Some more questions on Haag's theorem

Question

This is a follow-up question to my previous question How to understand the success of perturbation theory in QFT, despite Haag's theorem?. In that post a very intuitive heuristic proof for Haag's theorem is given: In the Schroedinger wavefunctional picture, an interaction term would alter the vacuum functional on each point in space. Suppose on one point the overlap between the interacting vacuum functional and the free vacuum wavefunctional is some $p<1$, then by translational invariance the overlap on every point will be $p$. Hence $\langle 0_\text{free}|0_\text{int}\rangle=p^\infty=0$, that is, the free and interacting vacua are orthogonal, and in fact it's in a superselection sense since no local change can break the orthogonality.

Here are some questions after some more thought:

(1) Why is "superselection" equivalent to "inequivalent representation"?

This question is not limited to the context of Haag's theorem but to superselection in general (say spontaneous symmetry breaking). I often see "superselection" and "inequivalent representation" being used interchangeably in the literature. I can more or less accept "inequivalent representation" $\implies$ "superselection", but I fail to see how the opposite direction is true: we would still have a superselection if we declare that our physics is described by a direct sum of two copies of the same representation of the observable algebra, wouldn't we?

I suspect it's just a matter of confusing semantics, but I can't be sure.

(2) What's a concrete model of Galilean QFT that is not plagued by Haag's theorem?

I see statements claiming that Galilean QFT does not necessarily suffer from inequivalent representations for nontrivial interactions (not that inequivalent representations are impossible). I'd like to see a concrete model of this sort.

The proof outlined at the beginning of this post relies only on the assumptions of translational invariance and the fact that the degrees of freedom are infinite, which are shared by Galilean QFT. Hence the only escape seems to be to engineer a nontrivial interaction that does not alter the vacuum wavefunctional at all, and I think this basically kills all the local interactions. Luckily Galilean QFT allows nonlocal interactions, I'm pushing along this direction but haven't got anything concrete. (At the moment I'm considering a classical coupled oscillator model, and trying to add a potential between different oscillators that is 0 when oscillators are in equilibrium positions and otherwise repulsive. Still, the thought is far from concrete.) (Update: Nonlocality is not needed, see the discussion with Ron Maimon)

Comments

In addition to the provided answers, there is at least another non-perturbatively renormalizable (partially) non-relativistic QFT, where the Hilbert space does not change after renormalization (I do not know if I would call it Galilean, it describes non-relativistic particles interacting with a relativistic field). It is called the Nelson model ( http://scitation.aip.org/content/aip/journal/jmp/5/9/10.1063/1.1704225 ). It allows creation of field excitations from the vacuum, but the number of non-relativistic particles is conserved.

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@yuggib, thanks!

Answer 1

(1): The physical reason why there are no multiple superselection sectors with the same representation is that there is no way to distinguish the two sectors, and superpositions are not allowed. - "we would still have a superselection if we declare that our physics is described by a direct sum of two copies of the same representation of the observable algebra, wouldn't we?" Yes, but then we'd also have two vacuum sectors. Uniqueness of the vacuum is responsible for the 1-1 correspondence. 

(2): In Galilean QFT, the Hamiltonian is a well-defined self-adjoint operator on Fock space (semibounded quadratic form), hence time evolution stays in it. In particular, any multiparticle quantum mechanics with a (sufficiently nice) translation invariant pair potential and an arbitrary number of particles is equivalent to a Galilean QFT in which each $N$-particle sector is conserved, hence ordinary QM applies. But there are also other examples in which particle number is not conserved. See http://projecteuclid.org/download/pdf_1/euclid.cmp/1103839847

Comments

Thanks and +1, you said 

 Uniqueness of the vacuum is responsible for the 1-1 correspondence. 

1-1 correspondence between what and what?

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@JiaYiyang: between superselection sectors and unitary irreps.

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By the way, what's the precise definition of "algebra of observables" in QFT?  To interpret Haag's theorem it seems $H_\text{int}$ and $H_\text{free}$ should not be put into the same set of algebra?

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In algebraic QFT, the "algebra of observables" is the algebra generated by the bounded functions of smeared field operators of a Wightman QFT. This includes the time translation operator $U(t)=e^{itH/\hbar}$, where $H$ is the Hamiltonian. Informally, unbounded limits (hence $H$ itself) are also included. The Wightman axioms themselves provide the vacuum representation of this algebra.

For interacting theories, nonconstant functions of $H_{free}$ are - because of Haag's theorem - not among the observables.

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Ok, thanks, then for now there seems to be no gain for me to adopt this way of thinking...it seems to convey no more physics than the more intuitive "infinitely shifted vacuum" picture.

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@JiaYiyang: There is more to inequivalent representations of the CCR than just the "infinitely shifted vacuum". Bogoliubov transformations such as those employed in superconductivity also represent the CCR in a nonstandard way. 

On the other hand, the superselection structure is not tied to inequivalent representations of the CCR but to inequivalent representations of the observable algebra.

independent of Haag's theorem, a fuller picture of the superselection structure of QFTs can be seen in 2 dimensions where many problems can be solved exactly, and the superselection structure is often known explicitly. In these case, it is typically described by a quantum group, more precisely by a weak Hopf algebra and related modular categories. In particular, one can see here that the superselection sectors cover each unitary irrep of the observable algebra (but not each unitary irrep of the CCR). But explaining this would need a separate thread. 

Answer 2

To clarify the nonrigorous physics version of the argument--- the "overlap" of vacua at one point independent of other points only fully makes sense if the vacuum is a product state over separate points, i.e. if the field theory is ultra-local, meaning it has no derivative terms. But the principle is exactly the same when you have derivative terms, as any local change in the vacuum is amplified by translation invariance to a zero inner product with the original vacuum.

The same argument works in momentum space to establish that two free theories with different mass have orthogonal vacua in infinite volume. You can work out the inner product of the Gaussian states with each other at each k, and the inner product is the product over all k of these inner product. Knowing that there is some k region where the inner product is less than 1, and knowing that in the infinite volume limit, the number of k-variables in any region diverges, you conclude that the inner product, which is the product over all k of the inner product at k, goes to zero, and there's nothing more to it.

But in this example, you got an inequivalent vacuum because you changed the mass. The Haag thing is talking about turning on an interaction which is renormalized so as to keep the vacuum energy and the mass of the single particle excitations exactly constant as you turn it on. The Haag theorem then shows that the inner product is still zero, for the same trivial reason--- there is a local difference in the vacuum, and whatever the local difference is, it multiplies out to zero inner product due to infinite volume.

It helps for the purpose of illustration to state the Haag principle in a path integral, rather than in a rigorous formalism. The inner product of two vacua with two different values of the coupling (on a lattice, so that there are no renormalization issues), may be written down as a path integral in the sudden approximation: you have a boundary at t=0 where $\lambda$ has one value on one side, and another value on the other. The question "What is the probability amplitude for the vacuum to remain a vacuum under this sudden perturbation?" is the value of the path integral with vacuum boundary conditions on either side.

The obvious answer, whether the perturbation is sudden, or soft, whether it is for a strong interaction or weak, is always zero, because there is some nonzero probability per unit spatial volume of creating excitations from any perturbation, which means that the probability of having no excitations at the end of the perturbation is zero, so there is zero overlap. Again, it is only a question of infinite volume, and there is nothing extraordinary about it. The only notable thing about it is that it is possible to make it rigorous relatively easily given the methods of the 1960s.

Arnold Neumaier answered question 1 completely, and gave a specific model in a reference for question 2, but for question 2, you wanted a specific example model, and here's how you get all of them.

The key point is that you don't have a Haag theorem exactly when there is no particle production in the sudden approximation, which is tantamount to saying that the interaction cann't produce particles in vacuum. Then the sudden approximation does nothing at all, per unit area, or over the whole space, and the vacuum is unchanged. If there is no particle creation/annihilation from vacuum due to the interaction in the nonrelativistic theory, there is no particle annihilation into vacuum either (by Hermiticity of H) and there are no vacuum bubbles, the interacting vacuum is exactly the same as the noninteracting vacuum.

A simple example of a local nonrelativistic field theory with no vacuum bubbles, and still with local interactions (in whatever spatial dimensions), is

$$S = \int \bar\psi ( i\partial_t - {\nabla^2 \over 2m})\psi + \lambda \bar\psi \bar\psi \psi\psi $$.

This is the case of conserved particle number and delta-function interactions. It makes sense without any coupling renormalization in 1+1 d, but in 2+1 d it requires logarithmic renormalization to take a continuum limit, and in higher dimensions, it requires a power-law renormalization, but in any case, consider this on a lattice and in infinite volume. This interactions preserves the particle number N in all states, so it never produces particles starting in a vacuum.

Another example, this time which does not conserve N, so that there is particle creation and annihilation in some states, is the two species model (i ranges from 1 to 2)

$$ S = \int \sum_i \bar\psi_i (i\partial_t - {\nabla^2\over 2m})\psi_i - \sum_i \bar\psi_i\bar\psi_i\psi_i\psi_i -g( \bar\psi_1 \psi^2 \psi^1 + \bar\psi_1 \bar\psi_2\psi_1)$$

The key point is that the creation of type 2 particles always requires a type 1 particle to seed it, and the type 1 particle is conserved. This is what's going on in the Lee model, where N type particles can create C type particles, but the number of N type particles isn't modified in the process. Again, even though the type 1 particles create type 2 particles, nothing happens starting in vacuum. You can cook up a similar example where particles of type 2 create particles of type 1 also, and still nothing happens in vacuum, but N is not conserved (aside from N=0).

These examples are extremely easy to cook up nonrelativistically because the nonrelativistic field separates the creation and annihilation into independent fields $\psi$ and $\bar\psi$, so that $\psi$ is pure annihilation and $\bar\psi$ is pure creation, and in this Hamiltonian, I didn't put in any $\bar\psi$ terms without $\psi$, so that there is no creation starting from vacuum. If you do put it in, there are vacuum bubbles, and you get Haag's theorem again, again from the infinite volume limit.

An argument analogous to the informal Haag theorem argument works for broken symmetry too, like for nonrelativistic phonons. If you move a solid the moved solid is outside the Hilbert space, because suddenly moving the solid creates infinitely many phonons relative to the old position, because the creation amplitude is per-unit-volume.

Relativistically, all local fields include both creation and annihilation operators, so you can't introduce a local interaction without particle creation, without vacuum bubbles. If you decide to normal order the interaction, and then throw away just the terms in $:H:$ which are pure creation operators, the ones which escape the vacuum, so that you make sure the vacuum is invariant by hand, the interaction term becomes hideously nonlocal and non-Lorentz invariant, because you can't separate $\phi$ into the creation and annihilation part, $\phi^+$ and $\phi^-$ locally in a Lorentz invariant way. All vacuum bubbles contain a divergent $\delta(0)$ momentum-space delta function term, which is just the infinite volume factor due to their translation invariance.

The reason that you need to mix creation and annihilation in relativity is obvious even in pure potential scattering--- when you scatter any particle off a potential V(x) coupling to a local field, the amplitude has to be crossing invariant, because the particle propagation is not purely causal, the propagator is nonzero outside the light-cone. There is a nice popular exposition of this principle in a lecture by Feynman from 1986, which you can find online.

Comments

Thanks and +1. You are right, not long after composing this question I also realized the key is to have an interaction where there is an annihilation operator on the rightmost position, which makes the construction rather easy in Galilean case. I was stuck because I was thinking in terms of solving the vacuum wavefunctionals, however, if I insist on viewing it in wavefunctional picture, it still seems quite miraculous that the interaction is not changing the wavefunctional at all. For example an even simpler construction would be adding a $\lambda \bar{\psi}\psi$ term which annihilates the vacuum, but in the wavefunctional picture, this is like changing the the stiffness of the spring of the oscillators, which should definitely change the wavefunctional on the first look.

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@JiaYiyang: Nonrelativistic fields are not like relativistic fields, the fields $\bar\psi$ and $\psi$ are canonically conjugate, look at the time derivative term, so adding $\bar\psi\psi$ to H is like adding $pq$, not $q^2$. If you want a nonrelativistic analog to relativistic fields, you can separate $\psi$ into real and imaginary fields, and integrate out the imaginary part. The standard nonrelativistic action is a phase space path integral, or "coherent state path integral" (I don't like this name, but it's due to the eigenstates of annihilation operator $\psi$ being coherent states, and these are the integration variables. But this thing is just the original Dirac/Feynman path integral over alternating p and q with action $\int p\dot{q} -H$ where H is commuted to have all p's to the left and q's to the right, it is nothing extraordinary). So your intuition regarding $\bar\psi\psi$ being a stiffness is not right. This term is a chemical potential, or an energy per particle. It doesn't change the vacuum properties at all. Likewise other such terms always involve both "p" and "q", so that both them and their conjugates (which must appear by Hermiticity) don't create particles out of the vacuum. So the modification of the Hamiltonian always comes in field wavefunctional interpretation with a differential operator that annihilates the original free vacuum.

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It would be better to write $a$ for $\psi$ and $a^*$ for $\bar\psi$ in your argument. then it fits the standard terminology of second quantization, as commonly used in statistical mechanics. 

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@ArnoldNeumaier: I have not seen this convention. Perhaps it is used in math literature? I have always seen the field called "psi", and the equation of motion called the "nonlinear Schrodinger equation", but I believe you. I learned Schrodinger Fields from a little book by Harris called "A Pedestrian Approach to Quantum Field Theory", and the Wikipedia article "Schrodinger Field" telegraphs the notation conventions here, and shows the imaginary part integration recovers a second order form of the path integral.

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Yes, I just noticed that nobody uses it in the position representation, though it is very natural, and commonly used in the momentum representation. I can only recommend its use in the position representation!

I learnt second quantization from the statistical physics book by Reichl. She explains it in the Appendix, using $a(p)$ for the annihilation operator in the momentum representation, from which one gets the position version by Fourier transform, $a(x)=\int dp e^{ix\cdot p}a(p)$. This operator annihilates position states, and using it gives a very transparent interpretation for the terms of the action, both in the nonrelativistic and in the relativistic case. 

The difficulty in the relativistic case is that in order to guarantee causal commutation rules a neutral scalar field must be given by $\phi(x)=a(x)+a^*(x)$. Note that Weinberg (Chapter 5.2) uses $\phi^+(x)$ for $a(x)$ and $\phi^-(x)$ for $a^*(x)$; but they are annihilation and creation operators, and all formulas speak much clearer when written that way!

Inserting $\phi(x)=a(x)+a^*(x)$ into the covariant action gives (after expansion and normal ordering) interaction terms that correspond to the various vertices of the associated Feynman diagram machinery. The fact that there are product terms in which the number of creation and annihilation operators do not match causes all the renormalization trouble. 

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@ArnoldNeumaier: ok, that's just old-fasioned pre-Feynman perturbation theory.

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I meant:

Independent of perturbation theory, I strongly recommend the use of the notation $a(x)$ in the position representation, both in the nonrelativistic (rather than $\psi(x)$) and the relativistic case (rather than $\phi^+(x)$).

Indeed, they are in both cases annihilation operators. The mathematics doesn't change, of course, but things look much more alike and rightly so. The difference between  the nonrelativistic and the relativistic case is just that in the former case, $a(x)$ itself is a good field, while in the relativistic case it it not causal, so that one needs a new symbol $\phi$ for the causal field.

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@ArnoldNeumaier: I see! It's a notation quibble! Ok, maybe. Is it such a big deal?

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For me, it made a lot of difference when I noticed that this relates the nonrelativistic and the relativistic case. It made the similarities and differences crystal clear! (For me, good notation is half of good understanding.)

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@ArnoldNeumaier: You're welcome to your convention, and it is clear and unified, but it's not my preferred choice. I make the exact opposite convension for mode expansions, also clear and also unified in the same way. instead of calling every creation operator $a^\dagger$ regardless of the field, I call the relativistically normalized creation operator $\phi^\dagger(P) = \sqrt{(2\pi)^32\omega_p}a^\dagger(P)$, and likewise for annihilation. In relativistic normalization, all the field expansion formulas are manifestly invariant and completely transparent:

$$\phi(x) = \int e^{-ipx} \phi^\dagger(P) + e^{ipx}\phi(P) dP$$

where P is spatial mode number (spatial momentum) $dP$ means the invariant measure on the mass shell, $d^3p \over (2\pi)^3 2\omega_p$, and $\phi^\dagger$ and $\phi$ are creation operators for relativistically normalized states. This for me is the "notationally correct" relativistic expansion the way I see it because it's trivial, no factors.

The canonical commutation relation is also properly relativistic now:

$$[\phi^\dagger(P),\phi(P')] = \delta(P-P') = (2\pi)^3 2\omega_p \delta^3(p-p')$$

so it can exactly pop a covariant mass-shell integral dP or dP' with no factors floating around afterwards. This is the easiest way for me to do bookkeeping on asymptotic state expansions, like in unitarity sums or cross sections, I can't keep track of the factors in the usual normalization convention which is not relativistic.

The natural nonrelativistic limit of this notation is to call the creation and annihilation operators by the name of their field (and include the $\sqrt{(2\pi)^3}$ in the a's but not the $\sqrt{2m}$ of rest energy--- this is a good convention, it is used in Dirac somewhere), so you end up calling the annihilation fields $\psi(P)$. not $a(P)$. There are two issues in the notation, the use of the same name "a" and "a-dagger" for different fields (I hate that) and also the more serious one of relativistically normalized states. I like my convention, so I use it universally, even though nobody else uses it as far as I know. (this is a really minor issue, I have no disagreement with you on content).

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So your intuition regarding $\bar{\psi}\psi$ being a stiffness is not right. This term is a chemical potential, or an energy per particle. It doesn't change the vacuum properties at all. 

True. Here are the follow-up questions:

(1) Indeed this is a chemical potential, and I now see this because it's proportional to the number operator, but it seems you infer this from the fact it is of the form of $pq$, what's your gist?

(2)Still it doesn't quite answer the mystery in the wavefunctional picture. Forget about $\bar{\psi}\psi$, let's go back to $\bar{\psi}\bar{\psi}\psi\psi$, now I guess you cannot quite get away with it by forming an easy physical intuition, how would you like explain Haag's theorem away this time? My point is, generically thinking in the wavefunctional picture, relativistic or not, any modification of the Hamiltonian would at least alter the functional at local points a little bit, which makes Haag's theorem again applicable. Isn't it a reasonable intuition? Why does it fail so miserably? 

I don't know if my insistence on getting an accurate wavefunctional intuition is leading anywhere, but if possible I think it'll help me internalize these ideas.

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I use $a_e(x)$ and $a_e(p)$ for electrons, $a_\gamma(x)$ and $a_\gamma(p)$ for photons, etc. This is more flexible and preserves my and your advantages.

For a nonrelativistic limit, you also need to make a Foldy-Wouthuysen transformation!

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@JiaYiyang: Just do it for a 1-d HO, the operator a is ${\partial + x}$, and annihilates the vacuum exactly, and deformations proportional to it don't change the vacuum.

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@ArnoldNeumaier: If you're old fasioned and insist on doing everything by operators, but that stuff is so musty to me. You can just take the limit in the path integral directly, although it amounts to the same thing. The FW transformation is just subtracting the rest mass from the operators and separating the creation and annihilation bits into separate fields.

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In my understanding, I want to be mathematically rigorous as far as possible, hence cannot use the path integral.

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 If you decide to normal order the interaction, and then throw away just the terms in :H: which are pure creation operators, the ones which escape the vacuum, so that you make sure the vacuum is invariant by hand, the interaction term becomes hideously nonlocal and non-Lorentz invariant,

@RonMaimon, but we do normal order the interactions in relativistic QFT all the time don't we? But the resulting perturbation theory seems covariant?

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@JiaYiyang: Ordinary covariant perurbation theory based on normal ordering is OK because we then do not throw away the part that does not annihilate the vacuum such as (Hermitian) terms like $a^*a^*a^*+aaa$. Thus the bare vacuum of the Fock space is not the vacuum of the interacting theory - a fact responsible for the divergences of naive perturbation theory, accounted for by the renormalization procedure.

In the similarity renormalization view of QFT,  renormalization consists of making (in the cutoff theory) a unitary transform of the Fock space that moves the interacting vacuum and the 1-particle states (actually all bound states of the interacting theory) to the bare vacuum and bare 1-particle states, then taking the cutoff to infinity (assuming that this is possible, which is the case perturbatively, but most likely not nonperturbatively). This transforms the a Hamiltonian to one in which the incriminated terms are absent. How this is done in QED can be seen from the book by Stefanovich (Chapters 11 and 14 on similarity renormalization are ok; ignore the weird stuff on superluminality). One can see the agreement with the standard approach.

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@ArnoldNeumaier, this is perhaps a notational issue that has been confusing me, doesn't normal ordering symbol by definition map $aa^*a^*$ to $a^*a^*a$? But then the interaction term obviously loses Hermiticity, this has been confusing me for a while. If this is not how one does normal ordering in relativistic QFT, what exactly is the definition of normal ordering?

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@JiaYiyang: Sorry, you are right. I corrected my wrong example. (But normal ordering preserves Hermiticity; the adjoint of $a^*a^*a$ is $a^*aa$, which is again normally ordered.)

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@ArnoldNeumaier,

Yes you are right, I messed up some easy algebra.

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@JiaYiyang:  In general $:F(a^*,a)^*:=:F(a^*,a):^*$, as you can prove by looking at monomials.

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@ArnoldNeumaier,

Actually what I had in mind about losing manifest covariance is this: consider a bilinear  current say $J^\mu=\bar{\psi}\gamma^\mu\psi$, then $:J^\mu:=J^\mu-\langle 0| J^\mu|0\rangle$, but then if you consider acting on both sides $U(\Lambda)^{-1}\ldots U(\Lambda)$, obviously $:J^\mu:$ and $J^\mu$ can't be both covariant since $\langle 0| J^\mu|0\rangle$ is just a c-number hence invariant under the similarity transformation. 

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@JiaYiyang:  Simce the vacuum is translation invariant, $⟨0|J^μ(x)|0⟩$ is independent of $x$ but transforms as a 4-vector. On the other hand, $J^μ(x)$ transforms under a Lorentz trnasformation into $(U(\Lambda)j)^\mu(\Lambda x)$. Thus everything is alright.

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@ArnoldNeumaier,

What I meant is $U(\Lambda)^{-1}:J^\mu:U(\Lambda)=U(\Lambda)^{-1}J^\mu U(\Lambda)- U(\Lambda)^{-1}\langle  0|J^\mu|0\rangle U(\Lambda) \\ =U(\Lambda)^{-1}J^\mu U(\Lambda)- \langle 0|J^\mu|0\rangle.$

Note the last term is invariant instead of covariant since it's a c-number, that's where I find it paradoxical.

@dimension10, the texts got truncated again, I still don't know how to fix it.

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@JiaYiyang:  But  $U^{−1}J^μU$ transforms for every linear operator $U$ as a vector, just as $J^\mu$; so everything is covariant. Writing $U(\Lambda)$ in place of $U$ doesn't change this. 

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@ArnoldNeumaier,

I don't understand your last comment, my point is $\langle J^\mu \rangle$ is a collection of c-numbers, then for any linear operator U we have $U^{-1}\langle J^\mu \rangle U= \langle J^\mu \rangle$

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@JiaYiyang: I still don't get what your problem is; in any case, this should be a separate question as it is unrelated to the remainder of the present thread.

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@ArnoldNeumaier, maybe I will start a separate post later. Meanwhile if you like you can move the discussions that deviate from the thread to chat.