Rigorous Quantum Field Theory.

[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?

 

[Bob_for_short]

I would like to avoid "putting words in strangerep's mouth" again.

I guess I am right. "Ill-defined" things occur in practical calculations and make them impossible. Otherwise nobody would care.

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.

I call "ill defined" expressions divergent, I am not shy. Renormalizations are discarding perturbative corrections to the masses and charge. (I am not shy to call the things as they are.) In most cases in QFT this prescription does not work.

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.

I know that. Imagine, in my Hamiltonian they come in such combinations that the vacuum and one-particle (electronium) states are stable.

 

[DarMM]

Please note that your interaction $V \propto a^{*} + a$ is called "bad" in the language of "dressed particle" theory.

Why is it bad?

In particular, your "dressed particles" (eigenstates of the interacting Hamiltonian) would be no different from "bare" particles (eigenstates of the free Hamiltonian).

I don't understand this, even in regular quantum mechanics the eigenstates of the interacting Hamiltonian are different to the eigenstates of the free Hamiltonian. I thought this would be easily agreed on.

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 $\pm 1$.

It doesn't matter. You can prove these operators have a Fock representation, hence there is a Hilbert space where they move the number of particles by $$\pm 1$$

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:

They are orthogonal to every single free state, so at the very list the Hilbert space is a direct sum,
$$\mathcal{F}_{I} \oplus \mathcal{F}$$. However you can show that such a thing is not true using representation theory, so they truly do live in different Hilbert spaces.

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.

Can I ask why Fock space is so important? In another thread you said that relativistic covariance could be given up to ensure that the interacting theory remained in Fock space. In this model we do give that property up and yet we still have a different Hilbert space. Now you want to change the concept of Hilbert space as used in QM, just so that we remain in Fock space? Why? It's just a Hilbert space, why change QM and relativity to avoid this fact?
In my opinion your doing the reverse of what you claim, giving up physics we know (no lorentz covariance, superluminal signaling) in order to keep a mathematical structure (Fock space).

 

[Bob_for_short]

And my questions about the new spectrum and the total Hamiltonian, please!

 

[meopemuk]

Please note that your interaction $V \propto a^{*} + a$ is called "bad" in the language of "dressed particle" theory.

Why is it bad?

By definition, "good" operators in the normally-ordered form (all annihilation operators on the right, all creation operators on the left) must have at least two creation operators and at least two annihilation operators. The simplest example of a "good" operator is $a^*a^*aa$. Their importance stems from the fact that they yield zero when acting on any 1-particle state or on the vacuum. Another important property is that the product or commutator of any number of "good" operators is also "good".

There is also a class of operators that I call "renorm". They are either $a^*a$ or multiplication by a constant. The free Hamiltonian is one example of a "renorm" operator.

In simple theories that we are discussing, all other operators are "bad". Your operator $V \propto a^{*} + a$ is "bad" in this classification. Its unpleasant property is that (normally ordered) products of such operators contain "renorm" terms. If $V$ happens to be in the interaction Hamiltonian, then the S-operator expansion $S \propto 1 + V + VV + VVV +...$ contains "renorm" terms (in addition to the first "1"), which signify the presence of self-interaction and self-scattering in the vacuum and 1-particle states. This leads immediately to the necessity of renormalization and other unpleasant effects.

On the other hand if interaction $V$ contained only "good" terms, then all multiple products of $V$ would yield zero when acting on the vacuum and 1-particle states, which agrees with the intuitive understanding that vacuum and single particle cannot scatter off themselves. There is no need for renormalization if $V$ is "good".

I don't understand this, even in regular quantum mechanics the eigenstates of the interacting Hamiltonian are different to the eigenstates of the free Hamiltonian. I thought this would be easily agreed on.

Yes, generally the interacting and free Hamiltonians have different eigenstates. However, the important issue is about the vacuum and 1-particle eigenstates. In your Hamiltonian $H_0 +a^* + a$, the interaction acts non-trivially on the free vacuum and 1-particle states. Therefore, 0-particle and 1-particle eigenstates of this Hamiltonian are different from 0-particle and 1-particle eigenstates of the free Hamiltonian. Your "dressed" particles are different from "bare" particles. This is why the (mass) renormalization is needed.

On the other hand, if I have a Hamiltonian with a "good" interaction, for example $H_0 + a^*a^*aa$, then its 0-particle and 1-particle eigenvectors are exactly the same as the free vacuum and free particles, respectively. There is no need for the (mass) renormalization. Both free and interacting theories live in the same Fock space. 0-particle and 1-particle sectors of the two theories are exactly the same. On the other hand, 2-particle, 3-particle, etc. sectors of the two theories are quite different. In the interacting theory, 2-particles are allowed to have a non-trivial scattering, form bound states, etc. This is consistent with the intuitive understanding that interactions can occur only if there are 2 or more particles. By choosing "good" interactions only we eliminate unphysical self-interactions in the vacuum and 1-particle states. At the same time we avoid unphysical renormalizations and divergences.

Can I ask why Fock space is so important? In another thread you said that relativistic covariance could be given up to ensure that the interacting theory remained in Fock space. In this model we do give that property up and yet we still have a different Hilbert space. Now you want to change the concept of Hilbert space as used in QM, just so that we remain in Fock space? Why? It's just a Hilbert space, why change QM and relativity to avoid this fact?
In my opinion your doing the reverse of what you claim, giving up physics we know (no lorentz covariance, superluminal signaling) in order to keep a mathematical structure (Fock space).

I like Fock space, because it is a natural consequence of two physically transparent statements:

(1) Particle numbers are valid physical observables (hence there are corresponding Hermitian operators);
(2) Numbers of particles of different types are compatible observables (hence their operators commute)

Please note that I make a distinction between "Lorentz covariance" and "relativistic invariance":

"Lorentz covariance" is the assumption that certain quantities (like space-time coordinates of events or components of quantum fields) have simple (e.g., linear) transformation properties with respect to boosts.

"Relativistic invariance" is the requirement that the theory is invariant with respect to the Poincare group.

In my opinion, the relativistic invariance is the most important physical law. It can be never compromised. On the other hand, Lorentz covariance cannot be derived from this law directly (some additional dubious assumptions are needed for this "derivation"). So, Lorentz covariance is just an approximate property, which makes sense for non-interacting or weakly-interacting systems only. Superluminal propagation of interactions is definitely in conflict with Lorentz covariance, but it is fully consistent with relativistic invariance. The conflict with causality is resolved by the fact that boost transformations of particle observables must be non-linear and interaction-dependent. This follows from the interaction-dependence of the total boost operators, as explained in Weinberg's book.

Note also that I don't want to modify quantum mechanics in any significant way. I just want to broaden the definiton of the Hilbert space, so that eigenvectors of unbounded operators (like momentum) can live there.

Eugene.

 

[strangerep]

Hmmm. So once again, while I slept, an avalanche of posts in this thread has
overwhelmed me. Although, curiously, a couple of people seemed to be
whispering in my dreams, debating about what I "really" meant but not actually
asking me. Quite bizarre. Anyway, you guys are probably all in bed by now,
so I can have a some peace to read carefully and respond... :-)

First, DarMM's posting of the "external field" example...

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.

I'm guessing that $\tilde{j}(k)}$ is a c-number, right? I.e., it commutes with everything?
(I'll proceed on this assumption, but if it's wrong, please tell me what commutation relations it satisfies.)

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.

OK, so we want to diagonalize the full Hamiltonian in terms of new a/c ops
$A^{*}(k),\,A(k)$, and in this case it's fairly easy to guess what
they are in terms of the free a/c ops. Let me re-write the 2nd-last formula
above in a simpler form:
$$A(k) ~=~ a(k) ~+~ z(k)$$
where (hopefully) the definition of my $z(k)$ is obvious.

Now, you said above that the $A(k)$ don't satisfy the usual commutation relations.
I don't understand this. If $z(k)$ commutes with the free a/c ops $a(k)$, etc,
then the $A(k)$ do satisfy the canonical commutation relations, afaict.
Or did I miss something?

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

Something looks wrong with that expression.
Should the minus sign be inside the parentheses?

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:
$$\frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3}) ~~~~~~~~(1)$$

then everything is fine. [...]

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

However, if condition (1) is violated something interesting happens.
$$Z$$ vanishes.

Did you forget a minus sign in your definition of Z above? It looks like
it goes to infinity rather than zero when condition (1) is violated.

(Or did you perhaps mean that $\Omega$ becomes non-normalizable when $Z\to\infty$?)

- - - - - - - - - - - - - - - - -

I'm also guessing that (after correcting any errors) the example is really
just the well-known so-called "field displacement" transformation, i.e.,
$$A^*(k) ~:=~ a^*(k) ~+~ g\,\bar{z}(k) ~~;~~~~~~ A(k) ~:=~ a(k) ~+~ g\,z(k)$$
which alternatively can be expressed as a formally-unitary transformation
$$A(k) ~:=~ U[z] \, a(k) \, U^{-1}[z]$$
where U is of the form
$$U[z] ~:=~ exp\int\!\!dk\Big( \bar{z}(k)a(k) - z(k)a^*(k) \Big)$$
(where possibly I might have a sign wrong.)
This U[z] is essentially equivalent to the operator you used to go from the free vacuum $\Psi_0$ to the interacting vacuum $\Omega$.

The alert reader may have noticed that the form of U[z] is exactly the
same as that which generates ordinary (Glauber) coherent states in the
inf-dof case. I might say more about that later, depending on what else
Bob_for_short wrote (and if he doesn't badger me about it).

The point is that states generated by U[z] acting on the free vacuum $\Psi_0$ are only in the free Fock space if $z(k)$ is square-integrable.
(I have some more detailed latex notes on this calculation that I could possibly post if anyone cares. :-)

For mathematical literature on this model:
Reed, M. and Simon, B. Methods of Modern Mathematical Physics, Vols. II-III

If you have those volumes handy, could you possibly give a more precise reference?

Rigorous QFT divides into three areas
[...]

Thanks. I see now that I'm mostly interested in algebraic-constructive
stuff. (I.e., constructing under the more general algebraic umbrella.)

And... hmm... I've run out of time, and can't do any more posts today. :-(

(Eugene, I know there's some posts aimed in my general direction that I haven't
answered yet. I'll try again tomorrow. :-)

 

[Bob_for_short]

...What's the difference?...

In your case (an external or classical current/source) your Lagrangian density contains the term jφ. In case of an external filed V(x) it would contain φ²V and V would get into the field equation as a potential.

I think your H is just H_tot corresponding to the original equation. There is no modification in the problem but "modification" of H₀ to get the original, full equation.

I verified, A and A⁺ have the same CCR so the "new" excitation spectrum is the same. Although in terms of A the Hamiltonian is diagonalized, the problem solution remains a superposition of different eigenstates, i.e., a state without a certain energy (a la coherent states). The particluar for massive φ spectrum $$\\w(k)$$ (dispersion law with or without gap) is not important.

---> Strangerep. I did not mean to offend you or answer for you. We were chatting on-line and I expressed what I thought following good sense in order to advance in discussion. I think my answer was reasonable (although I did not mention vertices). If you had been on-line, you would have answered yourself, I guess. A live chat needs quick responses. I am sorry, Strangerep, forgive me. (Consider it as my politeness - I did not want to wake you up and bother with a minor question.)

As to our problems with practical calculations in QED and QFT, the example of DarMM shows that there is no mathematical and conceptual problems in case of a know source j(x). (I hope we all agree that it is not a "free" case.)

The problems arise when we couple the unknown current j(x) with unknown filed φ (or A in QED via jA). It is precisely here where the self-action is introduced.
I take the advantage to show how one can proceed in a more physical way. In case of a know current j_ext the charge motion is determined with a strong external field so the charge acceleration in j_ext can be expressed via the external force F_ext. So instead of j_ext we can substitute its expression via the external force. Then the original equation reads as excitation of quanta due to the external force: it is the external force that stands in the right-hand side of the original equation for the quanta being excited. Then it is quite natural to suppose that this charge is a part of oscillators, - perturbing a part of oscillator excites the oscillator, like I described in my publications. Then a self-consistent theory is built quite straightforwardly in terms of compound systems without self-action - it is a theory of interacting compound systems.

 

[Bob_for_short]

By definition, "good" operators in the normally-ordered form (all annihilation operators on the right, all creation operators on the left) must have at least two creation operators and at least two annihilation operators. The simplest example of a "good" operator is $a^*a^*aa$. Their importance stems from the fact that they yield zero when acting on any 1-particle state or on the vacuum. Another important property is that the product or commutator of any number of "good" operators is also "good".

Just in this example it is clearly seen that there is nothing bad or non-physical it the solution expressed via certain combinations of "bad" operators. Many-photon states, like coherent ones, are natural and accompany most of scattering processes.

 

[meopemuk]

Just in this example it is clearly seen that there is nothing bad or non-physical it the solution expressed via certain combinations of "bad" operators. Many-photon states, like coherent ones, are natural and accompany most of scattering processes.

Could you please be more specific what you mean by "solution"?

My point is that "bad" operators should not be present in the interaction Hamiltonian.

By the way, "bad" operators are not present in the S-operator of any traditional QFT. This is because the sets of particles created and annihilated by "bad" operators have different energies. (For example, in the case of $$a^*(p)$$, the energy of annihilated particles is 0 and the energy of created particles is $$E_p$$). Therefore, these terms are always "killed" by the presence of energy delta function.

However, the absence of "bad" terms in the S-operator does not forbid formation of many-photon states in scattering. For example, the "good" operator responsible for the emission of two photons in a collision of two electrons is $$a^*a^*c^*c^*aa$$ (electron operators are denoted by a; photon operators are denoted by c).

Eugene.

 

[Bob_for_short]

Could you please be more specific what you mean by "solution"?

My point is that "bad" operators should not be present in the interaction Hamiltonian.

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

$$\Psi = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!}\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0} = Z^{1/2} \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)\Psi_{0}$$.

Where $$\Psi_{0}$$ is the free vacuum.

The total Hamiltonian contains the c/a operators in a "bad" way, the exact evolution operator does not leave the vacuum and one-particle state stable, the exact solution can be expressed in Glauber's form explicitly containing a "bad" combination. In fact, such a solution gives a Poisson probability distribution determined with one parameter - the average number of quanta (photons in QED).

 

[meopemuk]

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

$$\Psi = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!}\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0} = Z^{1/2} \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)\Psi_{0}$$.

Where $$\Psi_{0}$$ is the free vacuum.

The total Hamiltonian contains the c/a operators in a "bad" way, the exact evolution operator does not leave the vacuum an one-particle state stable, the exact solution can be expressed in Glauber's form explicitly containing a "bad" combination. In fact, such a solution is a Poisson probability distribution determined with one parameter - the average number of photons.

Please correct me if I misunderstood. Your $$\Psi$$ is the lowest-energy eigenvector of the interacting Hamiltonian. This is the "dressed" vacuum vector, which is different from the "free" vacuum vector $$\Psi_{0}$$. Taken literally, your expression means that in the "dressed" (or "physical") vacuum there is a non-zero chance to find one or more photons. However, this prediction disagrees with experiments. If I place a photon detector in absolute vacuum I will never see photons there. This is exactly the reason why I think that "bad" terms should not be present in interactions and that "physical" vacuum should be a zero-particle state.

Eugene.

 

[Bob_for_short]

Please correct me if I misunderstood. Your $$\Psi$$ is the lowest-energy eigenvector of the interacting Hamiltonian. This is the "dressed" vacuum vector, which is different from the "free" vacuum vector $$\Psi_{0}$$.

No, no, no! It is not a vacuum at all. It is a solution of the original equation and it is many-photon (many-quanta φ) state. It describes a permanent flux of coherent photons from the source (from the variable current j). It is what lasers emit.

 

[strangerep]

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".

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.

Your last sentence above doesn't follow. The interacting Hamiltonian
is not necessarily defined on the same domain as the free Hamiltonian.

[...] 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"?

If they are distributions, then equal-point multiplication is ill-defined,
just as is the case for any distributions. E.g., the square of a Dirac delta
distribution is ill-defined, because such a distribution is a mapping from a vector
space to a scalar space. (For the product to be well-defined everywhere, it would
have to be a mapping from a vector space to the same vector space. i.e., an operator.)

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.

Ahem. That's precisely the purpose of rigged Hilbert space.

I get the feeling you didn't have time to read quant-ph/0502053 after
I mentioned it in another thread. Perhaps we should postpone further
discussion of this point until after you've had enough time to study it.
Maybe in another thread.

[...] $\Psi$ is the lowest-energy eigenvector of the interacting Hamiltonian. This is the
"dressed" vacuum vector, which is different from the "free" vacuum vector $\Psi_0$ .
Taken literally, [this] means that in the "dressed" (or "physical") vacuum there is a non-zero
chance to find one or more photons.

Let us say "bosons", not photons -- the mass in DarMM's example is nonzero.

But no, it doesn't mean what you said. The physical bosons are associated with
the A(k), not the a(k). The A(k) annihilate the physical vacuum.

(To reconcile this with the seemingly-contradictory stuff that Bob_for_short
wrote in post #101 requires a separate post.)

 

[meopemuk]

No, no, no! It is not a vacuum at all. It is a solution of the original equation and it is many-photon (many-quanta φ) state. It describes a permanent flux of coherent photons from the source (from the variable current j). It is what lasers emit.

I go to bed.

Surely, I misunderstood. Possibly because your example does not fit under the heading "Rigorous Quantrum Field Theory". It is just a crude model of a complex physical system, which is laser.

 

[Bob_for_short]

If they are distributions, then equal-point multiplication is ill-defined, just as is the case for any distributions. E.g., the square of a Dirac delta
distribution is ill-defined, because such a distribution is a mapping from a vector
space to a scalar space.

The square of a Dirac delta-function never appears without a good reason. Normally it is infinity after integration. It is defined as infinity and means it.

Let us say "bosons", not photons -- the mass in DarMM's example is nonzero.

No objection.

But no, it doesn't mean what you said. The physical bosons are associated with
the A(k), not the a(k). The A(k) annihilate the physical vacuum.

If you can distinguish the solution in terms of a and A, please, show us. I wonder what especially physical creates A⁺.

 

[meopemuk]

strangerep, I am not going to argue about the domains of operators and Rigged Hilbert spaces. In my (perhaps uneducated) opinion, these are just technical subtleties. I would be very surprised if they have any relevance to physical predictions.

[...] [tex]\Psi[/itex] is the lowest-energy eigenvector of the interacting Hamiltonian. This is the
"dressed" vacuum vector, which is different from the "free" vacuum vector $$\Psi_0$$ .
Taken literally, [this] means that in the "dressed" (or "physical") vacuum there is a non-zero
chance to find one or more photons.

Let us say "bosons", not photons -- the mass in DarMM's example is nonzero.

But no, it doesn't mean what you said. The physical bosons are associated with
the A(k), not the a(k). The A(k) annihilate the physical vacuum.

Great, we then agree that "bare" vacuum, "bare" particles and their a/c operators a(k), a*(k) are just phantoms, which have no relevance to the stuff observed in Nature. By the way, we took them quite seriously when we initially wrote down the Hamiltonian or Lagrangian of our QFT theory (e.g., the QED Hamiltonian, which is normally written in terms of "bare" particle operators). Now we conclude that they are actually useless. Do you notice a weird contradiction here? In my opinion, this is one of the reasons to say that traditional QFT is self-contradictory.

Anyway, let us now focus on the good physical stuff - the physical vacuum $$\Omega$$ and physical particles created by $$A^*(k)$$. These particles can induce a response in real detectors, and that's the only thing we are interested in in physics. We've agreed that "bare" states and operators are useless, so let us now forget about them completely, and promise never mention them again. Indeed, we should never need them. Even without them, we have everything we need to do physics: We have the vacuum state $$\Omega$$, a/c operators $$A(k), A^*(k)$$ with usual commutation relations, the particle number operator

$$\int d^3k A^*(k)A(k)$$,

etc. We can build a Fock space which spans vectors like $$(A^*(k))^n \Omega$$. Let us call it the "interacting Fock space", to distinguish it from the (irrelevant) "free Fock space" built by applying $$(a^*(k))^n$$ to the "bare" vacuum. We also have the interacting Hamiltonian $$H$$ expressed as a function of $$A(k), A^*(k)$$. It is not difficult to build the free Hamiltonian (i.e., the one describing non-interacting A-particles) as well by formula

$$H_0 = \int d^3k \omega(k)A^*(k)A(k)$$.

It is important to note that, by construction, 0-particle $$\Omega$$ and 1-particle $$A^*(k) \Omega$$ states are eigenstates of the interacting Hamiltonian, which means that interaction between A-particles does not involve "bad" terms. So, we are all set. In our "interacting Fock space" we can represent any state of any multi-particle system. We can study the time evolution of such states, bound states, scattering etc. We will never encounter divergences, and we will never need to do renormalization.

So, we have achieved a peaceful transition from the traditional QFT (with bare particles, renormalization, and other curiosities) to the "dressed particle" theory formulated exclusively in terms of really observable ingredients. This is called "unitary dressing transformation". In the "dressed particle" theory both Hamiltonians $$H_0$$ and $$H$$ co-exist in the same Hilbert space. Moreover, they exactly coincide in 0-particle and 1-particle sectors.

Eugene.