Interacting theory lives in a different Hilbert space [ ]

[strangerep]

In article #34 of a recent thread about Haag's theorem, i.e.,
https://www.physicsforums.com/showthread.php?t=334424&page=3
a point of view was mentioned which I'd like to discuss further.
Here's the context:

[...] The interacting theory lives in a different Hilbert space [...]

I often see this statement, but I am not sure about its validity. In my
opinion, this statement goes against basic postulates of quantum theory. Let
me explain why I think the Hilbert space used to describe a physical system
should be independent on whether the system is interacting or not.

Let us first ask why we use Hilbert spaces to describe physical systems (their
states and observables) in QM? The answer is given by "quantum logic". This
theory tells us that subspaces in the Hilbert space are representatives of
"yes-no experiments" or logical "propositions" or experimental "questions".
Meets, joins, and orthogonal complements of subspaces represent usual logical
operations OR, AND, and NOT. It seems reasonable to assume that the same
questions can be asked about interacting and non-interacting system. The
logical relationships between these questions should not depend on the
interaction as well. Therefore, the same Hilbert space (= logical
propositional system) should be applied to both interacting and
non-interacting system, if their particle content is the same. [...]

I think I see a flaw in the argument above.

Suppose I want to know the accelerations of the two particles (or
maybe just their relative acceleration wrt each other). In the
case of 2 free particles I can certainly ask the question, but
the answer is always 0. But for two interacting particles, the
answer is nonzero in general.

Expressing this in the language of quantum logic and yes-no
experiments, the question "is the acceleration 0" always yields "yes"
in the free case, but can yield "no" in the interacting case.
Similarly, the question "is the acceleration nonzero" always yields
"no" in the free case but might yield "yes" in the interacting case.

Denote the 2-free-particle Hilbert space as $H_0$ and the
2-interacting-particle Hilbert space as $H$.

Proceeding by contradiction, let's assume that $H_0$ and $H$ are unitarily
equivalent, i.e., that any basis of $H_0$ also spans $H$. Assume as well that the
dynamical variable known as "acceleration" corresponds to (densely
defined) self-adjoint operators in $H_0$ and $H$, and that the two
operators are equivalent to each other up to a unitary transformation.

Every state in $H_0$ is a trivial eigenstate of the acceleration
operator, with eigenvalue 0. Expressed differently, the acceleration
operator annihilates every state in $H_0$. However, we expect to find
states in $H$ corresponding to nonzero accelerations, i.e., states which
the acceleration operator does not annihilate.

This implies that the basis states of $H_0$ cannot span H, contradicting
the initial assumption. Conclusion: $H_0$ and $H$ are not unitarily
equivalent.

So it's not enough that the same logical propositions (questions)
can be asked in both spaces. The spectrum of the corresponding
operator must also be considered, and whether both spaces
accommodate the full spectrum.

Or am I missing something?

 

[meopemuk]

Hi strangerep,

I must admit that my argument was not correct. Indeed, operators of observables (such as acceleration) have different properties and different commutation relations in the interacting and non-interacting cases. However, still I don't see the reason why these operators cannot coexist in the same Hilbert space. In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system. What is fundamentally different in the relativistic case and in QFT?

 

[Bob_for_short]

However, still I don't see the reason why these operators cannot coexist in the same Hilbert space. In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system.

That is a good point. Indeed, let us consider two Hamiltonians: one for a couple of free particles and another - with a repulsive potential. Are these Hamiltonians related with a "unitary" transformation? No. (Or, maybe, yes? Consider one-particle case). Yet, there is no problem with the perturbation theory (scattering) as well as with expansion of the exact solution of interacting case in series of free solutions. We have the same Hilbert space but two different basises to span this space.

In many cases it is so. I can also refer to the simplest 1D Sturm-Liouville problem considered in my publications where the perturbation theory may even give divergent matrix elements whereas the exact solutions are finite, physical, and they co-exist in the same Hilbert space.

 

[Bob_for_short]

What is fundamentally different in the relativistic case and in QFT?

In QFT there is an additional subtraction prescription. There is a principal difference before and after subtraction. Subtraction (discarding some corrections) changes the solution. The latter corresponds now to another Hamiltonian - a Hamiltonian without self-action. After that the rest is similar to a non-relativistic case.

 

[strangerep]

Indeed, operators of observables (such as acceleration) have different properties and different commutation relations in the interacting and non-interacting cases.

Hi meopemuk.

Thanks for your answer.

However, still I don't see the reason why these operators cannot coexist in the same Hilbert space.

As I mentioned in my original post, it's not so much about whether such an operator can
be represented in the Hilbert space(s), but whether the Hilbert spaces in question
account for the same (subset of) the full spectrum of the operator.

For finite degrees of freedom, one is saved by well-known theorems about isomorphisms
between Hilbert spaces, the Stone-von Neumann theorem, and all that. For infinite
degrees of freedom, continuous spectra, etc, these theorems fail.

In ordinary non-relativistic quantum mechanics we use the same Hilbert space for both interacting and non-interacting system. What is fundamentally different in the relativistic case and in QFT?

Afaict, it's all about how infinite degrees of freedom leads to unbounded operators.

 

[strangerep]

[...] let us consider two Hamiltonians: one for a couple of free particles and another - with a repulsive potential.

That sounds like a scenario with finite degrees of freedom. (?)

Are these Hamiltonians related with a "unitary" transformation? No. (Or, maybe, yes? Consider one-particle case).

If the answer was "yes", both Hamiltonians would have the same spectrum, so that
seems not correct.

In QFT there is an additional detraction prescription. There is a principal
difference before and after detraction. [...]

I'm not familiar with the word "detraction" in this context.
Did you mean "renormalization"?

 

[meopemuk]

Afaict, it's all about how infinite degrees of freedom leads to unbounded operators.

I don't see any big problem with unbounded operators. Physically, most relevant operators of observables are bound to be unbounded (pardon the pun). For example, the operator of position is unbounded, because the universe is infinite. The operator of momentum is unbounded, because there is no limit on the value of momentum.

If some mathematical formalism (functional analysis, theory of Hilbert spaces, etc.) has a problem with unbounded operators, then I consider this a problem of math rather than physics. Perhaps our math is build on unrealistic axioms (out of convenience). Perhaps, we need to be more creative and consider more general structures such as non-separable Hilbert spaces or non-standard Hilbert spaces, or something like that. In this particular case (unbounded operators) I don't think that math troubles indicate some new physics. I believe that physical considerations must take the lead, and math must follow.

Another point is that in the "dressed particle" approach to QFT (which is formulated in terms of particles rather than fields) the number of relevant degrees of freedom is always finite, because any realistic physical system is made of a finite number of particles.

 

[Bob_for_short]

I'm not familiar with the word "detraction" in this context.
Did you mean "renormalization"?

Yes, they call it "renormalization" although it is subtraction or discarding, to tell the truth .
I thought you had read my argumentations in "RiR".

P.S. I meant "subtraction", of course. My poor English...

 

[DrFaustus]

strangerep & meopemuk-> One crucial flaw of that argument is a purely logical one. While it is true that we can ask the same questions in both an interacting and a free theory, this does not imply at all that the underlying mathematical structure must be the same. Indeed, I can ask the same questions , say about particle scattering, in a classical, non quantum theory. Of course, physically the answers will be incorrect, but I can still ask those questions and obtain answers. But also, to an interacting theory I can ask questions a free theory simply cannot answer. For example, considering an out of equilibrium gas of particles, I could ask "How long will it take for the system to reach thermal equilibrium?". Clearly, if the particles cannot interact, an out of equilibrium system will never thermalize.

meopemuk -> strangerep nailed the answer: it's about the number of degrees of freedom. In ordinary QM you always have a finite nr of degrees of freedom and the Stone - Von Neumann theorem tells you, more or less, that all the representations of the canonical commutation relations are unitarily equivalent. That's why no one bothers talking about a different Hilbert space when dealing with QM other than the usual $$L^2(\mathbb{R}^n)$$. It's not so easy in QFT. If the nr of degrees of freedom is infinite, as it is when you deal with fields, the theorem no longer applies and it is in fact a physically relevant question as to what the appropriate Hilbert space is. And this holds for "simple" free theories. So the fundamental difference between QM and QFT is the number of degrees of freedom, not "additional subtraction procedures".

The problem with unbounded operators is precisely that they can have infinite expectation value in certain states. Sure, a purely mathematical problem. But if I'm to extract physics from some maths, then my maths better yield unique and finite answers or I'm in trouble. And it's not difficult to see how manipulating infinities can give you "paradoxes". That's the problem with unbounded operators: their maths is not clear yet we need them for physics.

And to say "consider a Hamiltonian with two particles" you must really specify if you're talking about QM or QFT. Actually, you don't need to: QFT is a multiparticle theory from the very beginning. So you're talking about QM, where things are "mathematically easy", so to say.

Finally, will now give you two more physical argument as to why the free field theory and the interacting one live in different Hilbert spaces. (Am still kinda struggling with the mathematical arguments myself :) )

Let's first clear what the Hilbert space of a (vacuum) free theory is. It's the usual Fock space, and the vectors of this space are multiparticle states. It is important however to observe now that these particles do not interact. So in the usual Fock space there is no space for interaction! Think about a self interacting particle propagating through space. The self-interaction will sure modify the one particle state of free theory and this state is in no sense a linear combination of multiparticle states of the same Fock state as all the "virtual particles" are also interacting whereas the Fock space ones are not.

Second, consider a system in thermal equilibrium at some finite temperature T. Such a state cannot be obtained in any way from the multiparticle states of the vacuum Fock space. Thermal states are not excitations of the vacuum state. Think about the quark-gluon plasma in the early Universe. The particle density in such states is non zero whereas the vacuum multiparticle states are zero density states. So even for free theories, the vacuum Hilbert space and a finite temperature Hilbert space will be different.

Yep, things are complicated in QFT - it's why you get $1m to prove the mass gap conjecture for Yang-Mills theories.

bob_for_short -> You insist on saying there are "problems" with renormalization because it involves subtraction of infinities. Read the book by Bogoliubov to see QFT without subtractions of infinities but with renormalization. Or read Scharf's book on "Finite QED" - as the title suggests, there is not a single infinity in that book. Or check out Epstein and Glaser papers about causal perturbation theory, where again renormalization is done without any subtraction of infinities. Renormalization is a (fairly) well understood issue in QFT and it's related to the mathematical nature of the various objects encountered in QFT, i.e. distributions.

 

[Bob_for_short]

...Read the book by Bogoliubov to see QFT without subtractions of infinities but with renormalization.

I read it in 1979.

Renormalization is a (fairly) well understood issue in QFT and it's related to the mathematical nature of the various objects encountered in QFT, i.e. distributions.

And I am a re-distribution master.

 

[meopemuk]

Let's first clear what the Hilbert space of a (vacuum) free theory is. It's the usual Fock space, and the vectors of this space are multiparticle states. It is important however to observe now that these particles do not interact. So in the usual Fock space there is no space for interaction! Think about a self interacting particle propagating through space. The self-interaction will sure modify the one particle state of free theory and this state is in no sense a linear combination of multiparticle states of the same Fock state as all the "virtual particles" are also interacting whereas the Fock space ones are not.

Hi DrFaustus,

I agree that in the traditional formulation of QFT particles do self-interact; "physical" particles of interacting theory are different from "bare" particles of free theory. I can also agree that this self-interaction makes definition of the interacting Hilbert (Fock) space problematic.

However, the more I think about the idea of "self-interacting particles" the less attractive it looks. The idea of "self-interacting QFT vacuum" is even less attractive.

Fortunately, there exists an alternative formulation of QFT, which I find more acceptable philosophically. In this formulation interactions do not modify 0-particle and 1-particle states of the theory. There is no distinction between "bare" and "physical" states. The interacting theory can happily occupy the Fock space of the free theory. This is called the "dressed particle" approach, which stems from the old work

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378.

There are only few researchers who take this approach seriously, but I hope this number will grow. The "dressed particle" approach can yield exactly the same renormalized S-matrix as traditional QFT. One side benefit is that calculations of the "dressed particle" S-matrix do not involve divergences.

 

[Bob_for_short]

...Think about a self interacting particle propagating through space. ...

Drunk, you mean?

...Think about the quark-gluon plasma in the early Universe. ...

I don't remember anything from that time.

 

[Fredrik]

DarMM wrote a couple of interesting posts about the Hilbert space of the interacting theory here. See #50 and #54. If I understand him correctly, the usual Fock space of non-interacting states is equivalent to (the completion of?) the vector space of square-integrable distributions, and the only thing that changes about that space when we introduce interactions is the measure we use to define the integrals.

I don't really understand any of this myself, so I don't have much to contribute here, but I hope to learn this stuff some day.

 

[Fredrik]

So in the usual Fock space there is no space for interaction!

I think this is a mistake. Any time evolution operator that doesn't leave every n-particle subspace invariant would by definition be an interaction, and you can certainly define such operators.

 

[Bob_for_short]

Let me summarize: Interacting theory lives just in a different Hilbert space, so its Hamiltonian is alive and kicking.

 

[strangerep]

If some mathematical formalism (functional analysis, theory of Hilbert spaces, etc.) has a problem with unbounded operators, then I consider this a problem of math rather than physics. Perhaps our math is build on unrealistic axioms (out of convenience).

Yes.

Perhaps, we need to be more creative and consider more general structures such as non-separable Hilbert spaces or non-standard Hilbert spaces, or something like that.

IIUC, that's why people use rigged Hilbert space -- which is larger than ordinary
Hilbert space. In the space of tempered distributions (of which ordinary Hilbert space
is a subspace), one can define a distribution-valued inner product like
$$\langle p | q \rangle ~=~ \delta(p-q)$$
and also obtain a usable generalized spectral theorem.

Other people prefer the C*-algebra approach which can also be regarded
as "larger" than the usual Hilbert space methods (but it would take ages
for me to make that statement precise. :-)

Maybe someone will get that $1M prize one day when they
figure out which maths is right.

 

[strangerep]

Yes, they call it "renormalization" although it is subtraction or discarding, to tell the truth .
I thought you had read my argumentations in "RiR".

Yes, but it was some time ago and I didn't remember
the word "detraction". Since you intended "subtraction",
I now understand what you meant. :-)

 

[strangerep]

Another point is that in the "dressed particle" approach to QFT (which is formulated in terms of particles rather than fields) the number of relevant degrees of freedom is always finite, because any realistic physical system is made of a finite number of particles.

I think that point could only be valid if the full (non-perturbative) Hamiltonian of the
dressed theory turned out to be a finite polynomial in the a/c operators. But if one
needs ever-longer more complicated terms at each perturbation order, one ends up
with an inf-dim theory, afaict.

 

[Bob_for_short]

I think that point could only be valid if the full (non-perturbative) Hamiltonian of the dressed theory turned out to be a finite polynomial in the a/c operators. But if one needs ever-longer more complicated terms at each perturbation order, one ends up with an inf-dim theory, afaict.

An "inf-dim theory" is not a problem per se if the series can be summed up into a certain function, as in my electronium potential. The a/c operators are in the denominator and in such a combination (Q) that is integrable in the perturbation theory: V(r,Q) = 1/|r + εeQ|.

 

[strangerep]

Any time evolution operator that doesn't leave every n-particle subspace invariant would by definition be an interaction, and you can certainly define such operators.

Examples?

You've got to make sure that the full Hamiltonian still participates correctly
in a representation of the Poincare algebra, and that the Hamiltonian is
densely well-defined on the multiparticle Fock space.

 

[strangerep]

An "inf-dim theory" is not a problem if the series can be summed up into a certain function, as in my electronium potential. The a/c operators are in the denominator and in such a combination (Q) that is integrable in the perturbation theory.

I wait patiently for your forthcoming paper giving a fully relativistic,
fully multiparticle treatment. :-)

 

[Bob_for_short]

Better tell me how to get a grant or at least a part-time research position to carry out this "a fully relativistic, fully multiparticle treatment".

 

[meopemuk]

I think that point could only be valid if the full (non-perturbative) Hamiltonian of the
dressed theory turned out to be a finite polynomial in the a/c operators. But if one
needs ever-longer more complicated terms at each perturbation order, one ends up
with an inf-dim theory, afaict.

I agree about that. There is no fully rigorous non-perturbative formulation of the "dressed particle" theory in a closed form. But this seems to be a common drawback of many modern theories. Attempts to put them on a solid mathematical footing do not bring much progress. A good case in point is axiomatic QFT, which hasn't produced any tangible results, AFAICT. In my opinion, if we want to move forward, we should often replace mathematical rigor with physical intuition.

 

[DrFaustus]

meopemuk -> I'll have to defend rigorous (axiomatic, algebraic) QFT a bit. It's not true that it hasn't produced any tangible result. Interacting models have been rigorously constructed in both 2D and 3D. It's 4D that is the problem. In fact, it is essentially the only big open problem that the rigorous approach has yet to overcome. (Tightly related to the $1m prize.) But if models in 2D and 3D don't seem tangible, one tangible result of rigorous QFT is the proper justification of scattering theory in the vacuum as we know it today. In particular, the use of the Gell-Mann and Low formula to compute the amplitudes.

As for the "dressed particle" approach, I don't know it and won't comment on it. But I want to point out why is it that people use fields in the first place. (Trust me if I say that people would be more than happy to throw them out of the window if a valid alternative would be found.) Fields, when governed by hyperbolic PDEs, have the mathematical property that perturbations to the system will propagate at finite speed. And it is essentially this the main reason to use fields - relativistic causality. And this is also the meaning of the commutation relations at the quantum level - measurements at space like separations cannot influence each other. And QFT is a theory of fields that on some occasions admits a particle interpretation. Which is precisely when we use it for scattering. But it is a theory of fields.

Bob_for_short -> Recipe to get a grant or research position.

1. Re-read Bogloiubov's book to find, say in the conclusions on pag. 644

A method has thus been developed for the actual construction of the operator functions $$S_n$$ free from divergences. It has turned out that this method is equivalent to the usual subtraction procedure. In this way the latter, which until now had the nature purely of a set of prescriptions, has been given a rigorous mathematical foundation within the framework of perturbation theory.

And, on pag.647, last paragraph of the book

Moreover, independently of the nature of possible changes in the mathematical methods and in the fundamental assumptions of local field theory, because of the singular nature of the local commutation and causal functions the principal quantities in terms of which the theory will have to be formulated will have to be generalized functions i.e., some modification of the subtraction formalism associated with the multiplication of generalized functions will be an unavoidable attribute of any future local theory. The question of the existence of divergences in a possible future non local theory of course remains open for the time being.

2. Stop wasting time on the internet ranting about renormalization and, instead, write a couple of interesting papers and publish them.

3. Find 3 academics that could understand and evaluate your papers and ask them for reference letters.

4. Enter the battlefield for the not many academic jobs.

Fredrik -> That claim was not meant to be perfectly rigorous, but more an a posteriori heuristic interpretation. But you should ask DarMM about more details - he is indeed the man to talk to when it comes to rigorous QFT. I have barely scratched the surface of it and not really delved into the details.

 

[Bob_for_short]

...If the answer was "yes", both Hamiltonians would have the same spectrum, so that seems not correct.

See Appendix 4 in http://arxiv.org/abs/0906.3504 where the same spectrum is obtained for two different Hamiltonians. Of course, the latter are related with a unitary ("dressing", if you like) transformation.

Bob_for_short -> Recipe to get a grant or research position.
1. Re-read Bogloiubov's book to find, say in the conclusions on pag. 644 And, on pag.647, last paragraph of the book...

What to do if I went in some respects farther than Bogoliubov? What if I have my own successful experience?

2. Stop wasting time on the internet ranting about renormalization and, instead, write a couple of interesting papers and publish them.

Done. By the way, I like mixing with clever guys.

3. Find 3 academics that could understand and evaluate your papers and ask them for reference letters.

Done. By the way, don't you want to read my papers? No? Anybody else?

4. Enter the battlefield for the not many academic jobs.

You know what they say? First you fulfil all calculations at your own account, and then we will see. It's a world upside down.

 

[meopemuk]

As for the "dressed particle" approach, I don't know it and won't comment on it. But I want to point out why is it that people use fields in the first place. (Trust me if I say that people would be more than happy to throw them out of the window if a valid alternative would be found.) Fields, when governed by hyperbolic PDEs, have the mathematical property that perturbations to the system will propagate at finite speed. And it is essentially this the main reason to use fields - relativistic causality. And this is also the meaning of the commutation relations at the quantum level - measurements at space like separations cannot influence each other. And QFT is a theory of fields that on some occasions admits a particle interpretation. Which is precisely when we use it for scattering. But it is a theory of fields.

I would like to call two witnesses who are going to support my view that the role of quantum fields is more formal than fundamental.

One of them is S. Weinberg, who explained the reasons to introduce quantum fields in his vol. 1. He introduces fields as a formal device to satisfy physical requirements of relativistic invariance and cluster separability. It appears that if one builds the interaction Hamiltonian as a polynomial in quantum fields, then these two conditions are satisfied. However, nobody's proved that quantum fields is the only possibility. There are examples of acceptable field-less models in the literature. In Weinberg's logic quantum fields appear as formal mathematical constructs rather than real physical entities. He even says that field commutators have nothing to do with measurability and causality. Also he doesn't see any physical relevance of the Dirac equation.

Just yesterday I "met" my other witness

N. D. Mermin, "What's bad about this habit", Physics Today, May (2009), 8.

http://www.ehu.es/aitor/irakas/mes/Reference/mermin.pdf

In particular, Mermin writes:

"But what is the ontological status of those quantum fields
that quantum field theory describes? Does reality consist of a
four-dimensional spacetime at every point of which there is a
collection of operators on an infinite-dimensional Hilbert space?
... But I hope you will agree that YOU are not a continuous
field of operators on an infinite-dimensional Hilbert space. Nor,
for that matter, is the page you are reading or the chair you are
sitting in. Quantum fields are useful mathematical tools. They
enable us to calculate things."

 

[Bob_for_short]

...Quantum fields are useful mathematical tools. They
enable us to calculate things."

OK, we will use fields.

 

[strangerep]

... (various complaints about the cruel real world) ...

I had previously skimmed Bob's paper:

V. Kalitvianski, "On Perturbation Theory for the
Sturm-Liouville Problem with variable coefficients and its applications."
(Originally from around 1991-1994 in Russian.)
Available as arXiv: 0906.3504

and I figured I should read it a bit more carefully before continuing this
thread. In this post I'll summarize what I think Bob's main point is,
and then (in my next post) relate/compare it to another more recent paper:

J.R.Klauder, "Rethinking Renormalization",
Available as arXiv:0904.2869 [hep-th]

(J.R.Klauder is a renowned mathematical physicist, perhaps
best known for his vast body of work on coherent states.)

First to Bob's paper, which I'll present in a different sequence
from the paper itself (for reasons which I'll explain later)...

Consider a Schrodinger-like equation
$$\left( \frac{d^2}{dz^2} + \lambda - U(z) \right) \Phi(z) ~=~ 0 ~~~~~~(A)$$
where $\lambda$ is an eigenvalue (to be found), and
$U(z)$ is a potential defined in terms of another function
$r(z)$. We take a specific $U$ of the form:
$$U(z) ~=~ \frac{(r^{1/4})''}{r^{1/4}} ~=~ -\frac{3}{16r^2}\left(\frac{dr}{dz}\right)^2 ~+~ \frac{1}{4r} \frac{d^2r}{dz^2} ~.$$
(The primes denote derivatives wrt z, as usual.)

Now consider the case where $r(z)$ is a step function (hence U
contains the square of a delta function) and is rather nasty. We wish to
find all the eigenvalues (ie values of $\lambda$ by perturbation
about a "free" theory where U is 0.

Denote the eigenvalues and eigenfunctions of the free theory with a "(0)"
superscript, e.g., $$\Phi^{(0)}_n$$ and $$\lambda^{(0)}_n$$
denote the n'th eigenfunction and eigenvalue, respectively, for the
free theory. Similarly, denote the eigenquantities of the perturbed
theory using a "PT" superscript.

A standard iterative procedure (assuming r changes slowly wrt z) yields
the perturbation series:
$$\Phi^{PT}_n ~=~ \Phi^{(0)}_n ~+~ \sum_{m\ne n} \frac{U_{mn}}{\lambda^{(0)}_n - \lambda^{(0)}_m} \; \Phi^{(0)}_m ~+~ \dots$$
$$\lambda^{PT}_n ~=~ \Phi^{(0)}_n ~+~ U_{nn} ~+~ \sum_{m\ne n} \frac{U_{nm}U_{mn}}{\lambda^{(0)}_n - \lambda^{(0)}_m} ~+~ \dots$$
where
$$U_{mn} ~=~ \left(\Phi^{(0)}_m , U \Phi^{(0)}_m\right) ~=~ \int dz \Phi^{(0)}_m U \Phi^{(0)}_m$$
which are divergent because of the delta fn squared in U. In other words,
U is not an everywhere-defined operator on the Hilbert space made from
the free eigenfunctions $\Phi^{(0)}_n$. Letting the step in
$r(z)$ occur at $z_1$, being a jump from $r_1$
to $r_2$, further evaluation of the above gives
$$U_{mn} ~=~ -\frac{1}{2} \Phi^{(0)}_n(z_1) \,{\Phi'}^{(0)}_n(z_1) \, ln(r_2/r_1) ~-~ \frac{1}{16}\int dz \, \left[(ln r)' \right]^2 \left(\Phi^{(0)}_n\right)^2$$
As we decrease the step size, i.e., let $r_2 \to r_1$, the
first term approaches 0, but the second remains infinite no matter how
close $r_2$ gets to $r_1$.

Thus we have an example of what's known as a "discontinuous perturbation"
problem: the full solution does not approach the free solution continuously
in any sense. Rather, there's a discontinuous jump. This is another way of
seeing how/why the two theories "live in different Hilbert spaces". :-)

In his paper, Bob does not start from the Schrodinger-like equation (A) where
I began. Rather, he starts from a Sturm-Liouville problem:
$$\left( \frac{d^2}{dx^2} + \lambda r(x) \right) \psi(x) ~=~ 0 ~~~~~~(B)$$
(plus some straightforward boundary conditions), and he notes that after
a change of variables $x \to z$ equivalent to:
$$\frac{dz}{dx} ~=~ \sqrt{r(x)} ~~;~~~~ \Phi(z) ~=~ r^{1/4}(x(z)) \; \psi(z(z)) ~~,$$
one gets the Schrodinger-like equation (A) I started with above.
(Work it out if you don't believe me. :-)

BTW, the relevance of all this "Sturm-Liouville" stuff is that a great
many applications in physics can be cast in Sturm-Liouville form, and
there's a large body of theory one can call on to analyze this stuff,
including a functional-analytic Hilbert space setting. See the Wiki entry
http://en.wikipedia.org/wiki/Sturm-liouville
for a bit more background. Hence, understanding a problem in the
more general S-L context has wider potential applicability than one
might naively think.

In Bob's paper, he then applies a different change of variable to
the Sturm-Liouville equation (B) above, similar to the previous one
but now the new dependent variable is
$$\phi(z) ~=~ \psi(x(z))$$
yielding the equation:
$$\left( \frac{d^2}{dz^2} + \lambda - V(z) \right) \phi(z) ~=~ 0 ~~~~~~(C)$$
where now the potential operator is
$$V(z) ~=~ -\left(ln \sqrt{r(z)}\right)' \; \frac{d}{dz} ~.$$
The derivative of the logarithm turns out to be proportional to a delta
function (ie not involving a delta function squared), and the matrix elements
$V_{nm}$ are finite. For the $r(z)$ step-function case
considered above, they turn out to be
$$V_{nm} ~=~ -\frac{1}{2} \, {\Phi}^{(0)}_n(z_1) \, {\Phi}^{(0)}_n(z_1) \, ln(r_2/r1) ~.$$
and the perturbation procedure for iterating the eigenfunctions and eigenvalues
now remains finite.

Bob's paper goes on to analyze the eigenvalue behaviour in more detail,
and for some other cases, but the above is sufficient for my purposes
in this thread. I'll continue the story (relating to Klauder's work) in
the next post, so please don't post any comments yet until I've
completed it.

 

[strangerep]

(Continuation of previous posting...)

The paper of J.R. Klauder that I want to discuss in this post is:

J.R.Klauder, "Rethinking Renormalization",
Available as arXiv:0904.2869 [hep-th]

It summarizes, and builds on a lot of earlier work. Some of his earlier
ideas can be found in textbook form as

J.R.Klauder, "Beyond Conventional Quantization",
Cambridge University Press, 2000 & 2005, ISBN 0-521-25884-7.

In "Rethinking Renormalization", Klauder considers a 1D example with
classical Hamiltonian
$$H_\lambda(p,q) ~=~ \frac{1}{2}\left(p^2 + q^2 \right) ~+~ \frac{\lambda}{|q|^\alpha}$$
where $\lambda$ is considered as a coupling constant. Clearly
the potential is singular at the origin q=0, so the particle can never reach
the origin for any $\lambda > 0$. The classical solutions to
the free problem (with $\lambda=0$) are just those of the
harmonic oscillator, proportional to $cos(t-a)$. However, the
solutions of the "interacting" theory with infinitesimal (but nonzero)
$\lambda$ are "rectified" waveforms, ie involving
$\pm|cos(t-a)|$. Klauder calls this the "pseudofree" theory,
since it clearly doesn't coincide with the free theory. One finds that
solutions to the full theory are indeed continuously connected to those
of the pseudofree theory. Klauder notes on p5 that:

If one were contemplating a perturbation series representation of the
interacting solution, that power series should not be about the free
theory (to which the interacting solutions are not continuously
connected!) but rather about the pseudofree theory.

Klauder then goes to analyze what happens when one quantizes the theory
for various values of $\alpha$. He takes a regularized form
of the potential:
$$V_\epsilon ~=~ \frac{\lambda}{(|q| + \epsilon)^\alpha}$$
and finds that for all $\alpha < 2$ one can construct a finite
series of counterterms with diminishing divergences, reminiscent of
super-renormalizable QFTs. For $\alpha = 2$, he finds that an
infinite series of counterterms is needed, all of equal divergence, like
a strictly renormalizable QFT. But for $\alpha > 2$ the
term-by-term divergences increase, reminiscent of nonrenormalizable QFTs.

In different words, for $\alpha \le 2$ it is possible to modify
the theory (using counterterms) such that it becomes continuously connected
to the free theory. But for $\alpha > 2$, Klauder finds that:

For $\alpha > 2$, ..., there is no modification of the theory that
can be made to prevent the theory from passing to a pseudofree theory as the
parameter $\lambda \to 0$. In other words, for
$\alpha > 2$ the interacting quantum theory passes to a pseudofree
theory with a set of eigenfunctions and eigenvalues that are generally
different from those that characterize the free theory.

Hopefully, the attentive reader can now see the parallels I'm trying to
draw between Bob's work, and the ideas of J.R.Klauder.

Even Klauder acknowledges that taking such experience, and hoping to apply
it to QFT, is still a "leap of faith" (or as someone else recently said of
Bob's ideas, "wishful thinking"). However, he then takes his ideas a little
further, attempting to characterize the singular nature of the interaction
that leads "to either a continuous connection with the original free theory
or instead leads to a continuous connection with a pseudofree theory". He
calls the latter a "hard-core interaction", and develops the idea in terms
of a functional integral approach. The distinction involves projecting
out all those paths in the free harmonic oscillator that reach or cross
the origin, and does indeed have its seeds in the classical theory in which
the feature of discontinuous perturbation can already be seen.

Klauder develops these ideas a lot further, however. He uses the example
of so-called "ultralocal" scalar QFT, which has been rigorously solved
nonperturbatively. IIUC, he shows that it is necessarily continuously
connected to a pseudofree theory distinct from the free theory. He remarks
on p23 that

... the solution obtained from a rigorous viewpoint is identical to the one
obtained by the supremely simple prescription of choosing a suitable
pseudofree model ...

We shall see that this breathtakingly elementary procedure, coupled with
a judicious choice of further details of the pseudofree model, will provide
a divergence-free formulation of additional examples of nonrenormalizable
models, formulations that would be difficult to arrive at by any other means.

Hopefully, the similarities and difference between Bob's and Klauder's
hopes and dreams should now be clearer to the attentive reader. :-)

However, Klauder takes it even further. He then looks at more
relativistic models, specifically covariant quartic self-interacting
scalar fields and finds a pseudofree model suitable for that case. This
is treated more fully in an earlier paper:

J.R. Klauder. “Taming Nonrenormalizability”,
Available as: arXiv:0811.3386.

which I haven't yet studied properly.

In the conclusion, Klauder says

For covariant scalar nonrenormalizable quantum field models, we have shown
that the choice of a nonconventional counterterm, but one that is still non-
classical, leads to a formulation for which a perturbation analysis of both
the mass term and the nonlinear interaction term, expanded about the
appropriate pseudofree model, are term-by-term finite.

- - - - - - - - - - -
Finally, I should mention that my intent in writing these lengthy posts was
to introduce into this thread the distinction between renormalizable theories
(which many people find boring, regarding them as tractable), and
nonrenormalizable theories (which are much more difficult). The question still
remains open whether renormalizability is an essential guide in finding
physically-relevant theories, or just the mathematically easier ones. No doubt
this would become considerably more interesting if the Higgs is not found,
and a nonrenormalizable Higgs-less electroweak theory must be confronted.

 

[meopemuk]

Hi strangerep,

in my opinion, your examples have little relevance to the problem of renormalization in QFT. Your examples are related to ordinary quantum mechanics, where the number of particles is conserved (actually, this number is 1). But renormalization in QFT is required exactly because its interaction does not conserve the number of particles. More specifically, a QFT model requires renormalization if two conditions are satisfied:

1. The interaction operator can change the number of particles (the Hamiltonian does not commute with particle number operators).

2. The interaction operator contains terms, which act non-trivially on the vacuum and 1-particle states. Examples of such terms are (in the creation-annihilation operator notation)$aaa, a^{\dag}aa, a^{\dag}a^{\dag}a, a^{\dag}a^{\dag}a^{\dag}$.

As long as you have a theory with conditions 1. and 2. you are guaranteed to have renormalization difficulties. Even if the interaction is well-behaving and loop integrals are convergent, the vacuum will be "polarized" and particles will "self-interact". Hamiltonians with interactions 2. are simply physically unacceptable. This fact is tacitly admitted in the standard QFT by adding renormalization counterterms to such Hamiltonians. (By the way, I don't see much difference between renormalizable and non-renormalizable QFT's. The former ones have finite number of types of counterterms. The latter ones have infinite number of types. Not a spectacular difference.)

Condition 1. above is physically justifiable, because we know that in the real world the number of particles changes all the time. However, condition 2. does not follow from any physical observation. This condition is bad and evil. So, if you want a physically acceptable theory without renormalization, make sure that your Hamiltonian satisfies condition 1. and does not contain terms of the type 2.

 

[strangerep]

in my opinion, your examples have little relevance to the problem of renormalization in QFT.

Well, they are not "my" examples. :-)

Your examples are related to ordinary quantum mechanics, where the number of particles is conserved (actually, this number is 1).

I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

 

[meopemuk]

I'm pretty sure Klauder's stuff with quartic interaction qualifies as a QFT.

Then we use different terminologies. In my opinion, the characteristic feature of QFT is the presence of interactions changing the number of particles. So, QFT requires the Fock space, in which the number of particles is allowed to vary from 0 to infinity. (Even the presence of "quantum fields" is not essential, as I tried to explain in one of my previous posts.).

If the number of particles does not change, then we have ordinary quantum mechanics, which can be formulated in an N-particle Hilbert space. The examples of Bob and Klauber belong to the latter category.

 

[DrFaustus]

When I quoted Bogoliubov in one of my previous posts I was really trying to emphasize the following point. There is one simple question everyone discussing QFT should answer and which is crucial to the whole renormalizability issue.

What is the mathematical nature of a quantum field? In other words, what kind of mathematical object is a quantum field?

You might try to say that it's an operator, but that is not correct. It cannot bean operator because with operators you simply cannot satisfy the canonical commutation relations. Up to now, no one has found a better way of describing quantum fields mathematically than using the concept of "operator valued distribution". This means that if $$\varphi(x)$$ is a quantum field, then $$\int \textrm{d} x \varphi(x) f(x)$$ is a well defined operator on some Hilbert space (with $$f(x)$$ a compactly supported function). And that's the "generalized function" that Bogoliubov talks about (see my previous post). Simply stated, quantum field = operator valued distribution. And the emphasis here is on "distribution".

Now, no matter how much you dislike renormalization, the problem with distributions is that pointwise multiplication is in general ill defined. (Convince yourself by multiplying two Dirac deltas by approximating them with Gaussians and taking the limit - you get infinity.) And this is the true origin of divergences in QFT - pointwise multiplication of distributions. In other words, an object as simple as $$\varphi^2(x)$$ is ill defined and needs an actual proper defnition. (A mathematically correct definition for the square is $$:\varphi^2(x): = \lim_{x \to y}\Big[ \varphi(x) \varphi(y) - \langle \varphi(x)\varphi(y)\rangle_0 \Big]$$ , where the colon denotes normal ordering and the expectation value is taken in the vacuum.) Such an object is a smooth function of x and the reason for this is that $$\varphi(x)\varphi(y)$$ and $$\langle\varphi(x)\varphi(y) \rangle_0$$ have the same "singularity structure". In other words, by subracting them you are removing the singularity in $$\varphi(x)\varphi(y)$$ and afterwards you can take the limit. But note that this is already a first example of renormalization! In this sense you could talk about renormalizing a squared Dirac delta. (As an aside, strangerep and Bob_for_short, if in that paper appears a squared delta, then the equation is mathematically ill defined and all the conclusions that one might draw are dubious to say the least. It's similar to dividing by zero - you simply cannot do it without some modification, like considering a limit.) And obviously, when you start multiplying Feynman diagrams, which are distributions themselves, things cannot be any better and in fact things actually get worse. For Christ's sake, they wouldn't be giving you a million if properly defining a QFT would be that easy.

Bottom line, as long as quantum fields are to be regarded as operator valued distributions, and they have to be some sort of nontrivial mathematical object if we want them to satisfy canonical commutation relations, then renormalization is unavoidable. That's the meaning of Bogoliubov's words.

Now, please find a way around that and I'll be happy to listen. But first answer to my question above. Because that'll be the starting point.

 

[Bob_for_short]

Thank you, Strangerep, for mentioning my paper. It shows that I did manage to reformulate and find a finite perturbative solution of that particular problem. I would also add that the the precision of my solution is very good and quite sufficient for practical goals.Dear Dr. Faustus,

Nobody argues about the nature of fields in QFTs. Nobody argues that they come in product in the perturbation theory. You yourself say that taking a finite (Gaussian or so) distribution makes the product well defined. What I found is a natural mechanism of keeping the product well defined due to quantum mechanical smearing. There is no need to "remove it". It follows from a new initial approximation and a new interaction Hamiltonian. It is natural rather than artificial. No corrections to the fundamental constants appear and the matrix elements are finite and small. The perturbation series thus are different from what you imply as inevitable.

 

[Bob_for_short]

When I quoted Bogoliubov in one of my previous posts I was really trying to emphasize the following point. There is one simple question everyone discussing QFT should answer and which is crucial to the whole renormalizability issue.

What is the mathematical nature of a quantum field? In other words, what kind of mathematical object is a quantum field?

...Up to now, no one has found a better way of describing quantum fields mathematically than using the concept of "operator valued distribution".

What we encounter in QFTs is not just a distribution product but a T-ordered product, i.e., a propagator, i.e., a solution of equation with a point-like source which is reduced to the Coulomb field in a non-relativistic case, i.e., a banal particle interaction. Make this interaction naturally smeared and no ill-definiteness arise. Roughly speaking, interaction of charge with the quantized EMF, taken into account in the first turn, brings this smearing in a natural way. The interaction reminder for the perturbation theory is then not so dangerous.