Rigorous Quantum Field Theory.

In summary, strangerep and DarMM are discussing rigorous issues in quantum field theory. Strangerep says that the Epstein-Glaser approach is not more ad hoc then solving LaTeX Code: a x^2 + b x +c =0. DarMM says that the upshot is that at the end you've constructed the perturbative expansion for S(g) (the S-matrix in finite volume) in a completely rigorous way. Modern work on the Epstein-Glaser approach tries to take the limit g \rightarrow 1, to go to infinite volume, although it has proven extremely difficult. They agree that renormalized QFT (such as QED) can calculate the S-matrix (i.
  • #141
meopemuk said:
Once you got the Hamiltonian H, finding solutions for wave functions is just a technical task. ...

Thank you, Eugene, for your answer. But let Strangerep and DarMM answer. Maybe they will write something specific for the problem solution.

As I said previously, apart from equations, there are also some "boundary" conditions that fix the linear superposition coefficients. I know the spectrum of operators A. I want the problem solution in terms of As and their vacuums. How many A-quanta are present in the solution, what is the average energy, etc.? I need an explicit solution to calculate all that. A formula like the original one for Ψ but in terms of A and their vacuums. Isn't it the formula given in my post #110? Anyway, I am very sure that our rigorous mathematicians can find and write this problem solution.
...In this particular case the solution is trivial: The Hamiltonian (in terms of physical a/c operators) has a non-interacting form, so physical particles propagate free, without interactions.
The exact solution in terms of a-operators does not contain any interaction between different a-modes either. They remain decoupled.
In the general case, it is a much more complicated task to express physical states as linear combinations of "bare" states. But, as I stressed a few times already, such expressions have no physical meaning, so we should not bother.
I consider the a-operators as physical. They carry energy-momentum, spin, etc., as dictated by their antenna j(r,t) You all blame them for nothing, in my opinion.

But let us wait for the answer.
 
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  • #142
Several words in favour of a-operators.

...what it shows is that they live in different Hilbert spaces,
which is what we've been talking about.
Let us see. The operator a has many eigenvectors called coherent states:
a|b> = b|b> where b is a complex number. The shifted operator A = a + z has the shifted eigenvalues in this basis: A|b> = (b+z)|b>. So do the momentum, coordinate, and many other operators in QM. I do not see any reason to build different Hilbert spaces because of this.

In particular, the eigenvector |-z> of a-operator is the vacuum vector for A. The operator A lives well in the space of eigenvectors of the a.
...there are no coherent states.
1) The operator's a and A algebras do not depend on mass m. In this respect they are similar to photon c/a operators.

2) If j(k) = jδ(k-k0) (the simplest monochromatic source), the exact solution |ψ> is an eigenstate of the operator a(k0) which is a coherent state by definition.

The dispersion law contains the mass m but it is a rather secondary thing in our discussion. It just means that for a given current j the heavier bosons will have a smaller average number in the given coherent state. It is directly seen from the exact solution and it is physically well comprehensible.
 
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  • #143
Bob_for_short said:
The exact solution in terms of a-operators does not contain any interaction between different a-modes either. They remain decoupled.

Let me remind you that the DarMM's Hamiltonian in terms of a-operators is

[tex]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)}[/tex]

The first term is the usual Hamiltonian of free a-particles, and the second term must be regarded as interaction. This interaction is very peculiar, because it leads to processes (described by the [tex]a^*[/tex] term) which spontaneously create a-particles out of vacuum. It also describes processes (described by the [tex]a[/tex] term) in which an a-particle gets annihilated. This means that given an initial state with just one a-particle this state evolves in time in a complicated linear combination of many-particle a-states. This doesn't look like a free non-interacting time evolution to me.

Note that I do not consider [tex]\tilde{j}(k)[/tex] as a "source" or "antenna". I consider it just as a numerical factor in the interaction Hamiltonian. So, in my interpretation there is no any external force. The above Hamiltonian describes an isolated system of (any number of) a-particles.

Bob_for_short said:
I consider the a-operators as physical. They carry energy-momentum, spin, etc., as dictated by their antenna j(r,t) You all blame them for nothing, in my opinion.

Are you then considering both (bare) a-particles and (dressed) A-particles as being physical? Are you saying that both of them can be seen in experiments? Is it then true that both bare and dressed electrons of QED can be observed? But we see only one type of electrons in nature. I am lost.

Eugene.
 
  • #144
meopemuk said:
... the DarMM's Hamiltonian in terms of a-operators is

[tex]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)}[/tex]

The first term is the usual Hamiltonian of free a-particles, and the second term must be regarded as interaction.
The second one is an interaction. And indeed, it is a specific case - it is one of two cases when the interaction does not bring problems. This is the case of a given external source for quanta ak. Then the problem is solved exactly.

This interaction is very peculiar, because it leads to processes (described by the [tex]a^*[/tex] term) which spontaneously create a-particles out of vacuum. It also describes processes (described by the [tex]a[/tex] term) in which an a-particle gets annihilated. This means that given an initial state with just one a-particle this state evolves in time in a complicated linear combination of many-particle a-states. This doesn't look like a free non-interacting time evolution to me.
You concentrate too much on a and vacuum. As I said previously, here the operators a and a+ come in such a combination that does not bring any problem. Everything is OK here: the particles are created indeed by the source. As soon as we work with a+, it creates particles "from vacuum". The solution, however, can be written in another, maybe better for your perception way:

|z> = exp(za+ - z*a)|0> = exp(-|z|2/2) eza+|0> = exp(-|z|2/2)∑n=0(zn/√n!)|n>.

The latter expression does not have any operators. It is a superposition of different states with different numbers of quanta a.

Note that I do not consider [tex]\tilde{j}(k)[/tex] as a "source" or "antenna". I consider it just as a numerical factor in the interaction Hamiltonian. So, in my interpretation there is no any external force. The above Hamiltonian describes an isolated system of (any number of) a-particles.
The system is not isolated because it is fed by a source. The evolution, however, is physical, without UV and IR divergences. There is no self-action in this interaction, that is why everything is OK.
Are you then considering both (bare) a-particles and (dressed) A-particles as being physical? Are you saying that both of them can be seen in experiments? Is it then true that both bare and dressed electrons of QED can be observed? But we see only one type of electrons in nature. I am lost.
In this particular problem the a- and A-operators are physical. I have already wrote the solution in their terms with explanations. They are not bare nor dressed. From physical and experimental point of view they are indistinguishable. They are alike.
I will write later about electrons and photons. It is a slightly different case. Besides, I do not want to get banned once more.
 
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  • #145
Bob_for_short said:
The system is not isolated because it is fed by a source.

Well, then we have a completely different physical interpretation of the Hamiltonian. You and I are talking about different physical systems. I am talking about an isolated system of a-particles with (self)interaction. You are talking about a-particles being emitted/absorbed by an external source or antenna.

If we want to consider this example as a model for quantum field theories without external potentials (like QED) then, I think, my interpretation is more appropriate.

Eugene.
 
  • #146
meopemuk said:
... I am talking about an isolated system of a-particles with (self)interaction.
But there is no self-interaction here!
You are talking about a-particles being emitted/absorbed by an external source or antenna.
Because the equation says so.
If we want to consider this example as a model for quantum field theories without external potentials (like QED) then, I think, my interpretation is more appropriate.
You are nearly right and I am right exactly. In fact, this situation is typical for soft radiation when the electron recoil is neglected and the accelerated electron motion is considered as known, given. Then the self-action term is neglected and the radiated filed coincides with the classical radiation. It is, of course, physical.
 
  • #147
Bob_for_short said:
Because the equation says so.

The same equation can describe (or model) different physical situations. Of course, you can apply DarMM's Hamiltonian to the emission of photons by antenna or laser. Then you make a crucial approximation of replacing a complex system of interacting particles (antenna or laser) with simple function j(k). Yes, you can do that, but then you are not talking about "Rigorous Quantum Field Theory".

In my opinion, any rigorous theory must deal with isolated systems only. So, I am suggesting to consider the DarMM's Hamiltonian as a description of a collection of (arbitrary number of isolated) a-particles. In this Hamiltonian, the [tex] a^*a[/tex] term is the free part and the [tex](a^*+a)[/tex] term is the interaction.

Eugene.
 
  • #148
Interaction of what with what? What species gives away its energy-momentum and what species gets it? The answer has been given above.

Exactly solvable problems belong to the Rigorous QFT too.

It's a chapter with the soft radiation treatment.
 
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  • #149
Bob_for_short said:
Interaction of what with what? What species gives away its energy-momentum and what species gets it?

This is a good question, and this is exactly why DarMM's Hamiltonian (where one particle is allowed to be created or anihilated) is a good model for QED. In QED we also have interaction terms in which three particles (1 electron, 1 positron, 1 photon) are spontaneously created or annihilated together. In my opinion, DarMM's model allows us to study the effect of these "bad" interactions in a simplified setting.

The conclusion of such a study should be this:

1. Bare a-particles in both DarMM's model and in QED have nothing to do with real observable particles.

2. The Hamiltonian written in terms of bare a-particles is unphysical and almost useless (the only useful property is that the S-matrix calculated with this Hamiltonian agrees with experiment).

3. To save the theory we must change to the basis of physical A-particles. If the Hamiltonian is expressed in terms of physical a/c operators, then all "bad" interaction terms disappear, and it becomes possible to give reasonable physical interpretation to "good" interactions present in the A-Hamiltonian. Using this A-Hamiltonian we can do all routine quantum mechanical calculations (finding bound states by diagonalization, exploring the time evolution of states and observables, calculating the S-matrix) without divergences and renormalization.

Eugene.
 
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  • #150
Bob_for_short said:
So, starting from H(A[j]), please write down the problem solution [itex]\Psi[/itex].
I can't wait to compare the physical and "non-physical" expressions for [itex]\Psi[/itex].
Patience, like politeness, is a virtue.

But I would have written something similar to Eugene's post #140,
so I won't repeat it.

Bob_for_short said:
[...]A formula like the original one for Ψ but in
terms of A and their vacuums. Isn't it the formula given in my post #110?
Similar, but with some caveats as explained below.

I consider the a-operators as physical. They carry energy-momentum,
spin, etc., as dictated by their antenna j(r,t)
You keep assuming that j(x) is an electric current, or similar. But in
this problem j(x) is just a time-independent external background field.
No extra assumptions concerning its nature were made in the original
problem statement.

"Physical energy" is only defined by reference to the "physical" Hamiltonian.
The "physical" vacuum is defined as the eigenstate of the full Hamiltonian
with lowest eigenvalue, i.e.,

[tex]
|0\rangle_z ~:=~ exp(za^* - \bar{z}a)|0\rangle
~=~(?)~ exp(-|z|^2/2) exp(za^*)|0\rangle
[/tex]
The [itex]za^*[/itex] is shorthand for an integral like
[tex]
\int\!\! dk \; z(k) \, a^*(k) ~.
[/tex]

[itex]|0\rangle_z[/itex] is annihilated by the [itex]A(k)[/itex] operators.

The [itex]A^*(k)[/itex] act on [itex]|0\rangle_z}[/itex] to generate a Hilbert
space [itex]\mathcal{H}_z[/itex], just as the [itex]a^*(k)[/itex] act on
[itex]|0\rangle[/itex] to generate a Hilbert space [itex]\mathcal{H}_0[/itex].
The question is then whether [itex]\mathcal{H}_z[/itex] is unitarily equivalent to [itex]\mathcal{H}_0[/itex]. I.e., can any vector in
[itex]\mathcal{H}_z[/itex] be written as a linear combination of vectors in
[itex]\mathcal{H}_0[/itex]?

Let's recall the definition of z(k), i.e.,
[tex]
z(k) ~:=~ C \, \frac{j(k)}{\omega(k)^{3/2}}
[/tex]
(where several constants have been written simply as "C").

The point is this: if z(k) is not square-integrable but divergent, we have
[tex]
exp(-|z|^2/2) ~\to~ 0
[/tex]
from which it follows that the two vacua are orthogonal to each other.
Further tedious calculations also show that every vector in
[itex]\mathcal{H}_0[/itex] is orthogonal to [itex]|0\rangle_z[/itex].

Depending on the details of j(x), it may or may not be true that
z(k) is square-integrable.

If [itex]j(k) = j\,\delta(k-k_0)[/itex] (the simplest monochromatic source),
the exact solution [itex]|\Psi\rangle[/itex] is an eigenstate of the operator
[itex]a(k_0)[/itex] which is a coherent state by definition.

The z(k) corresponding your j(k) above is not square-integrable. We cannot even say
that [itex]|z|^2[/itex] is "divergent". Rather it's simply undefined, involving the square
of a delta distribution. Therefore the solution cannot be expressed as a linear
combination of vectors from [itex]\mathcal{H}_0[/itex]. Nevertheless, the solution
can be expressed in another unitarily-inequivalent Hilbert space,
which is the point of all this.
 
  • #151
DarMM said:
Okay, here is a model which is exactly solvable nonperturbatively and is under complete analytic control. Also virtually every aspect of this model is understood mathematically.

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

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

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

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
[tex]H = \int{dk \omega(k)a^{*}(k)a(k)}[/tex]
Where [tex]a^{*}(k)[/tex],[tex]a(k)[/tex] 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:
[tex]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)}[/tex]

So far, so good.

Now the normal mode creation and annihilation operators for this Hamiltonian are:
[tex]A(k) = a(k) + \frac{g}{(2\pi)^{3/2}}\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}[/tex]
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:
[tex]H = \int{dk E(k)A^{*}(k)A(k)}[/tex],
where [tex]E(k)[/tex] is a function describing the eigenspectrum of the full Hamiltonian.

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

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

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

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

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

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

This is what a meant by my previous comment:
The interacting Hilbert space is a non-Fock representation of the usual creation and annihilation operators
or
It is a Fock representation of unusual creation and annihilation operators.


I hope this post helps.

I agree that "interacting vacuum" [tex]\Omega[/tex] has zero overlap with each and every free state. However, I don't agree that this fact implies that [tex]\Omega[/tex] is outside the free Fock space.

This may sound as a paradox, but my point is that "even if all components of a vector are zero, the vector itself could be non-zero". The important thing is that the number of components is infinite. So, infinite number of (virtually) zero components can add up to a non-zero total value.

In fact, here we have an uncertainty of the type "zero"x"infinity". In order to resolve this uncertainty we need to take a proper limiting procedure. I.e., we should slowly move the interaction from "weak" to "strong" regime and look not only at individual components of the [tex]\Omega[/tex] vector, but also at the total sum of squares of these components (which is a measure of the overlap of [tex]\Omega[/tex] with the free Fock space). Then we will see that each particular component indeed tends to zero, but the total sum of squares remains constant (if the transformation is unitary). This means that [tex]\Omega[/tex] does not leave the free Fock space even if the interaction is "strong".

Eugene.
 
  • #152
strangerep said:
Patience, like politeness, is a virtue.
I agree. I would love to be a gentlemen.
But I would have written something similar to Eugene's post #140, so I won't repeat it.
No need to repeat it indeed. But in this particular problem one can go further than writing a symbolic solution of the Schroedinger equation. Eugene is a great guy, I like and respect him very much; he wrote a book on QED. The only chapter missing is the coherent states.

I have "Quantum Electrodynamics" by Akhiezer-Berestetski, the last russian edition (1981). There are though many previous editions with the same treatment of time evolution in QED and in particular the coherent states. I will give some formulae from it.

S-matrix as an N-ordered operator is [1]:

S = Tψ{exp[-(1/2)∫jµ(x')Dc(x' - x'')jµ(x'')d4x'd4x''] NAexp(i∫jν(x)Aν(x)d4x)}

Here Tψ is the chronological ordering of ψ-operators, NA is the normal ordering of A-operators, and Dc is the causal Green's function of d'Alambert equation.

If the electron motion can be considered as known (j is a given function of space-time, not an operator), then the chronological ordering of ψ-operators is inessential and the S-matrix becomes:

S = exp[-(1/2)∫jµ(x')Dc(x' - x'')jµ(x'')d4x'd4x''] NAexp(i∫jν(x)Aν(x)d4x)

Here only fields Aν are operators. Thus the solution is:

|Ψ> = exp(i∫jν(x)Aν(x)d4x)|0>

If you represent the operator Aν as an expansion in "plane waves" over k, then the integral will give you the Fourier images of the current and the photon c/a operators. The solution is hence a coherent state of the photon field.

In case of our scalar massive boson field φ, we obtain a similar result for the problem solution (given first by DarMM).

Now, we can choose the current according to our physical situation. No problems with integrability arise. We have just to write explicitly the normalizing volumes in fields and currents.

[1] S. Hori, Progr. Theor. Phys., V. 7, p. 578 (1952).
 
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  • #153
strangerep said:
You keep assuming that j(x) is an electric current, or similar. But in this problem j(x) is just a time-independent external background field. No extra assumptions concerning its nature were made in the original problem statement.

It is not only a physical current but also a time-dependent one. DarMM made a mistake. A time-independent current cannot create propagating quanta. It is a four-dimensional Fourier transform of j(x) which is involved.

I also find it quite misleading to call Ω a "physical vacuum" or "interacting ground state". It is not an eigenstate of the total Hamiltonian whatever representation you use. If the energy is uncertain, then no variable change can make it certain, and in particular zero. It is clear even without calculations. No basis vector change can modify the state itself. It is another mistake of DarMM's.

But let us see it closer:

The coherent state is (one-mode state):

|z> = exp(za+ - z*a)|0>, a|z> = z|z>.

The combination (za+ - z*a) is invariant in case of the "shift" variable change: a = A + z*, where z is a complex number. The coherent state |z*> is the A-operator vacuum. But what about the Hamiltonian?

A+A = a+a - (a+z* + az) + |z|2

It looks like the original Hamiltonian but we see that whatever z is, the term |z|2 is unavoidable. So the DarMM's variable change may not be correct - it does not contain the corresponding shifts in spectra.

a+a - (a+z* + az) = A+A - |z|2

What a rigorous QFT! We have to verify everything.

If we make the variable change: a = A + z, then the coherent state |z> is the A-operator vacuum but the spectrum is shifted anyway:

a+a - (a+z + az*) = A+A - |z|2.

It is correct since acting on the solution with H in any representation should not give zero.

In this respect I would like to precise that the Hamiltonians H(a) and H0(a) both are good to describe the exact solution. In the first case H gives the solution proceeding from the source as a boundary condition (known antenna current). In the second case we have to measure the free field state in order to define the superposition coefficients in it. In both cases the solution is expressed in terms of the eigenvectors |n> of H0(a):

|z> = exp(za+ - z*a)|0> = exp(-|z|2/2) eza+|0> = exp(-|z|2/2)∑n=0 (zn/√n!) |n>.

The Hamiltonian H0(a) solely, however, is not sufficient for that since it allows defining its eigenvectors |n> but not the coefficients in a particular filed state. It is in this sense I thought of a and A as of alike (equally good) operators.

But since the total Hamiltonian is in fact different from that of DarMM's (it is not H0(A)), the actual meaning of operators A is also different from what I supposed.

strangerep said:
... Nevertheless, the solution can be expressed in another unitarily-inequivalent Hilbert space, which is the point of all this.
Very interesting to see this solution (expressed in terms of physical vacuums solely? No A-quanta? Are they then physical if the solution never contains them?).
 
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  • #154
meopemuk -> My understanding is that your main problem with QFT as it is thought in conventional books is the unphysical presence of bare particles. Is that correct? If that's the case, then things can be settled pretty easily. Even when dealing with QFT non-rigorously, if you have an interacting theory, it does not describe bare particles. Bare particles, with bare masses and charges, are just a mathematical, artificial, non physical construct that arises in perturbation theory. The real problem is that we do not know how to compute quantities non perturbatively, even in those cases where the QFT is defined rigorously, like phi^4 in 2D. And the thing we do is perturbation theory. So bare particles are nothing but a crude approximation to real particles. Much like Earth's orbit computed with the sun's presence exclusively is an approximation to it's real orbit. But the real problem is that we don't know how to compute things so we use calculational tricks. So a bare particle is nothing but a crude approximation to a real particle - it's the free (unphysical) theory's field quantum.

Keep also in mind that when people say they want to construct QFT rigorously it does not mean they want to extract physical quantities non-perturbatively - in most cases it's probably impossible anyways. The goal is to reach a point where things are defined and understood like, say, classical QED, which is well defined because it's given in terms of well defined mathematical objects, classical fields, and the field equations are well posed. Once you've said that, you're still left with the task of computing various quantities of interest. Not an easy task at all.

Maybe someone could enlighten me a bit, but I seem to remember that even in classical field theory, when self forces are computed, one uses some sort of renormalization? Is that correct?

A question about the dressed particle approach. You said that a good interaction term should contain at least 2 creation and 2 annihilation operators, that are normal ordered. If that's so, how would you describe in this approach the decay of particles? Something like one unstable particle decaying into 2 lighter ones? Ordinarily this is achieved with a product of three operators.

bob_for_short -> At some point you say that a Dirac delta squared is infinity and that that's what it means. I do not agree, and this is not really "up for discussion" as it is in all the books about distributions. It's established mathematical knowledge. The square of a Dirac delta is simply mathematically ill defined, just like division by zero.
 
  • #155
DrFaustus said:
Maybe someone could enlighten me a bit, but I seem to remember that even in classical field theory, when self forces are computed, one uses some sort of renormalization? Is that correct?
Yes, that is correct. The electron mass "renormalization" appeared first in the Classical Electrodynamics of one electron interacting with the EMF. Even after the mass renormalization the new equation turned out to have non-physical exact solutions.
DrFaustus said:
bob_for_short -> At some point you say that a Dirac delta squared is infinity and that that's what it means. I do not agree, and this is not really "up for discussion" as it is in all the books about distributions. It's established mathematical knowledge. The square of a Dirac delta is simply mathematically ill defined, just like division by zero.
In my experience I met two types of delta-function squared treatment. One is very well known in QFT and concerns the global energy-momentum conservation law. There δ4(0) is replaced with the product TV (total time times total volume). Then one calculates the things per unit of time and per unit of volume.

The second type gives just infinity and I get rid of it by the problem reformulation in better terms. This is exactly the type that is encountered in loops and vertices. It is just infinity. To get rid of them one subtracts the infinite contributions to the masses (electron and photon mass renormalization) and the charge (equation coupling constant). This subtraction is discarding from expressions and is not reformulation in better terms. That is why this "prescription" does not work without fail.
 
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  • #156
DrFaustus said:
So bare particles are nothing but a crude approximation to real particles.

I don't think that bare particles can be regarded even as a "crude approximation" to real particles. They are completely different. Let's just say that infinite mass and charge (of bare particles) is not a reasonable approximation to the finite mass and charge of physical particles. Moreover, "bad" interactions between bare particles are completely different from "good" interactions in the physical world. So, the transition between the bare and dressed pictures is a complete change of the point of view, rather than a perturbative refinement.


DrFaustus said:
A question about the dressed particle approach. You said that a good interaction term should contain at least 2 creation and 2 annihilation operators, that are normal ordered. If that's so, how would you describe in this approach the decay of particles? Something like one unstable particle decaying into 2 lighter ones? Ordinarily this is achieved with a product of three operators.

That's a good question. For simplicity I was talking about simple theories (such as QED) in which basic particles are stable. However, decays can be taken into account as well. Then the classification should be adjusted slightly. In the complete classification, "good" operators also include products of 1 creation and N annihilation operators (or N creation and 1 annihilation operators), where N>1. In contrast to "bad" operators of the same structure, in these "good" operators the (free particle) energy of the 1 unstable species can be higher than the energy of the N decay products, so that the decay is energetically permitted.

In true "bad" terms [tex]a^*bc+b^*c^*a[/tex] the energy of the particle "a" is always lower than the sum of energies of "b" and "c".

In addition to the mentioned classes of interactions, there is another class, which completes the full classification. If interaction [tex]a^*b[/tex] does not violate conservation rules, then it describes oscillations between two particle types "a" and "b". For example, neutrino oscillations.

Eugene.
 
  • #157
meopemuk said:
I don't think that bare particles can be regarded even as a "crude approximation" to real particles. They are completely different. Let's just say that infinite mass and charge (of bare particles) is not a reasonable approximation to the finite mass and charge of physical particles.
Infinite bare values make indeed everything senseless. But at what stage we discover that we in fact deal with bare particles and where from it follows that their masses and charges are infinite? First, it is the perturbative corrections that are infinite, not the initial masses and charges. There are two ways of discarding them:

1) A simple discarding. Then the initial masses and charges remain intact and physical (observable values that in some units can be put equal to unity). But this is obviously wrong mathematically. P. Dirac was pointing out namely this weakness of mathematics.

2) Declare the initial masses and charges to be infinite in order to add the infinite corrections to them without discarding and declare the resulting sum to be physical. Then no discarding is involved but the physics is ruined: our constants that served well in the first Born approximation are declared not to be themselves! Our particles are non observable! What an absurd solution! Technically it is equivalent to discarding but now it is not mathematicians who are wrong but sorry constants. It is they who are guilty now. There is a further rubbish like dependence of bare parameters from physical ones and the cut-off. This dependence is invented just in order to repeat discarding in higher orders.

In both cases the violence of the good sense is obvious. No change of the Hilbert space helps here.
 
  • #158
Bob_for_short said:
It is not only a physical current but also a
time-dependent one. DarMM made a mistake. [...]
No he didn't. He showed one particular example, and you show a
different example. Let us therefore speak of "DarMM's example" and
"Bob's example" to avoid confusion between the two.
I am discussing DarMM's example.

I also find it quite misleading to call Ω a "physical vacuum" or "interacting
ground state". It is not an eigenstate of the total Hamiltonian whatever
representation you use.
Yes it is.

Proof:

Let
[tex]U[z] ~:=~ exp(za^* - \bar{z} a)[/tex]
and
[tex]|z\rangle ~:=~ U[z]\, |0\rangle[/tex]
where
[tex]a\,|0\rangle = 0 ~.[/tex]

Then the transformation
[tex]a ~\to~ A = a - z[/tex]
is equivalent to
[tex]a ~\to~ A = U[z]\,a \, U^{-1}[z] ~.[/tex]
(Proof of the last statement is left as an exercise for the reader.)

Therefore
[tex]
A \, |z\rangle ~=~ \left( U[z]\,a \, U^{-1}[z] \right) U[z]\, |0\rangle
~=~ U[z]\,a \, |0\rangle ~=~ 0 ~.
[/tex]
I.e., A annihilates [itex]|z\rangle[/itex]. Since the full Hamiltonian H is expressible
in the form [itex]A^*A[/itex], it also annihilates [itex]|z\rangle[/itex].
Thus, [itex]|z\rangle[/itex] is an eigenstate of H with eigenvalue 0,
and hence qualifies as a vacuum state for H.

[Edit: Per Bob_for_short's request below...
The vector space constructed by acting on [itex]|z\rangle[/itex] with
(polynomials of) the [itex]A^*(k)[/itex] (and completing as usual to
get a Hilbert space) forms the solution space for the problem.]

It looks like the original Hamiltonian but we see that whatever z is, the term
[itex]|z|^2[/itex] is unavoidable. So the DarMM's variable change may not be
correct - it does not contain the corresponding shifts in spectra.
The constant shift [itex]|z|^2[/itex] is just a redefinition of the zero value of energy,
which is physically acceptable because we only measure energy differences.
 
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  • #159
This is all that you can supply? Where is the problem solution? Is it a pure physical/interacting vacuum? Is its energy certain and equal to zero? How can you measure the energy difference if it always equals zero?
 
  • #160
meopemuk said:
I agree that "interacting vacuum" [tex]\Omega[/tex] has zero overlap with each and every free state. However, I don't agree that this fact implies that [tex]\Omega[/tex] is outside the free Fock space.

This may sound as a paradox, but my point is that "even if all components of a vector are zero, the vector itself could be non-zero". The important thing is that the number of components is infinite. So, infinite number of (virtually) zero components can add up to a non-zero total value.
There's a big difference between "zero" and "almost zero". An infinite number of exactly zero
components still adds up to zero, whereas an infinite number of "almost zero" components
could add up to anything.

In fact, here we have an uncertainty of the type "zero"x"infinity". In order to resolve this uncertainty we need to take a proper limiting procedure. I.e., we should slowly move the interaction from "weak" to "strong" regime and look not only at individual components of the [tex]\Omega[/tex] vector, but also at the total sum of squares of these components (which is a measure of the overlap of [tex]\Omega[/tex] with the free Fock space). Then we will see that each particular component indeed tends to zero, but the total sum of squares remains constant (if the transformation is unitary).
That's only true if the transformation doesn't involve an unbounded operator (in which
case it is only formally unitary).

This means that [tex]\Omega[/tex] does not leave the free Fock space even if the interaction is "strong".
That depends on the details of the theory. In some cases, even the theory with an infinitesimal
coupling constant is not in the free Fock space. (DarMM's example doesn't show that, because
distinct Hilbert spaces only arise there if z(p) is not square-integrable.)

I guess this means we need a nastier example (maybe superconductivity with Bogoliubov
transforms) which show that the Hilbert spaces are distinct even for infinitesimal
couplings. (sigh)
 
  • #161
Frankly, Strangerep, did the "Rigorous QFT" of DarMM's result in complete zero?
 
  • #162
Bob_for_short said:
This is all that you can supply? Where is the problem solution?
See my edit in post #158.
 
  • #163
strangerep said:
See my edit in post #158.
Edit: Per Bob_for_short's request below...
The vector space constructed by acting on LaTeX Code: |z\\rangle with
(polynomials of) the LaTeX Code: A^*(k) (and completing as usual to
get a Hilbert space) forms the solution space for the problem.
The solution space (basis) could be useful if the solution involved at least one excited state (not vacuum). But according to you, the solution is expressed only via vacuums (it is a product of vacuums, posts #150 and #158):

|Ψ> = |0>, H|Ψ> = 0.
strangerep said:
The constant shift [itex]|z|^2[/itex] is just a redefinition of the zero value of energy, which is physically acceptable because we only measure energy differences.
And how about the total field momentum? With your discarding the shifts it is also equal to zero.

You propose to consider the flux of energy from a laser to be zero, don't you? Do you have any sense of physics?
strangerep said:
No he didn't. He showed one particular example, and you show a different example. Let us therefore speak of "DarMM's example" and "Bob's example" to avoid confusion between the two. I am discussing DarMM's example.
A time-independent current does not have any frequency in its Fourier spectrum. The DarMM's example contains a time-dependent current.

I am really disappointed. I did not expect that from you, Strangerep.
 
Last edited:
  • #164
Bob_for_short said:
The solution space could be useful if the solution involved at least one excited state (not vacuum). But according to you, the solution is expressed only via vacuums.
The state
[tex]
A^*(k) \, |z\rangle
[/tex]
is one excited state. Similarly, higher-order polynomials in the A's acting on |z>
are other excited states.

Of course the solution is expressed "only via vacuums".
A Fock space (aka "representation") is constructed by choosing a distinguished
vacuum vector, and then acting on it with creation ops to generate other states.
 
  • #165
strangerep said:
There's a big difference between "zero" and "almost zero". An infinite number of exactly zero
components still adds up to zero, whereas an infinite number of "almost zero" components
could add up to anything.

This is exactly my point. We need to understand clearly, which quantities are "exactly zero" and which are "almost zero" (or "exactly infinite" and "almost infinite").

The same uncertainty arises in the treatment of plane waves in ordinary QM. In order to get a normalized plane wave one needs to multiply [tex]\exp(ipx)[/tex] by a factor 1/"square root of the volume of space". One can say that the volume of space is "exactly infinite", and that its inverse is "exactly zero". So, normalized plane waves formally do not exist (they are "exactly zero"). At least, they do not belong to the "normal" Hilbert space, and the operator of momentum is not a valid observable in this Hilbert space. This is the same kind of logic, which leads you to the conclusion that interacting states in QFT do not belong to the free Fock space, and this logic forces you to introduce a separate (orthogonal) interacting Hilbert space to accommodate the interacting states.

I am not saying that this logic is bad or wrong. I am saying that this logic seems unsatisfactory to me. Intuitively, I would like to have a theory in which both position operator and momentum operator (and their eigenfunctions) can coexist peacefully in the same space of states. Similarly, I would like to have QFT in which both non-interacting and interacting states can coexist. I think that such a theory would need to modify our (presently primitive) notions of "zero" and "infinity". I hope that non-standard analysis is a good candidate framework for such a theory. Currently this is not more that a pure hope without any supporting evidence.

Eugene.
 
  • #166
strangerep said:
There's a big difference between "zero" and "almost zero". An infinite number of exactly zero components still adds up to zero, whereas an infinite number of "almost zero" components could add up to anything.
Yes, and I had such an experience in my practice. It was an extreme sensitivity of a spectral sum to its terms (very slow convergence): infinitesimal changes of each spectral term gave a finite change of the whole spectral sum.
 
Last edited:
  • #167
meopemuk said:
I am not saying that this logic is bad or wrong. I am saying that this logic seems unsatisfactory to me. Intuitively, I would like to have a theory in which both position operator and momentum operator (and their eigenfunctions) can coexist peacefully in the same space of states. Similarly, I would like to have QFT in which both non-interacting and interacting states can coexist. I think that such a theory would need to modify our (presently primitive) notions of "zero" and "infinity". I hope that non-standard analysis is a good candidate framework for such a theory. Currently this is not more that a pure hope without any supporting evidence.
Unfortunately it isn't that easy. Any modification of the Hilbert space framework results in a theory quite different from QM and missing some of its properties. Remember a complex Hilbert space is quite a specific structure mathematically and not that flexible.

To take your plane-wave example, plane-waves aren't an element of the Hilbert space of a non-relativistic particle, so a prediction of standard Hilbert space QM would be that there are no states of definite momentum. If you modified things to allow them, then they would become a physically possible state, however nobody has ever seen a state of definite momentum, so why would we do this?

Even if you object to different Hilbert spaces for free bare particles and physical particles, different Hilbert spaces are needed for QFTs at different temperatures and this is indisputable. However I still think a different Hilbert space for bare particles and interacting particles makes perfect sense, since you cannot prepare a free bare state given a collection of physical states. It's impossible, so just like the plane-wave states I have no idea why you would modify QM to allow this.

Also I have no idea why a theory like QFT would modify our notions of infinity or why the current notions are primitive.
 
  • #168
DarMM said:
Unfortunately it isn't that easy. Any modification of the Hilbert space framework results in a theory quite different from QM and missing some of its properties. Remember a complex Hilbert space is quite a specific structure mathematically and not that flexible.

I agree that Hilbert space postulates (i.e., quantum logic) are untouchable. However, I see a chance to generalize these postulates by using non-standard analysis (see below).

DarMM said:
To take your plane-wave example, plane-waves aren't an element of the Hilbert space of a non-relativistic particle, so a prediction of standard Hilbert space QM would be that there are no states of definite momentum. If you modified things to allow them, then they would become a physically possible state, however nobody has ever seen a state of definite momentum, so why would we do this?

By the same token, nobody has ever seen states with definite position. Shall we then say that position is not a good observable too?

DarMM said:
Even if you object to different Hilbert spaces for free bare particles and physical particles, different Hilbert spaces are needed for QFTs at different temperatures and this is indisputable.

I don't know much about QFT at nonzero temperatures. I am still trying to understand simple QFT with 1-2-3 particles, where temperature does not play any role.

DarMM said:
Also I have no idea why a theory like QFT would modify our notions of infinity or why the current notions are primitive.

Are you familiar with the idea of non-standard analysis?

http://en.wikipedia.org/wiki/Non-standard_analysis

It attempts to enrich the mathematical notions of zero and infinity. In my opinion, QM/QFT can benefit from such generalized approach. My guess is that this could be a more acceptable alternative to "non-equivalent representations of CCR".

Eugene.
 
  • #169
meopemuk said:
By the same token, nobody has ever seen states with definite position. Shall we then say that position is not a good observable too?
Yes, nobody has ever seen states of definite position. However I don't understand why this would make position a bad observable. When you model what's going on in an experiment the observable we are looking at is one of the bounded projections of position and these are perfectly good observables, whose statistical spread of observations matches results.

meopemuk said:
Are you familiar with the idea of non-standard analysis?

http://en.wikipedia.org/wiki/Non-standard_analysis

It attempts to enrich the mathematical notions of zero and infinity. In my opinion, QM/QFT can benefit from such generalized approach. My guess is that this could be a more acceptable alternative to "non-equivalent representations of CCR".
Well first of all, results in non-standard analysis on Hilbert spaces produce nothing that can't really be done with the standard methods. An example is provided in the article you link to.

Also even when QM or QFT are done with non-standard analysis the conclusions reached aren't the same as yours. Take the work of C. E. Francis, he shows that you can include the momentum eigenstates in the Hilbert space, however they aren't part of the physical Hilbert space. That is the physical subspace of his nonstandard analysis Hilbert space is the Hilbert space of standard analysis.

Also I should mention that nonstandard analysis is not really that well accepted in mathematics. Not that people disagree with it, it's just not certain if it adds anything new.

From my point of view I don't see why we should get rid of different Hilbert spaces. In the finite temperature and finite density case it is necessary, for example all Fock space states have zero density so you would need a new Hilbert space. Also Glimm and Jaffe's work (and the work of others) shows that this holds even in the case of zero temperature and density when you have interactions.

I don't agree that this should be replaced by a framework based on mathematics that hasn't shown any great utility. Especially because when it is applied what you describe doesn't happen, but instead the standard framework still holds and a few calculations are speeded up somewhat.

The rôle of different Hilbert spaces has a use in spontaneous symmetry breaking, finite temperature and density and different phases associated with phase transitions. Is it really that much of a stretch to find that it happens with interactions?
 
  • #170
DarMM,

apparently you know more about non-standard analysis than I do. So I should stop arguing.

Besides, the argument about different Hilbert spaces for bare and interacting particles looks purely academic to me. All particles existing in nature are interacting. Bare particles do not exist, so their Hilbert space should not be that interesting.

Eugene.
 
  • #171
meopemuk said:
DarMM,

apparently you know more about non-standard analysis than I do. So I should stop arguing.

Besides, the argument about different Hilbert spaces for bare and interacting particles looks purely academic to me. All particles existing in nature are interacting. Bare particles do not exist, so their Hilbert space should not be that interesting.

Eugene.
Yes, true. Of course that is not the point. All I'm stating is that they have different Hilbert spaces, you can find this interesting or boring if you wish. There are a few reasons why one may find it physically interesting.
For example in QM you can prepare a anharmonic eigenstate from a harmonic system, since all QM systems live in the same Hilbert space, but this procedure cannot be carried out in QFT.

However the focus of this thread is mathematical rigour. The reason why it becomes important then is because if the interacting and bare theories live in different Hilbert spaces then perturbation theory is harder to justify and you can't appeal to the same theorems to show that it works or analytically estimate its convergence.
For example in QED when you work to order [tex]\mathcal{O}(e^{4})[/tex], you ignore the higher terms because they are "small" compared to the ones you are interested in. As a physicist this needs no further comment, however mathematically you might ask "how do you know they are smaller?". To answer this question and to show that the terms you've included are significant you have to have analytic control of perturbation theory. In QM this is easy, a few theorems of Kato and some methods from Reed and Simon and you're done. For QFT it's much harder and to do it you must keep this different Hilbert space issue under control. Again if you've no interest in rigour, no problem. However this is rigorous QFT, putting everything on a sound mathematical footing.

Also if you want to prove a quantum field theory exists, since the only theory we can solve analytically and gain analytic control over is the free bare Hilbert space, you have to start from there to construct nonperturbatively the interacting model and this different Hilbert space stuff is a crucial issue in the construction. However, once again, the mathematical existence of a QFT is a question of rigour. One you might not be interested in, however the thread is about rigorous quantum field theories.

I must say that the interest in the bare Hilbert space is not for itself, but rather it is the first solid step on the way to rigorously demonstrating things about the interacting theory.
 
  • #172
DarMM said:
Also if you want to prove a quantum field theory exists, since the only theory we can solve analytically and gain analytic control over is the free bare Hilbert space, you have to start from there to construct nonperturbatively the interacting model and this different Hilbert space stuff is a crucial issue in the construction. However, once again, the mathematical existence of a QFT is a question of rigour. One you might not be interested in, however the thread is about rigorous quantum field theories.

I must say that the interest in the bare Hilbert space is not for itself, but rather it is the first solid step on the way to rigorously demonstrating things about the interacting theory.

I think that the connection between bare and interacting theories depends very much on what kind of interaction is assumed to exist.

In most quantum field theories (for example, in QED where interaction has the form [tex]jA[/tex]) it is assumed that "bad" interaction Hamiltonians (those which allow creation of multi-bare-particle states from one-bare-particle states or from vacuum) are OK. This leads to the situation in which bare vacuum and 1-particle states are different from physical vacuum and 1-particle states. This in turn leads to a host of problems associated with different Hilbert spaces needed for the the free and interacting theories, with renormalization, etc. In my opinion, this is a wrong path, and the problems we meet on this path are artificial. I think that "bad" interactions are not present in nature and their theoretical investigation is not useful.

Alternatively, if we limit ourselves to "good" interactions only, then there is no difference between bare and interacting vacuum and 1-particle states. The situation is very similar to the one we have in ordinary QM. As you said, many issues (like existence of the perturbation expansion) can be solved easily. We don't need to worry about different Hilbert spaces for the free and interacting theories. We don't need to worry about renormalization as well. By throwing out "bad" interactions we are not diminishing the predictive power of the theory. One can show that a "good" theory can reproduce exactly the S-matrix of a renormalized "bad" theory.

Eugene.
 
  • #173
meopemuk said:
This may sound as a paradox, but my point is that "even if all components of a vector are zero, the vector itself could be non-zero". The important thing is that the number of components is infinite. So, infinite number of (virtually) zero components can add up to a non-zero total value.

Eugene.

The little I know from axiomatic field theory seem to exclude exactly this. In the definition of the usual Fock space one starts from those vectors, for which only a finite number of components are different from zero, the others being exactly zero. Then one actually includes also the limit points of sequences in this space, but also for these elements the sum of the components is well defined. Hence, e.g. states of nonzero temperature, which contain an infinite number of electron hole excitations from the very beginning, do not lie in the Fock space, but form a separate Hilbert space.
 
  • #174
DrDu said:
In the definition of the usual Fock space one starts from those vectors, for which only a finite number of components are different from zero, the others being exactly zero.

Why would you make such an artificial assumption? Is there any physical reason? or just calculations become easier?
 
  • #175
I would argue like this: The Fock space is a separable Hilbert space and all separable Hilbert spaces are isomorphic. If you choose a representation with infinite dimensional vectors, then all vectors in H have to fulfill this criterion in order that the scalar product is defined.
 

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