Baryogenesis, finally
Sakharov's third condition for baryogenesis [A.D. Sakharov, Zh. Eksp. Teor. Fiz. Pis'ma 5, 32 (1967); JETP Lett. 91B, 24 (1967)] is usually stated as: departure from thermal equilibrium or interactions out of thermal equilibrium or some such. I've always had some trouble with the textbook arguments for it, but I think I've now had a personal epiphany.
The textbook argument, e.g., Trodden's notes, considers the expected baryon number in equilibrium: $$ \langle B \rangle = \text{Tr}\left(e^{-\beta H} B\right) $$ Inserting an identity, \( (\text{CPT}) (\text{CPT})^{-1} \), using the cyclicity of the trace and the invariance of the Hamiltonian under CPT: $$ \begin{align} \langle B \rangle =& \text{Tr}\left((\text{CPT}) (\text{CPT})^{-1} e^{-\beta H} B\right)\\ =& \text{Tr}\left(e^{-\beta H} (\text{CPT})^{-1} B (\text{CPT}) \right)\\ =& - \text{Tr}\left(e^{-\beta H} B\right)\\ =& -\langle B \rangle \end{align} $$ and thus \( \langle B \rangle = 0 \) in equilibrium.
The thing that confused me was that it seemed that we almost begged the question. We started by assuming that \(H\) was the only relevant macroscopic parameter. We didn't include any chemical potential for baryon number in our ensemble. Suppose we took: $$ \langle B \rangle = \text{Tr}\left(e^{-\beta H + \mu_B B} B\right) $$ The partition function now no longer commutes with CPT and we cannot conclude that \(\langle B \rangle = 0\). Try it. The above argument won't go through. Can we do that? I don't think so. Sakharov's first condition for baryogenesis was that \(B\) was not conserved. In which case, there is no chemical potential in the equilibrium ensemble.
But what if we can give a chemical potential to some other conserved quantity? What? Well, \(B - L\) is a natural candidate, as unlike individual baryon and lepton number, it is conserved. In that case, we could write the expectation of baryon number as, $$ \langle B \rangle = \text{Tr}\left(e^{-\beta H + \mu_{B - L} (B - L)} B\right) $$ Because of the chemical potential, we cannot conclude that \(\langle B \rangle = 0\). Is this a loophole to Sakharov's conditions? No, we've in fact rediscovered leptogenesis! We give a chemical potential for \(B - L\) and equilibration redistributes the charges such that \( \langle B \rangle \neq 0\). We convert a \(B- L\) asymmetry to a baryon asymmetry.
Did this require a departure from thermal equilibrium? Well, under standard cosmology, there are no chemical potentials after reheating, as the inflaton decays into a hot thermal bath with no chemical potentials. Creating a chemical potential for \(B-L\) from scratch requires an out-of-equilibrium process. Thus Sakharov's third condition holds.
What if we don't assume that there are no chemical potentials after inflation? Maybe inflation can produce a chemical potential for \(B-L\)? Is that a loophole? No, we've in fact discovered inflationary baryogenesis and changed the era in which baryogenesis occurs.
What if we don't assume inflation at all? In that case, indeed, we can input what we like as an initial condition, e.g., an equilibrium state with non-zero baryon number.
Tags: baryogenesis, statistical-mechanics, physics