the_pulp said:
The leptons are complex as the higgs right? Shouldnt you multiply by 2 like you did with the higgs?
The Dirac fields are all complex, which you might expect to double the number of degrees of freedom, but spinor fields obey the Dirac equation, which cuts this in half, so it's a wash. In detail:
Consider a complex scalar field ##\phi##. It represents two particles: a scalar and its antiparticle. We can see this when we write out the Fourier expansion of a complex scalar field in terms of creation and annihilation operators:
##\phi(x) = \int d^4 p \left ( a(p) e^{i p x} + b(p)^\dagger e^{-i p x} \right )##
Here ##a^\dagger## creates a particle and ##b^\dagger## creates an antiparticle. So indeed ##\phi## represents two particles.
Now consider a 4-component Dirac spinor ##\psi_\alpha##. It represents four particles: a spin-up fermion, a spin-down fermion, a spin-up antifermion, and a spin-down antifermion. But since ##\psi_\alpha## is complex in the sense that ##\psi_\alpha^\dagger \neq \psi_\alpha##, why does it not represent eight particles, two per component? If we wrote down the most general four-component complex field, it would indeed represent eight particles. The most general four-component complex field can be written
##\psi_\alpha(x) = \int d^4 p \sum_{i = 1}^4 \left ( u^{(i)}_\alpha(p) a^{(i)}(p) e^{i p x} + \bar{v}^{(i)}_\alpha(p)b^{(i)}(p)^\dagger e^{-i p x} \right )##
where the ##u^{(i)}_\alpha##, i = 1, 2, 3, 4 are four linearly independent vectors in Dirac space. There are eight creation operators here, ##{a^{(i)}}^\dagger## and ##{b^{(i)}}^\dagger## for i = 1, 2, 3, 4.
This looks like the usual mode expansion of a Dirac field except we seem to be summing over four spins. This is because two of the spins go away when we impose the Dirac equation
##(i \gamma^\mu_{\alpha\beta} \partial_\mu - m \delta_{\alpha\beta}) \psi_\beta(x) = 0##
I've written out the Dirac indices explicitly in the Dirac equation to emphasize that in addition to being a differential equation, it is also a matrix equation in Dirac space. When you apply this to the Fourier-expanded version of the field above, you get some conditions on ##u## and ##v##:
##(\gamma^\mu_{\alpha \beta} p_\mu + m \delta_{\alpha \beta}) u^{(i)}_{\beta}(p) = 0##
##(\gamma^\mu_{\alpha \beta} p_\mu - m \delta_{\alpha \beta}) v^{(i)}_{\beta}(p) = 0##
It turns out that each of these matrix equations only has two linearly independent solutions. So the index ##i## can only run from 1 to 2. We've lost half of our original eight degrees of freedom to the Dirac equation, which is why the Dirac field ##\psi_\alpha## only represents four degrees of freedom.
Of course, for the counting I did earlier we have to remember that before electroweak symmetry breaking the left- and right-handed components of each Dirac spinor are separate fields, each carrying two of the degrees of freedom.
the_pulp said:
Why the right handed quarks count as Six And the left ones as 3? Why does that not happen with the leptons?
There are equal numbers of left- and right-handed quarks, it's just that the left-handed fields are organized into SU(2) doublets but the right-handed ones aren't. The left-handed quarks form three SU(2) doublets [(u, d), (c, s), (t, b)] while the right-handed quarks are six separate SU(2) singlets [u, d, c, s, t, b]. You'll note that we end up with equal numbers of degrees of freedom for the left-handed and for the right-handed quarks (36 in each case), as we must.
The same thing happens with leptons, except that before neutrinos were known to have mass you could get away with claiming that there were no such things as right-handed neutrinos. That eliminates half the right-handed leptons. However, if neutrinos have Dirac masses then there must be right-handed neutrinos. I think if neutrinos have Majorana masses then there still might be no right-handed neutrinos?