Transverse polarizations of a massless spin 1 particle

  • #1
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Physical polarization vectors are transverse, ##p\cdot{\epsilon}=0##, where ##p## is the momentum of a photon and ##\epsilon## is a polarization vector.



Physical polarization vectors are unchanged under a gauge transformation ##\epsilon + a\cdot{p}=\epsilon##, where ##a## is some arbitrary constant.



1. Why are physical polarization vectors transverse?

2. Why is ##p\cdot{\epsilon}=0## the condition for the transverseness of ##\epsilon##?

2. How is ##\epsilon + a\cdot{p}=\epsilon## a gauge transformation? The gauge transformations I know are of the form ##A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda##.
 

Answers and Replies

  • #2
dextercioby
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Your text is ambiguous (if not plain wrong) to use both classical and quantum terminology. The polarization vector is assigned to a classical em wave, while p is the photon's momentum (3 or 4-momentum, it's not clear from what you wrote). The quantum analogue to the classical wave's polarization vector is called the helicity operator. Leaving this aside, for 3. I can say that the ##\epsilon## is not unique which can be (comparing to the standard gauge field theory) thought of a gauge transformation. For 2. the answer should be obvious (hint: why are the E and B fields called transverse?). The answer to 2. automatically sheds light (pun intended!) on 1.
 
  • #3
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Thank you for the excellent answer.

I have another question.

Assume that, in the centre of mass frame, photon 1 moves in the positive ##z##-direction so that photon 2 moves in the negative ##z##-direction.

Then, the helicity states of the two photons ##1## and ##2## are given by ##(\epsilon_{\mu}^{\pm})^{1}=\frac{1}{\sqrt{2}}(0,1,\pm i,0)## and ##(\epsilon_{\mu}^{\pm})^{2}=\frac{1}{\sqrt{2}}(0,1,\mp i,0)##.

1. Why is there a relative negative sign for the helicity state of each photon?
2. I know that these are so-called circular helicity states. But, how are these the components of these helicity states determined?
3. Is it a common terminology to call ##\epsilon^{+}## the positive helicity state and ##\epsilon^{-}## the negative helicity state?
 
  • #4
vanhees71
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There are two helicity states ##\epsilon^{\pm}##. What you've written down is the standard basis for the radiation gauge ##A^0=0##, ##\vec{\nabla} \cdot \vec{A}=0## for a photon with momentum in ##3##-direction. Of course, only either the upper or the lower choice of signs gives a complete set. The one basis is just interchanging the basis vectors of the other.
 

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