Rigorous Quantum Field Theory.

[DarMM]

Hey everybody, since the previous thread got locked I thought I would open this thread as a place to discuss rigorous issues in quantum field theory, be it on the constructive or axiomatic side of things.

I apologize if one is not supposed to start a discussion with posts from old threads, but I thought it would be a nice way to get the discussion going.

In the last thread:

strangerep & DarMM -> As I understand it, the renormalization procedure a la Epstein & Glaser (and hence the "infinite subtraction" one) is not more ad hoc then, say, solving LaTeX Code: a x^2 + b x +c =0 . Let me elaborate this a bit. I'm trying to obtain an answer from my (perturbation) theory, and to get that answer I must extend my product of distributions to coinciding points where it is in general ill defined. Of course, I'll have to satisfy some conditions (causality and locality), but in principle this is "just" another mathematical problem one has to solve. Much like trying to get an answer about some physical problem which would involve the solution of the above (simple) equation. So in this sense, it does not appear more ad hoc than every other problem in physics. Comments?

Yes, that is essentially my understanding. The upshot is that at the end you've constructed the perturbative expansion for S(g) (the S-matrix in finite volume) in a completely rigorous way. Modern work on the Epstein-Glaser approach tries to take the limit g \rightarrow 1, to go to infinite volume, although it has proven extremely difficult.

It is undisputable that renormalized QFT (such as QED) can calculate the S-matrix (i.e., the result of time evolution between - and + infinite times) very accurately. I think we also agreed that this theory does not have a well-defined finite Hamiltonian. Without a Hamiltonian it is impossible to calculate the finite time evolution. That's the major inconsistency I am talking about.

Yes, the Hamiltonian is not a well-defined finite operator on Fock space, but in most QFTs it can be shown to be a well-defined finite operator on another Hilbert space. Do you view this as a problem?

The investigation of the propogation of effects in QFT was performed by Segal and Guenin in the 1960s for specific models. Also Haag's algebraic apporach allows it to be treated in general, where you can show that effects do propogate at the speed of light.

I would appreciate exact references. Though, I am rather sceptical, for the same reason: the lack of a well-defined finite Hamiltonian in renormalized QFT.

Perhaps you are talking about 2D models? I admit, I know almost nothing about them.

For the two-dimensional models the best reference is:

Segal, I. Notes towards the construction of nonlinear relativistic quantum elds
I: The Hamiltonian in two spacetime dimensions as the generator of a C* -
automorphism group", Proc. Nat. Acad. Sci. U.S.A. 57, 1178-1183.

For the proof that general field theories have signals propogating at the speed of light a good read might be the treatise on algebraic quantum field theory:
Araki, H. Mathematical Theory of Quantum Fields, Oxford, 2000.

For several nonperturbatively constructed field theories there are actual calculations showing this to be the case. If you want references I can provide them.
I understand you are skeptical because there isn't a well-defined finite Hamiltonian on Fock space, however there is one on the correct Hilbert space.

 

[meopemuk]

Yes, the Hamiltonian is not a well-defined finite operator on Fock space, but in most QFTs it can be shown to be a well-defined finite operator on another Hilbert space. Do you view this as a problem?

Yes, I view this as a problem (though this could be a problem of my ignorance).

As far as I understand, when doing routine S-matrix calculation in QFT, we are working in the normal Fock space. We calculate matrix elements of the S-operator with states like a^{\dag}a^{\dag} |0 \rangle where |0 \rangle is the free vacuum and a^{\dag} are creation operators for free particles. These calculations are perfectly successful and agree with experiment.

Now, we seem to agree that renormalized QED does not provide a finite (Fock space) Hamiltonian, that is consistent with the S-matrix above. In order to define a Hamiltonian you are suggesting to go to another Hilbert space with different vacuum and different creation operators A^{\dag}. There, presumably, we will be able to calculate the finite time evolution of states and observables. If we can do that, then why don't we extend the time interval to infinity and get the S-matrix as well? If our finite time evolution is unitary and physically reasonable, then the S-matrix should come out OK too. Then why don't we forget about the original Fock space and move our entire formalism in your new Hilbert space? This Hilbert space has all we need: a finite Hamiltonian and a finite and accurate S-matrix. Then we will avoid the confusing necessity of dealing with two Hilbert (Fock) spaces.

Why this one-Hilbert-space description is not possible? Is it because asymptotic particles (at large distances where the interaction can be neglected) live in the non-interacting Fock space, while the same particles at short distances "migrate" to the interacting Hilbert space? How this "migration" occurs? Shouldn't we also consider some "intermediate" Hilbert spaces? Does the Hilbert space of interacting system change with time?

I just can't comprehend the overall picture.

I find it much easier to understand a theory in which entire life of an interacting system of particles is spent in the same fixed Hilbert (or Fock) space, where (just as in ordinary quantum mechanics) we have a fixed Hamiltonian and other operators of observables; the time evolution operator and the S-operator are related to each other by usual QM formulas. Such a theory can easily accommodate interactions which change the number of particles. The only limitation is that interaction should not contain "bad" terms like aaa + a^{\dag}aa +a^{\dag} a^{\dag} a + a^{\dag}a^{\dag}a^{\dag}. But the absence of such terms in the Hamiltonian is perfectly justifiable, because nobody has seen a physical process in which, e.g., three particle spontaneously appear out of vacuum. Without these "bad" terms, there is no distinction between "bare" and "dressed" (or "physical" or "interacting") particles. So there is no need to struggle with (at least) two Hilbert spaces.

Eugene.

P.S. Thank you for the references, I will check them out.

 

[meopemuk]

DarMM,

by the way, thanks for opening this new thread. I really appreciate the discussion.

Eugene.

 

[meopemuk]

Me too.

Just one modest question: whatever momentum q is transferred to the electron, no soft radiation appears on the tree level in charge-from-charge scattering (Rutherford, Mott, Moeller cross sections). Is it physical? "Do you view this as a problem?"

Bob_for_short,

I see your point: When two charges collide, many soft photons must be emitted, and the probability of scattering without such an emission should be negligibly small. On the other hand, the lowest order QED predicts a sizeable probability (cross section) for no-photons events. I don't know how to answer this question. I hope it is handled satisfactorily, when IR divergences are discussed, but I am not good at those.

Eugene.

 

[strangerep]

Just one modest question: whatever momentum q is transferred to the electron, no soft radiation appears on the tree level in charge-from-charge scattering (Rutherford, Mott, Moeller cross sections). Is it physical? "Do you view this as a problem?"

(Sigh.) Bob, you're committing the same impoliteness again: slowly hijacking someone
else's thread off in a tangential direction. DarMM clearly intended this thread to
concentrate on constructive/axiomatic/rigorous QFT.

Please move your question into your own new thread (quoting context from this
thread if necessary) so that any followup answers can appear there, with
redirections to your thread in the IR forum. That would be more polite to others.

 

[strangerep]

Hey everybody, since the previous thread got locked I thought I would open this thread as a place to discuss rigorous issues in quantum field theory, be it on the constructive or axiomatic side of things.

Thank you. I was indeed a bit disappointed that, while I slept in my part of the world,
my other thread became derailed enough to get it shut down.

I apologize if one is not supposed to start a discussion with posts from old threads,
but I thought it would be a nice way to get the discussion going.

I think it is a good, productive way to focus on particular items arising in another thread.

Yes, the Hamiltonian is not a well-defined finite operator on Fock space, but in most
QFTs it can be shown to be a well-defined finite operator on another Hilbert space.

What exactly do you mean here by "most QFTs"? Those that rigorously exist, or
other stuff like \phi^4, QED, etc (in 4D)?

Araki, H. Mathematical Theory of Quantum Fields, Oxford, 2000.

I see on Amazon that a much cheaper paperback version was released recently (July 2009).
I have ordered it, although the book is apparently for "mathematics graduates".
I'm hoping I now know enough topology and functional analysis to get by.

 

[Vanadium 50]

Strangrep is absolutely right. If this thread is hijacked too, it too will be closed. Please stick to the subject.

 

[DarMM]

What exactly do you mean here by "most QFTs"? Those that rigorously exist, or
other stuff like \phi^4, QED, etc (in 4D)?

I mean the ones which have either been completely rigorously constructed, or at least constructed far enough to have their Hamiltonian defined.
\phi^4 in 4d is one such theory, the only question left is if it is trivial or not, not if it exists. If it is not trivial, then it can be shown to have a Hamiltonian defined on another Hilbert space.
Virtually all theories in 2d and 3d have been shown to have this property.

I see on Amazon that a much cheaper paperback version was released recently (July 2009).
I have ordered it, although the book is apparently for "mathematics graduates".
I'm hoping I now know enough topology and functional analysis to get by.

Believe it or not the book is not that difficult, especially considering it is a translation of a Japanese original. Dr. Carow-Watamura must be congratulated for an excellent translation. Araki introduces the terminology when relevant and the appendicies are very well written. The only difficult mathematics in the book is operator algebras, but the book basically doubles as an intro to operator algebra so this doesn't matter too much.

For the causal properties of QFTs you should look at sections four and six.

 

[Avodyne]

\phi^4 in 4d is one such theory, the only question left is if it is trivial or not, not if it exists.

\phi^4 in 4d has been proved to be trivial: http://arxiv.org/abs/0808.0082

 

[DarMM]

\phi^4 in 4d has been proved to be trivial: http://arxiv.org/abs/0808.0082

This is not a proof in the sense that I mean. For instance it is difficult to analytically control the renormalization group in a mathematically rigorous way. Basically most treatments ignore something which is called the large field problem, which is what makes the renormalization group so difficult to treat rigorously. The large field problem is essentially that the renormalization group, even after a very short flow will produce a nonpolynomial action whose behaviour at very large fields \phi is not known and hence may not be integrable and the flow will halt. Papers such as the above ignore this problem.

There is nothing wrong with such ways of demonstrating the triviality of \phi^{4}, Luscher and Weisz and Wolff have all given excellent numerical demonstrations. However it still does not constitute a mathematical proof of triviality.

 

[DarMM]

As far as I understand, when doing routine S-matrix calculation in QFT, we are working in the normal Fock space. We calculate matrix elements of the S-operator with states like a^{\dag}a^{\dag} |0 \rangle where |0 \rangle is the free vacuum and a^{\dag} are creation operators for free particles. These calculations are perfectly successful and agree with experiment.

Yes, that is correct.

Now, we seem to agree that renormalized QED does not provide a finite (Fock space) Hamiltonian, that is consistent with the S-matrix above.

Indeed, however I shall stick to QFT in general rather than just QED in my answers below.

In order to define a Hamiltonian you are suggesting to go to another Hilbert space with different vacuum and different creation operators A^{\dag}.

Yes, another Hilbert space which is a different representation of the creation/annhilation operator algebra or canonical commutation relations, which ever you prefer.

There, presumably, we will be able to calculate the finite time evolution of states and observables. If we can do that, then why don't we extend the time interval to infinity and get the S-matrix as well? If our finite time evolution is unitary and physically reasonable, then the S-matrix should come out OK too.

It does indeed. We can take the time interval to infinity and obtain a well-defined S-matrix which agrees with the one from usual caculations. References on this available if you wish.

Then why don't we forget about the original Fock space and move our entire formalism in your new Hilbert space?

Now we come to the main issue. In the correct Hilbert space everything is defined, the Hamiltonian, the S-matrix, finite time evolution, e.t.c. The problem is that without very high powered mathematics, particularly the representation theory of operator algebras and functional integration, you cannot explicitly construct the correct Hilbert space. Fock space is the only rep which is easy to find/construct. Constructive Quantum Field Theory is basically the theory of constructing the interacting Hilbert space.
So once we find the correct Hilbert space everything can be done in one single space, however the kilometer high wall of advanced mathematics standing in the way of finding this Hilbert space basically forces this to be a concern only of mathematical physics.
Also you can show that the perturbative expansions of the S-matrix on the true Hilbert space can be written using Fock space expectation values, or to put it another way Fock space can simulate the real Hilbert space well enough for perturbative calculations. So standard perturbation theory has a mathematical backing.

By the way if anybody thinks this is too canonical, all this can be translated in path integral language:
Free Hilbert space is different to interacting Hilbert space
becomes
Free field measure on field path space is mutually singular with respect to the interacting measure.

Why this one-Hilbert-space description is not possible?

Hopefully my answer above has explained things. It is possible, just prohibitively difficult. The Clay mathematics institute Yang-Mills problem is basically asking you to construct/find the correct Hilbert space for 4D Yang-Mills theory.

 

[meopemuk]

Now we come to the main issue. In the correct Hilbert space everything is defined, the Hamiltonian, the S-matrix, finite time evolution, e.t.c. The problem is that without very high powered mathematics, particularly the representation theory of operator algebras and functional integration, you cannot explicitly construct the correct Hilbert space.

If you don't mind I would like to learn a bit more about the interacting Hilbert space. Let me know where my logic fails.

Let us assume that we have constructed this Hilbert space somehow. If it represent physics correctly, then it should have some specific properties. For example, it should have a particle interpretation, i.e., we can identify 0-particle, 1-particle, 2-particle, etc. states there. It seems natural that N-particle states are orthogonal to M-particle states. Therefore the Hilbert space can be divided into orthogonal N-particle sectors. Also, there must be an unitary representation of the Poincare group. I presume that both 0-particle and 1-particle sectors are invariant with respect to this representation. This is because in experiments we don't see spontaneous creation of particles from the vacuum and 1-particle states (suppose that we are talking about stable particles only). Furthermore, based on this structure, we should be able to define particle creation and annihilation operators, which change the number of particles by +1 or -1. Finally, the Hamiltonian (and other operators) should be expressable as a polynomial in these creation and annihilation operators.

In other words, we get all relationships the same as in the usual Fock space. (These relationships are forced upon us by physics. If some of them fails, there should be observable consequences.) Where is the difference? Why I am not allowed to do the same stuff in my favorite Fock space? Why can't I say that the interacting Hiolbert space and the Fock space are simply equivalent?

Eugene.

 

[DarMM]

If you don't mind I would like to learn a bit more about the interacting Hilbert space. Let me know where my logic fails.

Let us assume that we have constructed this Hilbert space somehow. If it represent physics correctly, then it should have some specific properties. For example, it should have a particle interpretation, i.e., we can identify 0-particle, 1-particle, 2-particle, etc. states there. It seems natural that N-particle states are orthogonal to M-particle states.

Remember in an interacting theory a one particle state could decay into two particles, so there is no reason to expect them to be orthogonal. Of course you mention later that we are discussing an interacting theory with only stable particles, so this doesn't matter too much.

Also, there must be an unitary representation of the Poincare group. I presume that both 0-particle and 1-particle sectors are invariant with respect to this representation.

Yes this is the same, although I think the one-particle space is covariant.

Furthermore, based on this structure, we should be able to define particle creation and annihilation operators, which change the number of particles by +1 or -1. Finally, the Hamiltonian (and other operators) should be expressable as a polynomial in these creation and annihilation operators.

This is where you run into a problem. There is no state which is annihilated by all annihilation operators. There is still a vacuum, i.e. a state with no energy or momentum which is invariant under Poincaré transformations, but there is no "lowest state on the ladder" for creation and annihilation operators.
You can define creation and annihilation operators for which this isn't a problem, but they have a different algebra. So you can say it two ways:
The interacting Hilbert space is a non-Fock representation of the usual creation and annihilation operators
or
It is a Fock representation of unusual creation and annihilation operators.

 

[meopemuk]

This is where you run into a problem. There is no state which is annihilated by all annihilation operators. There is still a vacuum, i.e. a state with no energy or momentum which is invariant under Poincaré transformations, but there is no "lowest state on the ladder" for creation and annihilation operators.
You can define creation and annihilation operators for which this isn't a problem, but they have a different algebra.

This is what I don't understand.

Suppose that I have divided my Hilbert space into N-particle sectors. For each sector I have a basis of vectors with defined particle momenta. For example, in the 1-particle sector the basis is |p1\rangle. In the 2-particle sector the basis is |p1, p2 \rangle etc. Then I can simply define my creation and annihilation operators by going through all basis vectors and postulating how the operators act on these vectors. For example, the annihilation operator a(p1) is *defined* by the following relationships

a(p1) |0 \rangle = 0
a(p1) |p1\rangle = |0 \rangle
a(p1) |p2\rangle = 0
a(p1) |p1, p2\rangle = |p1 \rangle
...

The action on vectors not belonging to the basis is extended by linearity. Furthermore, I can simply check that thus defined operators satisfy usual (anti)commutation relations. Where is my mistake?

Eugene.

 

[Avodyne]

Suppose that I have divided my Hilbert space into N-particle sectors.

You cannot do this, because the "number of particles" is not a well-defined concept. Even in a theory like \varphi^4 with a single type of particle that is stable, you can have pair creation, e.g. a process in which 2 high-energy particles scatter into 4.

Consider (for simplicity) states with zero 3-momentum. What are the allowed energies? Answer: E=0, E=m, and E>=2m. The E=0 state is the vacuum (no particles), the E=m state is one particle (at rest, because we have assumed zero 3-momentum), and the states with E>=2m form the multi-particle continuum. Consider, for example, a state with E=4m. This could be 4 particles at rest, or it could be 2 particles with equal and opposite 3-momenta such that their energies add up to 4m. But there is no reason for these states to be orthogonal, because we know one can scatter into the other. So what are the right basis states in the multiparticle continuum? This is the hard part of the problem.

But the theories that make sense perturbatively and have physical particles like those in the corrsponding free theory (\varphi^4, QED) are almost certainly trivial, which almost certainly means that they can't be rigorously constructed (at least IMO).

 

[meopemuk]

You cannot do this, because the "number of particles" is not a well-defined concept. Even in a theory like \varphi^4 with a single type of particle that is stable, you can have pair creation, e.g. a process in which 2 high-energy particles scatter into 4.

Pair creation does not mean that the particle number operators do not exist. It simply means that these operators do not commute with the total Hamiltonian, i.e., particle numbers are not conserved.

Consider (for simplicity) states with zero 3-momentum. What are the allowed energies? Answer: E=0, E=m, and E>=2m. The E=0 state is the vacuum (no particles), the E=m state is one particle (at rest, because we have assumed zero 3-momentum), and the states with E>=2m form the multi-particle continuum. Consider, for example, a state with E=4m. This could be 4 particles at rest, or it could be 2 particles with equal and opposite 3-momenta such that their energies add up to 4m.

Yes, I agree with this.

But there is no reason for these states to be orthogonal,

Here we disagree. In experiments the number of particles behaves as any any other observable. So, there should exist an Hermitian operator corresponding to this observable. Different eigenvalues of this operator (i.e., different numbers of particles) must correspond to orthogonal eigen-subspaces.

because we know one can scatter into the other. So what are the right basis states in the multiparticle continuum? This is the hard part of the problem.

The scattering matrix element between two orthogonal states is generally non-zero. So, two states can be orthogonal and still one can scatter into the other. So, your explanation is not convincing.

From physical requirements, N-particle sectors should be constructed as tensor products of N 1-particle subspaces. This also determines the basis in such sectors. Physics does not leave much frredom in the construction of N-particle states.

Eugene.

 

[Dmitry67]

Here we disagree. In experiments the number of particles behaves as any any other observable.

So, how many particles proton consists of?
I've heard about something called "scale problem" where the number of particles in QCD is different on different scale.

+Unruh effect, where the number of particles depend on frame.

 

[meopemuk]

So, how many particles proton consists of?
I've heard about something called "scale problem" where the number of particles in QCD is different on different scale.

+Unruh effect, where the number of particles depend on frame.

In ordinary QM the number of particles is regarded as a valid observable, i.e., the corresponding operator N is Hermitian (with real eigenvalues 0,1,2,3,...) and its eigensubspaces are orthogonal. The (expectation value of the) number of particles can depend on time (e.g., in decays). This simply means that the operator N does not commute with the Hamiltonian. The proton (which is an eigenstate of the interacting Hamiltonian) may be not an eigenstate of N. This is not unusual. The operator N may also not commute with generators of boosts, so the number of particles could be frame-dependent as well. This is not unusual also.

If I understand Avodyne correctly, then in rigorous 2D QFT models the particle number operator either does not exist or is not Hermitian. This seems rather strange and unphysical to me. I would like to learn more about it.

 

[DarMM]

In ordinary QM the number of particles is regarded as a valid observable

I'm not sure what is meant here, in Quantum Mechanics isn't the total number of particles fixed?

Suppose that I have divided my Hilbert space into N-particle sectors.

Let's start here. Given the Hilbert space of a relativistic quantum field theory, how would you go about dividing it into N-particle sectors? That is, how would you build the number operator?

 

[Bob_for_short]

I'm not sure what is meant here, in Quantum Mechanics isn't the total number of particles fixed?

Any particular atomic state ψ_n can be considered as a different "particle" in a narrow sense. After scattering one has a superposition of all ψ_n' allowed by the conservation laws. One can say (in a narrow sense) that in atom-atomic scattering the "particles" transform into each other. If the transferred energy is sufficient for ionization, then the number of different possible final n' is infinite.

 

[meopemuk]

I'm not sure what is meant here, in Quantum Mechanics isn't the total number of particles fixed?

I am sorry for deviating from my own terminology. Of course, "quantum mechanics" implies fixed number of particles. I should have used name "interacting theory in the Fock space".

Let's start here. Given the Hilbert space of a relativistic quantum field theory, how would you go about dividing it into N-particle sectors? That is, how would you build the number operator?

If I am allowed to do this job, I wouldn't start from fields, field equations, Lagrangians, etc. In my opinion, these things have little relevance to the construction of "quantum theory of systems with variable number of particles". I will first postulate (from physical considerations) that particle numbers are Hermitian observables, and that the number of photons is compatible (i.e., commutes) with the number of electrons, etc.. This implies that the full Hilbert space must be divided into orthogonal n-particle sectors. Then I will postulate that 1-particle sector is the Hilbert space of an irreducible unitary representation of the Poincare group, as was built by Wigner. I will also say that 2-particle sectors are tensor products of 1-particle subspaces, etc. In other words, I am going to obtain the usual Fock space structure from physical considerations.

The next step would be to construct a non-trivial unitary representation of the Poincare group in this Fock space, i.e., to specify relativistic interactions between particles. But this is a different story.

My views cannot be called revolutionary, because this approach was advocated in Weinberg's vol. 1. Unfortunately, Weinberg omits discussion of Fock spaces, but I hope I got the letter and spirit of his book right.

Eugene.

 

[Dmitry67]

In ordinary QM the number of particles is regarded as a valid observable, i.e., the corresponding operator N is Hermitian (with real eigenvalues 0,1,2,3,...) and its eigensubspaces are orthogonal. The (expectation value of the) number of particles can depend on time (e.g., in decays). This simply means that the operator N does not commute with the Hamiltonian.

Thank you, I did not know that.
Then how do we know that proton contains exactly 3 quarks? So how do we all agree on the number of quarks but not on the number of gluons? What is a difference?

 

[meopemuk]

Then how do we know that proton contains exactly 3 quarks? So how do we all agree on the number of quarks but not on the number of gluons? What is a difference?

I don't know QCD so well, but I believe that the most general proton wave function is a linear combination of states "3 quarks" + "3 quarks+1 quark-antiquark pair" + "3 quarks+2 quark-antiquark pairs" + ... In addition, any number of gluons should be also thrown into the mix.

In fact, even the hydrogen atom cannot be described exactly as a 2-particle state. It also has some admixture of states like "1 electron+1 proton+n photons+m electron-positron pairs+..."

Eugene.

 

[DarMM]

Okay meopemuk, let's take this statement:

I will first postulate (from physical considerations) that particle numbers are Hermitian observables

This is basically key to your whole construction, but under what conditions is particle number a Hermitian observable? That is when can you measure particle number?

Would you agree that particle number is a well-defined observable in the limit when the particles are spread far apart and are non-interacting. That is, it is well defined for in/out states?

 

[meopemuk]

This is basically key to your whole construction, but under what conditions is particle number a Hermitian observable? That is when can you measure particle number?

Would you agree that particle number is a well-defined observable in the limit when the particles are spread far apart and are non-interacting. That is, it is well defined for in/out states?

No, I don't agree with this statement. I would like to say that particle number can be measured for any state. It shouldn't matter whether particles are separated and free or close to each other and interacting. In usual quantum mechanics, operators of observables are fixed, and they do not depend on the state (wave function) of the system. Expectation values of observables do depend on the state, but operators themselves are independent on the state. So, if I say that the number of particles is a valid observable, I should also admit that this observable can be measured equally well in any state.


Eugene.

 

[DarMM]

No, I don't agree with this statement.

Let me rephrase this statement, you agree that particle number is a well-defined observable for in/out states? I don't mean that just in/out states have a well-defined particle number, but I'm asking do you consider in/out states to be realistic states that can have a well-defined particle number?

 

[meopemuk]

Let me rephrase this statement, you agree that particle number is a well-defined observable for in/out states? I don't mean that just in/out states have a well-defined particle number, but I'm asking do you consider in/out states to be realistic states that can have a well-defined particle number?

Yes, I agree with that. In my understanding in-out states are not different from any other (e.g., interacting) states. Simply in in-out states particles are far from each other and their interaction can be neglected. So yes, particle number is a well-defined operator for such states.

 

[DarMM]

Okay, perfect.
Now obviously we have two types of such asymptotic states. In states in the far past and out states in the far future. Obviously we can construct a Fock space for these in/out states. There is an In-Fock space and an Out-Fock space, and In/Out creation and annihilation operators. Now I'm sure you will agree that the in-vacuum is the same as the out-vacuum. However if the creation and annihilation operators satisfy exactly the same algebra, would this not imply that any n-particle in state is the same as any n-particle out state?

 

[meopemuk]

Okay, perfect.
Now obviously we have two types of such asymptotic states. In states in the far past and out states in the far future. Obviously we can construct a Fock space for these in/out states. There is an In-Fock space and an Out-Fock space, and In/Out creation and annihilation operators. Now I'm sure you will agree that the in-vacuum is the same as the out-vacuum. However if the creation and annihilation operators satisfy exactly the same algebra, would this not imply that any n-particle in state is the same as any n-particle out state?

Well, I will say even more. Both in and out states (as well as interacting states) live in the same Fock space with the same set of creation and annihilation operators. The in-state (at time t=-infinity) is such that wave packets of asymptotically free particles move toward each other, so that they are destined to collide at time t=0. In the out state (at t=+infinity)
wave packets move away from the collision point.

 

[DarMM]

Well, I will say even more. Both in and out states (as well as interacting states) live in the same Fock space with the same set of creation and annihilation operators. The in-state (at time t=-infinity) is such that wave packets of asymptotically free particles move toward each other, so that they are destined to collide at time t=0. In the out state (at t=+infinity)
wave packets move away from the collision point.

Let me say this again.
The two Fock spaces are basically the same Hilbert space, because they are unitarily related to each other, I'm not disputing that. However, the in-creation operator could not really be the same thing as the out-creation operator, since one creates a particle with certain properties at t = - \infty and the other a particle with the same properties at t = \infty. These are not the same state.
However since the two sets of creation/annihilation operators have the same vacuum, having the same algebra for the two sets of creation and annihilation operators would imply they were the same state.
Do you agree with this?

 

[meopemuk]

Let me say this again.
The two Fock spaces are basically the same Hilbert space, because they are unitarily related to each other, I'm not disputing that. However, the in-creation operator could not really be the same thing as the out-creation operator, since one creates a particle with certain properties at t = - \infty and the other a particle with the same properties at t = \infty. These are not the same state.
However since the two sets of creation/annihilation operators have the same vacuum, having the same algebra for the two sets of creation and annihilation operators would imply they were the same state.
Do you agree with this?

Annihilation and creation operators don't have time label, so they create states irrespective of time. For example the state

a^{\dag}|0 \rangle......(1)

is simply a 1-particle plane wave.

When studying simple 2-particle scattering we first create 2 separated wave packets out of basis states (1) at t = - \infty. We make sure that if the time evolution is applied to these wave packets, then at time t = 0 their paths cross, and the collision occurs. Continuing the time evolution to positive times we find that at t = \infty different free-propagating wave packets are formed (the number of particles may change). The wave function at t = \infty can be also expanded in states (1). These new expansion coefficients are related to the old expansion coefficient by means of the S-matrix.

So, I don't see any reason to introduce separate in- and out- creation/annihilation operators. The entire time evolution occurs in one Hilbert (Fock) space with one set of creation/annihilation operators. The role of these operators is to provide a basis set (1) in this Hilbert space. They also provide a convenient notation for writing operators of observables in the Fock space.

Eugene.

 

[DarMM]

Annihilation and creation operators don't have time label, so they create states irrespective of time.

I don't understand this.
What states are the creation and annihilation operators creating and annihilating then? A two particle state at t=\infty is not the same state as a two particle state at t=-\infty. So two different operators are necessary to create them from the vacuum.
In/Out states form two completely different basis of the Hilbert space which are related to each other through a unitary transformation known as the S-matrix. So if what you were saying is true, then the S-matrix would be the identity and the theory would be trivial.
This isn't even an aspect of rigorous field theory, it's in usual field theory textbooks like Srednicki.
If you object to this aspect of QFT, you are basically saying all theories are free.

 

[Bob_for_short]

...However, the in-creation operator could not really be the same thing as the out-creation operator, since one creates a particle with certain properties at t = - \infty and the other a particle with the same properties at t = \infty. These are not the same state.

In practice it would mean the electrons/photons (whatever) of today are different from yesterday's ones. It would destroy all attempts of scientific description since it would make the experiments non-repeatable and explicitly time-dependent.

What happens in reality is changing populations of states due to scattering. If one constructs a theory correctly, this should be the only scattering effect.

Do not tell me that it is impossible to do due to distribution character of fields. You may observe and describe difficulties in current constructions but they do not exclude more successful attempts.

 

[DarMM]

In practice it would mean the electrons/photons (whatever) of today are different from yesterday's ones. It would destroy all attempts of scientific description since it would make the experiments non-repeatable and explicitly time-dependent.

No it doesn't. It just means that a state which is composed of n separated particles in the far past is different to a state composed of n separated particles in the far future. Obviously one does not have to unitarily evolve into the other, since that would mean that the theory is trivial. I must say I'm kind of surprised to be debating this since it is introductory QFT stuff.

 

[Bob_for_short]

The confusion may originate from Schroedinger and Heisenberg pictures for operators and wave-functions (states). In order to avoid it, let us consider the interaction picture. Then In- and Out-states are the same lists of possible states. Any trivial or non-interacting theory corresponds to zero (or constant) interaction operator.

There is no physical reasons to imply that In- and Out-states are different. The Fock space, as a basis, is the same. What evolves in course of interaction is coefficients, amplitudes, or populations of states with certain number of particles (S-matrix).

If you speak of difference of In- and Out-states due to renormalizations, dressing, etc., it is another question: In-particles are bare, Out-particles are real in current theories. Do you speak of this?

 

[meopemuk]

I don't understand this.
What states are the creation and annihilation operators creating and annihilating then? A two particle state at t=\infty is not the same state as a two particle state at t=-\infty. So two different operators are necessary to create them from the vacuum.

Which time corresponds to the plane wave N\exp(ipx)? You cannot assign any time label to it. This is just a single state vector in the Hilbert space. Creation operators create only such plane wave states (or their linear combinations) without any reference to time. You are right when you say: "A two particle state at t=\infty is not the same state as a two particle state at t=-\infty". This is simply because the state vector created at t=\infty has evolved through time into another state vector. But both "infinite past" and "infinite future" state vectors can be build by applying (linear combinations of) the same creation operators to the same vacuum vector.

In/Out states form two completely different basis of the Hilbert space which are related to each other through a unitary transformation known as the S-matrix. So if what you were saying is true, then the S-matrix would be the identity and the theory would be trivial.
This isn't even an aspect of rigorous field theory, it's in usual field theory textbooks like Srednicki.
If you object to this aspect of QFT, you are basically saying all theories are free.

I've never understood what's the idea about separation of the in and out states. Suppose that I've shown you a wave function or a vector in the Hilbert space. Would you be able to tell if this is an "in" state or an "out" state? Or something else? They all look pretty similar to me. The difference is that in "in" states particles are moving toward each other (before the collision), while in "out" states particles are moving away from each other (after the collision). But nobody can forbid me to prepare a state in which particles are moving away from each other at time t = - \infty.

Excluding some pathological cases, the S-matrix is not trivial as long as there is a non-trivial interaction between particles. So, most interacting theories are not free.

Eugene.

 

[Bob_for_short]

I completely agree with Eugene.

Scattering result is just a redistribution of the initial energy/momentum over another population of states in the same basis. This was meant at the beginning of QED and other QFTs, and this has been finally achieved after renormalizations. The initial and final particles are of the same nature.

 

[DarMM]

Which time corresponds to the plane wave N\exp(ipx)? You cannot assign any time label to it. This is just a single state vector in the Hilbert space. Creation operators create only such plane wave states (or their linear combinations) without any reference to time. You are right when you say: "A two particle state at t=\infty is not the same state as a two particle state at t=-\infty". This is simply because the state vector created at t=\infty has evolved through time into another state vector. But both "infinite past" and "infinite future" state vectors can be build by applying (linear combinations of) the same creation operators to the same vacuum vector.

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple. Plus an in state has different expectation values for certain observables than an out state, so their difference can be characterised.

You guys have ideas on scattering theory that I've never really seen anywhere else. I mean the LSZ formalism and its more rigorous version of Haag-Ruelle theory uses the fact that the in and out basis are different. Srednicki uses it in his introductory textbook, as does Weinberg.

I don't really understand what is objectionable about there being different creation and annihilation operators for in and out states. It doesn't imply they live in a different Hilbert space. It's just like having ladder operators for the harmonic oscillators states and having different ladder operators for the anharmonic oscillators states. States of one can still be written as a superposition of the others. It's just a matter of them being different states, so they have different operators to create them. Sure using a combination of in-creation operators you can create an out state and vice-versa, but they are still different operators.

Do you guys object to this? If so why? I must admit I don't fully understand since it is in most QFT books.

 

[Bob_for_short]

We can use plane waves in the In-states and bound states in Out-states if we describe, for example, recombination an other reactions, no problem.

We do not understand how you distinguish In- and Out-electrons if they are free in both states. Similarly for photons.

Question of mine: does this necessity come from renormalizations? Use QED as an example, please. (I am afraid this "difference" of operators has no physical meaning and is explained exclusively with difficulties in constructing a working interaction model.)

 

[DarMM]

We do not understand how you distinguish In- and Out-electrons if they are free in both states. Similarly for photons.

Question of mine: does this necessity come from renormalizations? Use QED as an example, please. (I am afraid this "difference" of operators has no physical meaning and is explained exclusively with difficulties in constructing a working interaction model.)

Alright maybe the misunderstanding is the picture being used. Surely you can see that in the Heisenberg picture they must different states. In the Schrodinger picture where states are labeled only by their properties and not their properties at a time, there is of course no real difference because the language of in/out states would not arise.

Surely you can see how, in the Heisenberg picture, in/out states with the same properties are different states and hence would need different creation and annihilation operators?

Now if you guys would prefer I carry out the argument in the Schrodinger picture, that's fine. Although it is a little less natural for QFT. I'm assuming this is the only problem and that you do not disagree with the difference of in/out states in the Heisenberg picture.

 

[Bob_for_short]

In the Heisenberg picture the operators depend on time. It is like momentum dependence on time in CM p(t): in the initial state it is p, in the final one it is p'. Do you speak of this "difference"? It is not a problem in question. At least for me. What is a principal difference in QFT (QED)?

 

[meopemuk]

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple.

This is exactly the source of our disagreement. I hold the opinion that there is no fundamental difference between QM and QFT. The only difference is that interaction in QFT can change the number of particles, while in QM this number is fixed. If I had Weinberg's book with me right now, I could find a quote there to support this view.

 

[Bob_for_short]

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple.

It is not necessary to write down the exact wave-functions in QFT. It is sufficient to choose a good initial/final approximations taking into account an essential part of interaction. The interaction reminder can be considered perturbatively, in my humble but not groundless opinion.

 

[meopemuk]

Hi DarMM,

I think I understood which in-out operators you are talking about. Let me explain it in my own words and see whether you'd agree or not.

Let me define a set of creation operators a^{\dag}_{in}(p, -\infty). These operator create from vacuum 1-particle plane waves at time t = -\infty. If we apply the interacting time evolution operator to these plane waves, then we'll find that at time t = +\infty they do not look like plane waves at all. This is because scattering has occurred at times around zero. These time-evolved states can be obtained by using time-evolved in-creation operators

a^{\dag}_{in}(p, +\infty) = \exp(-iHT) a^{\dag}_{in}(p, -\infty) \exp(iHt)

where H is the total Hamiltonian and by T I formaly denoted an infinite time interval between t = -\infty and t = +\infty.

As I said, these new creation operators do not create plane wave states at t = +\infty. If we still want to create plane wave states at t = +\infty, we can define a new set of creation operators a^{\dag}_{out}(p, +\infty).

Both sets of operators a^{\dag}_{in}(p, -\infty) and a^{\dag}_{out}(p, +\infty) create the same plane wave states. However the former set creates them in the infinite past, while the latter set creates them in the infinite future.

Operators a^{\dag}_{in}(p, +\infty) can be expressed as linear combinations of a^{\dag}_{out}(p, +\infty). The expansion coefficients are elements of the S-matrix.

I think I can agree that scattering theory can be built on these principles, though this is not my favorite approach. Returning to our previous discussion, I would like to say that both in- and out-operators live in the same Hilbert space, and I still don't see a good reason why interacting and non-interacting theories must inhabit different Hilbert spaces.

Eugene.

 

[DarMM]

This is exactly the source of our disagreement. I hold the opinion that there is no fundamental difference between QM and QFT. The only difference is that interaction in QFT can change the number of particles, while in QM this number is fixed. If I had Weinberg's book with me right now, I could find a quote there to support this view.

Yes, but in QFT it is almost impossible to write the wavefunctions down since they are square integrable functions on an infinite dimensional space. Only in a few simple cases can you actually write down what the wavefunction is, which is why I said "explicitly right down wavefunctions". I wouldn't see this as a source of disagreement, since it isn't really related to what we are discussing and surely, for instance, you couldn't write down the wavefunction \psi\left[A_{\mu},t\right) for a proton. Yes there is no "fundamental" difference in some sense, but QFT is a lot harder, especially with regard to getting explicit results. Of course there are other ways in which QFT is quite different from QM, the presence of Lorentz invariance and locality imply strong analytic constraints on correlation functions in QFT that aren't there in QM. For instance one obtains crossing symmetry and various other properties of the S-matrix, where as in QM the S-matrix can be much more general. Also one has the spin and statistics theorem, e.t.c.

I'm sure Weinberg would agree, since he used to say that imposing analyticity, locality, e.t.c. and keeping the rules of quantum theory is literally what QFT is. Of course since QFT is a quantum theory it will have a lot in common with nonrelativistic QM, but it is also very different in some ways.

I think I can agree that scattering theory can be built on these principles, though this is not my favorite approach. Returning to our previous discussion, I would like to say that both in- and out-operators live in the same Hilbert space, and I still don't see a good reason why interacting and non-interacting theories must inhabit different Hilbert spaces.

Well the creation and annihilation operators weren't supposed to explain why, they were merely meant to be a step on the way there. However I can see that scattering is probably not the way to go, considering the difficulty in agreeing on the basics of the subject. So instead I'll write a post concerning an exactly soluable model in four dimensions where the presence of different Hilbert spaces becomes obvious, perhaps that will clear things up.

 

[meopemuk]

Yes, but in QFT it is almost impossible to write the wavefunctions down since they are square integrable functions on an infinite dimensional space. Only in a few simple cases can you actually write down what the wavefunction is, which is why I said "explicitly right down wavefunctions". I wouldn't see this as a source of disagreement, since it isn't really related to what we are discussing and surely, for instance, you couldn't write down the wavefunction \psi\left[A_{\mu},t\right) for a proton. Yes there is no "fundamental" difference in some sense, but QFT is a lot harder, especially with regard to getting explicit results. Of course there are other ways in which QFT is quite different from QM, the presence of Lorentz invariance and locality imply strong analytic constraints on correlation functions in QFT that aren't there in QM. For instance one obtains crossing symmetry and various other properties of the S-matrix, where as in QM the S-matrix can be much more general. Also one has the spin and statistics theorem, e.t.c.

I'm sure Weinberg would agree, since he used to say that imposing analyticity, locality, e.t.c. and keeping the rules of quantum theory is literally what QFT is. Of course since QFT is a quantum theory it will have a lot in common with nonrelativistic QM, but it is also very different in some ways.

I am glad that you've mentioned Weinberg, because his book presents the only approach to QFT that makes sense to me.

On page 49 he writes:

"First, some good news: quantum field theory is based on the same quantum mechanics that was invented by Schroedinger,..."

I don't think he ever considers the necessity to introduce a special Hilbert space for interacting theory. He also does not question the validity of Hermitian particle number operators. Although, I should admit that he doesn't bother to describe in any detail the Hilbert space he is working in. (the term "Fock space" is not even in the index). However, my guess is that he is doing "quantum mechanics in the Fock space", as I described earlier.

The crucial point of Weinberg's approach is indicated on page 200, where he expalins how it is different from the traditional QFT:

"Traditionally in quantum field theory one begins with such field equations or with the Lagrangian from which they are derived, and then uses them to derive the expansion of the fields in terms of one-particle annihilation and creation operators. In the approach followed here, we start with the particles, and derive the fields according to the dictates of Lorentz invariance, with the field equations arising almost accidentally as a byproduct of this construction."

On page 191 he explains why he needs to consider fields at all. The physical requirements of Poincare invariance and cluster separability are key here. Somewhat earlier he showed that interacting theory can be made Poincare-invariant if the interaction operator is represented as a space integral of "interaction density", which is a scalar with respect to the non-interacting representation of the Poincare group. The easiest way to construct such a scalar is to make a product of N "quantum fields". In Weinberg's interpretation, quantum fields appear as auxiliary technical constructs, whose only role is to ensure the Poincare invariance of interaction. Another benefit is that building interactions as products of quantum fields ensures cluster separability (or, as Weinberg calls it - the cluster decomposition principle). The obvious downside of such field theories is the inevitable need for renormalization: even single particle experiences unphysical self-interaction and self-scattering.

Note that Weinberg cannot prove that building interactions from fields is the only possible way to provide Poincare invariance, cluster separability, and changing number of particles. As an example of a failed alternative approach he mentions Bakamjian-Thomas theory. However, he doesn't tell that there are recent developments of this theory (see works by W.N. Polyzou), which show progress in both cluster separability and particle-number-changing interactions.

Weinberg doesn't mention yet another alternative - the "dressed particle" approach - which can easily accommodate Poincare invariance, cluster separability, and particle-number-changing interactions. Its main advantage is that renormalization is never needed. I consider this as a major sign of physical/mathematical consistency. The main disadvantage was mentioned by strangerep in one of recent threads: Since the "dressed particle" approach does not use fields, it does not accommodate the idea of "gauge invariance", so, unlike in QFT, there is no (relatively) simple and regular way to construct interaction Hamiltonians that immediately match observations. Currently we must either fit "dressed" Hamiltonians to experiment or use complicated tricks, like "unitary dressing". Though, this may be not a high price for getting rid of renormalization and divergences.

Eugene.

 

[Bob_for_short]

... the "dressed particle" approach - which can easily accommodate Poincare invariance, cluster separability, and particle-number-changing interactions. Its main advantage is that renormalization is never needed. I consider this as a major sign of physical/mathematical consistency. The main disadvantage was mentioned by strangerep in one of recent threads: Since the "dressed particle" approach does not use fields, it does not accommodate the idea of "gauge invariance", so, unlike in QFT, there is no (relatively) simple and regular way to construct interaction Hamiltonians that immediately match observations.

I do not remember mentioning such a "disadvantage" by Strangerep. I don not consider this as a "disadvantage". On the contrary, gauge-invariant formulation that uses tensions (transverse polarizations) involves only physical degrees of freedom that can carry and exchange the energy-momentum with charges.

IMHO there is a simple way to construct interaction Hamiltonians in the "dressed" particle approach but I am not sure of its unambiguity. Anyway, the final word is after experiment. There may be many trial Hamiltonians to test against the experimental data.

 

[meopemuk]

I do not remember mentioning such a "disadvantage" by Strangerep.

Post #66 in https://www.physicsforums.com/showthread.php?t=348911&page=5

He didn't mention local gauge invariance used as a tool for formulating interacting QFT Hamiltonians, but I guess he meant exactly this, because AFAIK currently there is no other means to obtain realistic interactions.

Eugene.

 

[strangerep]

Post #66 in https://www.physicsforums.com/showthread.php?t=348911&page=5

He didn't mention local gauge invariance used as a tool for formulating interacting QFT Hamiltonians, but I guess he meant exactly this, because AFAIK currently there is no other means to obtain realistic interactions.

No, I wasn't thinking about local gauge invariance in that post.
(You're probably on safer ground if you don't put words in my mouth. :-)

But let us not derail DarMM's thread off onto such tangents. I'm very interested to
discuss the exactly solvable 4D model he promises to write a post about...

 

[DrFaustus]

meopemuk -> There is one crucial difference between QM and QFT, which is related to the question of Hilbert spaces, the number of degrees of freedom. In particular, I'm talking about the Stone - Von Neumann theorem (see Wald - QFT in Curved Spacetime and Black Hole Thermodynamics, or simply Wikipedia). In short, the theorem states that in the case of finitely many degrees of freedom, all the representations of Weyl relations, i.e. of the canonical commutation relations, are unitarlity equivalent. And this is the reason why for QM you don't need to consider any other Hilbert space than the usual L^2(R). But this is not so in QFT, when the theorem breaks down. And there are, in fact, infinitely many representations of the CCR relations that are unitarily inequivalent (!), hence yielding different physical predictions. And one has to use physics to pick the appropriate Hilbert space. In other words, having different Hilbert spaces is ordinary in QFT. This is why you cannot draw consequences about QFT starting from QM. At best you can seek guidance, but you cannot, in general, just adopt arguments from QM to QFT.

An immediate example of the necessity of a different Hilbert space comes from QFT at finite temperature. A KMS state (or finite temperature state) is in no sense an excitation of the vacuum, and hence does not inhabit Fock space. An example of a KMS state is quark gluon plasma at some temperature. You cannot describe this state in any way by using states of Fock space. And this is only the tip of the iceberg. I think I've read somewhere that multiparticle states are a set of measure zero in the set of all the (physically relevant) states. And this brings me to the next point.

Weinberg does like his particles indeed. And sure in Minkowski spacetime, while dealing with particle scattering it leads to a consistent theory, this is not so when considering a generic Lorentzian spacetime. The very concept of particle is not at all clear. Even by staying in Minkowski space, and attempting to quantize a field in Rindler space (a wedge of Minkowski space), one ends up with a different particle interpretation. (See again book by Wald - chapter 5.1, The Unruh effect in flat space - or S.Fulling - Nonuniqueness of canonical field quantization in Riemannian spacetime, Phys.Rev.D 7, (1973)). The particle aspects of QFT are just a part of what QFT is. Sure, you can start with particles and eventually arrive to fields. But it is conceptually more elegant to start from a generic concept, the field, that has validity beyond the particle aspect.

When you talk about the dressed particle approach, what I'd like to know is how does such an approach cope with field theory at finite temperature. Or in condensed matter. Or on curved spaces. (In particluar I'd like to know about the latter as on a generic curved background you will not even be able to write down a mode expansion. And if you define some creation and annihilation operators you will still need to tell me why is that specific choice the physically correct one, given that there is no preferred vacuum state. Actually, there is no vacuum state at all.) Oh, and I don't want to turn this into a discussion about the dressed particle approach... or any other approach for that matter!

My bottom line is basically the following. One, you cannot draw parallels between QM and QFT because there's a world of difference between the two when it comes to the formal, mathematical aspects of the theories. Two, there is MUCH MORE to QFT than just particle scattering in Minkowski space. Finite temperature states, finite density states, QFT on curved spaces are just some examples. And any approach at modifying QFT will have to be able to deal with at least these scenarios, as QFT does. And each of these requires a different Hilbert space.

(I know this is not an argument that free theory andinteracting theory live in different Hilbert spaces, but just a suggestion that one should not be too surprised about it. Waiting for DarMM's post...)