About Gauge invariance - again

[izh-21251]

First of all, let me remind about an older thread on this topic:
https://www.physicsforums.com/showthread.php?t=330517
Here I'd like to thank again to everybody, who participated in that discussion.

However, I still find myself at a deadlock with some questions about Gauge Invariance (GI) principle. Thus, I decided to start a new thread about GI.

Before I come up to my problems, let me start with refreshing what I derived from books and what has been discussed previously.
So, the principle of GI arise from the fact, that a 4-vector quantum field cannot be built from
creation/annihilation opeartors of spin 1 massless particles (say, photons) (refer to S. Weinberg, "The Quantum Theory of Fields", Cambridge University Press, 1995, vol.1, p.246-250)
Namely, the coefficient functions of a desired field appear to transform not like 4-vectors under LORENTZ transformations! Instead,
$$D^{\mu}_{\nu}(W(\Theta,\alpha,\beta))e^{\nu}(k,\pm1)=e^{\pm i\Theta}\left\{e^{\mu}(k,\pm1)+\frac{\alpha\pm i\beta}{\sqrt{2}k}k^{\mu}\right\}$$

where $$e^{\mu}(k,\pm1)$$ are the coefficient functions of vector field for helicities +1 and -1, $$\alpha, \beta, \Theta$$ - arbitrary parameters of Lorentz-transformation.
We see - an additional term appeared, proportional to particle momentum $$k^{\mu}$$. This means, that photon field is not a 4-vector field.
However, the way out to build Lorentz-invariant S-matrix was suggested. It consists in coupling "vector field" to conserved currents only, so that additional term $$k^{\mu}$$ does not affect Lorentz-invariance. Different additional terms $$k^{\mu}$$ correspond to different gauges. As long as "vector potential" is coupled with conserved currents, one is free to choose any gauge, and it will not affect the observable results.

But - here begins my misunderstanding - what happens, when we build Poincare generators?
The Hamiltonian operator does not remain unchanged, if we added a term $$k^{\mu}$$ to a photonic coefficient function.
In other words, if we apply a LORENTZ transformation to Hamiltonian it will vary due to the fact, that coefficients $$e^{\mu}(k,\pm1)$$ will not transform as 4-vectors, but acquire a $$k^{\mu}$$-term. Stress, Hamiltonians of different gauges are connected with each other by LORENTZ transformations!

So far, my first question is - AM I RIGHT with my logic up to this point?

Ok, if I am right, Hamiltonian is not a Lorentz-scalar any more... However, the S-matrix still seems to be the same for all gauges, because, as I understand, all these different Hamiltonians are scattering-equivalent... (see Ekstein H. Equivalent Hamiltonians in scattering theory // Phys. Rev.,1960. Vol. 117. P. 1590 – 1595).

What about other Poincare generators? To ensure relativistic invariance of the theory, they must obey the set of commutation relations - the Lee algebra. For example - in instant form of relativistic dynamics, another generator, that carries interaction is the boost operator. Thus, the interaction in boost-operator will vary when one moves to another Lorentz-frame - in the same way as interaction in Hamiltonian did. No doubts - the whole set of Poincare-Lee commutation relations must remain unchanged.
So, we've come to a point, where in each Lorentz-frame we have unique set of Poincare generators.

My second question again - am I right here? And does it have any consequences for the theory?
Especially, if anybody is familiar with the so-called "clothing-transformation" method in QFT, what all this will mean for this approach??

Thanks.
Ivan.

 

[strangerep]

However, the way out to build Lorentz-invariant S-matrix was suggested. It consists in coupling "vector field" to conserved currents only, so that additional term $$k^{\mu}$$ does not affect Lorentz-invariance. Different additional terms $$k^{\mu}$$ correspond to different gauges.

Right -- that's effectively the same as adding a gradient to the EM potential.

As long as "vector potential" is coupled with conserved currents, one is free to choose any gauge, and it will not affect the observable results.

But - here begins my misunderstanding - what happens, when we build Poincare generators?
The Hamiltonian operator does not remain unchanged, if we added a term $$k^{\mu}$$ to a photonic coefficient function.

One must construct the "interacting representation" of the Poincare generators.
You begin to mention this later when you talk about forms of dynamics, etc, such
as "instant form". In that form there's an extra interaction term in the Hamiltonian,
and a corresponding one in the boost generators, such that the overall Poincare
Lie algebra remains satisfied. (BTW, it's "Lie", not "Lee". :-)

Constructed in this way, the physical observables are gauge-invariant, as they must be.

[...]
So, we've come to a point, where in each Lorentz-frame we have unique set
of Poincare generators.

I don't know how you came to that conclusion. There's only the interacting
representation of the Poincare generators appropriate to the form of dynamics
chosen, in this case the "instant form".

Especially, if anybody is familiar with the so-called "clothing-transformation" method
in QFT, what all this will mean for this approach??

The "clothing-transformation" is distinct from Poincare transformations, and
there's a number of variations on the theme. You'll need to be more specific...

I'm sure Meopemuk will have more to say if he stops by in this thread, but in
the meantime, have a look at the book http://arxiv.org/abs/physics/0504062
which talks about a lot of this stuff in more detail.

 

[meopemuk]

Ivan,

why do you want to build Hamiltonians in different gauges?

I think it is perfectly reasonable to build the Hamiltonian in the Coulomb gauge, where there are no uphysical (longitudinal) photon degrees of freeedom. Once you get this Hamiltonian and the corresponding boost operator you have all necessary ingredients for the theory. In particular, you can make a transition to the "clothed particle" picture.

Possibly, you can do the same work in other gauges, but then you'll need to deal with unphysical degrees of freedom, and calculations become messier. I guess that in the end you may obtain scattering-equivalent Hamiltonians. I am not sure about that, but, personally, I don't think that's important. In general, I don't consider GI an important physical principle. Yes, this principle played an important role in derivations of modern QFT theories, such as electroweak and QCD. But I suspect this role was heuristic rather than fundamentally physical.

Eugene.

 

[izh-21251]

I am not interested particularly in considering Hamiltonian in different gauges.
I understand, how GI-principle works in FIELD theories and agree to all things said.

But in PARTICLE theory, as it seems to me now, there is no any GI-invariance (or equivalence)... There is only a set of scattering-equivalent Hamiltonians, corresponding to different Lorentz-frames.
Please, correct me, if I'm wrong.

However, I cannot get on with the fact, that interaction operator in QED Hamiltonian is not a Lorentz scalar and in each Lorentz-frame we have new Hamiltonian - and new whole set of Poincare generators... What it means? :\\ That is what really interesting!

Further (it's rather a note, not a question) - can not it be possible to construct a Lorentz-scalar 3-linear operator, using massless spin-1 particle operators and avoiding the concept of fields at all??

What about non-abelian gauge theories? What if one tried to avoid the concept of fields there?..
How, in particle formalism, will gauge self interactions arise and will they be the same as in field theoreical picture?

Thanks.
Ivan

 

[meopemuk]

But in PARTICLE theory, as it seems to me now, there is no any GI-invariance (or equivalence)... There is only a set of scattering-equivalent Hamiltonians, corresponding to different Lorentz-frames.
Please, correct me, if I'm wrong.

However, I cannot get on with the fact, that interaction operator in QED Hamiltonian is not a Lorentz scalar and in each Lorentz-frame we have new Hamiltonian - and new whole set of Poincare generators... What it means? :\\ That is what really interesting!

The Hamiltonians in different Lorentz frames are related by a simple formula

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

where $$P(0)$$ is the total momentum operator. This relationship is universal and independent on any gauge invariance assumptions.

Further (it's rather a note, not a question) - can not it be possible to construct a Lorentz-scalar 3-linear operator, using massless spin-1 particle operators and avoiding the concept of fields at all??

I am not sure what you are trying to say. You can always expand field Hamiltonians in particle creation and annihilation operators. Then you get an equivalent "particle picture" Hamiltonian. Unfortunately, in this notation you lose the visibility of gauge invariance. As far as I know, in the particle picture there is no principle (either physical or heuristic) that can guide you to the acceptable form of interaction. I don't think that formulating the GI principle in the particle notation will lead you anywhere.

 

[izh-21251]

Sorry for a VERY late responce ((

Let me stress - I am not concerned with the Gauge Invariance itself.

What I am actually trying to understand, is how to apply the recently developing Unitary Clothing (Dressing) Transformations (UCT) approach to the gauge theories - QED, weak interactions and QCD.

And first of all - considering the idea of formulating theory without the concept of fields at all... I think it is not such crazy or leading nowhere...
Such approach have been developed recently by Shebeko et al. (not published yet as I know, please, refer to http://www.ccsem.infn.it/issp2010/newtalents/FROLOV.pdf )
As I understand their results, it is possible to ensure the Relativistic Invariance of the theory on the (algebraic) level of particle operators, avoiding the Lagrangian formalism and the concept of fields...
It could mean, that you can build Hamiltonian from particle operators (how to choose the proper form is another question...) and then ensure relativistic invariance of the theory, building Poincare generators and checking out the commutation relations..

The one problem - I haven't seen yet any implications of Clothing (Dressing) technique to gauge theories, to QED for example, as the simplest one... Are you aware of any?
Direct ('dumm') application of the UCT approach to the QED Hamiltonian led me to the problems, which I cannot resolve yet.
As I mentioned ealier, as the QED Hamiltonian has different forms in different gauges, thus I obtained different physical interactions between clothed particles, different mass shifts... and thus the new clothed particle operators are different if I start from different gauge..
Moreover, when I transform from one gauge to another... clothed interactions and mass counterterms loose their explicit Lorentz covariance...

Yet, I can not decide how to relate to such situation... This is really what troubling.

 

[meopemuk]

izh-21251,

You can find discussion of the dressed/clothed particle approach applied to QED in

https://www.physicsforums.com/showthread.php?t=474666

I don't think you should worry about the gauge invariance. There are no fields in this approach (there are only particles and their direct interactions), so the issue of gauge invariance simply doesn't appear there.

It is convenient to apply the unitary dressing transformation to the QED field Hamiltonian expressed in the Coulomb gauge, simply because in this gauge there are no unphysical longitudinal photon degrees of freedom. I am not sure what would happen if you choose to start with a QED field Hamiltonian in any other gauge. My guess is that you'll end up with a different (but scattering equivalent) set of particle potentials.

Eugene.

 

[izh-21251]

Thank you.
Tomorrow will plunge into reading it )))