Rigorous Quantum Field Theory.

[Bob_for_short]

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

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

 

[strangerep]

The difficult part is to define an interacting representation of the Poincare
group, which satisfies cluster separability and permits changes in the number
of particles. This is the place where quantum fields (= certain formal linear
combinations of creation and annihilation operators) come handy. We simply
notice that if the interaction Hamiltonian (= the generator of time
translations) and interacting boost operators are build as integrals of
products of fields at the same "space-time points", then all physical
conditions are satisfied automatically. In this approach, there is no need to
worry about different Hilbert spaces for the non-interacting and interacting
theories. [...]

The Hilbert space of "QFT2" is still infinite-dimensional -- so any
mathematical issues arising as a consequence of infinite degrees of freedom
still lurk here, just as they lurk in "QFT1".

I agree with that, but the Hilbert space of QFT2 is built *before* any
interaction is introduced. So, the same Hilbert space is used in both
non-interacting and interacting theories.

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

[...] the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations
of them are also operator-valued distributions. Hence QFT2 suffers exactly the
same "ill-defined equal-point multiplication of distributions" mathematical
problem as QFT1.

Could you give an example in which a product of a/c operators or quantum
fields is "ill-defined"?

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

Consider simple 1-particle quantum mechanics. Eigenstates of the momentum
operator are usual plane waves in the position representation. If we want
these eigenstates to be normalized, we must multiply them by a normalization
factor that is effectively zero (but not exactly zero!). So, it would be
tempting to conclude that normalized plane waves are orthogonal to all
"normal" states in the 1-particle Hilbert space. Do they belong to some other
orthogonal Hilbert space? Some people try to resolve this problem by
introducing "rigged Hilbert spaces", "Gelfand triples", etc. Personally, I
don't like these ideas. My opinion is that we are using too narrow a
definition of the Hilbert space. We should use a broader definition of Hilbert
spaces, i.e., such that eigenvectors of unbounded operators (like momentum)
can find their place there.

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

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

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

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

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

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

 

[meopemuk]

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

I go to bed.

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

 

[Bob_for_short]

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

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

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

No objection.

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

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

 

[meopemuk]

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

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

<br /> <br /> <blockquote data-attributes="" data-quote="strangerep" data-source="post: 2451873" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> strangerep said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Let us say &quot;bosons&quot;, not photons -- the mass in DarMM&#039;s example is nonzero.<br /> <br /> But no, it doesn&#039;t mean what you said. The physical bosons are associated with<br /> the A(k), not the a(k). The A(k) annihilate the physical vacuum. </div> </div> </blockquote><br /> Great, we then agree that &quot;bare&quot; vacuum, &quot;bare&quot; particles and their a/c operators a(k), a*(k) are just phantoms, which have no relevance to the stuff observed in Nature. By the way, we took them quite seriously when we initially wrote down the Hamiltonian or Lagrangian of our QFT theory (e.g., the QED Hamiltonian, which is normally written in terms of &quot;bare&quot; particle operators). Now we conclude that they are actually useless. Do you notice a weird contradiction here? In my opinion, this is one of the reasons to say that traditional QFT is self-contradictory.<br /> <br /> Anyway, let us now focus on the good physical stuff - the physical vacuum \Omega and physical particles created by A^*(k). These particles can induce a response in real detectors, and that&#039;s the only thing we are interested in in physics. We&#039;ve agreed that &quot;bare&quot; states and operators are useless, so let us now forget about them completely, and promise never mention them again. Indeed, we should never need them. Even without them, we have everything we need to do physics: We have the vacuum state \Omega, a/c operators A(k), A^*(k) with usual commutation relations, the particle number operator <br /> <br /> \int d^3k A^*(k)A(k), <br /> <br /> etc. We can build a Fock space which spans vectors like (A^*(k))^n \Omega. Let us call it the &quot;interacting Fock space&quot;, to distinguish it from the (irrelevant) &quot;free Fock space&quot; built by applying (a^*(k))^n to the &quot;bare&quot; vacuum. We also have the interacting Hamiltonian H expressed as a function of A(k), A^*(k). It is not difficult to build the free Hamiltonian (i.e., the one describing non-interacting A-particles) as well by formula <br /> <br /> H_0 = \int d^3k \omega(k)A^*(k)A(k). <br /> <br /> It is important to note that, by construction, 0-particle \Omega and 1-particle A^*(k) \Omega states are eigenstates of the interacting Hamiltonian, which means that interaction between A-particles does not involve &quot;bad&quot; terms. So, we are all set. In our &quot;interacting Fock space&quot; we can represent any state of any multi-particle system. We can study the time evolution of such states, bound states, scattering etc. We will never encounter divergences, and we will never need to do renormalization. <br /> <br /> So, we have achieved a peaceful transition from the traditional QFT (with bare particles, renormalization, and other curiosities) to the &quot;dressed particle&quot; theory formulated exclusively in terms of really observable ingredients. This is called &quot;unitary dressing transformation&quot;. In the &quot;dressed particle&quot; theory both Hamiltonians H_0 and H co-exist in the same Hilbert space. Moreover, they exactly coincide in 0-particle and 1-particle sectors. <br /> <br /> Eugene.

 

[strangerep]

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

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

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

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

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

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

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

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

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

 

[meopemuk]

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

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

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

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

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

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

Can we agree about that?

Eugene.

 

[strangerep]

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

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

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

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

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

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

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

Can we agree about that?

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

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

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

 

[meopemuk]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Eugene.

 

[Bob_for_short]

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

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

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

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

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

<br /> \Psi = \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)\Psi_{0,a} = \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}A^{*}(k)}\right)\Psi_{0,A}<br />

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

I have another proof:

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

|z> = exp(za⁺ - z^*a)|0> =

= exp(-|z|²/2) e^za+|0>.

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

Thus, we may determine the solution in two ways:

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

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

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

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

 

[Bob_for_short]

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

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

 

[strangerep]

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

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

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

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

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

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

 

[meopemuk]

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

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

Cheers.
Eugene.

 

[Bob_for_short]

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

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

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

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

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

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

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

4) Operators A and A⁺ have different CCR.

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

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

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

|Ψ⁽¹⁾> ≈ [1 - g∫dk(...)] |0>

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

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

 

[Neo_Anderson]

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

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

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

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

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

4) Operators A and A⁺ have different CCR.

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

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

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

|Ψ⁽¹⁾> ≈ [1 - g∫dk(...)] |0>

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

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

LOL...

 

[DarMM]

Yes, generally the interacting and free Hamiltonians have different eigenstates. However, the important issue is about the vacuum and 1-particle eigenstates. In your Hamiltonian H_0 +a^* + a, the interaction acts non-trivially on the free vacuum and 1-particle states. Therefore, 0-particle and 1-particle eigenstates of this Hamiltonian are different from 0-particle and 1-particle eigenstates of the free Hamiltonian. Your "dressed" particles are different from "bare" particles. This is why the (mass) renormalization is needed.
..., then its 0-particle and 1-particle eigenvectors are exactly the same as the free vacuum and free particles, respectively. There is no need for the (mass) renormalization. Both free and interacting theories live in the same Fock space. 0-particle and 1-particle sectors of the two theories are exactly the same.

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

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

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

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

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

 

[DarMM]

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

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

Also in an earlier post of yours:

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

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

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

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

 

[DarMM]

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

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

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

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

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

 

[DarMM]

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

It does, there are no coherent states.

 

[meopemuk]

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

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

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

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

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

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

Thanks.
Eugene.

 

[meopemuk]

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

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

OK, let us agree that theories with different masses and/or different coupling constants live in different Hilbert spaces (though I have my reservations about the true meaning of this statement). In Nature we have just one set of masses and one set of coupling constants. So, everything relevant to the physical world happens in just one Hilbert space. What interesting can be learned form studying other (non-physical) Hilbert spaces?

Eugene.

 

[DarMM]

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

I really don't understand this. I can't think of a single pair of Hamiltonians in nonrelativistic QM where the ground states and first excited states coincide. This is nonrelativistic QM with no renormalizations. So why would it be true in any QFT?

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

I'm not talking about gauge invariance. I'm talking about global symmetries and the group transformations associated with them. This can apply even without gauge invariance.
You have the interacting theory invariant under less or more global symmetries than the free theory, so if it has more (or less) symmetries how can its one particle states coincide with the free one particle states? The interacting particles will literally have more or less charges.

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

That's a big question and answering it would amount to an explanation of Haag-Ruelle theory. Haag's monograph "Local Quantum physics" has a good explanation of your particle questions on pages 75-83. The basic result is that there are indeed n-particle sectors. I can't say what their properties are because that really depends on the theory you are talking about. It's not even necessarily true that they can't be represented as eigensubspaces of an Hermitian operator. In general, no they don't have to be orthogonal, for instance in theories with unstable particles.

 

[DarMM]

OK, let us agree that theories with different masses and/or different coupling constants live in different Hilbert spaces (though I have my reservations about the true meaning of this statement). In Nature we have just one set of masses and one set of coupling constants. So, everything relevant to the physical world happens in just one Hilbert space. What interesting can be learned form studying other (non-physical) Hilbert spaces?

I think you're coming at this from the wrong angle. The general aim is to prove QFT is mathematically rigorous, for a few reasons:
(i)Intellectual satisfaction, we would like our best physical theories to be mathematically consistent.
(ii)Incorporation of QFT in mathematics to help with the analysis, geometry and topology of spaces of functions. See how Supersymmetric theories have helped Morse theory.
(iii)The increased mathematical rigour has allowed the discovery of various previously unknown features of QFT.
(iv)Nonperturbative knowledge of QFT seems to come mainly from (a)Lattice Field Theory or (b)Rigorous Field Theory.

The reason you want to learn about other Hilbert spaces are then several fold.

Also some examples:
There is a different Hilbert space for every single temperture in finite temperture field theory. You need to know about them to discuss thermal properties of quantum fields.

In general however, your question would be the equivalent of asking:
"Why, if our spacetime is one exact solution of Einstein's field equations, are we interested in other solutions?"
For instance I might be looking at QCD on its own, in which case I want to find out about its Hilbert space, not the entire standard model.

 

[meopemuk]

I really don't understand this. I can't think of a single pair of Hamiltonians in nonrelativistic QM where the ground states and first excited states coincide. This is nonrelativistic QM with no renormalizations. So why would it be true in any QFT?

QM and QFT are fundamentally different in only one respect: QM is a theory with a fixed number of particles, while this number is allowed to change in QFT. The ground and first excited state in QM have the same number of particles as all other states. In QFT the "ground" and the "first excited states" are 0-particle and 1-particle states respectively. So, your analogy does not look good to me. Renormalization is a feature specific to QFT. It does not have true QM analogs. The reason for the (mass) renormalization is the (undesirable) presence of interactions which affect the 0-particle and 1-particle states.

I'm not talking about gauge invariance. I'm talking about global symmetries and the group transformations associated with them. This can apply even without gauge invariance.
You have the interacting theory invariant under less or more global symmetries than the free theory, so if it has more (or less) symmetries how can its one particle states coincide with the free one particle states? The interacting particles will literally have more or less charges.

If you are not talking about gauge invariance, then I am not sure what's the meaning of U(1)? Why should I care about U(1) invariance when building interacting QFT models? The only true symmetry that I care about is the Poincare symmetry.

That's a big question and answering it would amount to an explanation of Haag-Ruelle theory. Haag's monograph "Local Quantum physics" has a good explanation of your particle questions on pages 75-83. The basic result is that there are indeed n-particle sectors. I can't say what their properties are because that really depends on the theory you are talking about. It's not even necessarily true that they can't be represented as eigensubspaces of an Hermitian operator. In general, no they don't have to be orthogonal, for instance in theories with unstable particles.

Thank you for the reference, I'll take a look when I have a chance. If we are not talking about unstable particles (which is a whole separate subject by itself), would you agree that different n-particle sectors must be orthogonal, which implies that there exists an Hermitian particle number operator?

Eugene.

 

[DarMM]

QM and QFT are fundamentally different in only one respect: QM is a theory with a fixed number of particles, while this number is allowed to change in QFT. The ground and first excited state in QM have the same number of particles as all other states. In QFT the "ground" and the "first excited states" are 0-particle and 1-particle states respectively. So, your analogy does not look good to me. Renormalization is a feature specific to QFT. It does not have true QM analogs. The reason for the (mass) renormalization is the (undesirable) presence of interactions which affect the 0-particle and 1-particle states.

First of all, you can have finite renormalizations in QM:
Zamastil J.; Czek J.; Skala L., "Renormalized Perturbation Theory for Quartic Anharmonic Oscillator", Annals of Physics, 276(1), p.39-63

Secondly you can have infinite renormalizations in QM:
Jackiw, R., "What good are quantum field theory infinities", pg.101-110 Mathematical Physics 2000, Ed. Kibble, T., World Scientic, Singapore.
Holstein, B., "Anomalies for Pedestrians", Am. J. Phys., 61, pg 142.

It's not specific to QFT, it's just generic in QFT. However there are QFTs which need no ultraviolet renormalizations.

Again I'm to have to come back to the fact that the interacting particles can have different quantum numbers (from global symmetry groups) than the free particles.

If you are not talking about gauge invariance, then I am not sure what's the meaning of U(1)? Why should I care about U(1) invariance when building interacting QFT models? The only true symmetry that I care about is the Poincare symmetry.

Well look forget about U(1). Although surely you understand that the global U(1) group is related to the conserved electric charge. That's the meaning of global U(1).
Also I can't believe you're saying why would you care about a group other than the Poincare group. Remember Noether's theorem, extra symmetries means extra conserved currents and charges. You want electrons to have electric charge U(1), quarks should carry color SU(3). Pion should come in a flavour octet SU(3)_{F}. The big insight of particle physics in the 1960s and 1970s, the use of symmetry. Do you also disagree with this?

For example take the linear sigma model. The free theory has O(N) symmetry, but this is spontaneously broken by the interacting theory, which has only O(N-1)
symmetry. Now we have one boson in the trivial rep and N others in the fundamental rep. Hence the charge structure of the free theory is different from the charge structure of the interacting theory, so how could their one particle states coincide?

There are examples without spontaneous symmetry breaking if you object to that. In general the interacting particle can have dfferent internal symmetries and hence different charges.

Thank you for the reference, I'll take a look when I have a chance. If we are not talking about unstable particles (which is a whole separate subject by itself), would you agree that different n-particle sectors must be orthogonal, which implies that there exists an Hermitian particle number operator?

The details of the analysis of Haag and Ruelle is quite involved. Essentially, I agree that states which are "particulate" will be orthogonal when the numbers of particles differ. Of course they can unitarily evolve overlaps. Of course this is only for Minkowski space, in curved spacetime things will not be as easy.

 

[meopemuk]

First of all, you can have finite renormalizations in QM:
Zamastil J.; Czek J.; Skala L., "Renormalized Perturbation Theory for Quartic Anharmonic Oscillator", Annals of Physics, 276(1), p.39-63

Secondly you can have infinite renormalizations in QM:
Jackiw, R., "What good are quantum field theory infinities", pg.101-110 Mathematical Physics 2000, Ed. Kibble, T., World Scientic, Singapore.
Holstein, B., "Anomalies for Pedestrians", Am. J. Phys., 61, pg 142.

It's not specific to QFT, it's just generic in QFT.

Thank you for the references. I remember reading similar articles about "renormalization" in QM. However, if I am not mistaken, all of them discuss renormalization as some modification of the strength of the interaction operator (e.g., making the coupling constant energy-dependent). Indeed, this can serve as an analog of the *charge* renormalization in QFT.

However, in QM (in contrast to QFT) 0-particle and 1-particle states don't live in the same Hilbert space with n-particle states. So, changing n-particle interactions has no effect on 0-particle and 1-particle states. So, the *mass* renormalization (which is my main interest) cannot be modeled in QM.

However there are QFTs which need no ultraviolet renormalizations.

Do you have a reference?

Again I'm to have to come back to the fact that the interacting particles can have different quantum numbers (from global symmetry groups) than the free particles.

But I can also say that charges and quantum numbers (like strangeness) are properties of interactions rather than particles themselves. If there is no interaction, then everything is conserved, and to talk about "conserved charges" is kind of redundant. If there is interaction, then conservation laws are guaranteed by the structure of the corresponding interaction operator. For example, the charge conservation in QED follows from the absence of interaction terms, which can change the "number of electrons minus the number of positrons". Then there is no need to assume that interaction has any effect on 1-particle properties.

I can admit that the "dressed particle" ideology has one important drawback. It does not use quantum fields. Therefore, it cannot entertain the idea of the group of gauge transformations between fields. In the traditional QFT this idea is the main heuristic tool for constructing realistic interaction Hamiltonians. On the other hand, "dressed particle" Hamiltonians can be obtained only by guessing, fitting to experiments, or applying the ugly "unitary dressing transformation" to the standard QFT. So, in the "dressed particle" approach group theory applications are limited, basically, to the Poincare group.

Eugene.

 

[meopemuk]

Jackiw, R., "What good are quantum field theory infinities", pg.101-110 Mathematical Physics 2000, Ed. Kibble, T., World Scientic, Singapore.

A remarkable passage at the end of the paper:

"Apparently the mathematical language with which we are describing Nature cannot account for all natural phenomena in a clear fashion. Recourse must be made to contradictory formulations involving *infinities*, which nevertheless lead to accurate descriptions of experimental facts in *finite* terms."

I could live with infinities, which nevertheless lead to "accurate description" of *all* experimental facts. However, traditional QFT does not fulfil even this limited promise. It does provide accurate description of scattering amplitudes (even with the formally divergent Hamiltonian). However, it fails to provide any reasonable description of the time evolution. In my understanding, nobody pays any attention to this obvious failure only because the time evolution is very difficult to measure in experiments.

 

[DarMM]

Do you have a reference?

Virtually every paper on N=4 Super Yang-Mills will mention that it is ultraviolet finite.

But I can also say that charges and quantum numbers (like strangeness) are properties of interactions rather than particles themselves.

I can't see how. Strangeness literally measures the amount of strange quarks present, surely that is a property of the particles, in particular strange quarks.

If there is no interaction, then everything is conserved, and to talk about "conserved charges" is kind of redundant.

Despite the fact that this isn't always true (interactions may add additional conservation laws), it isn't really about what is conserved, but more about the charges.

Again take the O(4) sigma model, the free theory has four particles in the vector rep of O(4). So they each have different O(4) charges. The interacting theory has one particle with O(3) charge of magnitude 0 and value (in 3-direction) 0 and three other particle with O(3) charge of magnitude 1 and value (in 3-direction) of -1,0,1.

So the interaction will have a direct effect on one-particle states. Not only will they have different values for their charges, but they will actually have a charge of a different type. I can not see there being any way in which their one particle states can coincide.

In the opposite direction take pure Yang-Mills, the free theory is conformally invariant, the interacting theory is not. The one particle states in the free theory (gluons) are massless, in the interacting theory they are massive (glueballs). So again, how could they coincide?

I can admit that the "dressed particle" ideology has one important drawback. It does not use quantum fields. Therefore, it cannot entertain the idea of the group of gauge transformations between fields. In the traditional QFT this idea is the main heuristic tool for constructing realistic interaction Hamiltonians. On the other hand, "dressed particle" Hamiltonians can be obtained only by guessing, fitting to experiments, or applying the ugly "unitary dressing transformation" to the standard QFT. So, in the "dressed particle" approach group theory applications are limited, basically, to the Poincare group.

What about global symmetry groups? Let's say I wanted to give the particles an internal SU(n) symmetry that was not gauge.

 

[DarMM]

I could live with infinities, which nevertheless lead to "accurate description" of *all* experimental facts. However, traditional QFT does not fulfil even this limited promise. It does provide accurate description of scattering amplitudes (even with the formally divergent Hamiltonian). However, it fails to provide any reasonable description of the time evolution. In my understanding, nobody pays any attention to this obvious failure only because the time evolution is very difficult to measure in experiments.

Well people have paid attention to this, since the 1950s. Glimm and Jaffe and others have constructed non-perturbatively defined interacting theories in d = 2, 3 dimensions with well-defined finite time evolution. Hence we do know of QFTs with finite time evolution.
As well as that there are several papers (in Perturbative Field Theory and Lattice field theory) which deal with finite time evolution in four dimensions, there's just nothing substantial known about finite time evolution nonperturbatively with no approximations in four dimensions. Even then there is knowledge about nonperturbative finite time evolution on the axiomatic level in 4d, it's just not at the level of specific models.

 

[meopemuk]

I can't see how. Strangeness literally measures the amount of strange quarks present, surely that is a property of the particles, in particular strange quarks.

I am looking at this from the point of view of an experimentalist. Looking at a particle's track he cannot say what is the particle's strangeness. There is no label attached. The only thing he can do is to collide different particles (i.e., force them to interact) and notice that some pairs of particles are often created together in these collisions. Then he can assign the "strangeness" labels as a sort of shorthand notation, which simply expresses some peculiar property of the inter-particle interaction. If there was no interaction, we wouldn't be able to assign charge, strangeness, and other labels to our particles. These labels are just meaningless in the absence of interactions.

What about global symmetry groups? Let's say I wanted to give the particles an internal SU(n) symmetry that was not gauge.

Yes, I think that isotopic symmetry and SU(n) can be still applied in the particle-based approach, because these symmetries do not rely on the field description, which is absolutely essential for the local gauge invariance.

Eugene.

 

[meopemuk]

Well people have paid attention to this, since the 1950s. Glimm and Jaffe and others have constructed non-perturbatively defined interacting theories in d = 2, 3 dimensions with well-defined finite time evolution. Hence we do know of QFTs with finite time evolution.
As well as that there are several papers (in Perturbative Field Theory and Lattice field theory) which deal with finite time evolution in four dimensions, there's just nothing substantial known about finite time evolution nonperturbatively with no approximations in four dimensions. Even then there is knowledge about nonperturbative finite time evolution on the axiomatic level in 4d, it's just not at the level of specific models.

OK, then I take back the "nobody pays any attention" claim with apologies. Though I still can't understand how this can be done (unless the idea of "dressed particles" is introduced). Perhaps you can straighten me up?

One thing that looks obvious to me is that considering the time evolution of "bare" particles cannot be good. For example, in QED I can write the full Hamiltonian H in the "bare" particle representation (see Weinberg's book). I can also form the time evolution operator \exp(i \hbar Ht) \approx 1 +i \hbar Ht and apply it to the 1-particle state a^{dag}|0 \rangle. Even if I close my eyes on the presence of infinite counterterms in H, the result of such "time evolution" will be totally unphysical: the initial single particle will immediately disintegrate into a complicated linear combination of many-particle states. It seems to me that even exact non-perturbative solution (if it can be found) will have the same properties.

I hope you would agree that this doesn't make sense. Stable single particles do not disintegrate in experiments.

It seems to me that the only sensible approach is to say that "bare" particles have no relationship to the real particles seen in experiment. We should actually study the time evolution of "physical" or "dressed" particles. Then the Hamiltonian should be rewritten in the "dressed" particle basis. Is this a fair description of what is done in the research that you've mentioned? Are these people studying the time evolution of "dressed" particles? Possibly using a different name for these entities? Then I can understand and agree.

I would appreciate if you can tell me where my logic is failing. This is rather important for me as I am trying to make sense of QFT.

Thanks.
Eugene.

 

[DarMM]

I am looking at this from the point of view of an experimentalist. Looking at a particle's track he cannot say what is the particle's strangeness. There is no label attached. The only thing he can do is to collide different particles (i.e., force them to interact) and notice that some pairs of particles are often created together in these collisions. Then he can assign the "strangeness" labels as a sort of shorthand notation, which simply expresses some peculiar property of the inter-particle interaction. If there was no interaction, we wouldn't be able to assign charge, strangeness, and other labels to our particles. These labels are just meaningless in the absence of interactions.

Okay, well what about Pure Yang-Mills or any other theory with dimensional transmutation/conformal symmetry breaking? In that case the free theory is massless, while the interacting theory is massive. That's something that an experimentalist could see, is it not? Surely in the case when the free theory is massless and the interacting theory is massive their one-particle states could not coincide?

 

[DarMM]

OK, then I take back the "nobody pays any attention" claim with apologies. Though I still can't understand how this can be done (unless the idea of "dressed particles" is introduced). Perhaps you can straighten me up?

One thing that looks obvious to me is that considering the time evolution of "bare" particles cannot be good. For example, in QED I can write the full Hamiltonian H in the "bare" particle representation (see Weinberg's book). I can also form the time evolution operator \exp(i \hbar Ht) \approx 1 +i \hbar Ht and apply it to the 1-particle state a^{dag}|0 \rangle. Even if I close my eyes on the presence of infinite counterterms in H, the result of such "time evolution" will be totally unphysical: the initial single particle will immediately disintegrate into a complicated linear combination of many-particle states. It seems to me that even exact non-perturbative solution (if it can be found) will have the same properties.

I hope you would agree that this doesn't make sense. Stable single particles do not disintegrate in experiments.

It seems to me that the only sensible approach is to say that "bare" particles have no relationship to the real particles seen in experiment. We should actually study the time evolution of "physical" or "dressed" particles. Then the Hamiltonian should be rewritten in the "dressed" particle basis. Is this a fair description of what is done in the research that you've mentioned? Are these people studying the time evolution of "dressed" particles? Possibly using a different name for these entities? Then I can understand and agree.

Essentially, yes. That is what they do. For example take one of the first papers on the subject for \phi^{4} in three dimensions:
James Glimm, "Boson fields with the :\phi^{4}: interaction in three dimensions", Comm. Math. Phys. 10(1) p.1-47.

The third section of his paper is literally called "The Dressing Transformation T". He moves to the "physical particle basis" using a completely non-perturbative dressing transformation and one can see that the Hamiltonian is completely well-defined in this basis. It is this basis which has well-defined finite time evolution.

However and this is the main point, the physical basis is totally disjoint from Fock space. It is another Hilbert space. To quote the abstract:
"Consequently the Hilbert space (of physical particles) in which H_{ren} acts is disjoint from the bare particle Fock space"

So dressing transformations move you to a new Hilbert space. Although you will not see this at any order in perturbation theory.

 

[meopemuk]

Okay, well what about Pure Yang-Mills or any other theory with dimensional transmutation/conformal symmetry breaking? In that case the free theory is massless, while the interacting theory is massive. That's something that an experimentalist could see, is it not? Surely in the case when the free theory is massless and the interacting theory is massive their one-particle states could not coincide?

I have a feeling that we are talking about different things, because we haven't accepted common terminology. Let me attempt to clarify this.

There are three types of particles present in our discussion. They are "bare" particles, "dressed" particles and "non-interacting dressed" particles.

The full Hamiltonian of QFT is usually formulated in terms of "bare" particle operators. Due to the presence of "bad" terms in the interaction, "bare" particles are not eigenstates of this Hamiltonian. "Bare" particles (even isolated) are subject to ever present self-interaction which changes their properties. In other words, "bare" particles are not physical objects, they cannot be seen in experiments, so we should not discuss them in physical terms.

One can form certain linear combinations of "bare" states and obtain "dressed" particles, which are eigenstates of the full Hamiltonian. By definition, there is no self-interaction of "dressed" particles. These are particles whose theoretical properties can be compared with what we see in experiments. The full Hamiltonian can be rewritten in terms of "dressed" particle a/c operators. This Hamiltonian will have the form H = H_0 + V where H_0 is the Hamiltonian of "non-interacting dressed" particles (it is completely different from the free "bare" particle Hamiltonian) and V is interaction. This interaction does not contain "bad" terms, so it has no effect on the "dressed" vacuum and "dressed" 1-particle states. So, even if we managed to "turn off" interaction V, the properties (e.g., masses) of "dressed" particles would not be affected.

I have a feeling that when you talk about the difference between particles in the free and interacting theories you compare "bare" particles and "dressed" particles. There is indeed a huge difference between them. On the other hand, I am comparing "dressed" particles with and without interaction V (or, what is the same, "dressed" particles close to each other, i.e., interacting, and far apart, i.e., non-interacting). And there is no difference between free and interacting "dressed" particles.

Can we resolve our argument simply by adopting common terminology?

 

[meopemuk]

Essentially, yes. That is what they do. For example take one of the first papers on the subject for \phi^{4} in three dimensions:
James Glimm, "Boson fields with the :\phi^{4}: interaction in three dimensions", Comm. Math. Phys. 10(1) p.1-47.

The third section of his paper is literally called "The Dressing Transformation T". He moves to the "physical particle basis" using a completely non-perturbative dressing transformation and one can see that the Hamiltonian is completely well-defined in this basis. It is this basis which has well-defined finite time evolution.

However and this is the main point, the physical basis is totally disjoint from Fock space. It is another Hilbert space. To quote the abstract:
"Consequently the Hilbert space (of physical particles) in which H_{ren} acts is disjoint from the bare particle Fock space"

So dressing transformations move you to a new Hilbert space. Although you will not see this at any order in perturbation theory.

Great! It is good to know that I am not completely off-track. I will need to pay more attention to the works on "Rigorous Quantum Field Theory" then.

However, as I already mentioned, the fact that bare and dressed particles live in different Hilbert spaces should not be disturbing (or even interesting). Bare particles and their Hilbert space have no physical relevance. So, we can simply disregard everything which carries the name "bare" and focus only on the properties of dressed or physical stuff. If interaction between dressed particles is "turned off", then we shouldn't see any effect on their properties or on their Hilbert space. Is this a reasonable point of view?

Eugene.

 

[meopemuk]

Essentially, yes. That is what they do. For example take one of the first papers on the subject for \phi^{4} in three dimensions:
James Glimm, "Boson fields with the :\phi^{4}: interaction in three dimensions", Comm. Math. Phys. 10(1) p.1-47.

The third section of his paper is literally called "The Dressing Transformation T". He moves to the "physical particle basis" using a completely non-perturbative dressing transformation and one can see that the Hamiltonian is completely well-defined in this basis. It is this basis which has well-defined finite time evolution.

However and this is the main point, the physical basis is totally disjoint from Fock space. It is another Hilbert space. To quote the abstract:
"Consequently the Hilbert space (of physical particles) in which H_{ren} acts is disjoint from the bare particle Fock space"

So dressing transformations move you to a new Hilbert space. Although you will not see this at any order in perturbation theory.

I've got this paper. It is a bit too mathematical for me, but I think I understood a couple of important points. Let me know if I got them wrong.

I was interested to see how Glimm's approach is related to the Greenberg-Schweber dressed particle approach. My main observation is that Glimm and G-S are talking about different "dressing transformations". G-S dressing transformation is designed to remove only "bad" interaction terms from the Hamiltonian. But Glimm's dressing transformation connects the full renormalized Hamiltonian H_{ren} with the free Hamiltonian H_0 (see eq. 1.2.1). So, this transformation basically "kills" all interaction terms, not just "bad" ones.

For this reason Glimm's dressing transformation cannot be unitary (H_{ren} and H_0 have different spectra for sure). Therefore, a/c operators of dressed particles do not satisfy usual (anti)commutation relations. This is how I understand Glimm's words on page 27:

"We remark that annihilation and creation operators act in a natural fashion in [the interacting Hilbert space] and that this representation appears to be inequivalent to the Fock representation."

In contrast, Greenberg-Schweber dressing transformation is unitary, and a/c operators of dressed particles have the same (anti)commutation relations as a/c operators of bare particles.

I think it is important to stress this distinction to make sure that we are using the same terminology.

Eugene.

 

[Bob_for_short]

...Let me first clarify something in the context of that example before continuing...

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

For a given j(x), we can indeed generate a Fock space by acting with
A^*(k) on \Omega. But that's all we can do in this model.

You have written that the total Hamiltonian H and operators A are physical. My question to DarMM and Strangerep: what is the problem solution in terms of operators A? Write it down explicitly, please.

 

[strangerep]

what is the problem solution in terms of operators A? Write it down explicitly, please.

Sorry, but I don't understand what precisely you want.
The A(k) are already given in terms of the a(k) and they diagonalize the
Hamiltonian.

But obviously you want something else. (?)

 

[Bob_for_short]

Sorry, but I don't understand what precisely you want.
The A(k) are already given in terms of the a(k) and they diagonalize the
Hamiltonian. But obviously you want something else. (?)

Yes, obviously I want the problem solution Ψ. Hamiltonian serves to find the problem solution.

So, starting from H(A[j]), please write down the problem solution Ψ. I can't wait to compare the physical and "non-physical" expressions for Ψ.

 

[meopemuk]

Yes, obviously I want the problem solution Ψ. Hamiltonian serves to find the problem solution.

So, starting from H(A[j]), please write down the problem solution Ψ. I can't wait to compare the physical and "non-physical" expressions for Ψ.

Once you got the Hamiltonian H, finding solutions for wave functions is just a technical task. You can find stationary states by diagonalizing this Hamiltonian, and you can find the time evolution of any initial state |\Psi(0)\rangle by applying the time evolution operator |\Psi(t)\rangle = exp(-iHt)|\Psi(0)\rangle.

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.

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.

Eugene.

 

[Bob_for_short]

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.

 

[Bob_for_short]

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.

 

[meopemuk]

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

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)}

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 a^* term) which spontaneously create a-particles out of vacuum. It also describes processes (described by the a 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 \tilde{j}(k) 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.

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.

 

[Bob_for_short]

... the DarMM's Hamiltonian in terms of a-operators is

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)}

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 a_k. Then the problem is solved exactly.

This interaction is very peculiar, because it leads to processes (described by the a^* term) which spontaneously create a-particles out of vacuum. It also describes processes (described by the a 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) e^za+|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 \tilde{j}(k) 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.

 

[meopemuk]

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.

 

[Bob_for_short]

... 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.

 

[meopemuk]

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 a^*a term is the free part and the (a^*+a) term is the interaction.

Eugene.

 

[Bob_for_short]

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.

 

[meopemuk]

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.

 

[strangerep]

So, starting from H(A[j]), please write down the problem solution \Psi.
I can't wait to compare the physical and "non-physical" expressions for \Psi.

Patience, like politeness, is a virtue.

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

[...]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.,

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

|0\rangle_z is annihilated by the A(k) operators.

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

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

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

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

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

The z(k) corresponding your j(k) above is not square-integrable. We cannot even say
that |z|^2 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 \mathcal{H}_0. Nevertheless, the solution
can be expressed in another unitarily-inequivalent Hilbert space,
which is the point of all this.