Rigorous Quantum Field Theory.

[strangerep]

QFT1 is taught in most QFT textbooks (excluding the Weinberg's one). [...]
we build the Fock space by applying creation operators to the vacuum many times, etc. etc. In this approach, it is not clear whether interacting theory lives in the same Hilbert (Fock) space as the non-interacting one. Are the creation/annihilation operators of the two theories different? Do they have different particle number operators? Possibly, it makes sense to build the interacting Hilbert space as a representation space of canonical (anti)commutation relations? Due to the "infinite number of degrees of freedom", it might even happen that mathematically the two Hilbert spaces are different or inequivalent.

On the other hand, QFT2 is a version presented in Weinberg's book. It postulates particles as primary physical objects, and uses quantum mechanics from the beginning (so, no need for "quantization"). First, the Hilbert (Fock) space is build as a direct sum of N-particle spaces. Then creation and annihilation operators are explicitly defined in this Fock space. The next step is to define dynamics (= an unitary representation of the Poincare group) in the Fock space. The non-interacting representation can be constructed trivially (as a direct sum of N-tensor products of irreducible representations). The difficult part is to define an interacting representation of the Poincare group, which satisfies cluster separability and permits changes in the number of particles. This is the place where quantum fields (= certain formal linear combinations of creation and annihilation operators) come handy. We simply notice that if the interaction Hamiltonian (= the generator of time translations) and interacting boost operators are build as integrals of products of fields at the same "space-time points", then all physical conditions are satisfied automatically. In this approach, there is no need to worry about different Hilbert spaces for the non-interacting and interacting theories. [...]

The Hilbert space of "QFT2" is still infinite-dimensional -- so any mathematical issues arising
as a consequence of infinite degrees of freedom still lurk here, just as they lurk in "QFT1".

 

[meopemuk]

The Hilbert space of "QFT2" is still infinite-dimensional -- so any mathematical issues arising
as a consequence of infinite degrees of freedom still lurk here, just as they lurk in "QFT1".

I agree with that, but the Hilbert space of QFT2 is built *before* any interaction is introduced. So, the same Hilbert space is used in both non-interacting and interacting theories.

 

[strangerep]

[...] the Hilbert space of QFT2 is built *before* any interaction is introduced. So, the same Hilbert space is used in both non-interacting and interacting theories.

Except that the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations of them
are also operator-valued distributions. Hence QFT2 suffers exactly the same "ill-defined
equal-point multiplication of distributions" mathematical problem as QFT1.

 

[meopemuk]

Except that the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations of them
are also operator-valued distributions. Hence QFT2 suffers exactly the same "ill-defined
equal-point multiplication of distributions" mathematical problem as QFT1.

Could you give an example in which a product of a/c operators or quantum fields is "ill-defined"?

 

[Bob_for_short]

I think Strangerep speaks of loops in practical calculations, not of something different.

I would say the loop expressions are not "ill-defined" but simply divergent. There are many "cut-off" approaches to make them temporarily finite but they are just infinite as they should be.

 

[DarMM]

Okay, here is a model which is exactly solvable nonperturbatively and is under complete analytic control. Also virtually every aspect of this model is understood mathematically.

Now in the is model, the field does not transform covariantly. The reason I'm using the model is to show that even with this property removed there are still different Hilbert spaces.

Firstly, the model is commonly known as the external field problem. It involves a massive scalar quantum field interacting with an external static field.

The equations of motion are:
\left( \Box + m^{2} \right)\phi\left(x\right) = gj\left(x\right)

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
H = \int{dk \omega(k)a^{*}(k)a(k)}
Where a^{*}(k),a(k) are the creation and annihilation operators for the Fock space particles.

In order for this to describe the local interactions with an external source, I would modify the Hamiltonian to be:
H = \int{dk \omega(k)a^{*}(k)a(k)} + \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) + a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)}

So far, so good.

Now the normal mode creation and annihilation operators for this Hamiltonian are:
A(k) = a(k) + \frac{g}{(2\pi)^{3/2}}\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}
A short calculation will show you that these operators have different commutation relations to the usual commutation relations.

They bring the Hamiltonian into the form:
H = \int{dk E(k)A^{*}(k)A(k)},
where E(k) is a function describing the eigenspectrum of the full Hamiltonian.

Now, if you use Rayleigh-Schrödinger perturbation theory you obtain the interacting ground state as a superposition of free states:
\Omega = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!} - \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0}.
Where \Psi_{0} is the free vacuum.
Also Z = exp\left[\int{\frac{g^{2}}{(2\pi)^{3}}\frac{|\tilde{j}(k)|^{2}}{\sqrt{2}\omega(k)^{3}}}\right]

Now for a field weak enough that:
\frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})
then everything is fine. I'll call this condition (1).

However if this condition is violated, by a strong external field, then we have some problems.

First of all A(k), A^{*}(k) are just creation and annihilation operators. They have a different commutation relations, but essentially I can still use them to create a Fock basis, since I can prove the exists a Hilbert space with a state annihilated by all A(k). Now this constructed Fock space always exists, no problem. Let's call this Fock space \mathcal{F}_{I}.

However, if condition (1) is violated something interesting happens. Z vanishes. Now the expansion for \Omega is a sum of terms expressing the overlap of \Omega with free states. If Z=0, then \Omega has no overlap with and hence is orthogonal to all free states. This can be shown for any interacting state.
So every single state in \mathcal{F}_{I} is completely orthognal to all states in \mathcal{F}, the free Fock space. Hence the two Hilbert spaces are disjoint.

So the Fock space for a(k),a^{*}(k) is not the same Hilbert space as the Fock space for A(k), A^{*}(k). They are still both Fock spaces, however A(k), A^{*}(k) has a different algebra, so it's the Fock representation of a new algebra. If one wanted to still use the a(k),a^{*}(k) and their algebra, you would need to use a non-Fock rep in order to be in the correct Hilbert space.

This is what a meant by my previous comment:
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.

I hope this post helps.

 

[DarMM]

I also want to say that in \mathcal{F}_{I}, the Hamiltonian and S-matrix are both finite. So one has unitary evolution in this space.

For mathematical literature on this model:
Reed, M. and Simon, B. Methods of Modern Mathematical Physics, Vols. II-III, New York: Academic Press.
Wightman's article in Partial Differential Equations edited by D. Spencer, Symposium in Pure Mathematics (American Mathematical Society, Providence), Vol. 23.

 

[Bob_for_short]

Thank you, DarMM, for this example. Before reading it, I would like to know if it is very different from "radiation of classical current", i.e., from coherent states (if m=0 and the field is the quantized EMF)? Or it is akin to the coherent states?

 

[DarMM]

What's not clear to me is whether rigorous QFT is really all about taming this uncountable
infinity of inequivalent representations, or something else entirely.

Rigorous QFT divides into three areas:

Axiomatic Field Theory
This is basically, as Dr. Faustus said, Functional analysis with the Wightman axioms on top. One tries to understand what type of mathematical object quantum fields are and what conditions they should obey, either in canonical or path integral form. This has been accomplished by Wightman, Osterwalder, Schrader, Frohlich, Nelson, Symanzik and others. Given these conditions (axioms), you then try to figure out properties of the quantum fields, such as:
Analyticity of the vertex functions.
Poles in correlation functions.
The connection between spin and statistics.
e.t.c.

Basically the study of what mathematical objects fields are and the consequences of this.

Algebraic Field Theory
Here one is being very general and simply concentrates on the properties of local quantum theories in Minkowski or other spacetimes, not specifically requiring there to be fields involved. This area can be seen as working out the general physical consequences of quantum theory and relativity. For instance this area provides the simplest treatment of Bell's inequalities in general spacetimes. It is this area that really tries to understand the uncountable infinity of inequivalent representations, since it is trying to understand the very general implications of quantum theory and relativity.

You could see Axiomatic field theory as a subset of Algebraic field theory, which focuses on Minkowski spacetime and assumes the theory is a field theory and is non-thermal, which leads to more detailed properties. However because of fields being involved, the mathematics and the focus of the two areas tend to be quite different.

For instance an algebraic field theory question might be "What are the general characteristics of thermal states that separate them from pure states? What are the different representations in which they live? Can we characterise them?"
Axiomatic Field Theory would ask "Where are the poles in the three-point vertex function? What physical information is contained in these poles?"

Constructive Field Theory
This area essentially tries to prove that the field theories that physicists work with actually belong to the catagories above. For instance in Axiomatic Field Theory you have the Wightman axioms, but how do you know there exists any mathematical object that actually satisfies them? Constructive field theory builds (constructs) such objects by nonperturbatively controlling actual quantum field theories.

So constructive field theory would ask something like:
"Does \phi^{4} in two dimensions exist as a well-defined mathematical entity?, If it does exist, does it satisfy the axioms from axiomatic field?"

 

[DarMM]

Thank you, DarMM, for this example. Before reading it, I would like to know if it is very different from "radiation of classical current", i.e., from coherent states (if m=0 and the field is the quantized EMF)? Or it is akin to the coherent states?

If I had choosen the scalar field to be massless there would be coherent states. I haven't choosen this specifically to avoid dealing with that issue. So the issues dealt with are quite different.

 

[Bob_for_short]

If I had choosen the scalar field to be massless there would be coherent states. I haven't choosen this specifically to avoid dealing with that issue. So the issues dealt with are quite different.

No, there is no any issue in case of quantized EMF radiated with a classical current (source).

In your case there is just a threshold for massive quanta but the external current can be sufficiently powerful to radiate even massive quanta, so I see it as a quite akin problem.

What energy can absorb/emit new quanta corresponding to operators A(k)? (You missed dk in two integrals but it is not essential.)

 

[DarMM]

No, there is no any issue in case of quantized EMF radiated with a classical current (source).

In your case there is just a threshold for massive quanta but the external current can be sufficiently powerful to radiate even massive quanta, so I see it as a quite akin problem.

Okay, but it isn't an akin problem. Having a mass gap to the emitted quanta fundamentally changes things. If I had massless quanta there would be two features, change of Hilbert space and coherent states. Since the quanta are massive there are no coherent states. A problem with no coherent states is not akin to a problem with coherent states.
Unless you are saying they are akin simply because in both the external field can create quanta, however that's no different than saying they are both external field problems. The main focus here is the change in Hilbert space, not coherent states or anything like them, because they are not present.

What energy can absorb/emit new quanta corresponding to operators A(k)?

I'm not sure what you are asking.

 

[Bob_for_short]

..A problem with no coherent states is not akin to a problem with coherent states. Unless you are saying they are akin simply because in both the external field can create quanta, however that's no different than saying they are both external field problems.

I see it as a problem with an external source, not a problem in an external filed. If j(x) is a known function, then I have an exact solution.

The old quanta are physical - they exchange with energy-momentum corresponding to experimental data (in case of photons). To what do correspond new quanta, if any?

 

[DarMM]

I see it as a problem with an external source, not a problem in an external filed.

What's the difference?

If j(x) is a known function, then I have an exact solutions.

So do I, so has everybody for the last fifty years. They are written above or are to be found in Reed and Simons books.

The old quanta are physical - they exchange with energy-momentum corresponding to experimental data (in case of photons). To what correspond new quanta, if any?

They are energy eigenstates, the eigenstates of the new Hamiltonian. That's why they diagonalize it, just like regular QM. The old quanta do not diagonalize the new Hamiltonian and so are not energy eigenstates. Hence they do not correspond to the energy measured in experiments. Just like regular QM.

 

[Fredrik]

First of all A(k), A^{*}(k) are just creation and annihilation operators. They have a different commutation relations, but essentially I can still use them to create a Fock basis, since I can prove the exists a Hilbert space with a state annihilated by all A(k). Now this constructed Fock space always exists, no problem. Let's call this Fock space \mathcal{F}_{I}.
...
So every single state in \mathcal{F}_{I} is completely orthognal to all states in \mathcal{F}, the free Fock space. Hence the two Hilbert spaces are disjoint.

You said that you're using "QFT2", which means that you start with a Hilbert space \mathcal H of one-particle states. That Hilbert space is the representation space of a massive spin-0 irreducible representation. The Fock space \mathcal F is (I assume) the Fock space constructed from \mathcal H. So you must be using that to define a(k). And then you define A(k)=a(k)+f(k)I where f is a complex-valued function and I is the identity operator. This means that A(k) is just another operator on \mathcal F.

How can any argument you make after that lead to the conclusion that A(k) is an operator on a Hilbert space that's disjoint with \mathcal F?

 

[Bob_for_short]

What's the difference?

An external source (current) emits photons in coherent states. The energy is not defined - there is no an eigenstate but a superposition (coherent states). There is an average energy and its dispersion, etc., since the number of photons is not certain but the wave phase is.

An external filed is understood for a particle, not for photons. It can lead to bound states different from free states. So the difference between these two cases is obvious - the subjects are different.

They are energy eigenstates, the eigenstates of the new Hamiltonian. That's why they diagonalize it, just like regular QM. The old quanta do not diagonalize the new Hamiltonian and so are not energy eigenstates. Hence they do not correspond to the energy measured in experiments. Just like regular QM.

A classical current (antenna) emits uncertain number of photons. It is not an eigentsate and it is known. It does not prevent the emitted photons from being the true quanta with E=hf.

As soon as your, new quanta, correspond to a certain energy (do they?) there is an obvious contradiction with the exact solution. That is why I am asking about them.

EDIT: We can take an exact solution in terms of old operators, make the variable change and express it in terms of the new operators. If the new operators correspond to new eigenstates (let us admit it for instance), then the original solution will be a superposition of new eigenstates too, so the solution is not transformed in a state with a certain energy (a pure eigenstate).

An analogue of such a variable change exists already in the usual QED if you transform the "linearly polarized" c/a operators into "circularly polarized" ones. The original solution remains a coherent state with uncertain energy.

 

[DarMM]

You said that you're using "QFT2", which means that you start with a Hilbert space \mathcal H of one-particle states. That Hilbert space is the representation space of a massive spin-0 irreducible representation. The Fock space \mathcal F is (I assume) the Fock space constructed from \mathcal H. So you must be using that to define a(k). And then you define A(k)=a(k)+f(k)I where f is a complex-valued function and I is the identity operator. This means that A(k) is just another operator on \mathcal F.

How can any argument you make after that lead to the conclusion that A(k) is an operator on a Hilbert space that's disjoint with \mathcal F?

Good question. Basically by staying in the abstract algebra of operators at all times. I have the operators for creating free massive spin-0 bosons, I can then express the "external-source" Hamiltonian as a function of these operators.
Of course I can also express it as a function of operators which diagonalize it.

So I have two different Hamiltonians, one diagonalized by one set of creation and annihilation and another by another set.
However I have not yet passed to a representation. If I do I can use the quantity Z to measure how disjoint the two representations are. If Z is non-zero, then the reps are unitarily equivalent, otherwise they are not.
Essentially A(k) is made a function of a(k), at the abstract operator level, before I pass to reps. I apoligise if this is unclear.

The reason this is an excellent question is that it leads naturally to the point of view of algebraic field theory. That is, the operators themselves and their algebra are fundamental. Things become clearer at the level of the operators in abstract, before we pass to a rep. This is an important insight in Rigorous Quantum Field Theory.

 

[DarMM]

An external source (current) emits photons in coherent states. The energy is not defined - there is no an eigenstate but a superposition (coherent states). There is an average energy and its dispersion, etc., since the number of photons is not certain but the wave phase is.

I'm not talking about photons. My example above discusses massive spin-0 bosons, not massless spin-1 bosons. The problems you are talking about concern massless particles. I'm aware of these problems, but I am not discussing massless particles, hence there is no need for coherent states.

 

[Bob_for_short]

I'm not talking about photons. My example above discusses massive spin-0 bosons, not massless spin-1 bosons. The problems you are talking about concern massless particles. I'm aware of these problems, but I am not discussing massless particles, hence there is no need for coherent states.

First of all there is no problem in case of the coherent radiation in QED, so I do not impose " a problem" to your model. Next, if your source (current j(x)) is sufficiently energetic (ω>>mc²), your case is not different from a massless case. That is why I have not noticed any particular difference in physics in these both cases.

 

[Bob_for_short]

Rigorous QFT divides into three areas:

Axiomatic Field Theory
...you then try to figure out properties of the quantum fields, such as:
Analyticity of the vertex functions. Poles in correlation functions. The connection between spin and statistics. e.t.c.

Basically the study of what mathematical objects fields are and the consequences of this.

Algebraic Field Theory
Here one is being very general and simply concentrates on the properties of local quantum theories in Minkowski or other spacetimes...

...Axiomatic Field Theory would ask "Where are the poles in the three-point vertex function? What physical information is contained in these poles?"

Constructive Field Theory
This area essentially tries to prove that the field theories that physicists work with actually belong to the catagories above. ...

So constructive field theory would ask something like:
"Does \phi^{4} in two dimensions exist as a well-defined mathematical entity?, If it does exist, does it satisfy the axioms from axiomatic field?"

Here a certain interaction Lagrangian/Hamiltonian is implcitely implied and this fact is crucial. It is not about "the most general" properties of fields but in certain frames imposed with the interaction term. An the the corresponding properties are crucially dependent on this term, on its physics. \phi^{4} ot jA imply a self-action.

 

[DarMM]

First of all there is no problem in case of the coherent radiation in QED, so I do not impose " a problem" to your model. Next, if your source (current j(x)) is sufficiently energetic (ω>>mc²), your case is not different from a massless case. That is why I have not noticed any particular difference in physics in these both cases.

There are some similarities in that both models have particles being created by an external source, but with the presence of a mass gap there is the fundamental difference that there is no coherent states. If the source is strong enough to create particles, then yes particles will be emitted. However there is no soft particles here, so I don't have to treat infrared issues such as coherent states. This is a big difference, the physics of massive and massless particles are not similar even if both are being created by the same mechanism. This is the importance of the mass gap.

 

[DarMM]

Here a certain interaction Lagrangian/Hamiltonian is implcitely implied and this fact is crucial. It is not about "the most general" properties of fields but in certain frames imposed with the interaction term. An the the corresponding properties are crucially dependent on this term, on its physics. \phi^{4} ot jA imply a self-action.

In Axiomatic Quantum Field and Algebraic Quantum Field theory no interaction term is implied or any properties of the interaction, besides that it obey relativity.
In Constructive Field Theory a certain interaction is explicitly chosen, because you are trying to show a specific theory is well-defined. So by showing Yukawa Theory exists a Yukawa interaction is obviously chosen.

In none of the cases is it implict though, it's either not assumed (Axiomatic and Algebraic) or explicitly chosen (Constructive).

 

[Bob_for_short]

There are some similarities in that both models have particles being created by an external source, but with the presence of a mass gap there is the fundamental difference that there is no coherent states. If the source is strong enough to create particles, then yes particles will be emitted. However there is no soft particles here, so I don't have to treat infrared issues such as coherent states. This is a big difference, the physics of massive and massless particles are not similar even if both are being created by the same mechanism. This is the importance of the mass gap.

Fortunately there is no IR problem in QED description of a classical current radiations. The exact solution exists and is physically meaningfull. Your fears are groundless and the solutions are physically very similar. So in both cases we have an exact solution with incertain energy. This is clear and there is no problem with it at all.

As soon as A(k), A(k)⁺ have different commutation relationships, what states do they create? Are their "energy levels" equidistant, etc.?

 

[Bob_for_short]

In Axiomatic Quantum Field and Algebraic Quantum Field theory no interaction term is implied or any properties of the interaction, besides that it obey relativity.

But any vertex implies a certain interaction, doesn't it?

 

[DarMM]

Fortunately there is no IR problem in QED description of a classical current radiations. The exact solution exists and is physically meaningfull. Your fears are groundless and the solutions are physically very similar. So in both cases we have an exact solution with incertain energy. This is clear and there is no problem with it at all.

Okay, I'm not talking about QED. I'm talking about the simple model posted in message #66. I have no fears about QED and I'm aware that there aren't any problems with the infrared part of the theory since those problems are solved by coherent states. However what I'm talking about is not QED, so I'm not going to talk about those things. It's not that I think QED has a problem, it is simply that I'm not talking about it and have chosen a model where coherent states are absent. I understand what you are talking about, but it doesn't occur here.

As soon as A(k), A(k)⁺ have different commutation relationships, what states do they create? Are their "energy levels" equidistant, etc.?

They create the physical eigenstates, as I have said.

 

[DarMM]

But any vertex implies a certain interaction, doesn't it?

Sure, but they don't treat field theory using diagrams. Particularly Algebraic Field Theory. They just study consequences of conditions on the field themselves.

 

[Bob_for_short]

Sure, but they don't treat field theory using diagrams. Particularly Algebraic Field Theory. They just study consequences of conditions on the field themselves.

OK, for me everything is clear. My message is that we can construct a QFT in a rigorous way similar to atom-atomic interaction mentioned above, i.e., with everyhting physically meaningful.

Thanks, DarMM, for your posts and replies.

Regards,

Vladimir.

 

[meopemuk]

DarMM, thank you for the good example. It will allow us to discuss a few important issues.

In order for this to describe the local interactions with an external source, I would modify the Hamiltonian to be:
H = \int{dk \omega(k)a^{*}(k)a(k)} + \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) + a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)}

Please note that your interaction V \propto a^{*} + a is called "bad" in the language of "dressed particle" theory. This theory explicitly forbids using such "bad" interactions. If you had selected a "good" interaction, then you wouldn't get all the problems that you've described below. In particular, your "dressed particles" (eigenstates of the interacting Hamiltonian) would be no different from "bare" particles (eigenstates of the free Hamiltonian).

On the other hand, such "bad" interactions are used very often in QFT (they will appear every time you build interactions as products of quantum fields), and we need to discuss how to make sense of them.

Now the normal mode creation and annihilation operators for this Hamiltonian are:
A(k) = a(k) + \frac{g}{(2\pi)^{3/2}}\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}
A short calculation will show you that these operators have different commutation relations to the usual commutation relations.

They bring the Hamiltonian into the form:
H = \int{dk E(k)A^{*}(k)A(k)},
where E(k) is a function describing the eigenspectrum of the full Hamiltonian.

Now, if you use Rayleigh-Schrödinger perturbation theory you obtain the interacting ground state as a superposition of free states:
\Omega = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!} - \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0}.
Where \Psi_{0} is the free vacuum.
Also Z = exp\left[\int{\frac{g^{2}}{(2\pi)^{3}}\frac{|\tilde{j}(k)|^{2}}{\sqrt{2}\omega(k)^{3}}}\right]



Now for a field weak enough that:
\frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})
then everything is fine. I'll call this condition (1).

However if this condition is violated, by a strong external field, then we have some problems.

First of all A(k), A^{*}(k) are just creation and annihilation operators. They have a different commutation relations, but essentially I can still use them to create a Fock basis, since I can prove the exists a Hilbert space with a state annihilated by all A(k). Now this constructed Fock space always exists, no problem. Let's call this Fock space \mathcal{F}_{I}.

This is exactly what other people call a "dressing transformation". The new a/c operators A^{*}, A are said to create/annihilate "dressed" particles. One troublesome point is that your operators A^{*}, A do not satisfy usual commutation relations. I think this is unphysical. The fact that there exists a vacuum vector annihilated by all A(k) is not sufficient to declare that A^{*}, A are valid a/c operators. The canonical form of commutation relations is important as well. This form ensures that a/c operators behave as they supposed to do, i.e., change the number of particles in the system by \pm 1. This condition can be satisfied by using unitary "dressing transformation". Apparently, your "dressing transformation" is not unitary. However, this is not such a big deal, and does not affect the issue of "different Hilbert spaces". I think your transformation can be made unitary without much trouble, if needed.

However, if condition (1) is violated something interesting happens. Z vanishes. Now the expansion for \Omega is a sum of terms expressing the overlap of \Omega with free states. If Z=0, then \Omega has no overlap with and hence is orthogonal to all free states. This can be shown for any interacting state.
So every single state in \mathcal{F}_{I} is completely orthognal to all states in \mathcal{F}, the free Fock space. Hence the two Hilbert spaces are disjoint.

So the Fock space for a(k),a^{*}(k) is not the same Hilbert space as the Fock space for A(k), A^{*}(k). They are still both Fock spaces, however A(k), A^{*}(k) has a different algebra, so it's the Fock representation of a new algebra. If one wanted to still use the a(k),a^{*}(k) and their algebra, you would need to use a non-Fock rep in order to be in the correct Hilbert space.

I think we should be careful before claiming that a(k), a^{*}(k) and A(k), A^{*}(k) give us orthogonal Hilbert spaces. I can agree that expansion coefficients of "dressed" states wrt "bare" states are zero. But this does not mean that they belong to different Hilbert spaces. Let me explain what I mean on this simple example:

Consider simple 1-particle quantum mechanics. Eigenstates of the momentum operator are usual plane waves in the position representation. If we want these eigenstates to be normalized, we must multiply them by a normalization factor that is effectively zero (but not exactly zero!). So, it would be tempting to conclude that normalized plane waves are orthogonal to all "normal" states in the 1-particle Hilbert space. Do they belong to some other orthogonal Hilbert space? Some people try to resolve this problem by introducing "rigged Hilbert spaces", "Gelfand triples", etc. Personally, I don't like these ideas. My opinion is that we are using too narrow a definition of the Hilbert space. We should use a broader definition of Hilbert spaces, i.e., such that eigenvectors of unbounded operators (like momentum) can find their place there. This requirement is dictated to us by physics, and our math must follow physical requirements, not the other way around. I am not sure exactly what mathematics should be used for this purpose. Perhaps the "non-standard analysis" of A. Robinson could help, but my math skills are too weak to go there.

My guess is that in a properly defined "broad" or "non-standard" Fock space there should be enough place for both "bare" and "dressed" particles.

Eugene.

 

[meopemuk]

I think Strangerep speaks of loops in practical calculations, not of something different.

I would say the loop expressions are not "ill-defined" but simply divergent. There are many "cut-off" approaches to make them temporarily finite but they are just infinite as they should be.

I would like to avoid "putting words in strangerep's mouth" again. If you are right and the mentioned "ill-definiteness" is related to loops (as we discussed already in the post #53 in https://www.physicsforums.com/showthread.php?t=348911&page=4), then I have a few comments.

If our interaction Hamiltonian is constructed as a product of quantum fields, then when we calculate the S-matrix in perturbation theory we must evaluate products of such products. These terms lead inevitably to the appearance of non-zero (in most cases even infinite) loop contributions to vacuum->vacuum and 1-particle->1-particle scattering amplitudes. This "ill-definiteness" is exactly what is cured by the renormalization prescription. You may not like this "cooking recipe", but it works well as far as practical applications are concerned.

There is however a different way to deal with the "ill-definiteness" of the products of fields. Just make sure that products of fields (and associated "bad" terms) never appear in your interactions. If your interaction is built from "good" terms only, then there is no need for renormalization, and there are no divergences.

Eugene.

 

[Bob_for_short]

The equations of motion are:
\left( \Box + m^{2} \right)\phi\left(x\right) = gj\left(x\right) ...(1)

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
H_0 = \int{dk \omega(k)a^{*}(k)a(k)}
Where a^{*}(k),a(k) are the creation and annihilation operators for the Fock space particles.

In order for this to describe the local interactions with an external source, I would modify the Hamiltonian to be:

H = \int{dk \omega(k)a^{*}(k)a(k)} + \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) + a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)} ...(2)

So far, so good.

I did not verify what Hamiltonian corresponds to the equation (1). Is (2) a modification of (1) or just the total Hamiltonian?

If (2) does not correspond to (1), then (2) is a new problem with its own dynamics different from what I discussed above. Is it indeed?