What controls/defines S=1 for Photons?

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what_are_electrons
[SOLVED] What controls/defines S=1 for Photons?

The spin for photons is S=1 for photons, but I do not know what controls or defines that assignment. Your help will be appreciated.
 

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  • #2
dextercioby
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Group theory...A photon is described by a vector field [itex] A_{\mu}(x) [/itex] which is a spin 1 (more correctly [itex] (1/2,1/2) [/itex]) representation of the restrained Lorentz group...

Daniel.
 
  • #3
what_are_electrons
dextercioby said:
Group theory...A photon is described by a vector field [itex] A_{\mu}(x) [/itex] which is a spin 1 (more correctly [itex] (1/2,1/2) [/itex]) representation of the restrained Lorentz group...

Daniel.
My understanding is that a spin = 1 implies +1, 0, -1.
 
  • #4
dextercioby
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Your understanding is not valid for real photons,but for hypothetical ones (so-called Proca photons).

Daniel.
 
  • #5
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what_are_electrons said:
My understanding is that a spin = 1 implies +1, 0, -1.
yes, that is correct. However QFT shows us that the [tex]s_{z} = 0[/tex]-state corresponds to a non-physical degree of freedom. Such photons will exhibit a negative probability-distribution. The reason for these troubles are the fact that a photon has zero-restmass.

Virtual photons however can exhibit the three spin-values...

regards
marlon
 
  • #6
dextercioby
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Marlon,it would be silly to include in the quantization the 2 nonphysical degrees of freedon for the em. field... :rolleyes:

Daniel.
 
  • #7
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dextercioby said:
Marlon,it would be silly to include in the quantization the 2 nonphysical degrees of freedon for the em. field... :rolleyes:

Daniel.
What exactly do you mean by these words ?

marlon
 
  • #8
dextercioby
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I meant the whole procedure of quantization of the em.field should be accomplished by realizing the fundamental Dirac brackets on the reduced phase space (which already includes only 2 classical degrees of freedom,the physical one, instead of 4,the initial ones) as quantum Lie brackets (commutators,if u prefer),in perfect agreement with the second postulate of QM.

Daniel.
 
  • #9
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Well, you need to add the fact that during the quantization of a massive spin 1 field, there is no problem. However in the case of a massless spin 1 field, things get more difficult for the known reasons. The massive spin 1 field has three spin states or, classically, polarization states after performing a Lorentz boost to bring it in its rest frame. We cannot do that in the case of a photon. We need gauge-fixing, Fadeev Poppov gohsts or Gupta-Bleuler-theory...

marlon
 
  • #10
dextercioby
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What i said is appliable entirely for the Proca Lagrangian.Proca theory has initially 4 degrees of freedom (just like the Maxwell field) and,after applying the Dirac bracket,the initial phase space,spanned by 8 coordinates,is reduced to a 6 dimensional one,which would correspond to 3 degrees of freedom.
Because second-class constrained systems are rather dull and simple to quantize using the Dirac bracket & the second principle,other more spectacular methods (BRST) were developed,as the transformation of a II-nd class system in a I-st class counterpart and applying the traditional BRST techniques.
I'm quite proud that the 2 teachers who contributed mostly to my scientifical knowledge are working in this domain. :approve:
Quite interesting... :approve:
 
  • #11
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Sounds very interesting indeed

marlon
 
  • #12
selfAdjoint
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Yes it does. References, Dexter?
 
  • #13
dextercioby
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The Bible of quantization:Marc Henneaux & Claudio Teitelboim:"Quantization of Gauge Systems",PUP,1992.
My 2 teachers wrote a book (unfortunately in Romanian,so it's not in your libraries) on "Quantization of Second-Class Theories" based on their numerous articles in Nucl.Phys.B and other peer-reviewed journals.Actually they wrote many books,but that's on the sbject i discussed earlier:transforming a second class system---->gauge system and quantizing it via BRST cohomological approach.

There are 3 working groups (on this specific subject) in Europe that i know of:the one at ULB with prof.Henneaux,the one at Craiova with prof.Bizdadea and the one at Moscow with Batalin,Fradkin,Vilkoviski,...(about the Russians,they must be really in their '60-s at least,they published at the end of the '70's,so they're probably retired,but surely they've left their places to someone else).

Daniel.
 
  • #14
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any references to articles in peer reviewed journals, dexter ?

marlon
 
  • #15
dextercioby
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C.Bizdadea,S.O.Saliu,Nucl.Phys.B456(1995),473.
C.Bizdadea,S.O.Saliu,Nucl.Phys.B469(1996),302.
C.Bizdadea,S.O.Saliu,Phys.Lett.B368(1996),202.
C.Bizdadea,Phys.Rev.D53(1996),7138.

And so on and so forth.

And a ton on arxiv.

Daniel.
 
  • #16
Haelfix
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The Proca Lagrangian is just the lagrangian for a general spin 1 particle. If we take the photon massless, this just reduces to classical EM field. By simple Fadeev Popov methods we see what Marlon said, that the spin zero state is eaten by the ghost term.

The three spin state photon must therefore have a small mass. But wait its not that easy. If we require local gauge invariance, one immediately notices the small mass term is not gauge invariant, which clashes horribly with what we know. The theory will be nonrenormalizable.

Am I missing something?
 
  • #17
dextercioby
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Proca Lagrangian,by construction,with nonzero mass term,is not gauge invariant,simply because gauge invariance is associated to first class systems,while Proca Lagrangian is second class...End of story.
If we build interaction theories (not necessarily in the SM) from the principle of minimal coupling,then matter fields will never couple to second class fields,because the latter don't have gauge invariance by definition.

Daniel.

P.S.Guys,this discussion is at classical level... :wink: Quantization is another matter.
 
  • #18
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Haelfix said:
The Proca Lagrangian is just the lagrangian for a general spin 1 particle. If we take the photon massless, this just reduces to classical EM field. By simple Fadeev Popov methods we see what Marlon said, that the spin zero state is eaten by the ghost term.

The three spin state photon must therefore have a small mass. But wait its not that easy. If we require local gauge invariance, one immediately notices the small mass term is not gauge invariant, which clashes horribly with what we know. The theory will be nonrenormalizable.

Am I missing something?
All you say is very true. However, the Higgs-mechanism incorporates zero mass spin 1 bosons (the photon that is). The local U(1) symmetry of EM-interactions is never broken after the continuous breakdown of symmetry. Therefore the photon does not interact with the Higgs-field. Before this symmetry-loss, all elementary particles are massless...

marlon
 
  • #19
Haelfix
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Right so, dare I ask, whats the draw of the theory of proca photons?

There is an infinite class of non gauge invariant, nonrenormalizable theories, what makes this attractive?
 
  • #20
dextercioby
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Who said it's "attractive"...?As i said,second class theories are rather dull and too easy to quantize.That's why new constraints are generated to turn them into I-st class.
In the end,it's purely a theoretical model.We'll have to change everything,if the photons will turn out to be massive...Which i hope will never happen.

Daniel.
 
  • #21
what_are_electrons
dextercioby said:
Group theory...A photon is described by a vector field [itex] A_{\mu}(x) [/itex] which is a spin 1 (more correctly [itex] (1/2,1/2) [/itex]) representation of the restrained Lorentz group...

Daniel.
Been reading Quantum Theory of Light (R. Loudon) and Quantum Chromodynamics (G.Schaefer) but have not come across any reference to a spin = 1 (using 1/2, 1/2) for photons. Where can I read about these 1/2, 1/2 states?
 
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  • #22
dextercioby
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Pierre Ramond:"Field Theory:A Modern Primer",L.H.Ryder:"Quantum Field Theory".

Daniel.
 
  • #23
Haelfix
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The idea is that photons are irreducible unitary representations of the poincare group SL(2,C). This group is the double cover of the proper Lorentz group, which can be decomposed as SU(2)*SU(2). You always have to be careful about the Z2 factors in there, but thats basically what you end up with.
 

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