Exploring the Possibilities of a New Relativistic Quantum Theory

[izh-21251]

Sorry, I edited my post too long, so you answered it earlier than I finished it. ))

Ivan

 

[izh-21251]

Exactly what I meant.
If you apply boost to the interaction operator (14.70), page 546 in the book, the photonic coefficient function won't stay unchanged - an additional term, proportional to $$\acute{k}^{\mu}$$ will occur.
So, instead of $$\epsilon_{i}(\acute{s},\acute{k})$$ (in your notations) one will get something like:

$$\epsilon_{i}(\acute{s},\acute{k}) + C(\theta, \acute{s}, \acute{k} )k_{i}$$,
where $$C(\theta, \acute{s}, \acute{k} )$$ is a certain complex function.

All this will then mean, that the operator under consideration is gauge-dependent (or, in other words, change its form under certain transformation)...

 

[izh-21251]

Basically, I am solving equations (10.27) - (10.30).

This procedure is equal to the case, when I take field-theoretical Hamiltonian (leading to the QED S-matrix) and apply dressing transformation to it, isn't it?
The two Hamiltonians are scattering-equivalent - the second (dressed) Hamiltonian lead to the same S-matrix.

 

[meopemuk]

I still believe, that in particle theory gauge transformations is nothing but Lorentz-boost in certain direction.

In my opinion, the significance of the so-called "gauge invariance" is blown out of proportion in modern physics. In my opinion, this is just a technical/heuristic detail.

Let me recall the logic, which Weinberg uses to build the interaction Hamiltonian $$J_{\mu}(x)A^{\mu}(x)$$ in QED. The basic idea is to first build certain linear combinations of particle c/a operators (=quantum fields), which have covariant transformation laws and vanishing space-like commutators. Then relativistic interaction "density" should be constructed simply as a product of such quantum fields.

This idea works fine when particles are massive. But for massless photons we meet the problem that the field $$A^{\mu}(x)$$ does not transform covariantly under boosts. See Appendix K.6 in the book. The transformation law (K.24) contains an additional term, which spoils the covariance. If you like, you can call this additional term the "gauge transformation", but usually, when people talk about gauge transformations in classical or quantum electrodynamics, they have in mind something more general than simply additional term in the bost transformation of the EM field.

The non-covariance of $$A^{\mu}(x)$$ means that we are not allowed to build our interaction as a product of $$A^{\mu}(x)$$ with arbitrary fermion fields. Such products will not obey the requirements of relativistic invariance. However, we can still satisfy this requirement if we chose to build our interaction as the specific product $$J_{\mu}(x)A^{\mu}(x)$$, where $$J_{\mu}(x)$$ is the so-called "conserved fermion current". Coupling our photon field $$A^{\mu}(x)$$ with the current operator $$J_{\mu}(x)$$ has a convenient property that the non-covariant term in the boost transformation law of $$A^{\mu}(x)$$ does not have any adverse effect on the relativistic invariance of the interaction operator $$J_{\mu}(x)A^{\mu}(x)$$. This non-trivial fact is proved in Appendix N.

All the above considerations are nothing but heuristic tricks, whose only justification is that the resulting interaction $$J_{\mu}(x)A^{\mu}(x)$$ works mysteriously well in all situations. In my opinion, these considerations do not justify raising the "gauge invariance" to the level of fundamental physical principle.

You show the electron-proton ($$ep\rightarrow ep$$) interaction in the 2-nd order in the Coulomb gauge... If I work in different gauge (or make a boost, which I believe to be equal actions) - this potential will be different. Thus - the dressed Hamiltonian IS gauge dependent.

Yes, you can make a boost transformation of the interaction operator V and you will get a different operator V', simply because usually V does not commute with the boost generator K in the Poincare algebra

$$V' = exp(iK \theta) Vexp(-iK \theta)$$

This happens in all theories, even when there are no massless particles (like photons) and no traces of gauge invariance.

Eugene.

 

[meopemuk]

Exactly what I meant.
If you apply boost to the interaction operator (14.70), page 546 in the book, the photonic coefficient function won't stay unchanged - an additional term, proportional to $$\acute{k}^{\mu}$$ will occur.
So, instead of $$\epsilon_{i}(\acute{s},\acute{k})$$ (in your notations) one will get something like:

$$\epsilon_{i}(\acute{s},\acute{k}) + C(\theta, \acute{s}, \acute{k} )k_{i}$$,
where $$C(\theta, \acute{s}, \acute{k} )$$ is a certain complex function.

All this will then mean, that the operator under consideration is gauge-dependent (or, in other words, change its form under certain transformation)...

Function $$\epsilon_{i}(\acute{s},\acute{k})$$ is a purely numerical coefficient defined in Appendix K.2. When you apply boost transformation to V_3 in (14.70)

$$V_3 \to \exp(iK \theta) V_3 \exp(-iK \theta)$$

this numerical coefficient remains unchanged. So, I don't think I understand your point.

Eugene.

 

[meopemuk]

This procedure is equal to the case, when I take field-theoretical Hamiltonian (leading to the QED S-matrix) and apply dressing transformation to it, isn't it?
The two Hamiltonians are scattering-equivalent - the second (dressed) Hamiltonian lead to the same S-matrix.

Yes, this is correct.

If I understand correctly, you are also asking what would happen if you apply the dressing transformation to field Hamiltonians in different gauges?
As I said at the end of post #210, I believe that you'll get non-equal but scattering-equivalent Hamiltonians. I am not 100% sure about that, because I haven't done these calculations myself.

In gauges other than the Coulomb gauge one has all those "non-physical" photon degrees of freedom, and they make me sick. So, I don't want to go down that route. Moreover, in my opinion, such calculations have only academic interest, because fields and gauges should not belong to a good physical theory.

Eugene.

 

[izh-21251]

In my opinion, the significance of the so-called "gauge invariance" is blown out of proportion in modern physics. In my opinion, this is just a technical/heuristic detail.

I think we are in total agreement here. There is no physical meaning in 'gauge invariance'.
I will try to avoid using term 'gauge invariance' in discussing particle theory... since it causes a lot of misundersatndings.

 

[izh-21251]

Function $$\epsilon_{i}(\acute{s},\acute{k})$$ is a purely numerical coefficient defined in Appendix K.2. When you apply boost transformation to V_3 in (14.70)

$$V_3 \to \exp(iK \theta) V_3 \exp(-iK \theta)$$

this numerical coefficient remains unchanged. So, I don't think I understand your point.

Yes, I probably was not accurate enough in my point.
I meant, that the product $$\epsilon_{i}(\acute{s},\acute{k})a(\acute{s},\acute{k})$$ (where $$a$$ is the photonic operator) changes it's form under any boost in the plane, orthogonal to $$\acute{\vec{k}}$$, so that additional term. proportional to $$\acute{k}^{\mu}$$ arises.

Weiberg showed this property on the p.250 of his book (formula (5.9.22)).
$$D^{\mu}_{\nu}(W(\Theta,\alpha,\beta))e^{\nu}(k,\pm1)=e^{\pm i\Theta}\left\{e^{\mu}(k,\pm1)+\frac{\alpha\pm i\beta}{\sqrt{2}k}k^{\mu}\right\}$$


$$\epsilon_{i}(\acute{s},\acute{k})a(\acute{s},\acute{k})$$ enters the operator of 'bremsstrahlung' (14.70) in your book.
Thus - under the boost in the $$\acute{\vec{k}}$$-orthogonal plane - in this operator appears new term proportional to $$\acute{k}^{\mu}$$

Ivan

 

[meopemuk]

izh-21251,

I agree that the interaction operator V_3 experiences some non-trivial transformation under boosts. Actually, the transformation law is even more complicated than that shown in Weinberg's book, because Weinberg's formula is written for a free field transforming under non-interacting representation of the Poincare group. In the presence of interaction the boost generator contains additional terms and the transformation law of V_3 becomes quite complicated.

But I don't see anything unusual in this situation. I guess I failed to understand why you consider this operator transformation as something remarkable? The interaction operator looks different for different moving observers. So what?

Eugene.

 

[izh-21251]

The interaction operator looks different for different moving observers. So what?

I suppose it's ok with respect to the S-matix - it is the same QED S-matrix for Hamiltonians in all Lorentz frames.

But what about time evolution? Won't these Hamiltonians (containing interactions which look different in different Lorentz frames) lead to different time-evolution laws?
-----------------------------------------------------------------------------------
Also - what about renorm operators? They also will look different in different Lorentz frames.

I was always thinking, that (for instance) mass-renorm operators correspond to the particle mass shifts.
For the case of mesodynamics (with massive bosons) Shebeko et al. proved, that mass shifts do not depend on particle momenta. They introduce mass counterterms and all mass-renorm operators in Hamiltonian simply contribute to the mass-shifts of clothed particles.
In other words, Shebeko-Shirokov's clothed particles have rest masses equal in all Lorentz frames.

In the case of QED Hamiltonian situation is different - particle mass shifts will depend on the momenta of particles..
If I apply Shebeko-Shirokov technique to the QED Hamiltonian - I will obtain rest masses of clothed particles different in different Lorentz frames..

Do you insert all renorm-operators (which then appear during dressing) from the very beginning into QED-Hamiltonian, so that they mutually cancel??

Thanks,
Ivan

 

[meopemuk]

I suppose it's ok with respect to the S-matix - it is the same QED S-matrix for Hamiltonians in all Lorentz frames.

But what about time evolution? Won't these Hamiltonians (containing interactions which look different in different Lorentz frames) lead to different time-evolution laws?

Hamiltonians in different frames are different, because H does not commute with the boost generator. Yes, this means that the time evolution looks different when watched from different frames. If I am looking from a moving train window I see that houses around are moving. However, a person on the ground would swear that houses stay at rest. We perceive the time evolution of the houses differently. This is hardly surprising.

In the case of QED Hamiltonian situation is different - particle mass shifts will depend on the momenta of particles..
Do you insert all renorm-operators (which then appear during dressing) from the very beginning into QED-Hamiltonian, so that they mutually cancel??

The QED interaction with renormalization counterterms is shown in eq. (9.38). The counterterms relevant for the mass renormalization are those with factors $$\Delta_2$$ and $$\Lambda_2$$. Obviously, there are renorm operators in these counterterms. The dressing transformation in section 10.2 is designed to cancel these renorm interactions exactly. So, the only remaining renorm part in the dressed Hamiltonian is the sum of free particle energies in H_0. This means that after renormalization and dressing we arrive to the same particles (with the same original constant masses) from which we started.

Eugene.

 

[izh-21251]

The QED interaction with renormalization counterterms is shown in eq. (9.38). The counterterms relevant for the mass renormalization are those with factors $$\Delta_2$$ and $$\Lambda_2$$. Obviously, there are renorm operators in these counterterms. The dressing transformation in section 10.2 is designed to cancel these renorm interactions exactly.

Let me be more specific..
What is the nature of these factors - $$\Delta_2$$ and $$\Lambda_2$$? Are they the C-numbers?

 

[izh-21251]

Hamiltonians in different frames are different, because H does not commute with the boost generator. Yes, this means that the time evolution looks different when watched from different frames. If I am looking from a moving train window I see that houses around are moving. However, a person on the ground would swear that houses stay at rest. We perceive the time evolution of the houses differently. This is hardly surprising.

Hamiltonian and Boost-generator do not commute. That I undestand. That's why moving objects look differently from static (have larger energy, show time delay, length shortening etc.)

What I tried to ask is, whether the UNUSUAL properties of interactions in Hamiltonians (consisting in that they are different in different frames due to photonic operators properties) affect time evolution laws...
If I understand you correctly, you are saying that there are no problems - and different forms of Hamiltonian in different Lorentz frames mean nothing special. Am I right?

Ivan

 

[meopemuk]

Let me be more specific..
What is the nature of these factors - $$\Delta_2$$ and $$\Lambda_2$$? Are they the C-numbers?

Yes, they are c-numbers defined in (9.24) and (9.25). In most textbooks they are denoted as $$\delta m$$ and $$Z_2 - 1$$ respectively. Perhaps I should also switch to this common notation. I'll think about that.

Eugene.

 

[meopemuk]

Hamiltonian and Boost-generator do not commute. That I undestand. That's why moving objects look differently from static (have larger energy, show time delay, length shortening etc.)

What I tried to ask is, whether the UNUSUAL properties of interactions in Hamiltonians (consisting in that they are different in different frames due to photonic operators properties) affect time evolution laws...
If I understand you correctly, you are saying that there are no problems - and different forms of Hamiltonian in different Lorentz frames mean nothing special. Am I right?

Ivan

Yes. No matter how complicated is the Hamiltonian, there is a universal simple relationship between Hamiltonians $$H(\theta)$$ and $$H(0)$$ in two moving frames

$$H(\theta) = H(0) \cosh \theta - P(0) c \sinh \theta$$


where P(0) is the total momentum operator. This relationship remains valid for all relativistic theories.

Eugene.

 

[izh-21251]

The counterterms relevant for the mass renormalization are those with factors $$\Delta_2$$ and $$\Lambda_2$$. Obviously, there are renorm operators in these counterterms. The dressing transformation in section 10.2 is designed to cancel these renorm interactions exactly.
[...]
Yes, they are c-numbers defined in (9.24) and (9.25).

When explicitely perfoming dressing transformation in QED Hamiltonian in Coulomb gauge, I faced the problem that 2-nd order renormalization operators do not cancel mass counterterms.

In other words, I obtain mass shifts (say, for electron) that depend on particle momentum and different in different Lorentz frames.
The result - it is not possible to define factors of mass counterterm as the C-numbers.

Namely for electron mass counterterm , we have:
$$M_{ren}=\int\frac{d\vec{p}}{E_{p}}\delta m(\vec{p})b^{+}_{p}b_{p}$$ (1)
where $$b$$ is the electron operator.
The corresponding renorm operator from 2-nd order dressed Hamiltonian is:
$$H_{2,ren}=\int\frac{d\vec{p}d\vec{k}}{E_{p}}F(\vec{p}, \vec{k})b^{+}_{p}b_{p};$$ (2)

For the QED Hamiltonian the result is that $$\delta m$$ depends on $$\vec{p}$$

Thus, one has to interoduce counterm in the form (1) to cancel all mass-renorm terms in dressed Hamiltonian... in this case $$\delta m$$ won't be the C-number... Otherwise, you will have Hamiltonian with remaining "renorm" terms in it.

Ivan

 

[meopemuk]

Ivan,

I am not exactly sure what is your complaint. The mass renormalization counterterm in QED is shown in eq. (9.20). If you expand the field operators there in particle c/a operators you'll obtain renorm terms whose coefficient functions are momentum-depend. I don't see anything wrong with that.

Eugene.

 

[izh-21251]

My point is that $$\Lambda_{2}$$ is not a C-number.

If I suppose it is a C-number, I cannot cancel the renorm operator of second order in Hamiltonian.

Ivan

 

[meopemuk]

My point is that $$\Lambda_{2}$$ is not a C-number.

If I suppose it is a C-number, I cannot cancel the renorm operator of second order in Hamiltonian.

Ivan

$$\Lambda_{2}$$ is a constant (independent on momentum) given in (9.25). In the mass renormalization counterterm (9.20) this constant is multiplied by a non-trivial (momentum-dependent) operator. The form of this operator was chosen specifically to cancel divergent loop integrals in the S-operator, in particular the electron self-energy integral shown in fig. 9.2. This is explained in subsections 9.2.1 and 9.2.2. Please let me know if this explanation is not clear, and I'll try to correct it.

Eugene.

 

[izh-21251]

Dear Eugene,
sorry for delayed reply. I unfortunately had a plenty of work lately.

I spared some time to revisit your book.
Concerning $$\Lambda_{2}$$, you are right, I apologize for a mistake.
This constant enters renormalization term in conventional S-matrix approach, so it is a C-number and I have no questions about it.

Interesting is the idea to restore Hamiltonian from the S-matrix, which is well-known for QED, and ensure the restored Hamiltonian has a set of "good" properties.
Does this method gives unique dressed Hamiltonian?
How this program will look provided the S-matrix is not known exactly? E.g., in the case of hadrodynamics?

What I was actually trying to talk about is Dressing (Clothing) in Hamiltonian, not restoring dressed Hamiltonian from the S-matrix.
I believe the two approaches (Shirokov-Shebeko and the one of yours) worth deeper comparison (though you say they are equivalent).
---------------------------------------------------------------
I questioned Shebeko about clothing in QED.
He agreed that the major peculiarity of QED (any gauge theory) is that it's Hamiltonian is not a Lorentz scalar and it can be divided into two parts: scalar and non-scalar.

He stressed that this division is important with respect to how to choose clothing transformation. One should consider applying Clothing only to the SCALAR part.

Well, I haven't grasped what these all mean yet... But I think this will give a shift to my undestanding...
----------------------------------------------------------------

 

[A. Neumaier]

I questioned Shebeko about clothing in QED.
He agreed that the major peculiarity of QED (any gauge theory) is that it's Hamiltonian is not a Lorentz scalar.

This is not peculiar at all, and has nothing to do with gauge theories.

No relativistic field theory can have a Lorentz invariant Hamiltonian, since H must transform as the 0-component of the 4-momentum vector.

 

[izh-21251]

Let me argue.

Usually, the Hamiltonian density has the property:
$$U(\Lambda, a)H(x)U^{-1}(\Lambda, a)=H(\Lambda x+a)$$ (1)
then one says that Hamiltonian density is a scalar.
Here U - is a Lorentz transformation.
Did you mean
$$U(\Lambda, a)H(x)U^{-1}(\Lambda, a)\neq H(x)$$ ??
This is of cause true.

Example of a theory with such Hamiltonian - a theory of massive spinless particles, interacting via Yukawa 3-linear coupling.

However, in some fileld-theoretical models (including couplings with derivatives or particles with spin >=1), the property (1) does not hold.
In this situation one can say that Hamiltonian is not a scalar.

The most common example of non-scalar Hamiltonian - the Hamiltonian of QED.
Such property of QED Hamiltonian is directly related with properties of photons - massless spin-1 particles.
You might know that QED Hamiltonian looks differently in different gauges - this is exactly the property of QED Hamiltonian, which I called "non-scalarity".

"Gauge equivalence principle" in field theory in particle theory should be referred to as "scattering equivalence" of non-scalar Hamiltonians.

Ivan

 

[izh-21251]

By the way - the same properties (scalarity or non-scalarity) one can observe in particle representation - via considering properties of particle operators due to Lorentz-transformations.

Refer to S.Weinberg, "The quantum theory of fields", Volume 1, Chapter 5.

 

[A. Neumaier]

Let me argue.

Usually,
the Hamiltonian density has the property:
$$U(\Lambda, a)H(x)U^{-1}(\Lambda, a)=H(\Lambda x+a)$$ (1)
then one says that Hamiltonian density is a scalar.

Before you were talking about the Hamiltonian, not the Hamiltonian density.

The density may indeed be a scalar, but in simple cases only; the conditions for it are discussed in Weinberg's book. But there is nothing peculiar about it - it is needed to make the Hamiltonian itself the 0-component of a covariant 4-vector.

 

[meopemuk]

Hi Ivan,

Interesting is the idea to restore Hamiltonian from the S-matrix, which is well-known for QED, and ensure the restored Hamiltonian has a set of "good" properties.
Does this method gives unique dressed Hamiltonian?

No, this method does not provide the unique Hamiltonian. Similarly, the dressing/clothing unitary transformation does not provide a unique solution as well. This uncertainty is explained in subsection 10.2.9. This is not suprising, because the same S-matrix can follow form many different (scattering-equivalent) Hamiltonians.

How this program will look provided the S-matrix is not known exactly? E.g., in the case of hadrodynamics?

If the S-matrix is approximate, then the calculated dressed Hamiltonian will be approximate too.

I questioned Shebeko about clothing in QED.
He agreed that the major peculiarity of QED (any gauge theory) is that it's Hamiltonian is not a Lorentz scalar and it can be divided into two parts: scalar and non-scalar.

He stressed that this division is important with respect to how to choose clothing transformation. One should consider applying Clothing only to the SCALAR part.

Please give my regards to Dr. Shebeko. I am not sure why he insists on separating scalar and non-scalar parts of the Hamiltonian (thanks to Dr. Neumaier for clarifying your terminology). I don't make this separation in the book and I apply dressing to the entire Hamiltonian. I start with the Coulomb-gauge QED Hamiltonian (8.10) - (8.13). Then I expand fields through annihilation-creation operators to obtain explicit expressions for interaction terms in Appendix L. Then renormalization counterterms should be calculated and added to this Hamiltonian as explained in chapter 9. This leads to the Hamiltonian H^c shown in eq. (10.9). From this point the dressing technique proceeds as described in subsections (10.2.4) - (10.2.6).

As you can see from Appendix L and chapter 9, the Hamiltonian H^c contains huge number of terms. The dressing procedure requires calculations of multiple commutators involving these terms. So, this path is enormously laborious and cumbersome. I wasn't patient enough to go beyond the 2nd perturbation order.

On the other hand, the alternative direct fitting of the dressed Hamiltonian to the S-matrix (as described in subsection 10.2.7) is much simpler. It requires only knowledge of renormalized QED scattering amplitudes in the desired perturbation order. They are denoted by F's in eqs. (10.27) - (10.30). And they can be taken from any regular QFT textbook.

Cheers.
Eugene.

 

[Dickfore]

I skimmed through the article and saw the discussion on Darwin Lagrangian. So, OP is the Darwin Lagrangian relativistically invariant?

 

[meopemuk]

I skimmed through the article and saw the discussion on Darwin Lagrangian. So, OP is the Darwin Lagrangian relativistically invariant?

I do not discuss Lagrangians in the book. I derive Darwin Hamiltonian as a part of corresponding interacting representation of the Poincare Lie algebra. This Hamiltonian is relativistically invariant, i.e., the commutation relations of the Poincare algebra are valid. They are verified in Appendix O of the book.

Eugene.

 

[Dickfore]

I do not discuss Lagrangians in the book. I derive Darwin Hamiltonian as a part of corresponding interacting representation of the Poincare Lie algebra. This Hamiltonian is relativistically invariant, i.e., the commutation relations of the Poincare algebra are valid. They are verified in Appendix O of the book.

Eugene.

In your 'proofs', I see a lot of $\approx$ signs. Does this mean that the commutators are only valid up to the same order as the Darwin Hamiltonain is?

 

[meopemuk]

In your 'proofs', I see a lot of $\approx$ signs. Does this mean that the commutators are only valid up to the same order as the Darwin Hamiltonain is?

Yes, this is correct.

Eugene.

 

[Dickfore]

So, then, you have not proven relativistic covariance exactly, have you?

 

[meopemuk]

So, then, you have not proven relativistic covariance exactly, have you?

This is correct. The whole approach in Appendix O is perturbative and expansion in powers of 1/c^2 was used. Only low-order terms were retained in the proof.

However, there is a good reason to believe that relativistic invariance will be valid in higher orders as well. The traditional field-based QED is relativistically invariant (see Appendix N). The dressed particle approach is obtained from QED by means of a unitary dressing transformation, which preserves Poincare commutators (see subsection 10.2.8). Therefore, the full interaction potential between dressed charged particles must be relativistically invariant as well. Unfortunately, this full non-perturbative potential is not known yet and explicit proof of its invariance is not possible.

Eugene.

 

[Dickfore]

This is correct. The whole approach in Appendix O is perturbative and expansion in powers of 1/c^2 was used. Only low-order terms were retained in the proof.

However, there is a good reason to believe that relativistic invariance will be valid in higher orders as well.
Eugene.

lol, this is not considered a scientific argument. Why don't you derive a Hamiltonian in the next order in perturbation theory?

 

[meopemuk]

Why don't you derive a Hamiltonian in the next order in perturbation theory?

This is exactly what I'm doing right now. The calculations are rather non-trivial as they involve loop integrals with their ultraviolet and infrared divergences. If I'm successful I'll have radiative corrections to the Darwin-Breit Hamiltonian, which describe, e.g., the anomalous magnetic moment of the electron and the Lamb shift. I will be happy to report my findings here when I'm done.

Eugene.