# How to properly count the SM particles degrees of freedom?

I'm studying the thermal history of the universe, and I found a bit difficult to properly identify the correct number of d.o.f. for each SM particle. I'm referring to the usual factor $$g_{*}(T)=\sum_{Bosons}g_{B}(\frac{T_{B}}{T})^{4}+\sum_{Fermions}g_{F}(\frac{T_{F}}{T})^{4}$$ To find each contribute $$g_{B,F}$$ one should use the following known facts (notice that it refers to the relativistic limit, where the particles can be considered massless):

$$G_{\mu}^{I}$$ $$W_{\mu}^{I}$$ $$B_{\mu}$$ Q $$u_{R}$$ $$d_{R}$$ E $$e_{R}$$ H
Families 1 1 1 3 3 3 3 3 1
Spin 2 2 2 2 2 2 2 2 2
Weak isospin 1 3 1 2 1 1 2 1 2
Colors 8 1 1 3 3 3 1 1 1
Statistics 1 1 1 $$\frac{7}{8}$$ $$\frac{7}{8}$$ $$\frac{7}{8}$$ $$\frac{7}{8}$$ $$\frac{7}{8}$$ 1
$$δg$$ 16 6 2 $$\frac{63}{2}$$ $$\frac{63}{4}$$ $$\frac{63}{4}$$ $$\frac{21}{2}$$ $$\frac{21}{4}$$ 4

Now, my question is the following:-Why do we put here a factor of 2 for each particle? For example the Gauge bosons have spin 1 so I'd have put a factor of 3, while for the Higgs boson (spin zero) a factor of 1. Why it is not the case? I want to stress that I'm still not very comfortable with SM, so please be patient and don't assume anything as obvious, cause for me it's not.

EDIT: If can be helpful, it's written that we can write (I assume when $$T=T_{i}$$) $$g_{*}(T)=\stackrel{Photons\hspace{0.2cm}pol.}{2}+\sum_{i\neq photons}δg_{*,i}$$ so for the photons we consider just their polarizations, but do the other particles have two polarizations as well?

• If you want to improve the table formatting, StackExchange supports Markdown tables. Here's a handy site that generates a blank table for you, which you can then fill in. Commented Jul 31, 2023 at 14:04
• a quote for the electron "the electron's three degrees of freedom (charge, spin, orbital) " from en.wikipedia.org/wiki/Elementary_particle . So the spin orientations do not count in this case. Reference for your particular table? to check what it refers as d,o,f,? Commented Jul 31, 2023 at 16:16
• The reference is in the book written by my professor, but since it's still not ready I can't share it. The degrees of freedom refers to those factors $g_{B,F}$ above, in particular they come from the expression for the number density $n_{i}$, the energy density $\rho_{i}$ and pressure $p_{i}$ of each specie labelled with index "i". It should be standard (quantum) Stat. Mec. I add a row to the table making it more explicit. Commented Jul 31, 2023 at 16:27
• I wonder whether the degree-of-freedom counting is being done under the assumption that you're above the electroweak symmetry breaking temperature. Above this temperature all the weak gauge bosons are massless and so they'd only have two spin states. Not sure what's happening with the Higgs, though — this isn't my field of expertise. Commented Jul 31, 2023 at 16:57
• That's something, indeed the table refers to degrees of freedom in the relativistic limit where $m_{i}\simeq0$, so this is remarkable. I've written that in the massless case there are less number of possible polarizations, should be because of that. Thank you btw! Commented Jul 31, 2023 at 17:44

The degrees of freedom of any quantum field have to do with the number of polarizations of said field. For example, the Higgs field $$\phi(x)$$ is mathematically a scalar field or in other words, a scalar number per point in space. Equivalently, you can think of it as being represented by one scalar number $$\phi_0$$ per wavevector $$\vec{k}$$ if you think in Fourier terms as a plain wave

$$$$\phi (x)=\phi_0 e^{i k \cdot x}$$$$

This is what we mean when we say that there's one degree of freedom in the Higgs field. The reason why this is the correct number we want to use for thermodynamics is because each plain wave is essentially a harmonic oscillator, so we are just using the equipartition theorem for very abstract springs.

If we move to the electromagnetic field $$A_\mu$$ one might think that it has 4 degrees of freedom, since it is a 4-vector and thus you need to specify 4 numbers per point in space. A generic one particle state with wavevector $$k$$ looks like

$$$$A_\mu (x)=\epsilon_\mu e^{i k \cdot x}$$$$

where $$\epsilon_\mu$$ is the polarization vector. However, using gauge symmetry it is possible to set $$\epsilon_0=0$$ and $$\epsilon_\mu p^\mu =0$$ so we are left with two independent polarizations. More intuitively, we know that if there's an EM wave travelling in the $$\hat{z}$$ direction, we only need to specify the electric field $$E_x$$ and the magnetic field $$B_y$$.

This analysis extends to all gauge fields, which also have two polarizations per excitation. Note however that you also need to count other degrees of freedom. For example, there are eight different gluon colors and therefore the total number of degrees of freedom in the gluon field should be sixteen, two polarizations per each color.

Finally, note that this doesn't hold for matter fields, since they are not gauge fields. However, in the ultra relativistic limit where we can consider them to be massless, they basically behave like gauge fields so there's also two polarizations per excitation.

• Thank you! I've read that there is a more rigorous proof of that using Poincarè irrepresentations... If you have some good references they are well accepted:) Commented Jul 31, 2023 at 21:25
• You are welcome! There's a bit of that in QFT in a nutshell by Zee. In the section "A photon can find no rest" if you want to check that out. It explains that fields need to be, mathematically, in some representation of the Poincaré group and how being massless (and therefore not being able to boost to the rest frame) change things. Commented Jul 31, 2023 at 23:17