Rigorous Quantum Field Theory.

[meopemuk]

The Fock space is a separable Hilbert space...

Again, why do you think that the Hilbert space of a physical system must be separable? Is there any physical reason for that? Or it simply makes your math easier?

From the point of view of physics, it seems that even the Hilbert space of one particle must be non-separable. There is an uncountable number of points in 3D space. A distinct position eigenvector can be associated with each such point. These eigenvectors would form an uncountable orthonormal basis in the 1-particle Hilbert space.

Eugene.

 

[strangerep]

From the point of view of physics, it seems that even the Hilbert space of one particle must be non-separable. There is an uncountable number of points in 3D space. A distinct position eigenvector can be associated with each such point. These eigenvectors would form an uncountable orthonormal basis in the 1-particle Hilbert space.

Except that you don't have a scalar-valued inner product, and therefore don't have
a Hilbert space in the strict sense of that word. Instead, I guess you're thinking of the
usual delta-distribution valued inner product, which is fine, but it's not a Hilbert space.
It's actually a rigged Hilbert space. (Oops! There's that phrase again that you don't
want to talk about. :-)

But true nonseparable Hilbert spaces are interesting too. In Kibble's work on finding
particular dressing transformations to deal with IR divergences in QED in a more
logically satisfactory way than the standard approaches, he indeed constructs a very large
nonseparable space and solves the dynamical problem therein.

 

[meopemuk]

Except that you don't have a scalar-valued inner product, and therefore don't have
a Hilbert space in the strict sense of that word. Instead, I guess you're thinking of the
usual delta-distribution valued inner product, which is fine, but it's not a Hilbert space.
It's actually a rigged Hilbert space. (Oops! There's that phrase again that you don't
want to talk about. :-)

I guess you are talking about the usual practice to represent the position-space wave function of a state localized at point $$a$$ by the delta function

$$\psi_a(x) = N\delta(x-a)$$

where $$N$$ is a normalization factor. If we adopt this rule and represent the inner product by integration, then the product of two localized states is

$$\langle a | b \rangle = N^2 \int dx \delta(x-a) \delta(x-b) = N^2 \delta(a-b)$$

Then in order to have a normalized localized state $$\langle a| a \rangle = 1$$, we need to set $$N = 0$$, which is nonsense.

I think we can avoid this controversy by choosing the "square root of the delta function" as the wave function for localized states. Then the norm of any such localized state is

$$\langle a | a \rangle = \int dx \sqrt{\delta(x-a)} \sqrt{\delta(x-a)} = \int dx \delta(a-x) = 1$$

as required in quantum mechanics.


Eugene.

 

[strangerep]

I guess you are talking about the usual practice to represent the position-space wave function of a state localized at point $$a$$ by the delta function

$$\psi_a(x) = N\delta(x-a)$$

where $$N$$ is a normalization factor. If we adopt this rule and represent the inner product by integration, then the product of two localized states is

$$\langle a | b \rangle = N^2 \int dx \delta(x-a) \delta(x-b) = N^2 \delta(a-b)$$

Then in order to have a normalized localized state $$\langle a| a \rangle = 1$$, we need to set $$N = 0$$, which is nonsense.

Yes, but in RHS we don't need the N at all. The inner product is distribution-valued.

I think we can avoid this controversy by choosing the "square root of the delta function" as the wave function for localized states. Then the norm of any such localized state is

$$\langle a | a \rangle = \int dx \sqrt{\delta(x-a)} \sqrt{\delta(x-a)} = \int dx \delta(a-x) = 1$$

as required in quantum mechanics.

Is this "square-root of delta function" thing your own idea?
If not, could you give some references, please?

What happens in the following case?

$$\langle a | b \rangle = \int dx \sqrt{\delta(x-a)} \sqrt{\delta(x-b)} ~=~ (?) \int dx \sqrt{\delta(x-a) \; \delta(x-b)} ~=~ (?)$$

If a and b are unequal, then the product under the square-root sign is
exactly zero everywhere. So how does one interpret the original integral
rigorously to end up with something like $$\delta(a-b)$$ ?

Edit: Oh, I just realized... you probably mean it's 0, right?
But if so, these states don't give a well-defined resolution of unity
of the form
$$\int\!dx |x\rangle\langle x|$$
because each $$\langle a | a \rangle$$ is only defined on a
set of measure zero, -- hence it's not good Lebesgue integral.

 

[meopemuk]

Is this "square-root of delta function" thing your own idea?
If not, could you give some references, please?

I haven't seen this idea in the literature.

What happens in the following case?

$$\langle a | b \rangle = \int dx \sqrt{\delta(x-a)} \sqrt{\delta(x-b)} ~=~ (?) \int dx \sqrt{\delta(x-a) \; \delta(x-b)} ~=~ (?)$$

If a and b are unequal, then the product under the square-root sign is
exactly zero everywhere. So how does one interpret the original integral
rigorously to end up with something like $$\delta(a-b)$$ ?

In the "square root" approach

$$\langle a | b \rangle = 1, \ \ \ \ if a=b$$
$$\langle a | b \rangle = 0, \ \ \ \ if a \neq b$$

so $$\langle a | b \rangle$$ can be interpreted as the *probability* amplitude of finding state a in the state b.

In the usual approach, where $$\langle a | b \rangle = \delta(a-b)$$, the inner product $$\langle a | b \rangle$$ should be interpreted as the *probability density* amplitude.

 

[DrFaustus]

meopemuk -> I think one should be very careful with the type of expression you wrote (I'm referring to your "square root of the delta"). The reason is again the problem with multiplication of distributions. First of all you'd have to define what kind of object is your square root of the delta, and I think you'd need it to be a distribution as well or you wouldn't be able to get the singularity structure correctly for its square, i.e. Dirac's delta. And then you'd have to take care about it's product. And to do so, you'd have to first define how it acts so one could study it's properties and then show that you can (a) multiply two such objects and (b) that the product is indeed the Dirac delta. I don't know if this can be done or not, but I suspect it cannot. So your expression would need some sort of correction.

 

[meopemuk]

DrFaustus,

I can always define the delta function as a limit of a sequence of "normal" functions whose integral is equal to 1 and whose support is shrinking around one point.

Similarly, I can define the "square root of the delta function" as a limit of a sequence of "normal" functions such that integrals of their squares are equal to 1 and supports are shrinking.

Eugene.

 

[Avodyne]

I can always define the delta function as a limit of a sequence of "normal" functions whose integral is equal to 1 and whose support is shrinking around one point.

Similarly, I can define the "square root of the delta function" as a limit of a sequence of "normal" functions such that integrals of their squares are equal to 1 and supports are shrinking.

You can do that, but the problem is that if you take the product of the "square root of the delta function" with any smooth square-integrable function, and integrate, the result is zero. So the "square root of the delta function" is equivalent to the null vector in Hilbert space of square-integrable functions.

 

[meopemuk]

You can do that, but the problem is that if you take the product of the "square root of the delta function" with any smooth square-integrable function, and integrate, the result is zero. So the "square root of the delta function" is equivalent to the null vector in Hilbert space of square-integrable functions.

This is correct if you limit yourself to "smooth" functions only. The "square root of delta" is not smooth, but it *is* square-integrable, and the inner product of this function with itself is equal to 1. So, this function is not a null vector in the Hilbert space of "both smooth and non-smooth" square-integrable functions.

I don't think that we should ban non-smooth functions from our Hilbert space. If we do that then eigenfunctions of position and momentum are not permitted, and our theory lacks the important observables of position and momentum. This is not acceptable, in my opinion.

Moreover, wave function is not a physical field or substance. It is just a mathematical measure of probability distribution. There is no law forbidding the probability density to change abruptly. So, nothing forbids existence of discontinuous or non-smooth wave functions.

Eugene.

 

[DarMM]

This is correct if you limit yourself to "smooth" functions only. The "square root of delta" is not smooth, but it *is* square-integrable, and the inner product of this function with itself is equal to 1. So, this function is not a null vector in the Hilbert space of "both smooth and non-smooth" square-integrable functions.

No, Avodyne is referring to how it behaves as a distribution. As a distribution it vanishes, since its integral against a smooth function vanishes. It's nothing to do with its own smoothness.

I don't think that we should ban non-smooth functions from our Hilbert space. If we do that then eigenfunctions of position and momentum are not permitted, and our theory lacks the important observables of position and momentum. This is not acceptable, in my opinion.

We don't. The Hilbert space of standard QM has several non-smooth functions in it.

Also you should stop repeating the claim that if two operators have no eigenfunctions in the Hilbert space then they don't exist as observables in the theory. To be observables they just have to be self-adjoint on the Hilbert space, which position and momentum are. If their eigenfunctions aren't elements of the Hilbert space it just means that they have no states with no uncertainty in the value of that observable, which certainly doesn't mean they aren't observables.

 

[meopemuk]

Also you should stop repeating the claim that if two operators have no eigenfunctions in the Hilbert space then they don't exist as observables in the theory. To be observables they just have to be self-adjoint on the Hilbert space, which position and momentum are. If their eigenfunctions aren't elements of the Hilbert space it just means that they have no states with no uncertainty in the value of that observable, which certainly doesn't mean they aren't observables.

In my definition, an observable exists if for each (eigen)value there are states in which this (eigen)value is measured with no uncertainty. This is, of course, a personal opinion, which cannot be proven neither by experiment nor by a deeper theory.

Eugene.

 

[strangerep]

I can always define the delta function as a limit of a sequence of "normal" functions whose integral is equal to 1 and whose support is shrinking around one point.

Similarly, I can define the "square root of the delta function" as a limit of a sequence of "normal" functions such that integrals of their squares are equal to 1 and supports are shrinking.

As others have indicated, this doesn't seem possible.
Could you please give a specific example of such a sequence of "normal
functions" that does indeed have the desired properties?

The obvious first attempt fails: consider the usual delta distribution
represented as the limit of a sequence of Gaussians:

$$\delta_\epsilon(x) ~:=~ \lim_{\epsilon\to 0} ~ \frac{1}{\epsilon\sqrt{\pi}} ~ \exp(-x^2/\epsilon^2)$$

Then the naive square-root of this is

$$\sqrt{\delta_\epsilon(x)} ~:=~ \lim_{\epsilon\to 0} ~ \frac{1}{\sqrt{\epsilon}\; \pi^{1/4}} ~ \exp(-x^2/2\epsilon^2)$$

but a short computation shows that

$$\lim_{\epsilon\to 0} \int\!dx \; \sqrt{\delta_\epsilon(x)} ~=~ 0$$

and similarly,

$$\lim_{\epsilon\to 0} \int\!dx \; x^n \sqrt{\delta_\epsilon(x)} ~=~ 0$$

for any non-negative n.

So defining the "square root of a delta distribution" in the above way
doesn't work usefully. It's equivalent to the trivial zero distribution,
hence a set of such "functions", indexed by a continuous parameter,
cannot serve as a basis for a nontrivial Hilbert space.

(BTW, this also means that "square-root" is a misleading name since we
normally think that if $$\sqrt{z}=0$$, then $$z=0$$,
which is not the case here.)

I suppose one could then say: ok, the square of a dirac delta is not a
distribution, but some other object type which I'll call "Rdist", with
properties (presumably) like the following:

Definition: An Rdist space V is a linear space over the complex
field, equipped with a symmetric bilinear product

$$\star : V\times V \to D$$

where D is a space of distributions, such that (for $$v,w\in V$$),

$$v\star v = \delta$$ and $$v\star w = 0$$ if $$v \ne w$$.

But then how does one define a resolution of unity on V? There needs to
be two different kinds of product (inner and outer, presumably).
And we need integration over the elements of V to form continuous
linear combinations. But the trivially-zero integrals above seem to
prohibit this.

So if there's a way to make this idea both rigorous and useful,
you need to show me what it is. I've failed to figure it out for myself.

 

[meopemuk]

As others have indicated, this doesn't seem possible.
Could you please give a specific example of such a sequence of "normal
functions" that does indeed have the desired properties?

The obvious first attempt fails: consider the usual delta distribution
represented as the limit of a sequence of Gaussians:

$$\delta_\epsilon(x) ~:=~ \lim_{\epsilon\to 0} ~ \frac{1}{\epsilon\sqrt{\pi}} ~ \exp(-x^2/\epsilon^2)$$

Then the naive square-root of this is

$$\sqrt{\delta_\epsilon(x)} ~:=~ \lim_{\epsilon\to 0} ~ \frac{1}{\sqrt{\epsilon}\; \pi^{1/4}} ~ \exp(-x^2/2\epsilon^2)$$

but a short computation shows that

$$\lim_{\epsilon\to 0} \int\!dx \; \sqrt{\delta_\epsilon(x)} ~=~ 0$$

and similarly,

$$\lim_{\epsilon\to 0} \int\!dx \; x^n \sqrt{\delta_\epsilon(x)} ~=~ 0$$

for any non-negative n.

So defining the "square root of a delta distribution" in the above way
doesn't work usefully. It's equivalent to the trivial zero distribution,
hence a set of such "functions", indexed by a continuous parameter,
cannot serve as a basis for a nontrivial Hilbert space.

Let me rephrase your example a little bit.

Let us take a "normal" square-integrable function $$\phi(x)$$, which is normalized to unity. For example, this can be a fixed-width gaussian. If "square root of delta" functions form a valid basis in the Hilbert space, then the sum of squares of projections of $$\phi(x)$$ on all these basis functions must be equal to one.

Next we can show (as you already did) that the inner product (the projection) of $$\phi(x)$$ with any "square root delta" centered at point $$a$$ tends to zero as $$\epsilon\to 0$$

$$\lim_{\epsilon\to 0} \int\!dx \; \phi(x) \sqrt{\delta_\epsilon(x-a)} ~=~ 0$$

This is definitely true. However, this does not mean that the sum of squares of all projections also tends to zero. As $$\epsilon\to 0$$ we should increase the number of $$\sqrt{\delta_\epsilon(x-a)$$ functions in the basis, so that in the limit there is one such function centered at each value of $$a$$. In this limit there are uncountably many basis functions and the total sum (of squares of projections) becomes an indefinite number of the type (zero)x(infinity).


My guess is that if we take this limit properly, then the result should be (zero)x(infinity) = 1, which means that "square root of delta" functions is a valid basis set in the Hilbert space of square-integrable functions.

Eugene.

 

[DarMM]

Hey, I don't know if bringing up old threads is bad here (Apologies if it is), I just wanted to bring this thread back up since I feel that, like its predecessor, it got derailed by Bob_for_short and meopemuk discussing their own views of what QFT should be like. I just thought I'd bring this back up in case anybody wants a good discussion or wants to ask questions.

 

[George Jones]

Although there is no hard-and-fast rule, this is somewhat frowned upon. In this case, however, I think its is a great idea. I neither know enough nor have enough time to participate actively in the discussion, but I will be watching with interest.

 

[DrFaustus]

DarMM -> I do have a question for you. It might be a bit off topic, but not much. Do you know of any references where I could find an explicit position space expression for the Feynman propagator on the cylinder (1+1 D spacetime $$\mathbb{R} \times \mathbb{S}$$ )? The momentum space representation is the same as on Minkowski space, but I basically don't know how to compute the inverse Fourier transform, which now involves a sum over the discrete momenta rather than an integral.

 

[DarMM]

DarMM -> I do have a question for you. It might be a bit off topic, but not much. Do you know of any references where I could find an explicit position space expression for the Feynman propagator on the cylinder (1+1 D spacetime $$\mathbb{R} \times \mathbb{S}$$ )? The momentum space representation is the same as on Minkowski space, but I basically don't know how to compute the inverse Fourier transform, which now involves a sum over the discrete momenta rather than an integral.

Hey, yes actually. If the infinite volume Feynman propagator is $$\Delta(x,t)$$, then the finite volume propagator you are looking for is given by:
$$\sum_{n}\Delta(x + nL,t) \qquad n \in \mathbb{Z}$$
where L is the size of the circle. (The "length of the world")
A proof can be found in Glimm and Jaffe's book, section 7.3, immediately following Proposition 7.3.1. This is also a routine type of calculation in Lattice Field theory so you could take a look at:
"Quantum Fields on the Lattice" by I. Montvay and G. Münster, Cambridge Monographs on Mathematical Physics, Cambridge University Press, 1994.

If you want a quick idea of the proof, you can change the sum in the Fourier series into an integral and obtain
$$\frac{1}{L} \sum_{\underline{n} \in \mathbb{Z}} \int{d\omega} = \int{d^{2}k} \left[\sum_{\underline{n} \in \mathbb{Z}}\delta\left(\underline{k} - \frac{2\pi\underline{n}}{L} \right)\right]$$
then use Córdoba's formula to replace the delta function:
$$\left[\sum_{n \in \mathbb{Z}}\delta\left(\underline{k} -\frac{2\pi\underline{n}}{L} \right)\right] = \sum_{n \in \mathbb{Z}} e^{i nkL} \end{equation}$$
And you'll see the origin of the terms.

 

[Ben Niehoff]

I appreciate the proof above that $\sqrt{\delta(x)}$ does not work as a basis element. I have often thought, "Why not just use the square root of delta?", and now I see why not. :)

 

[DrFaustus]

DarMM -> Thanks for the answer! It looks like the same answer I was considering, only in a slightly different way (didn't do all the computations, but it should be the same). The idea was to first perform the "time integral", which gives me $$\Delta(t,\vec{p})$$ in your notation. Then use Poisson's summation formula which basically gives the answer you wrote down. Only "problem" now is that the infinite volume propagator is written in terms of Bessel functions and I have no idea if that sum could be explicitly evaluated, perhaps in terms of some "exotic" function. Which is what I was actually hoping for and asking for. But I'm guessing that if in the book of Glimm and Jaffe it's left in that form, than that's pretty much the best we have. I will have a look at the references though. Oh, and it's good to know you're around again. :)

 

[strangerep]

I just wanted to bring this thread back up since I feel that,
like its predecessor, it got derailed [...]. I just thought I'd bring this back up in case
anybody wants a good discussion or wants to ask questions.

Thanks for coming back. I was quite disappointed about what happened before.
I wanted to get clear first how the very simple example of a time-independent external
field can give rise to unitarily inequivalent Hilbert spaces, before moving on to
the more interesting case of a time-dependent external field which Bob_for_short
was interested in, and then move on to more difficult cases. Unfortunately, he became
impatient and walked out in a huff too soon.

Anyway, I'm still interested in seeing a case where the interaction entails a
loss of the CCRs (which is not the case for the simple external field example).
If you have the time and patience to write a post about that I'd be grateful. :-)
Maybe a new separate thread on that specific subject would be best, in which
we stick to that sub-topic (and this current thread is already very long).

 

[DarMM]

Only "problem" now is that the infinite volume propagator is written in terms of Bessel functions and I have no idea if that sum could be explicitly evaluated, perhaps in terms of some "exotic" function. Which is what I was actually hoping for and asking for. But I'm guessing that if in the book of Glimm and Jaffe it's left in that form, than that's pretty much the best we have. I will have a look at the references though. Oh, and it's good to know you're around again. :)

Yeah, unfortunately not. Glimm and Jaffe leave it in that form because there's not much more you can do with it. I imagine you now have sums over modified Bessel functions of the second kind (only a guess, maybe you used other Bessel functions). However I can tell you that the series converges quite rapidly and so only the first few sums in the series are "large".
Oh, and thanks for welcoming me back!

 

[DarMM]

Canonical Commutation Relations

Anyway, I'm still interested in seeing a case where the interaction entails a
loss of the CCRs (which is not the case for the simple external field example).
If you have the time and patience to write a post about that I'd be grateful. :-)
Maybe a new separate thread on that specific subject would be best, in which
we stick to that sub-topic (and this current thread is already very long).

I'll certainly do a post on it. Essentially we've already covered step one with the external field, which is the interaction causing a change of representation, but only to another Fock rep. The second step would be the interaction causing moving things further to a non-Fock rep (this is associated with mass renormalization). Finally we could cover the case where the interactions cause a complete failure of the CCR, (associated with Wave-Function/Field Strength renormalization).

I'll try it here first, however if it's a bit cumbersome I could move it to another thread.
Or perhaps, if the moderators could tell me, would it be better to start a fresh thread altogether?

 

[DarMM]

Scattering Theory

Before I begin I wanted to clear up some possible confusion about the different types of particles "bare", "free" and "physical" and how these relate to the different Hilbert spaces issue.

In scattering theory in QM, if you have the usual properties of Asymptotic completeness, e.t.c. then as the physical state vector $$\psi$$ evolves past the time of interaction its evolution coincides with that of a free particle. Essentially:
$$U(t)\psi - U_{0}(t)\psi_{out} \rightarrow 0$$ as $$t \rightarrow \infty$$.
With $$U(t)$$ the evolution operator and $$U(t)_{0}$$ the free evolution operator.
Essentially the physical state eventually begins to look like a bunch of freely moving particles.

In QFT this is still the case, the state will eventually evolve into a collection of non-interacting particles, i.e. free particles. However Haag's theorem basically says that this evolution never tends to the evolution of an actual free theory. In other words, even though the particles do become non-interacting asymptotic scattering states their evolution is at no point similar to the evolution of particles from the free theory.

An example will make this less surprising. Take QED and set the electric charge to zero. This is a free theory consisting of free fermions and spin one bosons. The free fermions have charge $$0$$. Now take normal QED and imagine Coulomb scattering. After the scattering the electrons will move away from each other and no longer interact. They become free. However they never actually look like the fermions from the $$e = 0$$ theory, because they always have an electric charge.

In a more general way this is the basic difference between QM and QFT that Haag's theorem tells us about. In QM I can take an eigenstate of the harmonic oscillator and through a series of physical operations prepare an eigenstate of the anharmonic oscillator.

However, in QFT, I can never take a single electron with charge $$e$$ and from it prepare a free fermion with no charge. These two theories live in totally different worlds or in the language of QM, two different representations or Hilbert spaces.

This doesn't change anything about scattering theory fundamentally, I still have in/out states its just that they have nothing to do with the free Hamiltonian. You can even see this in regular (non-rigorous) QFT and QM. In QM the S-matrix is usually defined with things like Møller operators, however in QFT the S-matrix is defined as poles of the correlation functions. This is because in QFT Møller operators don't exist, since there is no link to an interacting theory. This is the whole reason for the LSZ formalism.

 

[DrFaustus]

DarMM -> Why do you say that in QFT Moeller operators don't exist? Staying in 2D and for a scalar field, I certainly can define an object like $$\exp[-i(t-s)H_0] \exp[i(t-s)H]$$. And this object is well defined and for finite s and t. My understanding is that problems occur when you take the limits $$s \to -\infty$$ and $$t \to +\infty$$. Or am I missing something here?

 

[strangerep]

In scattering theory in QM, if you have the usual properties of Asymptotic completeness, e.t.c.
then as the physical state vector $\psi$ evolves past the time of interaction
its evolution coincides with that of a free particle.
[...]
In QFT this is still the case, the state will eventually evolve into a
collection of non-interacting particles, i.e. free particles.

I've begun to be quite puzzled why QFT is still based on this notion,
since it's been known for many decades that even for the
non-relativistic Coulomb interaction the asymptotic dynamics does not
tend to the free dynamics. I wrote a summary post on this recently:

https://www.physicsforums.com/showpost.php?p=2560777&postcount=3

But it seems that, even if one tries to solve the full theory in
terms of the correct asymptotic dynamics (as found in the paper
by Kulish & Fadeev referenced in my post), one overcomes only the
IR divergence but UV problems persist.

But don't let me divert your train of thought... :-)

 

[DarMM]

DarMM -> Why do you say that in QFT Moeller operators don't exist? Staying in 2D and for a scalar field, I certainly can define an object like $$\exp[-i(t-s)H_0] \exp[i(t-s)H]$$. And this object is well defined and for finite s and t. My understanding is that problems occur when you take the limits $$s \to -\infty$$ and $$t \to +\infty$$. Or am I missing something here?

Believe it or not, it isn't well defined for finite s and t. There is no Hilbert space on which the two unitary groups $$\exp[-i(t-s)H_0]$$ and $$\exp[i(t-s)H]$$ are both well-defined. This is, essentially, Haag's theorem.

 

[DarMM]

I've begun to be quite puzzled why QFT is still based on this notion,
since it's been known for many decades that even for the
non-relativistic Coulomb interaction the asymptotic dynamics does not
tend to the free dynamics. I wrote a summary post on this recently:

https://www.physicsforums.com/showpost.php?p=2560777&postcount=3

But it seems that, even if one tries to solve the full theory in
terms of the correct asymptotic dynamics (as found in the paper
by Kulish & Fadeev referenced in my post), one overcomes only the
IR divergence but UV problems persist.

But don't let me divert your train of thought... :-)

There are a few issues here. First of all the infrared divergences in QED are a special case that doesn't occur in most other field theories. There, as you have said, the asymptotic states are really the electrons and their radiation fields (in old language "soft" photons) and "hard" photons. The rigorous version of this an be seen in the book by Steinmann "Perturbative quantum electrodynamics and axiomatic field theory" and the Symmetry breaking book by Strocchi. In those books you can see that the infrared problem is actually solved by using the correct asymptotic states.

In the case of massive field theories this will not happen. What I was discussing above is something else. Basically in massive field theories the dynamics do tend to free dynamics, but not free dynamics controlled by the non-interacting Hamiltonian. That is, the particles do become free and evolve as a non-interacting state. However unlike in QM, that free motion cannot be described by $$H_{0}$$. This is why the Møller operators do not exist and why there is no interaction picture.
Basically we are using the correct asymptotic states and they are free states. However they are not, as the usual approach assumes, actually states of the the theory with Hamiltonian $$H_{0}$$.

As you said, the UV divergences are another issue, which is the main issue I'll be covering.

 

[DrFaustus]

DarMM -> Consider the following. The Hamiltonian of my 2D scalar field with quartic interaction is the usual one and I split it as $$H = H_0(t) + V(t)$$ with
$$H_0(t) = \frac{1}{2}\int d x \Big[ \pi^2(t,x) + (\partial_x \varphi(t,x))^2 + m^2 \varphi^2(t,x) \Big]$$ and
$$V(t) = \frac{\lambda}{4!} \int \ud x \varphi^4(t,x)$$, and everything is to be considered normal ordered so it all makes sense. I also want to emphasize that both the "free hamiltonian" and the "potential" (according to my arbitrary split) are constructed out of interacting fields. No interaction picture fields around. Now, with these two objects, I define
$$S(s,t) := \exp\big[i (s-t) H_0(s) \big] \exp\big[- i (s-t) H\big]$$. Are you saying that this object is ill defined? That would confuse me a lot... :/

 

[strangerep]

DarMM -> Consider the following. The Hamiltonian of my 2D scalar field with quartic interaction is the usual one and I split it as $$H = H_0(t) + V(t)$$ with
$$H_0(t) = \frac{1}{2}\int d x \Big[ \pi^2(t,x) + (\partial_x \varphi(t,x))^2 + m^2 \varphi^2(t,x) \Big]$$ and
$$V(t) = \frac{\lambda}{4!} \int \ud x \varphi^4(t,x)$$, and everything is to be considered normal ordered so it all makes sense. I also want to emphasize that both the "free hamiltonian" and the "potential" (according to my arbitrary split) are constructed out of interacting fields. No interaction picture fields around. Now, with these two objects, I define
$$S(s,t) := \exp\big[i (s-t) H_0(s) \big] \exp\big[- i (s-t) H\big]$$. Are you saying that this object is ill defined? That would confuse me a lot... :/

Looks like DarMM doesn't have much spare time to devote to this thread, so I'll offer
my $0.02 worth (just to keep this thread alive)...

Typically, interacting Hamiltonians in fermion-boson theories, expressed via a/c ops,
contain terms like a*b*c* (where a,b are fermionic, and c is bosonic). In phi^4 theory
we have stuff like c*c*c*c*. Such operators are ill-defined on the usual Hilbert space,
and normal-ordering doesn't fix this for such terms.

Paraphrasing Dirac, the interaction is so "violent" that one is ejected
from the Hilbert space within an infinitesimal time. :-)

 

[meopemuk]

Typically, interacting Hamiltonians in fermion-boson theories, expressed via a/c ops,
contain terms like a*b*c* (where a,b are fermionic, and c is bosonic). In phi^4 theory
we have stuff like c*c*c*c*. Such operators are ill-defined on the usual Hilbert space,
and normal-ordering doesn't fix this for such terms.

Paraphrasing Dirac, the interaction is so "violent" that one is ejected
from the Hilbert space within an infinitesimal time. :-)

The idea of the "dressed particles" approach is to declare such violent interactions illegal. Only Hamiltonians without such terms should be considered physical. By introducing this limitation the dressed particle approach is not losing generality. It can be shown that the whole point of renormalization is to neutralize the disastrous effect of these "bad" interaction terms on the S-matrix.

Eugene.

 

[DrFaustus]

strangerep -> Thanks for the input. However, the reason why I sticked to $$\lambda \varphi^4$$ in 2D is precisely because in this case you "don't need much" to make the theory well defined. It is a result of Glimm and Jaffe that all you need in 2D is normal ordering, and temrs like a*a*a*a* in the interaction are normal ordered, so the resulting interaction term, actually the Hamiltonian, is well defined. And I think you also get to stay in the free particle Fock space (not totally sure about this tho'). So yeah, in 2D interactions are pretty *mild*, not even strong wnough to eject me fromFock space :) And this is essentially the reason why I'd be surprised if the (local) Moeller operators I have defined above would not be meaningful objects. And as I remarked above, there are no "interaction picture" fields around...

meopemuk -> You again started talking about the dressed particle approach. Please, if you want to discuss it, open a new thread and by all means let's talk about it. (As I replied to strangerep, the interaction terms you'd like to ban are perfectly acceptable in 2D.). But *please* do not disrupt this thread by ideas on "how should QFT be modified so it'd be nicer". I want to emphasize again that QFT is a well established *mathematical physical* framework (emphasis on mathematical) and as such it has it's own rules and axioms. Let us talk about QFT staying within QFT, shall we?

 

[meopemuk]

(emphasis on mathematical)

Sorry, I've thought mistakenly that we are talking about physics here.

Eugene.

 

[DarMM]

strangerep -> Thanks for the input. However, the reason why I sticked to $$\lambda \varphi^4$$ in 2D is precisely because in this case you "don't need much" to make the theory well defined. It is a result of Glimm and Jaffe that all you need in 2D is normal ordering, and temrs like a*a*a*a* in the interaction are normal ordered, so the resulting interaction term, actually the Hamiltonian, is well defined. And I think you also get to stay in the free particle Fock space (not totally sure about this tho'). So yeah, in 2D interactions are pretty *mild*, not even strong wnough to eject me fromFock space :) And this is essentially the reason why I'd be surprised if the (local) Moeller operators I have defined above would not be meaningful objects. And as I remarked above, there are no "interaction picture" fields around...

Hey, DrFaustus sorry for the delay, I now have the time to focus on this thread so I won't be absent for so long again (hopefully!)

Anyway to your question. The 2D case is a funny one, in that it is locally Fock. I'll explain what this means. If the interaction is $$\int{:\phi(x)^{4}:dx}$$ then you are ejected from the Fock space. However if the interaction is $$\int_{\Lambda}{:\phi(x)^{4}:dx}$$, then you are not. Basically provided the interaction is confined to a finite region, then things are not violent enough to push you out of Fock space.
Or, to put it another way, the ultraviolet singularities can be tamed in Fock space (by Wick Ordering), but the infrared ones cannot. A field theory which is like this is called "locally Fock". Another example of a locally Fock theory is Yukawa theory in two dimensions.

(By the way, these infrared singularities are unrelated to the infrared singularities associated with photons in QED. They've a different origin and meaning.)

However these infrared singularities are nonperturbative so you'll never see them in perturbation theory. Hence the Møller operators can't exist on Fock space when the volume cutoff is removed.

However they can exist in the real Hilbert space in a more generalised fashion, due to Haag-Ruelle theory. Haag-Ruelle theory states that if a quantum field theory has a mass gap, then the overlap between outgoing and ingoing states in the real Hilbert space can be expressed as a mapping in an abstract Fock space. This mapping is the Møller operators. This is how you justify scattering theory in axiomatic QFT. It's also why the mass gap is important for the Clay Mathematics prize.

 

[DarMM]

Sorry, I've thought mistakenly that we are talking about physics here.

Eugene.

Well, we are and we are not. In this thread I take the assumption that QFT as it stands is an acceptable framework because of its empirical success. Given that, I am discussing the mathematical foundation of QFT and the rigorous justification of some of its formulas. So I am discussing the mathematical nuts and bolts of one of our greatest physical theories and hopefully some misunderstandings of its inner workings will be cleared up.

 

[sheenashirley]

Also you can show that the perturbative expansions of the S-matrix on the true Hilbert space can be written using Fock space expectation values, or to put it another way Fock space can simulate the real Hilbert space well enough for perturbative calculations. So standard perturbation theory has a mathematical backing.

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