Renormalizable quantum field theories

[LaserMind]

In Quantum Field Theory, there are 'infinity' problems:

At extremely short distances the energy quanta increase so
much that infinities would occur. In order to overcome this
a cut off is imposed that postulates that quanta cannot possess
energy above some arbitary high value.
This works well for low energy calculations but not for high
energy interactions.

1) At what length does this cut off happen - approximately
2) How is this justified?

A reference if you don't know what I am talking about:
http://en.wikipedia.org/wiki/Quantum_field_theory#Renormalization

 

[Bob_for_short]

1) At what length does this cut off happen - approximately
2) How is this justified?

The necessity to make a cut-off, let us say, in QED for certainty, originates from the fact that without it the integrals diverge.
The cut of is an artificial trick, it is not related to some physics. Introducing a cut-off is called a regularization. There are many of them. Then, in the "renormalizable" theories, the divergent terms are grouped so that they are perturbative corrections to the initial mass and charge.
After that they say: "Let us discard these corrections", and the corrections are discarded. So no dependence of the cut-off parameter remains. Others say: "Our original mass and charge are not observable but with these big corrections they are observable, therefore they are equal to the finite m and e." Both ideologies give the same final "renormalized" finite expressions. They are cut-off independent.

This discarding prescription is not legitimate mathematically and reflects a too bad (too distant) initial approximation and bad interaction (self-action) used in the perturbation theory. The "renormalized" solutions are not solutions of the original theory but those of another one that I call a Novel QED.

Practically the renormalization justification is done by comparison with the experimental data. It is a good luck that sometimes this prescription works (see my "Reformulation instead of Renormalizations" by Vladimir Kalitvianski). In fact there is a possibility to reformulate QED in better terms and obtain convergent series automatically.

Bob_for_short.

 

[meopemuk]

In Quantum Field Theory, there are 'infinity' problems:

At extremely short distances the energy quanta increase so
much that infinities would occur. In order to overcome this
a cut off is imposed that postulates that quanta cannot possess
energy above some arbitary high value.
This works well for low energy calculations but not for high
energy interactions.

1) At what length does this cut off happen - approximately
2) How is this justified?

A reference if you don't know what I am talking about:
http://en.wikipedia.org/wiki/Quantum_field_theory#Renormalization

I basically agree with Bob_for_short's assessment. Just wanted to add a couple of points.

In QFT (really, I have QED in mind) we meet problems when calculating the S-matrix (e.g., scattering amplitudes). Some momentum integrals in these calculations appear to be divergent when the integration momentum tends to infinity. The fix suggested by the renormalization theory is two-fold:

1. Introduce momentum cutoff, so that integrals are forced to be finite.
2. Add certain (momentum-dependent) "counterterms" to the Hamiltonian of QED.

If you do these steps carefully, then you can find that calculated scattering amplitudes

1. Become finite and cutoff-independent
2. Agree perfectly well with experiment when involved particle momenta are below the cutoff.

All this is nice and good. The question is what if we want to study interactions at momenta higher than the cutoff? The problem is that we cannot take the cutoff momentum to infinity, because "counterterms" in the Hamiltonian are divergent in this limit and we would obtain an ill-defined Hamiltonian. The prevailing attitude is that we should not take the infinite cutoff limit, because this would mean probing interactions at such small distances that our ideas about smooth space-time are no longer applicable. It is assumed that at small distances (on the order of Planck length or whatever), some new effects take place (like space-time granularity or string-like nature of particles,...) which invalidate the use of standard QED. So, the suggested solution is to pick the momentum cutoff to be above the characteristic momentum in the considered physical problem and below the inverse Planck length.

There is also an alternative point of view, which is called the "dressed particle" approach. It says that the QED Hamiltonian with countertems is badly screwed up. It suggests to fix the divergences in counterterms by applying an unitary "dressing" transformation to the Hamiltonian. As a result of this we obtain:

1. New (cutoff-independent and finite) Hamiltonian of QED.
2. All scattering amplitudes computed with this Hamiltonian are exactly the same as in the traditional renormalized QED, i.e., agree with experiment very well.

 

[ExactlySolved]

In the classical field theory of electromagnetism it is necessary to do the same renormalization procedure in order to account for the radiation backreaction.

I just see renormalization as a 'problem' with Lorentz invariant theories involving point particles, the impossibility of having a finite energy lorentz-invariant near-field of an accelerating point particle being similar but obviously more subtle than the problem of having a lorentz invariant rigid body. But my point is that this issue does not only arise in QFT.

By the way, for those who do not like renormalization, one of the awesome properties about strings as opposed to particles is that their interactions are finite to all orders in perturbation theory, no renormalization required, the integrals converge!

 

[meopemuk]

By the way, for those who do not like renormalization, one of the awesome properties about strings as opposed to particles is that their interactions are finite to all orders in perturbation theory, no renormalization required, the integrals converge!

This is not unique to strings. The same is true in the "dressed particle" approach to QFT. All loop integrals are convergent. No regularization/renormalization required.

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139

 

[LaserMind]

This is not unique to strings. The same is true in the "dressed particle" approach to QFT. All loop integrals are convergent. No regularization/renormalization required.

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139

A unitary clothing transformation in which the Hamiltonian remains finite
in the limit of removed regularization because infinities present in theta
exactly cancel infinities in V (where a transformation 'e to the i theta'
-sorry LATEX is too hard for me - has been applied).


E. V. Stefanovich,
Quantum Field Theory without Infinities
http://www.geocities.com/meopemuk/AOPpaper.html

 

[Bob_for_short]

As soon as Eugene starts from the same QED Hamiltonian and from the same initial approximation (free particles), he is bound to reproduce the usual renormalized QED expansion however it is called - dressing or just renormalizations, whatever. In both approaches occurs the perturbative "dressing". By the way, this name was invented in the standard QED, before "dressing" transformation. Nobody has been able to explain how finally a dressed electron looks like. (At best, one speaks of infinite "vacuum polarization" around a point-like charge that screens it original (infinite) "bare" charge.)

In Eugene's approach the same banal infrared difficulties arise. It means too distant initial approximation (the strong photon-electron coupling neglected).

As I said in my first reply, the fundamental constant renormalizations are equivalent to discarding the correction to the known, phenomenological constants. This reveals their non legitimate mathematical character. Renormalizations remove perturbatively the self-action interaction term. What does remain after renormalization? A potential interaction of compound systems, not self-action. That is why in the Novel QED I start directly from another Hamiltonian - without the self-interaction term and with electroniums as the initial approximations. No wonder I obtain immediately mathematically finite and physically correct results.

In order to show that there is no problem at short distances, let as consider an atom. It is practically unknown but true that the positive charge in atoms is quantum mechanically smeared, just like the negative (electron) charge. The positive charge cloud size is much smaller than the negative charge cloud size, but it is finite and is of the same nature (turning aroung the atomic center of inertia). So the effective ("dressed" if you like) potential is not as singular as the Coulomb one (1/r) but is much softer, for example, U_eff(r) ≈ 1/{r²+[(m_e/M_A)a₀]²}^1/2 for Hydrogen. This potential gives correct amplitude of elastic scattering at large angles (it gives the positive charge elastic form-factor that serves as a natural regularization factor in integrals). This potential does not tend to infinity when r →0 but remains constant. An of course, the cut-off size is much much larger than the Plank's length.

The same is valid for an electron permanently coupled to the quantized electromagnetic field: it charge is quantum mechanically smeared, so no singularity of the effective potential appears in calculations if one takes the coupling exactly in the initial approximation. Perturbative taking into account leads to the infrared divergence. It is easy to understand: the potential (1/r) is infinitely "far" from U_eff(r) at short distances and the corresponding integrals diverge. That is why it is necessary to start from better initial approximation where photon-electron coupling is taken into account exactly.

Pushing the bound nucleus in atom or pushing the bound electron in electronium excites the internal degrees of freedom of the corresponding compound system, so the inelastic processes (atom exciting or photon oscillator exciting) happens automatically in the first Born approximation. So no infrared divergence arises. The inclusive cross section give well known classical results - the Rutherford cross section, as if the target charge were point-like ans situated at the center of inertia of the compound system.

Details of physics and mathematics of the Novel QED are reported in "Atom as a "Dressed" Nucleus" and in "Reformulation instead of Renormalizations" by Vladimir Kalitvianski (available on arXiv).

Bob_for_short.

 

[LaserMind]

bose condensate & renormalizable quantum field theories

1) Are we assuming that the wavefunction region is a limit of energy quanta following the 1/r laws? So, as we approach the centre of a 'wavefunction region' the energy quanta diminish to zero instead of diverging to infinity?

2) How does this cut off affect an Einstein Bose condensate where wavefunctions are merged?

 

[Bob_for_short]

I could not understand you first question. Apparently we think differently.

I could not understand your second question: Bose-Einstein condensate of what? Of electrons? I am sorry, I cannot reply.

Bob.

 

[Bob_for_short]

In the classical field theory of electromagnetism it is necessary to do the same renormalization procedure in order to account for the radiation backreaction.

I just see renormalization as a 'problem' with Lorentz invariant theories involving point particles, the impossibility of having a finite energy lorentz-invariant near-field of an accelerating point particle being similar but obviously more subtle than the problem of having a lorentz invariant rigid body. But my point is that this issue does not only arise in QFT.

By the way, for those who do not like renormalization, one of the awesome properties about strings as opposed to particles is that their interactions are finite to all orders in perturbation theory, no renormalization required, the integrals converge!

I start my article "Reformulation instead of Renormalizations" from analysis of H. Lorentz ansatz about self-action and point-likeness of the electron. I show that the mass renormalization is just discarding corrections (perturbative or exact) to the phenomenological electron mass. In other words, it is postulating new equation for the electron dynamics.
The same discarding is made in QED. It is good luck that such a prescription works. Normally it does not work - the number of non-renormalizable theories is much larger.

I show also how the energy-momentum conservation law can be preserved without self-action ansatz.

Finally, I show that in compound systems there is always a natural cut-off mechanism so no necessity to invent strings or other "grained space-time" appear.

Starting from compound systems gives naturally soft radiation which is not case in Eugene dressing transformation or in the standard QED approaches. I propose physically and mathematically justified approach. I do not rely on good luck.

Bob_for_short.

 

[malawi_glenn]

bob, read the forum ruels; don't adress answers to "problems" from your own, non-published, ideas/theories.

I start my article "Reformulation instead of Renormalizations" from analysis of H. Lorentz ansatz about self-action and point-likeness of the electron. I show that the mass renormalization is just discarding corrections (perturbative or exact) to the phenomenological electron mass.

I show also how the energy-momentum conservation law can be preserved without self-action ansatz.

Finally, I show that in compound systems there is always a natural cut-off mechanism so no necessity to invent strings or other "grained space-time" appear.

Starting from compound systems gives naturally soft radiation which is not case in Eugene dressing transformation or in the standard QED approaches.

Bob_for_short.

 

[Bob_for_short]

bob, read the forum ruels; don't adress answers to "problems" from your own, non-published, ideas/theories.

My works were published. For example, "Atom as a "dressed" nucleus" has been published in the Central European Journal of Physics, V. 7, N. 1, pp. 1-11 (2009). I consider there the same problem. As well it was published long ago in the USSR (1990-93). My "RiR" is available on arXiv as a prepublication and I refer to it since it answers the questions of this thread. They are not problems of my own but the "eternal" problems of interacting fields. I, as many of us, tried to resolve them. None has found an error in my works so far. What is my own is my opinion based on my results. It is what this forum is made for - an exchange of opinions. Those who have no their own results refer to somebody else's opinions.

Bob.

 

[malawi_glenn]

oh it has been published recently? That is better :-)

Well we tend to discuss accepted opinions here, if you read the forum rules you will understand.

have fune

 

[Bob_for_short]

oh it has been published recently? That is better :-)

No, I published it long ago. I propose you to take part in my poll about positive charge atomic form-factor, please.

Well we tend to discuss accepted opinions here, if you read the forum rules you will understand.

I read somewhere your opinion about people who are not happy with renormalizations. You consider them to be crackpots.

I am sure you are happy with renormalizations and I am sure you know the whole universe history from the Big Bang till the end. You feel so high, you even touch the sky. That is why you take a liberty to lecture me.

have fune

Learn hard and be open-minded.

Regards,

Bob_for_short.

 

[LaserMind]

I could not understand you first question. Apparently we think differently.

I could not understand your second question: Bose-Einstein condensate of what? Of electrons? I am sorry, I cannot reply.

Bob.

Bob, I got this from a physics blog (I believe its yours??):
http://vladimirkalitvianski.wordpress.com/

“It is also described with an atomic (positive charge or “second”) form-factor, so the positive charge in an atom is not “point-like”. The positive charge “cloud” in atoms is small but finite. It gives a natural “cut-off” or regularization factor in calculations”
also in blog:
"It is practically unknown but true that the positive (nucleus electric) charge in an atom is quantum mechanically smeared, just like the negative (electron) charge. "

I assume by +ve 'charge cloud' you are referring to some sort of wavefunction region?
Its interesting to me to know what's going on here - a wavefunction region seems a good reason for capping energies - or maybe I am on the wrong path here?

 

[George Jones]

Let me remind everyone that Physics Forums rules to which everyone who registers agrees,

https://www.physicsforums.com/showthread.php?t=5374,

in part, state

Overly Speculative Posts: One of the main goals of PF is to help students learn the current status of physics as practiced by the scientific community; accordingly, Physicsforums.com strives to maintain high standards of academic integrity. There are many open questions in physics, and we welcome discussion on those subjects provided the discussion remains intellectually sound. It is against our Posting Guidelines to discuss, in most of the PF forums, new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion. Posts deleted under this rule will be accompanied by a private message from a Staff member, and, if appropriate, an invitation to resubmit the post in accordance with our Independent Research Guidelines. Poorly formulated personal theories, unfounded challenges of mainstream science, and overt crackpottery will not be tolerated anywhere on the site.

I'm locking this thread. Any discussion of Bob_for_short's ideas should take place in the appropriate thread in the Independent Research Forum,

https://www.physicsforums.com/showthread.php?t=307642.

 

[LaserMind]

Is there a discussion of renormalization and dressing anywhere else?
I am interested in this topic, but cannot get a discussion going or join
one.

 

[Bob_for_short]

I think you may discuss the generally accepted issues here. I will not participate any more in order not to have the thread locked.

If you are interested in my personal findings, you can read my articles available on arXiv and discus them in the independent research group, the thread https://www.physicsforums.com/showthread.php?t=307642.

Bob.

 

[DarMM]

If you do these steps carefully, then you can find that calculated scattering amplitudes

1. Become finite and cutoff-independent
2. Agree perfectly well with experiment when involved particle momenta are below the cutoff.

All this is nice and good. The question is what if we want to study interactions at momenta higher than the cutoff? The problem is that we cannot take the cutoff momentum to infinity, because "counterterms" in the Hamiltonian are divergent in this limit and we would obtain an ill-defined Hamiltonian.

I just wanted to mention that this is not correct. The addition of counterterms to the Hamiltonian makes it finite as the cutoff is removed. There are several field theories where this has proven to be the case, see Glimm and Jaffe's book.
In fact to be totally accurate without renormalizations the Green's functions of theories are zero, not infinite. So the addition of infinite counterterms makes the theory non-zero and finite.

 

[Bob_for_short]

In fact to be totally accurate without renormalizations the Green's functions of theories are zero, not infinite. So the addition of infinite counter-terms makes the theory non-zero and finite.

Dear DarMM and George Jones,

Don't you think that starting from "bare" (non physical, wrong) Hamiltonian furnished with non physical counter-terms is less overt crackpottery than starting from physical (well defined, everything is known experimentally) Hamiltonian with physical interaction?

Could you also participate in my poll on the positive charge atomic form-factor, please?

Bob (for short).

 

[meopemuk]

The addition of counterterms to the Hamiltonian makes it finite as the cutoff is removed. There are several field theories where this has proven to be the case, see Glimm and Jaffe's book.

What about QED? The counterterms in the QED Hamiltonian are explicitly cutoff-dependent and they diverge in the limit of removed cutoff. Or I'm missing something?

 

[DarMM]

What about QED? The counterterms in the QED Hamiltonian are explicitly cutoff-dependent and they diverge in the limit of removed cutoff. Or I'm missing something?

In all theories with renormalizations "the counterterms are explicitly cutoff-dependent and diverge in the limit of the removed cutoff". However the Hamiltonian with the counterterms included converges in the limit where the cutoffs are removed. I know it's pretty strange, so if you want details please ask.

 

[Bob_for_short]

In all theories with renormalizations "the counterterms are explicitly cutoff-dependent and diverge in the limit of the removed cutoff". However the Hamiltonian with the counterterms included converges in the limit where the cutoffs are removed.

Not always. There are non-renormalizable theories where such a prescription fails. So when it "works", there is nothing behind it but a good luck.

Bob.

 

[meopemuk]

In all theories with renormalizations "the counterterms are explicitly cutoff-dependent and diverge in the limit of the removed cutoff". However the Hamiltonian with the counterterms included converges in the limit where the cutoffs are removed. I know it's pretty strange, so if you want details please ask.

Yes, this sounds pretty strange, and I would appreciate if you can provide details.

To be specific, let me refer to S. Weinberg's "The quantum theory of fields" vol. 1 (Let me know if you want to discuss any other textbook. I have quite a few of them, though not Glimm & Jaffe's, unfortunately). In eqs. (11.1.6) - (11.1.9) Weinberg writes the Lagrangian of renormalized QED (the expression for the Hamiltonian should be essentially similar). The L_2 terms in (11.1.9) are counterterms. They have factors like deltam, Z_3 and Z_2, which (as shown later in the chapter) are expressed through divergent integrals. Weinberg writes explicit 1-loop formulas for these divergent factors. As I understand, in all higher orders, contributions to these factors are divergent too.

If I understand what you're saying, then the "counterterm" part of the Hamiltonian can be convergent only if divergent contributions from different orders somehow cancel each other. Is it what you mean?

 

[DarMM]

Not always. There are non-renormalizable theories where such a prescription fails. So when it "works", there is nothing behind it but a good luck.

Bob.

Of course it fails for non-renormalizable theories, that's why they are non-renormalizable. However it's not random luck that's behind the situations where it works, there are detailed physical reasons and mathematical reasons for why it works.
Mathematically its related to probability theory and the theory of random walks e.t.c.

 

[Avodyne]

Coefficients in the hamiltonian are in general cutoff dependent, and become infinite when the cutoff is removed. The reason this does not cause problems is that, in quantum mechanics, coefficients in the hamiltonian are not directly measureable.

Consider an anharmonic oscillator (with mass m=1),

H = {\textstyle{1\over2}}p^2 + {\textstyle{1\over2}}\omega^2 x^2 + {\textstyle{1\over24}}gx^4.

Classically, we can measure \omega by looking at the frequency of small oscillations. For very small amplitude A, we have x(t)=A\cos(\omega t).

But in the quantum theory, we cannot measure \omega directly. The smallest oscillation we can get comes from taking a linear combination of the ground state and the first excited state, |\psi(0)\rangle = c_1|0\rangle+c_2|1\rangle, and then we find

\langle \psi(t)|x|\psi(t)\rangle \propto \cos(\omega_{10}t),

where \omega_{10}=(E_1-E_0)/\hbar, which works out to be

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

So we can directly measure some complicated functions of \omega and g, but not \omega itself.

In field theory, we want the measureable quantities to come out cutoff independent, and this turns out to be possible only if the coefficients in the hamiltonian are cutoff dependent. But that's OK, because we can't directly measure them. We only think we can because we're used to using intuition from classical theory.

 

[DarMM]

Yes, this sounds pretty strange, and I would appreciate if you can provide details.

To be specific, let me refer to S. Weinberg's "The quantum theory of fields" vol. 1 (Let me know if you want to discuss any other textbook. I have quite a few of them, though not Glimm & Jaffe's, unfortunately). In eqs. (11.1.6) - (11.1.9) Weinberg writes the Lagrangian of renormalized QED (the expression for the Hamiltonian should be essentially similar). The L_2 terms in (11.1.9) are counterterms. They have factors like deltam, Z_3 and Z_2, which (as shown later in the chapter) are expressed through divergent integrals. Weinberg writes explicit 1-loop formulas for these divergent factors. As I understand, in all higher orders, contributions to these factors are divergent too.

If I understand what you're saying, then the "counterterm" part of the Hamiltonian can be convergent only if divergent contributions from different orders somehow cancel each other. Is it what you mean?

Let me see if I can explain this.
Take H as the original unrenormalized Hamiltonian. When I introduce cutoffs I create H^{\Lambda}, the cutoff Hamiltonian. To this I add the counterterm part of the Hamiltonian with cutoffs, let's call it H^{\Lambda}_{c.t.}.
Now H^{\Lambda} and H^{\Lambda}_{c.t.} are both totally divergent, non-self-adjoint, unbounded operators, in fact they are so badly behaved that they're not even operators.

However the renormalized Hamiltonian which is:
H_{ren} = \lim_{\Lambda \rightarrow \infty} H^{\Lambda} + H^{\Lambda}_{c.t.}
is a well-defined self-adjoint and semi-bounded operator. Although the counterterms diverge, (there are no cancellations between different orders in perturbation theory), they exactly cancel the divergences coming from powering operator-valued distributions like \phi(x) or A_{\mu}(x), so H_{ren} is well-defined.<br /> <br /> Does that make sense?

 

[Bob_for_short]

Coefficients in the hamiltonian are in general cutoff dependent, and become infinite when the cutoff is removed. The reason this does not cause problems is that, in quantum mechanics, coefficients in the hamiltonian are not directly measureable.

Consider an anharmonic oscillator (with mass m=1),

H = {\textstyle{1\over2}}p^2 + {\textstyle{1\over2}}\omega^2 x^2 + {\textstyle{1\over24}}gx^4.

Classically, we can measure \omega by looking at the frequency of small oscillations. For very small amplitude A, we have x(t)=A\cos(\omega t).

But in the quantum theory, we cannot measure \omega directly. The smallest oscillation we can get comes from taking a linear combination of the ground state and the first excited state, |\psi(0)\rangle = c_1|0\rangle+c_2|1\rangle, and then we find

\langle \psi(t)|x|\psi(t)\rangle \propto \cos(\omega_{10}t),

where \omega_{10}=(E_1-E_0)/\hbar, which works out to be

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

So we can directly measure some complicated functions of \omega and g, but not \omega itself.

In field theory, we want the measureable quantities to come out cutoff independent, and this turns out to be possible only if the coefficients in the hamiltonian are cutoff dependent. But that's OK, because we can't directly measure them. We only think we can because we're used to using intuition from classical theory.

The transition frequency for harmonic and slightly anharmonic oscillators (g is small) is still

\omega_{10}=\omega,

so it is as observable as the classical frequency. Your example is not convincing. See my "Reformulation instead of Renormalizations" for a more convincing example.

Bob.

 

[meopemuk]

Let me see if I can explain this.
Take H as the original unrenormalized Hamiltonian. When I introduce cutoffs I create H^{\Lambda}, the cutoff Hamiltonian. To this I add the counterterm part of the Hamiltonian with cutoffs, let's call it H^{\Lambda}_{c.t.}.
Now H^{\Lambda} and H^{\Lambda}_{c.t.} are both totally divergent, non-self-adjoint, unbounded operators, in fact they are so badly behaved that they're not even operators.

However the renormalized Hamiltonian which is:
H_{ren} = \lim_{\Lambda \rightarrow \infty} H^{\Lambda} + H^{\Lambda}_{c.t.}
is a well-defined self-adjoint and semi-bounded operator. Although the counterterms diverge, (there are no cancellations between different orders in perturbation theory), they exactly cancel the divergences coming from powering operator-valued distributions like \phi(x) or A_{\mu}(x), so H_{ren} is well-defined.<br /> <br /> Does that make sense?

<br /> <br /> I see your point, but it doesn&#039;t make sense to me.<br /> <br /> If we return to the Weinberg&#039;s Lagrangian, then what you call &quot;the original non-renormalized Hamiltonian&quot; is equivalent to L_0 and L_1 in (11.1.7) and (11.1.8). These are finite operators, and their definition does not involve loop integrals. So, they are independent on the cutoff.<br /> <br /> The cutoff-dependent divergent part is L_2. So, the sum L_0 + L_1 + L_2 is still cutoff-dependent and divergent. I don&#039;t see where the cancelation comes from.<br /> <br /> What am I missing here?

 

[Avodyne]

Although the counterterms diverge, (there are no cancellations between different orders in perturbation theory), they exactly cancel the divergences coming from powering operator-valued distributions

What you mean is that divergences cancel when you calculate observable quantities (or certain generalizations, such as renormalized Green's functions, which are not directly observable but still come out finite). The renormalized hamiltonian can then be given an abstract definition such that it has finite matrix elements between certain classes of states. But if you try to write it down explicitly, it will still have cutoff-dependent coefficients in it.

 

[Bob_for_short]

Let me see if I can explain this.
Take H as the original unrenormalized Hamiltonian. When I introduce cutoffs I create H^{\Lambda}, the cutoff Hamiltonian. To this I add the counterterm part of the Hamiltonian with cutoffs, let's call it H^{\Lambda}_{c.t.}.
Now H^{\Lambda} and H^{\Lambda}_{c.t.} are both totally divergent, non-self-adjoint, unbounded operators, in fact they are so badly behaved that they're not even operators.

However the renormalized Hamiltonian which is:
H_{ren} = \lim_{\Lambda \rightarrow \infty} H^{\Lambda} + H^{\Lambda}_{c.t.}
is a well-defined self-adjoint and semi-bounded operator. Although the counterterms diverge, (there are no cancellations between different orders in perturbation theory), they exactly cancel the divergences coming from powering operator-valued distributions like \phi(x) or A_{\mu}(x), so H_{ren} is well-defined.<br /> <br /> Does that make sense?

<br /> <br /> No, it does not make sense, in my opinion. This &quot;counter-term&quot; ideology starts from a wrong idea that the original Hamiltonian contains non-observable masses and charges. In fact, it is the badly guessed interaction (self-action) that brings corrections (divergent or not) to the fundamental (known) constants. The interaction Hamiltonian should be changed from the very beginnig rather than perturbatively - namely one can reformulate the theory without self-action. Then the perturbation corrections are finite and small, and the theory becomes physical - its initial approximations are physical and it describes correctly the soft radiation and other effects quite naturally, without appealing to not physical counter-terms. The counter-terms detract, order by order, the contributions of the self-action term of the total Hamiltonian. There is a short-cut based on good physics rather than on good luck.<br /> <br /> Bob.

 

[Avodyne]

The transition frequency for harmonic and slightly anharmonic oscillators (g is small) is still \omega_{10}=\omega

No, it isn't, it's

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

It doesn't matter how small g is (unless it's exactly zero), you are still only able to measure \omega_{10}, which is not the same as \omega. And to find out experimentally how small g is, you have to do some other measurements. And these won't measure g directly either, but some other complicated function of g and \omega. Then, knowing the theory, you could compute g and \omega. But you can't measure them. And this is why it's OK, in field theory, for hamiltonian coefficients to be cutoff dependent.

 

[meopemuk]

What you mean is that divergences cancel when you calculate observable quantities (or certain generalizations, such as renormalized Green's functions, which are not directly observable but still come out finite). The renormalized hamiltonian can then be given an abstract definition such that it has finite matrix elements between certain classes of states. But if you try to write it down explicitly, it will still have cutoff-dependent coefficients in it.

Yes, this is my understanding too. The full Lagrangian/Hamiltonian of the renormalized theory L_0 + L_1 + L_2 is cutoff-dependent and divergent. However, S-matrix elements (scattering amplitudes) computed with this Hamiltonian are cutoff-independent, finite, and agree with experiment perfectly well. That's the great achievement of the renormalization theory.

If you care only about scattering theory (as in all QFT textbooks), then the ill-defined Hamiltonian does not pose a serious problem. However, if you want to study the time evolution of states and observables, then the cutoff-dependence and divergences in the Hamiltonian is a show-stopper.

 

[Bob_for_short]

No, it isn't, it's

\omega_{10}=\omega+{\textstyle{1\over8}}g\hbar^2\omega^{-2}+O(g^2).

It doesn't matter how small g is (unless it's exactly zero), you are still only able to measure \omega_{10}, which is not the same as \omega. And to find out experimentally how small g is, you have to do some other measurements. And these won't measure g directly either, but some other complicated function of g and \omega. Then, knowing the theory, you could compute g and \omega. But you can't measure them. And this is why it's OK, in field theory, for hamiltonian coefficients to be cutoff dependent.

Even when g is not small, the anharmonic oscillator has a discrete spectrum which is directly measurable. You have chosen a bad example to show "complexity" of the QFT problems.

The Hamiltonian coefficients - masses and charges are not cutoff dependent if one uses a better Hamiltonian - without self-action. These fundamental constants are observable and constant. What is the energy dependent is not the "running" charge but the exact scattering amplitude.

Bob.

 

[Bob_for_short]

Yes, this is my understanding too. The full Lagrangian/Hamiltonian of the renormalized theory L_0 + L_1 + L_2 is cutoff-dependent and divergent. However, S-matrix elements (scattering amplitudes) computed with this Hamiltonian are cutoff-independent, finite, and agree with experiment perfectly well. That's the great achievement of the renormalization theory.

Eugene, you give too much flattery to the renormalized S-matrix. It is known/shown that the elastic processes are impossible - the S-matrix elements are identically equal to zero! It is the inclusive cross sections that are different from zero. So even after the renormalizations you have to work hard with the IR problem.

Bob.

 

[meopemuk]

Eugene, you give too much flattery to the renormalized S-matrix. It is known/shown that the elastic processes are impossible - the S-matrix elements are identically equal to zero! It is the inclusive cross sections that are different from zero. So even after the renormalizations you have to work hard with the IR problem.

Bob.

Yes, I know that. There are ultraviolet divergences and there are infrared divergences. But let us solve one problem at a time. Otherwise the whole thing becomes too confusing. I am interested to solve the ultraviolet problem first. Temporarily, we can assign a small non-zero mass to the photon, and thus avoid the annoying issues with "soft" photons and exclusive/inclusive cross sections.

 

[Bob_for_short]

Yes, I know that. There are ultraviolet divergences and there are infrared divergences. But let us solve one problem at a time. Otherwise the whole thing becomes too confusing. I am interested to solve the ultraviolet problem first. Temporarily, we can assign a small non-zero mass to the photon, and thus avoid the annoying issues with "soft" photons and exclusive/inclusive cross sections.

It turns out the the both problems can be solved at once: writing a physical Hamiltonian without self-action but with a potential interaction of coupled charges (electroniums) eliminates both mathematical problems and makes the theory physical: the soft radiation in present automatically as excitations of the internal (or relative) degrees of freedom of compound systems - electroniums. The inclusive consideration is now obligatory, otherwise the elastic S-matrix contains only zero-valued elements. I would love to work out the Novel QED with you. You have everything for that. Just take may ideas and solutions and make calculations. If anything seems to you strange or non-comprehensible - I will stand by. You always can count on me. I have outlined how the Lamb shift and the anomalous magnetic moment can be calculated (straightforwardly, without artificial cut-offs). We can start tomorrow, if you like.

Vladimir Kalitvianski.

 

[ytuab]

It turns out the the both problems can be solved at once: writing a physical Hamiltonian without self-action but with a potential interaction of coupled charges (electroniums) eliminates both mathematical problems and makes the theory physical:ted (straightforwardly, without artificial cut-offs). We can start tomorrow, if you like.
Vladimir Kalitvianski.

Sorry. I have not read your paper completely.

Is your theory relativistic or non-relativistic?
Does your theory keep Lorentz invariant?

 

[Bob_for_short]

Sorry. I have not read your paper completely.
Is your theory relativistic or non-relativistic?
Does your theory keep Lorentz invariant?

Yes, relativistic and Lorentz invariant. If you like to discuss details, I invite you to the Independent Research forum where I posted a thread on this subject.

Bob.

 

[DarMM]

I see your point, but it doesn't make sense to me.

If we return to the Weinberg's Lagrangian, then what you call "the original non-renormalized Hamiltonian" is equivalent to L_0 and L_1 in (11.1.7) and (11.1.8). These are finite operators, and their definition does not involve loop integrals. So, they are independent on the cutoff.

Okay here is the first difficulty. In Weinberg L_{0} in (11.1.7) is a well defined self-adjoint operator. L_{1} in (11.1.8) is not. It isn't self-adjoint and it's not semi-bounded. Hence L_{0} + L_{1} is not a well defined operator. I can sketch a proof if you wish, it involves some functional analysis.

The cutoff-dependent divergent part is L_2. So, the sum L_0 + L_1 + L_2 is still cutoff-dependent and divergent. I don't see where the cancelation comes from.

What am I missing here?

Let me describe the procedure.
Okay, so L_{0} is a well behaved operator so let's leave it alone. L_{1} is not well defined, so let's make it well defined by intorducing a cutoff. Taking the cutoff to infinity would also mean L_{2} is undefined so let's leave it cutoff dependant.
Now let's add the cutoff L^{\Lambda}_{1} and L^{\Lambda}_{2} to obtain:
L^{\Lambda}_{int} = L^{\Lambda}_{1} + L^{\Lambda}_{2}.

The \Lambda \rightarrow \infty limit of L^{\Lambda}_{int} is a well-defined self adjoint operator with no cutoff dependence. We can then add it to L_{0} (although by "adding" it you have to deal with certain technical difficulties but I'm being lose), to obtain a well-defined cutoff independant theory with well-defined finite time evolution.

 

[DarMM]

What you mean is that divergences cancel when you calculate observable quantities

I hope my previous post makes things clearer, but what I'm saying is that you can remove all cutoff dependence from the Hamiltonian and have it finite after you have performed renormalization.

But if you try to write it down explicitly, it will still have cutoff-dependent coefficients in it.

No it doesn't. It has no cutoff dependence once you take the limit and it is finite in that limit.

 

[Bob_for_short]

Dear DarMM,

What would you prefer: to get rid of UV and IR infinities perturbatively or to do with a physically correct theory without physical and mathematical difficulties?

Bob.

 

[ytuab]

Yes, relativistic and Lorentz invariant. If you like to discuss details, I invite you to the Independent Research forum where I posted a thread on this subject.

Bob.

I have now read " Atom as a Dressed Nucleus".

"Positive charge cloud" theory is very interesting, and I think your idea is true.

But I'm sorry if I make mistakes.

I think your paper is not relavent to the divergent problems of QFT ( is only relevant to 1/r?).
Your paper is nonrelativistic and does not keep Lorentz invariant?
It is almost impossible to solve the all divergent problems of QFT keeping Lorentz invariant.

How do you think about it?

 

[meopemuk]

L_{1} is not well defined, so let's make it well defined by intorducing a cutoff.

I don't understand why you are saying that L_1 is not well defined and cutoff-dependent. First note that L_1 is interaction "density", so in order to obtain the interaction operator, L_1 must be integrated on d^3x. So, in total there are 4 integrations in this expression: one on d^3x and 3 momentum integrals (which come from definitions of quantum fields). Quantum fields also supply exponential factors like exp(ipx). Their integration on d^3x results in one momentum delta function (which simply expresses the fact that our interaction conserves the total momentum). As a result we obtain a sum of terms. Each term has 3 momentum integrals whose integrands contain 1 momentum delta function and a product of 3 creation/annihilation operators. This is exactly the standard form of any interaction operator as shown in Weinberg's eq. (4.4.1). There is no cutoff in this equation.

 

[DarMM]

I don't understand why you are saying that L_1 is not well defined

I'm saying it because it's not. It's not self-adjoint or semi-bounded, there are proofs of this fact in a few books. The basic problem is that \bar\psi\gamma^{\mu}\psi is an operator valued distribution, but in more than two dimensions...

First note that L_1 is interaction "density", so in order to obtain the interaction operator, L_1 must be integrated on d^3x. So, in total there are 4 integrations in this expression: one on d^3x and 3 momentum integrals (which come from definitions of quantum fields). Quantum fields also supply exponential factors like exp(ipx). Their integration on d^3x results in one momentum delta function (which simply expresses the fact that our interaction conserves the total momentum). As a result we obtain a sum of terms. Each term has 3 momentum integrals whose integrands contain 1 momentum delta function and a product of 3 creation/annihilation operators. This is exactly the standard form of any interaction operator as shown in Weinberg's eq. (4.4.1).

...after performing integration it's not a well-defined operator. This was proven by Wightman in the 1960s. I can provide you with some references. So \int{\bar\psi\gamma^{\mu}\psi}A_{\mu} is not a well defined operator, this is rigorous mathematical fact.

and cutoff-dependent.

It's not cutoff-dependent, I'm just doing that so I can add it to the counterterm part without getting into the rigours of distribution theory. I'm introducing a cutoff in order to turn it into an operator which is well-defined and then adding it to the cutoff counterterm part obtaining another operator which is well-defined when the cutoff is removed.

There is no cutoff in this equation.

Of course, but the problem is that equation doesn't give you an operator unless it is cutoff.
Think about it, if the integral of L_{1} was an operator why would you even need renormalization? If it was well-defined there would be no loop divergences.

 

[meopemuk]

...if the integral of L_{1} was an operator why would you even need renormalization? If it was well-defined there would be no loop divergences.

In my opinion, L_1 is a well defined operator. In any case, whatever subtle irregularities were found in it by Wightman, they are no match for the explicitly divergent constants deltam, Z_2 and Z_3 in L_2. So, I don't see how any cancelation of divergences is possible in L_1 + L_2.

My understanding of the origin of loop divergences is different from yours. In my view the major problem is that L_1 (when expressed in terms of creation/annihilation operators) contains "trilinear" terms, like

L_1 = a*ac + a*c*a + ...

where a is annihilation operator for electrons and c is annihilation operator for photons. When you calculate the 2nd order S-operator with interaction L_1 you need to take the product of two copies of L_1

S = (a*ac + a*c*a + ...)(a*ac + a*c*a + ...)

After normal ordering you may notice that there is a non-zero term of the type

(loop integral) a*a

This term describes some kind of "scattering of the electron on itself", i.e., self-interaction. One effect produced by this term is that the electron mass in the interacting theory is different from the electron mass in the free theory. That's why the electron mass renormalization is needed. In QED things are even worse: the momentum dependence of the trilinear interactions is such that the loop integral is divergent, so the electron mass correction is infinite.

As long as you have trilinear interaction terms in your Hamiltonian you'll always have renormalization problems. In the "dressed particle" approach these trilinear interaction terms are called "bad". The idea of this approach is to change the Hamiltonian so that these bad terms are not present. All interactions must be written in terms of "good" operators only. An example of such a "good" operator is

a*a*aa

You may notice that if you take a product of two such operators and normal-order this product, you'll never get terms of the type a*a. So, there can be no "corrections" to the electron mass. No renormalization is needed.

The question is: can we get rid of the "bad" terms in the Hamiltonian and still obtain the same accurate S-matrix as we know it from renormalized QED? The answer is "yes", and the "dressed particle" approach shows how this can be done.

 

[Bob_for_short]

I have now read " Atom as a Dressed Nucleus".
"Positive charge cloud" theory is very interesting, and I think your idea is true.
But I'm sorry if I make mistakes.
I think your paper is not relavent to the divergent problems of QFT ( is only relevant to 1/r?).
Your paper is nonrelativistic and does not keep Lorentz invariant?
It is almost impossible to solve the all divergent problems of QFT keeping Lorentz invariant.
How do you think about it?

Hamiltonian formulation of QED looks like non Lorentz invariant but it is in fact. It's been proven many times. I use the Hamiltonian formulation in the gauge invariant (Dirac's) variables (known also as a Coulomb gauge). I build the relativistic Hamiltonian basing on the physics of the quantum mechanical charge smearing outlined in the first part of the article. The resulting Hamiltonian (see also "Reformulation instead of Renormalizations", formula (60)) is relativistic but free from the self-action. This, new formulation, is free from non-physical entities and describes the right physics analogous to the atomic scattering description. Preliminary non relativistic estimations show that it is right. I have not presented the detailed relativistic calculations but it is clear that they only bring some numerical corrections to the right physics obtained already in the non-relativistic approximation.

Bob.

 

[DarMM]

In my opinion, L_1 is a well defined operator.

It's not though and this isn't some subtlety, it is the origin of the divergences in quantum field theories. Operating on any vector in the Hilbert space twice with L_{1} (which is second order in perturbation theory) maps to a vector outside Fock space, hence the divergences.

In any case, whatever subtle irregularities were found in it by Wightman, they are no match for the explicitly divergent constants deltam, Z_2 and Z_3 in L_2.

Yes, in fact they are. You can prove that they are matches for these counterterms, are you contesting the proofs? If you are I suggest we start with the quartic scalar case.

So, I don't see how any cancelation of divergences is possible in L_1 + L_2.

Maybe so, but that doesn't change the fact that these cancellations indeed occur. Take a look at Glimm's "Boson fields with the \Phi^{4}interaction in three dimensions", to see it at work in the case of quartic scalar theory in three dimensions.

My understanding of the origin of loop divergences is different from yours. In my view the major problem is that L_1 (when expressed in terms of creation/annihilation operators) contains "trilinear" terms, like

L_1 = a*ac + a*c*a + ...

where a is annihilation operator for electrons and c is annihilation operator for photons. When you calculate the 2nd order S-operator with interaction L_1 you need to take the product of two copies of L_1

S = (a*ac + a*c*a + ...)(a*ac + a*c*a + ...)

After normal ordering you may notice that there is a non-zero term of the type

(loop integral) a*a

This term describes some kind of "scattering of the electron on itself", i.e., self-interaction. One effect produced by this term is that the electron mass in the interacting theory is different from the electron mass in the free theory. That's why the electron mass renormalization is needed. In QED things are even worse: the momentum dependence of the trilinear interactions is such that the loop integral is divergent, so the electron mass correction is infinite.

As long as you have trilinear interaction terms in your Hamiltonian you'll always have renormalization problems. In the "dressed particle" approach these trilinear interaction terms are called "bad". The idea of this approach is to change the Hamiltonian so that these bad terms are not present. All interactions must be written in terms of "good" operators only. An example of such a "good" operator is

a*a*aa

You may notice that if you take a product of two such operators and normal-order this product, you'll never get terms of the type a*a. So, there can be no "corrections" to the electron mass. No renormalization is needed.

The question is: can we get rid of the "bad" terms in the Hamiltonian and still obtain the same accurate S-matrix as we know it from renormalized QED? The answer is "yes", and the "dressed particle" approach shows how this can be done.

Remember that by Haag's theorem, interacting theories do not live in Fock space and hence cannot in general be written using creation and annhilation operators.
Anyway if you look at any of the literature on rigorous field theory you will see that the origin of ultraviolet divergences has been known for a long time as being due to two facts:
1. The producting of operator valued distributions resulting in ill-defined powers which will not not give rise to operators when integrated.
2. The non-Fock representations needed for interacting theories.

These are the origins of the ultraviolet difficulties. They are the reason L_{1} is divergent. In fact L_{1} is a tremendously divergent object, it even has what physicists would call nonperutrbative infrared divergences. To say it's well-defined up to some subtleties is totally provably wrong.

See the paper of Glimm above for an example of how badly divergent L_{1} is even in the case of a scalar field theory.

 

[meopemuk]

There can be no cancelation between L_1 and L_2 for the simple reason that L_1 is first order in the coupling constant (1st power in e) and L_2 is composed of 2nd, 3rd, 4th etc. order terms.

Operating on any vector in the Hilbert space twice with L_{1} (which is second order in perturbation theory) maps to a vector outside Fock space, hence the divergences.

This agrees with what I was saying: the (2nd order) product L_1 * L_1 is divergent (due to loop integrals). This divergence is compensated by the divergent 2nd order term in L_2. However, as I said above, there is no cancelation of divergences in the sum L_1 + L_2.

 

[DarMM]

There can be no cancelation between L_1 and L_2 for the simple reason that L_1 is first order in the coupling constant (1st power in e) and L_2 is composed of 2nd, 3rd, 4th etc. order terms.

This stuff is strange and perhaps I've been explaining badly. Think of what L_{1} is meant to be a function of, it's a function of the interacting fields \psi and A_{\mu} which are themselves functions of e. So L_{1} does contain terms of higher order in e through its dependence on the interacting fields.

Does this make sense?

However I must say that the best way of thinking of this probably isn't through the use of perturbative additive renormalizations, but with nonperturbative multiplicative renormalizations where you can this whole thing become a problem related to the theory of distributions.
At the end of the day however it has been proven that cancellations do take place between L_{1} and L_{2}.