A Polarization of gauge bosons and gauge choice

spaghetti3451

Consider the following facts:

1. For a particle with momentum $k$, the two transverse polarization vectors $\epsilon({\bf k}, \lambda_{1})$ and $\epsilon({\bf k}, \lambda_{1})$ are purely spatial and orthogonal to $\bf k$, that is,
$\epsilon^{0}({\bf k}, \lambda_{1}) = 0,$
$\epsilon({\bf k}, \lambda_{1})\cdot{k} = 0,$
$\epsilon^{0}({\bf k}, \lambda_{2}) = 0,$
$\epsilon({\bf k}, \lambda_{2})\cdot{k} = 0.$
2. The third, longitudinal, polarization vector $\epsilon({\bf k}, \lambda_{3})$}, for a particle with momentum $k$, is timelike positive, orthogonal to $k$ as well as the transverse polarization vectors, and has unit negative norm, that is,
$\epsilon^{0}({\bf k}, \lambda_{3}) > 0,$
$\epsilon({\bf k}, \lambda_{3})\cdot{k} = 0,$
$\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{1}) = 0,$
$\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{2}) = 0,$
$\epsilon({\bf k}, \lambda_{3})\cdot{\epsilon({\bf k}}, \lambda_{3}) = -1.$

3. We can infer from the orthogonality of the polarization vectors that the longitudinal polarization vector of a particle points in the direction of momentum of the particle.

--------------------------------------------------------------------------------------------------------------------------------------

Are these facts true for any gauge choice of the electromagnetic four-vector $A^{\mu}$?

Related Quantum Physics News on Phys.org

PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

PeterDonis

Mentor
For a particle with momentum $k$,
Do you mean a photon with momentum $k$? Your notation and your question seem to imply that.

spaghetti3451

Do you mean a photon with momentum $k$? Your notation and your question seem to imply that.
There is a longitudinal polarization. This is for a W or Z boson.

PeterDonis

Mentor
This is for a W or Z boson.
But you refer to a gauge choice for the electromagnetic 4-vector. That's not the same as a W or Z boson. Which are you asking about?

vanhees71

Gold Member
In the electroweak standard model (as for any Higgsed gauge theory) the unitary gauge reveals the particle content explicitly. It's drawback is that the proper vertex functions are not explicitly renormalizable in this gauge, but to get the physical degrees of freedom, it's great. The great discovery by 't Hooft and Veltman in 1971 was thas one can choose a renormalizable gauge (called $R_{\xi}$ gauges), where the proper vertex functions are manifestly renormalizable since the power counting works as for scalar fields, i.e., the gauge-boson propogator falls qudratically with momentum in these gauges. Together with dim. reg. that enabled them to prove both the renormalizability of non-abelian Higgsed gauge theories and the unitarity and physicality of the S-matrix. It makes clear that both the Faddeev-Popov ghosts and the "would-be Goldstone bosons" conspire with the unphysical polarization degree of freedom of the gauge bosons to cancel these unphysical degrees out of the S-matrix (at any order in perturbation theory), and this makes the Standard Model a physically consistent QFT of the electroweak interaction.

"Polarization of gauge bosons and gauge choice"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving