to physical excitations over the vacuum. That's what everyone does today anyway, without you telling them. That's the point of the Haag-Ruelle scatterng formalism, and later the LSZ formalism that subsumed it. Nobody needs to reference bare states to describe scattering anymore, you just look at correlation functions.

The question here is whether it makes sense to do perturbation theory Dyson's way, with a bare vacuum, an interacting vacuum, and an explicit description of the interacting vacuum as an expansion in a series of bare vacuum states. Does it make sense mathematically.

It makes sense in finite volume, where the description works. The coefficients in the description don't converge to a limit in infinite volume, so it doesn't work if you take the infinite volume limit too early. That's it, done, question answered, problem solved. There's nothing more to say, and all this bloviating is wasting everybody's time.

By the way, all your calculations are trivial and did not need to be shown. I think that showing this trivial stuff is in bad taste, as it gives the impression that you think someone didn't already know this.

I am glad that you agree with my trivial observation that we always deal with the dressed states.

Concerning their representation with bare states, the quality of such a representation depends indeed on "proximity" of the bare states to the dressed ones. With choosing another bare basis, we can get a better or a worse representation. Thus, one has to implement the main features of the dressed states into the bare basis functions, and then the perturbative coefficients will be reasonably small. This is another trivial observation of mine and I propose in my papers to creatively choose a better initial approximation instead of keeping to a bad one and saying everything is already OK.

Yes, this too is an issue--- you want to choose the best "bare" state to compare to. This is the reasoning behind the Gell-Mann Low version of the renormalization group--- they choose the subtraction point so that the leading order approximation is closest-fitting for the particular scattering in question.

The question is the essence of renormalization group equations, and the method of the renormalization group optimizes the choice of the split between free and counterterm Lagrangian for the problem. These issues have been solved already a long time ago, and there is no point in discussing them more unless you have an actual new idea. I don't see one above.

There is no one above, Ron. They are in my papers. But forget it.

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):
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:
p$\hbar$ysics$\varnothing$verflow
Then 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.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...