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 A_{μ}:

ℒ=-¼F^{μν}F_{μν}

EoM: ∂_{μ}F^{μν}=0 ⇒ □A^{ν}-∂^{ν}(∂_{μ}A^{μ})=0

We know that ℒ is invariant under following Gauge Transformation (GT) where λ is a gauge parameter:

GT: A'^{μ}(x)=A^{μ}(x)–∂^{μ}λ(x)

We can use GT to change EoM: Given A^{μ}(x) that solves EoM, construct λ(x) such that □λ=∂_{μ}A^{μ }which implies ∂_{μ}A'^{μ}=0. A thing to note is that A'^{μ}(x) still has same F^{μν }and EoM, i.e. A'^{μ} is physically equivalent to A^{μ}.

Now we gauge fix the Lagrangian, i.e. give up manifest gauge invariance but keep all the physically distinct solutions:

ℒ=-¼F^{μν}F_{μν}–½α(∂_{μ}A^{μ})^{2}

EoM: □A^{ν}-(1-α)∂^{ν}(∂_{μ}A^{μ})=0

where α is a (real & non-zero) gauge fixing parameter.

Acting with ∂_{ν} on EoM gives α□(∂_{μ}A^{μ})=0 ⇒ ∂_{μ}A^{μ}=0.

Now we are left with D EoM & 1 Lorentz condition:

□A^{μ}=0 & ∂_{μ}A^{μ}=0.

Solving the first equation gives:

A^{μ}(x)=a^{μ}e^{-ik.x} where k^{2}=0 with k^{0}≡ω_{k}≡|**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: a^{μ}=Ck^{μ}+ε^{μ }where ε^{μ} should satisfy ε^{0}=0 & **ε**.**k**=0 (as is easily checked!) This already suggests 2 DoF have vanished; lets make it concrete by gauging A^{μ} using

λ(x)=iCe^{-ik.x }⇒ A'^{μ}(x) = A^{μ}(x)–∂^{μ}λ(x) = ε^{μ}e^{-ik.x}.

Now, it is obvious that a vector gauge field has only (D-2) DoF since ε^{0}=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=-¼F^{μν}F_{μν}+½m^{2}A_{μ}A^{μ}^{ }(Not gauge invariant!)

EoM: □A^{ν}-∂^{ν}(∂_{μ}A^{μ})+m^{2}A^{ν}=0

∂_{ν} EoM ⇒ m^{2}(∂_{ν}A^{ν})=0

Here also we are left with D EoM and 1 Lorentz condition:

(□+m^{2})A^{μ}=0 & ∂_{μ}A^{μ}=0.

But that is as far as we can go since no GT can reduce DoF any further (basically due to m^{2 }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!

thanks, that settles the dust.

ReplyDeleteand the font was nice.

-PJ

George's link takes me to one of your web page instead of his :)

ReplyDeletequite concise and elegant!

ReplyDeleteI am commenting now having read some of the relevant literature and in a better position to appreciate the conciseness.

Does e.k = 0 still hold if I use a generalization of Lorentz gauge fixing condition [say R-\xi gauge such as one used in Spontaneous SB where dA = xi x(some fields)]

ReplyDelete