Rigorous Quantum Field Theory.

[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)} +<br /> \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) +<br /> 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) +<br /> \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:
<br /> A(k) ~=~ a(k) ~+~ z(k)<br />
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!} -<br /> \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n}<br /> \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 =<br /> 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:
<br /> \frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})<br /> ~~~~~~~~(1)<br />

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.,
<br /> A^*(k) ~:=~ a^*(k) ~+~ g\,\bar{z}(k) ~~;~~~~~~<br /> A(k) ~:=~ a(k) ~+~ g\,z(k)<br />
which alternatively can be expressed as a formally-unitary transformation
<br /> A(k) ~:=~ U[z] \, a(k) \, U^{-1}[z]<br />
where U is of the form
<br /> U[z] ~:=~ exp\int\!\!dk\Big( \bar{z}(k)a(k) - z(k)a^*(k) \Big)<br />
(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.

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

<br /> <br /> <blockquote data-attributes="" data-quote="strangerep" data-source="post: 2451873" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> strangerep said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Let us say &quot;bosons&quot;, not photons -- the mass in DarMM&#039;s example is nonzero.<br /> <br /> But no, it doesn&#039;t mean what you said. The physical bosons are associated with<br /> the A(k), not the a(k). The A(k) annihilate the physical vacuum. </div> </div> </blockquote><br /> Great, we then agree that &quot;bare&quot; vacuum, &quot;bare&quot; 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 &quot;bare&quot; 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.<br /> <br /> 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&#039;s the only thing we are interested in in physics. We&#039;ve agreed that &quot;bare&quot; 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 <br /> <br /> \int d^3k A^*(k)A(k), <br /> <br /> etc. We can build a Fock space which spans vectors like (A^*(k))^n \Omega. Let us call it the &quot;interacting Fock space&quot;, to distinguish it from the (irrelevant) &quot;free Fock space&quot; built by applying (a^*(k))^n to the &quot;bare&quot; 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 <br /> <br /> H_0 = \int d^3k \omega(k)A^*(k)A(k). <br /> <br /> 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 &quot;bad&quot; terms. So, we are all set. In our &quot;interacting Fock space&quot; 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. <br /> <br /> So, we have achieved a peaceful transition from the traditional QFT (with bare particles, renormalization, and other curiosities) to the &quot;dressed particle&quot; theory formulated exclusively in terms of really observable ingredients. This is called &quot;unitary dressing transformation&quot;. In the &quot;dressed particle&quot; 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. <br /> <br /> Eugene.

 

[strangerep]

[...] 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.

You seem to have jumped from DarMM's example involving a quantized
boson field interacting with a classical external field j(x).
Let me first clarify something in the context of that example before
continuing...

From my post #95,
<br /> A(k) ~:=~ a(k) ~+~ z(k)<br />
where
<br /> z(k) ~:=~ \frac{g}{(2\pi)^{3/2}} ~ \frac{j(k)}{\sqrt{2}\,w(k)^{3/2}}<br />
The A(k) diagonalize the full Hamiltonian H:
<br /> H ~=~ \int\!\!dk\, E(k) A^*(k) A(k)<br />
not the free Hamiltonian H_0 which corresponds to the
case when j(x) is 0.

For a given j(x), we can indeed generate a Fock space by acting with
A^*(k) on \Omega. But that's all we can do in this
model. In fact, we should probably change the notation from A(k)
to A[j](k) to show that the new a/c ops have a functional
dependence on j. We should also write \Omega[j] and
z[j](k) for similar reasons.

If we change the external field j(x)\to h(x) then we have a
different set of a/c ops A[h](k). Clearly, these two external
fields corresponds to distinct physical situations, and the question is
then whether both can be represented in the same Hilbert space. I.e., we
have two Hilbert spaces: H[j] generated by applying the A^*[j](k)
to \Omega[j] and another one, H[h] generated by applying the
A^*[h](k) to \Omega[h].
We enquire whether these two Hilbert spaces in fact coincide,
i.e., whether any vector in one of them exists as a superposition of basis vectors of the other.
It turns out that this is only the case if z[j] - z[h] is square-integrable. In particular,
when h=0 (the free case where there's no external field), the Hilbert space
H[j] only coincides with H[0] if z[j] is square integrable (which is what
DarMM called a "weak external source"). But if our physical situation is
such that z[j] is not square-integrable, H[j] and H[0] are unitarily
inequivalent Hilbert spaces. That's the central point of DarMM's example, iiuc.

But in general, we cannot use H[j] as the Hilbert space for other arbitrary
choices of external field, but only for other external fields h which are
"close enough" to j, in the sense that (z[j]-z[h]) is a square-integrable
function.

(DarMM: please correct me if I've got anything wrong above about what
you really intended. :-)

If you can distinguish the solution in terms of a and A, please, show us.

My answer to Eugene above is essentially also an answer to this.
Sure, one can formally express one set of solutions in terms of the other,
but the mathematical difficulties arise when one tries to construct
a Hilbert space from the solutions. Then one finds distinct Hilbert
spaces, in general, in the way I explained above.

 

[meopemuk]

You seem to have jumped from DarMM's example involving a quantized
boson field interacting with a classical external field j(x).

I thought (perhaps incorrectly) that DarMM's example is valuable only as a platform for discussing more general (and more realistic) quantum field theories. So, I thought it would be appropriate to make some general statements, even not related to the example itself.

I mostly agree with what you wrote about Hilbert spaces H[j] for different sources j. However, I would like to make a couple of comments.

1. The claimed "inequivalence" or "orthogonality" of H[j] and H[h] is of somewhat peculiar nature. Using this logic we can also claim that plane waves (eigenvectors of momentum) are orthogonal to "normal" state vectors in the Hilbert space. (I've discussed this example in an earlier post). Perhaps, this is a valid mathematical point of view. But I still think that this is just a mathematical subtlety, without real physical consequences. Usually, we can do ordinary QM calculations by simply ignoring peculiar properties of plane waves (or approximate them by some wavepackets, or put our system in a large box, or whatever). Similarly, I think it would be appropriate not to give too much relevance to the "orthogonality" of H[j] and H[h].

2. In the DarMM's model we are free to change the source j. However, in more realistic theories (like QED) the interacting Hamiltonian is rigidly fixed. So, we have only two different Fock spaces: The free Fock space H[0] is built on bare a/c operators a, a*. The interacting Fock space H[j] is build on dressed a/c operators A, A*. I think we have agreed that bare operators a, a* are not physically relevant (particles created/annihilated by them even cannot be observed). So, I have absolutely no interest in knowing whether H[0] and H[j] are orthogonal or not. H[0] is just irrelevant. All physically relevant calculations are performed in H[j] with operators of observables constructed from A, A*. The Hamiltonian for free A-particles can be defined in this Fock space by simple formula

H_0 = \int d^3k \omega(k) A^{*}(k)a(k)

Can we agree about that?

Eugene.

 

[strangerep]

Usually, we can do ordinary QM calculations by simply ignoring peculiar properties of plane waves (or approximate them by some wavepackets, or put our system in a large box, or whatever).

These techniques only work in cases where the approximated theory is continuously
connected to the full theory.

(In the following I've changed the font so that Hilbert spaces have a different symbol from
the Hamiltonians.)

Similarly, I think it would be appropriate not to give too much relevance to the "orthogonality" of \mathcal{H}[j] and \mathcal{H}[h] .

Except that elements of \mathcal{H}[h] cannot be meaningfully approximated by
elements of \mathcal{H}[j].
In other language, a "phase transition" is involved in passing from j to h, analogous to
the phase transitions in condensed matter at a critical temperature. Each phase must
be described by a distinct Hilbert space.

I have absolutely no interest in knowing whether \mathcal{H}[0] and
\mathcal{H}[j] are orthogonal or not. \mathcal{H}[0] is just irrelevant.
All physically relevant calculations are performed in \mathcal{H}[j] with operators
of observables constructed from A, A*. The Hamiltonian for free A-particles can be defined in this Fock space by simple formula

H_0 = \int d^3k \omega(k) A^{*}(k)a(k)

Can we agree about that?

Apart from the obvious typo above, I don't what you intended by "H_0".
All we have so far is:

H[j] = \int\!\! d^3k \, \omega(k) \, A^{*}[j](k) \, A[j](k)

and that the A[j](k) form an irreducible set (in the rigged Hilbert space containing
\mathcal{H}[j], -- but you said you don't want to talk about such things).

 

[meopemuk]

Except that elements of \mathcal{H}[h] cannot be meaningfully approximated by
elements of \mathcal{H}[j].
In other language, a "phase transition" is involved in passing from j to h, analogous to
the phase transitions in condensed matter at a critical temperature. Each phase must
be described by a distinct Hilbert space.

Let me ask you this: do you think that study of these "phase transitions" would result in some new physical insight (like prediction of new effects), or these are just some mathematical niceties?

Apart from the obvious typo above, I don't what you intended by "H_0".
All we have so far is:

H[j] = \int\!\! d^3k \, \omega(k) \, A^{*}[j](k) \, A[j](k)....(1)

and that the A[j](k) form an irreducible set (in the rigged Hilbert space containing
\mathcal{H}[j], -- but you said you don't want to talk about such things).

Perhaps I wasn't clear enough. To be more general, in my talking points below I am referring to the full-fledged QFT (such as QED) rather than to DarMM's simplified model.

1. The non-interacting Fock space \mathcal{H}[0] is not interesting from the physical point of view, because electromagnetic interaction cannot be turned off in Nature.

2. For the same reason, the connection between \mathcal{H}[0] and the interacting Fock space \mathcal{H}[j] is not relevant to physics. (here I use symbol j borrowed from DarMM's model; it could be more appropriate to use the QED's coupling constant \alpha.)

3. All interesting physical stuff happens in \mathcal{H}[j], where operators A[j](k) are defined.

4. DarMM's example shows us how to define the interacting Hamiltonian (we call it H(j)) in the interacting Fock space \mathcal{H}[j]. In DarMM's example this Hamiltonian resembles the non-interacting Hamiltonian (see your formula (1)). In QED, this Hamiltonian has rather complicated form.

5. At the same time we can also *define* the non-interacting Hamiltonian H_0(j) in \mathcal{H}[j] by formula

H_0[j] = \int\!\! d^3k \, \omega(k) \, A^{*}[j](k) \, A[j](k)

6. As soon as we have all these ingredients, we can do all physical calculations (like S-matrix) in the interacting Fock space \mathcal{H}[j] without ever asking ourselves how this Fock space is related to \mathcal{H}[0].

This is why I still don't understand what physically relevant can be learned from "phase transitions" between \mathcal{H}[j] and \mathcal{H}[0]?

Eugene.

 

[Bob_for_short]

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.

I like this passage! It made clear that it is not a traditional QFT which is self-contradictory but researchers that contradict themselves.

Let me say several words about this story. The original quanta φ or photons in QED are observable. An external source (antenna with current j) emits them and it this is quite physical. The filed solution has some amplitude and phase; let me speak of the electromagnetic filed, in particular, of the tension E(r,t). It depends on its source j_ext but I do not label the filed, it is not necessary. It is implied and it is explicitly seen from the solution.

Now, this electric filed achieves a distant receiver antenna and gets into the equation motions of its charges as an external force eE(r,t). It induces a current. This current is detected. This is how a free EMF is observable. Moreover, this says us what the current j_ext in the emitter is.

Let me consider then your total Hamiltonian in terms of new c/a operators A. It looks as a free Hamiltonian. As soon as its spectrum is identical to that of the old non-perturbed Hamiltonian H₀(a,a+), the only way to distinguish the states corresponding to them is to measure experimentally the filed state with a distant receiver antenna. Doing so, we find that the field is in a coherent state with a given average number of photons. Both Hamiltonians are good for describing this state so we cannot distinguish the initial H₀ and the total H_tot expressed via a and A. We use a and a⁺ with j_ext to calculate the field state Ψ or we use H_tot(A,A+) and a receiver to measure the field state Ψ - both methods give the same state with the same average determined with the external source j_ext. Anyway, Ψ expressed via a or A contains j_ext in the same way!. It is not necessary to label the operators with j. Such are the physics and mathematics of this phenomena. I did not find any problem here.

<br /> \Psi = \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,a} = \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,A}<br />

I omitted Z since it is the same in case of normalized Ψ.

I have another proof:

A coherent state |z> can be represented in the following simple way: |z> =exp(za⁺ - z^*a)|0>. The combination (za⁺ - z^*a) is invariant in case of the "shift" variable change: a = A + z^*, where z is a complex number. (The average number of photons is <n_z> = |z|².) Although |0> is the vacuum state for a-operators (|0>_a), the expression for |z> in terms of A is the same so in this expression one can use the vacuum of A-operators (|0>_A).

|z> = exp(za⁺ - z^*a)|0> =

= exp(-|z|²/2) e^za+|0>.

The latter expression is what DarMM has written above for a given k. Now it is clear that solutions in terms of a and A coincide if the corresponding vacuums implied. There is no necessity to distinguish A(0) and A(j) and the corresponding vacuums since it does not bring anything useful.

Thus, we may determine the solution in two ways:

1) with solving the original equation with a known source j_ext, or

2) with observing the field state (a boundary condition).

In both cases it is sufficient to use operators a and a⁺ with their standard algebra since in both cases we obtain the same solution. We may label the operators with k but not with j. It is the field states Ψ who are labeled with j, h, etc., i.e. who are the source-dependent and obtained in this or another way. The basis |n> can be left intact.

As we know, the average number of photons is <n_z> = |z|². In case of k-dependence of z, its square-integrability is attained in all physical situations (any physical current j).

 

[Bob_for_short]

...This is why I still don't understand what physically relevant can be learned from "phase transitions" between \mathcal{H}[j] and \mathcal{H}[0]?

In my opinion, your feeling is absolutely correct. No variable change can change physics. It may help resolve the equations (with variable separation, for example) but it does not change the physics contained in the original problem. The physics can be changed with the problem change.

 

[strangerep]

In case of k-dependence of z, its square-integrability is attained in all physical situations (any physical current j).

Could you please provide a proof of this statement? (I'm guessing it rests on
how do you define "physical current" precisely?)

No variable change can change physics. It may help resolve the equations (with variable separation, for example) but it does not change the physics contained in the original problem.

Indeed. But this thread is concerned with "rigorous" QFT, which means "no naughty
fudging of the maths".

Let me ask you this: do you think that study of these "phase transitions" would result in some new physical insight (like prediction of new effects), or these are just some mathematical niceties?

Since this thread is about rigorous QFT, such personal speculations are off-topic.
I just want to understand the maths (and difficulties therein) involved in proving
whether theories are/aren't mathematically honest.

 

[meopemuk]

Since this thread is about rigorous QFT, such personal speculations are off-topic.
I just want to understand the maths (and difficulties therein) involved in proving
whether theories are/aren't mathematically honest.

At the risk of being off-topic I would like to express my personal opinion as well. I am most interested in applying DarMM's prescription (the transition from "bare" a/c operators a, a* to "physical" a/c operators A, A* and expressing the full Hamiltonian through A, A*) to realistic theories, like QED. I suspect that this exercise can teach us a few important and unexpected lessons about the nature of electromagnetic interactions.

Cheers.
Eugene.

 

[Bob_for_short]

Could you please provide a proof of this statement? (I'm guessing it rests on how do you define "physical current" precisely?)

I think I could. For example, I could take a periodic current j(x) in a wire, calculate its Fourier image and make sure it is OK for calculating Z(j).

...this thread is concerned with "rigorous" QFT, which means "no naughty fudging of the maths".

As we could see, the DarMM's example has no mathematical problems - all formulae are correct (we forget about typos). It is the perception of this example which is wrong. Let me list the wrong statements made about this example:

1) It is a problem in an "external filed".

2) H₀, operators a and a⁺, as well as their states are non-physical and non-observable.

3) Mass gap brings a principal difference with respect to a massless case.

4) Operators A and A⁺ have different CCR.

5) Operators A and A⁺ and the total Hamiltonian are more "physical".

This all made Eugene puzzle about something essential contained in these statements whereas they are just misleading and this fact is incredible.

There is only one instructive feature of this example: the perturbative solution and the exact one may "predict" different results. Indeed, the perturbative solution is:

|Ψ⁽¹⁾> ≈ [1 - g∫dk(...)] |0>

In the zeroth order it "predicts" that vacuum can stay vacuum <0||Ψ⁽⁰⁾> = <0|1|0> = 1. This is possible indeed if the current is rather weak and the mass gap is large, loosely speaking. In a general case the perturbation theory gives bad numerical predictions since the perturbation may not be small. In QED it is the case in any scattering process with free charges <0||Ψ⁽⁰⁾> = 1 but <0||Ψ^(exact)> = 0 !

On the other hand, the exact solution (post #99) is valid always, whatever j is. It is crucial in QED - to be able to correctly describe the soft radiation that always happens. This was one of motivations of my developments in QED reformulation. I obtain the soft radiation automatically in the zeroth order so no elastic processes are possible: <0||Ψ⁽⁰⁾> = 0. It is so because for the radiation my |Ψ⁽⁰⁾> is in fact |Ψ^(exact)>. The superscript n in my |Ψ⁽ⁿ⁾> relates to the order of inter-charge potential rather than to the charge-filed quanta coupling. The latter is taken into account exactly - by construction.

 

[Neo_Anderson]

I think I could. For example, I could take a periodic current j(x) in a wire, calculate its Fourier image and make sure it is OK for calculating Z(j).

As we could see, the DarMM's example has no mathematical problems - all formulae are correct (we forget about typos). It is the perception of this example which is wrong. Let me list the wrong statements made about this example:

1) It is a problem in an "external filed".

2) H₀, operators a and a⁺, as well as their states are non-physical and non-observable.

3) Mass gap brings a principal difference with respect to a massless case.

4) Operators A and A⁺ have different CCR.

5) Operators A and A⁺ and the total Hamiltonian are more "physical".

This all made Eugene puzzle about something essential contained in these statements whereas they are just misleading and this fact is incredible.

There is only one instructive feature of this example: the perturbative solution and the exact one may "predict" different results. Indeed, the perturbative solution is:

|Ψ⁽¹⁾> ≈ [1 - g∫dk(...)] |0>

In the zeroth order it "predicts" that vacuum can stay vacuum <0||Ψ⁽⁰⁾> = <0|1|0> = 1. This is possible indeed if the current is rather weak and the mass gap is large, loosely speaking. In a general case the perturbation theory gives bad numerical predictions since the perturbation may not be small. In QED it is the case in any scattering process with free charges <0||Ψ⁽⁰⁾> = 1 but <0||Ψ^(exact)> = 0 !

On the other hand, the exact solution (post #99) is valid always, whatever j is. It is crucial in QED - to be able to correctly describe the soft radiation that always happens. This was one of motivations of my developments in QED reformulation. I obtain the soft radiation automatically in the zeroth order so no elastic processes are possible: <0||Ψ⁽⁰⁾> = 0. It is so because for the radiation my |Ψ⁽⁰⁾> is in fact |Ψ^(exact)>. The superscript n in my |Ψ⁽ⁿ⁾> relates to the order of inter-charge potential rather than to the charge-filed quanta coupling. The latter is taken into account exactly - by construction.

LOL...

 

[DarMM]

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.
..., 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.

This doesn't make sense to me. The analogue of the vacuum and one particle state in QM is the ground state and first excited state. Now for two different Hamiltonians, even in nonrelativistic QM, the ground state and the first excited state do not coincide, so why should the vacuum and one-particles states of two theories coincide in QFT?

Secondly, taking QED as an example, the free particles have different properties to the interacting particles. For instance the free fermion particle does not transform under the U(1) group, it is invariant or in the trivial representation. The interacting electron does transform under U(1). So surely with these different properties they could not coincide.

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)

Particle number is not tied to the number operator. In a non-Fock space you can still have particles. For example a theory with non-zero field strength Z has a pole at a specific value of momentum for its two-point function. That value being the mass of a particle in the theory. So you have a one-particle sector. Similarly if you read about the Haag-Ruelle theory of scattering you can go on from here to obtain the n-particle sectors. The main difference is that in non-Fock spaces particles are not as neatly tied to operators obeying the canonical commutation relations.

 

[DarMM]

(DarMM: please correct me if I've got anything wrong above about what
you really intended. :-)

No, you have everything right. Also all your corrections to my original post on the external field model are correct. I must also explain an error in the original post. I said that A(k) and A^{*}(k) satisfy different commutation relations. This is incorrect, in more complicated examples such as a \phi^{4} theory it will be true, but it is not true here. My apologies for this error, I was thinking about the general case too much.

Also in an earlier post of yours:

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

Specifically check out:
Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, Volume III - Scattering Theory, pages 293-317.

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

Then I recommend Stephen Summers guide to exactly that to be found: http://www.math.ufl.edu/~sjs/construction.html".

 

[DarMM]

Let me ask you this: do you think that study of these "phase transitions" would result in some new physical insight (like prediction of new effects), or these are just some mathematical niceties?

Well, we are talking about the fact that the two models live in two different Hilbert spaces and this shows it. That is the importance.

This is why I still don't understand what physically relevant can be learned from "phase transitions" between \mathcal{H}[j] and \mathcal{H}[0]?

Again, what it shows is that they live in different Hilbert spaces, which is what we've been talking about. Quantum Field Theory uses different Hilbert spaces for different theories. Unlike non-relativistic QM, which uses the same Hilbert space always, as you can show through the Stone-Von Neumann theorem.

I must mention that two free scalar theories with different values of the mass also live in separate Hilbert spaces. They're both Fock spaces, but they are two different Fock spaces which are unitarily inequivalent.

 

[DarMM]

As we could see, the DarMM's example has no mathematical problems - all formulae are correct (we forget about typos). It is the perception of this example which is wrong. Let me list the wrong statements made about this example:
...
3) Mass gap brings a principal difference with respect to a massless case.

It does, there are no coherent states.

 

[meopemuk]

This doesn't make sense to me. The analogue of the vacuum and one particle state in QM is the ground state and first excited state. Now for two different Hamiltonians, even in nonrelativistic QM, the ground state and the first excited state do not coincide, so why should the vacuum and one-particles states of two theories coincide in QFT?

It is true that the most general QFT interaction does affect the vacuum and 1-particle states. You have your full rights to study such interactions. However, my point is that if we limit our attention only to interactions that do not change the vacuum and 1-particle states, then our life becomes much easier (e.g., no renormalization) and no important physics is lost.

Secondly, taking QED as an example, the free particles have different properties to the interacting particles. For instance the free fermion particle does not transform under the U(1) group, it is invariant or in the trivial representation. The interacting electron does transform under U(1). So surely with these different properties they could not coincide.

You are talking about gauge invariance, but I wouldn't like to open this yet another can of worms here. Can we stick to theories where the gauge invariance issue is not present?

Particle number is not tied to the number operator. In a non-Fock space you can still have particles. For example a theory with non-zero field strength Z has a pole at a specific value of momentum for its two-point function. That value being the mass of a particle in the theory. So you have a one-particle sector. Similarly if you read about the Haag-Ruelle theory of scattering you can go on from here to obtain the n-particle sectors. The main difference is that in non-Fock spaces particles are not as neatly tied to operators obeying the canonical commutation relations.

What are properties of n-particle sectors in "non-Fock spaces"? why can't they be represented as eigensubspaces of an Hermitian operator? are sectors with different "n" non-orthogonal?

Thanks.
Eugene.