# Polarization vectors of massive and massless particles

I read from Mandl & Shaw that when quantizing massless vector particles such as photons in Lorentz gauge, there are 4 linearly independent polarization vectors (2 of them being able to "gauged away"), while when quantizing massive vector particles such as W bosons, there are only 3 linearly independent space-like polarization vectors. It also says that when W and Z gauge bosons "eat" the Goldstone bosons from Higgs field, they become massive and acquire longitudinal polarization. So apparently there is some rule that says massive vector particles have longitudinal polarization, while massless particles don't. Why is that? From what principle is it derived? Their Euler-Lagrangian equations are essentially the same (both look like Klein-Gordon equation), just one of them has $m=0$. Is it the case that when $m=0$, it has four linearly-independent solutions while only three when $m\neq0$ ?

• This answer is great. – jinawee Jan 1 '14 at 20:18
• – DJBunk Jan 1 '14 at 21:36

So apparently there is some rule that says massive vector particles have longitudinal polarization, while massless particles don't. Why is that?

Massive vector may have longitudinal or transversal polarization. This is disscused in almost every QFT book, like Weinberg vol. 1.

Anyway, an intuitive explanation is that if $m=0$, logitudinal polarization is not possible since the particle would travel at $v<c$ or $v>c$ (due tu vibrations in it's direction's motion). While in the massive case, $v<c$ and longitudinal polarization is possible.

Mathematicaly, this can be explained in the context of Wigner's classification.

If we define $W_\mu=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}P^\nu M^{\rho\sigma}$ and $P_\mu=i\partial_\mu$. We can classify any particle in function of the eigenvalues of $W_\mu W^\mu$ and $P_\mu P^\mu$ .

In the massive case $P_\mu P^\mu >0$, the eigenvalues of $W_\mu W^\mu$ are $-m^2 s(s+1)$ and the spin $s$ can take $n/2$ values.

But if $m=0$, the eigenvalues of $W_\mu W^\mu$ are zero, since $P\mu W_\mu=0$. Which means that $P$ and $W$ are proportional, where the proportionality constant is the helicity (polarization).

• I have to say I'm not entirely satisfied with your answer. There must be a simpler, more intuitive answer which doesn't quote Weinberg or Wigner. – ashpool Jan 1 '14 at 20:09
• A particle at $v=c$ cannot go back and forth, since it would have to accelerate and decelerate (this is an intuitive explanation of polarization). That's why light has transversal polarization only. Is that simple enough? – jinawee Jan 1 '14 at 20:11
• In Mandl & Shaw, the photon field is originally given 4 polarization vectors, and the scalar and longitudinal polarizations being gauged away using a fairly sophisticated argument (which I think your answer is related to). On the other hand, for W boson field, it just has 3 polarization vectors to begin with. I was wondering why essentially the same Euler-Lagrangian equations give such different results. Is it the case that $(\square+m)W^\mu=0$ has three linearly independent solutions while $\square A^\mu=0$ has four linearly independent solutions? – ashpool Jan 1 '14 at 20:36

The solution of the Proca equation for massive spin-1 particles automatically satisfies Lorentz condition, eliminating one degree of freedom from the 4. This turns the polarization vector into space-like vector with three components.

For massless vector particles, gauge invariance eliminates one more degree of freedom, leaving only two polarization states. This fact for massless particles can also be explained intuitively by the fact that it has no 3-dimensional rest frame -- only 2 helicity states. Apparently Wigner proved this mathematically.