Many people have asked me and I’ve asked many people why a photon has only two polarizations/ degrees of freedom (DoF) and not four! The ‘qualitative’ answer which most people (including me) gave and with which I was satisfied for quite some time consists simply of two parts:
- one DoF reduced due to gauge invariance (4-1=3);
- another DoF removed due to masslessness of photon (3-1=2)!
But after sometime, hearing this mundane answer did not satisfy me or the others anymore (basically others but lets not digress)! So I took it upon myself to settle this once and for all. You may ask why go to such lengths to answer an age-old question of which everyone knows the answer which is ‘DoF for a massless vector gauge field in D dimensions is (D-2)’. Well, there are two reasons:
- As PvN stresses in his classes it is always good to have a concrete example worked out than always deal with generalities/ abstractions!
- In GS’s QFT class, I got to know the ‘answer’ which one of the many people (lets call him PJ) does not agree to happening… go figure!
There is one more (2+1=3) very important reason: after reading this page by WS, I got the urge to test the Unicode stuff myself. So I post here the answer starting ‘from scratch’. This will probably be the only place in which I present Math and Physics simultaneously and roughly in equal proportion. [As should be obvious from my second reason, the following is ‘shamelessly’ lifted from notes of GS’s QFT class. Though any mistakes below would be entirely mine due to the well-known phenomenon of Lost in Translation.]
Lets start from the Lagrangian and Equations of Motion (EoM) for the massless vector gauge field Aμ:
ℒ=-¼FμνFμν
EoM: ∂μFμν=0 ⇒ □Aν-∂ν(∂μAμ)=0
We know that ℒ is invariant under following Gauge Transformation (GT) where λ is a gauge parameter:
GT: A'μ(x)=Aμ(x)–∂μλ(x)
We can use GT to change EoM: Given Aμ(x) that solves EoM, construct λ(x) such that □λ=∂μAμ which implies ∂μA'μ=0. A thing to note is that A'μ(x) still has same Fμν and EoM, i.e. A'μ is physically equivalent to Aμ.
Now we gauge fix the Lagrangian, i.e. give up manifest gauge invariance but keep all the physically distinct solutions:
ℒ=-¼FμνFμν–½α(∂μAμ)2
EoM: □Aν-(1-α)∂ν(∂μAμ)=0
where α is a (real & non-zero) gauge fixing parameter.
Acting with ∂ν on EoM gives α□(∂μAμ)=0 ⇒ ∂μAμ=0.
Now we are left with D EoM & 1 Lorentz condition:
□Aμ=0 & ∂μAμ=0.
Solving the first equation gives:
Aμ(x)=aμe-ik.x where k2=0 with k0≡ωk≡|k|
(Notation: k is a D-vector and k is a (D-1)-vector!)
Lorentz condition enforces k.a=0. This suggests the general solution: aμ=Ckμ+εμ where εμ should satisfy ε0=0 & ε.k=0 (as is easily checked!) This already suggests 2 DoF have vanished; lets make it concrete by gauging Aμ using
λ(x)=iCe-ik.x ⇒ A'μ(x) = Aμ(x)–∂μλ(x) = εμe-ik.x.
Now, it is obvious that a vector gauge field has only (D-2) DoF since ε0=0 & ε.k=0. In other words, timelike and longitudinal components are not physical, only transverse components (usually called polarizations in 4-D and hence the symbol ‘remaining’ is ε!) are physical and count as DoF!
Thus, we see that loosing two DoF is not so straightforward as the argument at the beginning of this post suggested; there is a subtlety as to how gauge invariance and masslessness conspire to get rid of these DoF. If you are wondering where the masslessness was used(!), keep reading…
Finally, lets look at the massive vector field and observe how longitudinal component remains physical!
ℒ=-¼FμνFμν+½m2AμAμ (Not gauge invariant!)
EoM: □Aν-∂ν(∂μAμ)+m2Aν=0
∂ν EoM ⇒ m2(∂νAν)=0
Here also we are left with D EoM and 1 Lorentz condition:
(□+m2)Aμ=0 & ∂μAμ=0.
But that is as far as we can go since no GT can reduce DoF any further (basically due to m2 in EoM) and so we are left with one more DoF compared to massless case, i.e. D-1 DoF.
I hope this satisfies some (if not all) of the many people I referred to earlier. And if you (the reader) have a succinct/ concise/ elegant proof (but explicit as done here), you are welcome to post it in the comments or tell me via any other communication means you may find reasonable!
So, that is it for this month and if you have not read this ‘ground-breaking’ paper in fundamental mathematics; you are missing out something very important in your life!
