# QCD chPT chemical potential

I have recently been to a guest lecture about low energy QCD with a finite isospin chemical potential. I have two overall questions about this, since it wasn’t possible to ask questions at the given lecture, and i have to wait a while before future classes about this are available.

Question 1: It was mentioned that in low energy QCD the baryon chemical potential has no effect in chiral perturbation theory. Can anyone briefly explain why this is so?

Question 2: In general i would like to know how the isospin chemical potential enters the effective QCD Lagrangian, and what kind of changes it brings - i.e. what it actually does. We can just take the simpler case where the baryon chemical potential is set to zero: $\mu_B = 0$

I know it has something to do with promoting $SU(2)_L \times SU(2)_R$ to a local gauge symmetry, and that $\mu_I$ will enter the chiral Lagrangian through the covariant derivative as a zeroth component of a gauge potential. How can one see this?

I would be very interested in literature which can give me an introduction to isospin chemical potential in effective low energy QCD Lagrangians - especially with the connection to the chiral condensate and isospin density. But as far as this question goes, the focus is just to start with understanding the chemical potentials of QCD. I have spent a couple of hours reading everything i could find both on stackexchange and in various published papers. With this question, I am trying to boil everything down to a basic understanding.

Thank you very much.

The partition function is $$Z(\mu)={\rm Tr}[\exp(-\beta(H-\mu Q))]$$ where $H$ is the Hamiltonian, $Q$ is a conserved charge, $\mu$ is the conjugate chemical potential, and $\beta$ is the inverse temperature.

1) Consider baryon number $Q=B$ and $\mu=\mu_B$. QCD has states with non-zero baryon number (protons, neutrons, etc), but all the fields in the chiral lagrangian (in the meson sector) have $B=0$. Therefore, the partition function is independent of $\mu_B$. There is a subtlety here, because the chiral lagrangian (may) support topological field configurations (skyrmions) that acquire $B\neq 0$ from the Wess-Zumino term. However, these states are heavy and their contribution is relevant only for $\mu_B\simeq m_p$ (the mass of the proton), which is outside the regime of validity of chiral perturbation theory.

2) Now consider isospin, $Q=I_3$ and $\mu=\mu_I$. The chiral lagrangian contains states (pions etc) with non-zero isospin. Indeed, the conserved charge in the chiral lagrangian is $$Q = f_\pi^2{\rm Tr}[\tau^3i(\partial_0 U^\dagger U+U\partial_0U^\dagger)] + \ldots$$ where $U$ is the chiral field. This expression can be obtained using the Noether procedure, or, much more efficiently, by gauging the $SU(2)_L\times SU(2)_R$ symmetry.

2b) Under $SU(2)_L\times SU(2)_R$ we have $U\to LUR^\dagger$. We can gauge this symmetry by introducing the covariant derivative $$i\nabla_\mu U = i\partial_\mu U + l_\mu U - Ur_\mu\, .$$ A vector field corresponds to $v_\mu=l_\mu=r_\mu$. The corresponding charge is $$Q = \frac{\delta S}{\delta v_0},$$ Note that $v_0$ is conjugate to $Q$, so we can study matter at finite density by setting $v_0=\mu_I$.

3) The lightest state with non-zero $I_3$ is the charged pion, which has isospin $\pm 1$. This means that, depending on the sign of $\mu$, for $|\mu|\geq m_\pi$ there is no energetic cost to adding a positive (negative) pion to the system and charged pion condensation takes place. This happens within the regime of validity of chiral perturbation theory, but in real QCD the situation is a little more complicated because of the electromagnetic interaction (there would have to be a neutralizing background).

4) The ground state is most easily studied by gauging $SU(2)_V$, setting $V_0=\mu_I$, and then minimizing the ground state energy with respect to $U$ at fixed $\mu_I$. The standard reference is https://arxiv.org/abs/hep-ph/0005225 . This analysis shows, in particular, that interactions between pions (automatically included in the chiral lagrangian) do not modify the critical chemical potential for Bose condensation, but limit the growth of the condensate density, so that the onset transition is second order.

• Thank you so much for the answer! You mention the your second expression can more easily be obtained by gauging the $SU(2)_L \times SU(2)_R$ symmetry. Can you give some hints on how exactly - so what's the motivation behind that exact form given in eq. (5) in the article from Son and Stefanov? I heard it was because it couples to the timelike vector potential - but why only the timelike part? Also for some reason the $\partial_0 \Sigma$ can be ignored, and the rest is kept.
– Yuki
Jan 6 '17 at 7:03
• Great! I have one last remark. Isn't the external vector field defined as $v_{\mu} = l_{\mu} + r_{\mu}$ corresponding to the remaining $SU(2)_V = SU(2)_{L+R}$ symmetry? That is what my books say. I would think the covariant derivative would be something like $D_{\mu}U = \partial_{\mu}U + i[U,V_{\mu}]$ with $V_{\mu}=\frac{\mu_I}{2}v_{\mu}\tau_3$ so the vector field itself couples to $\tau_3$. I understand we couple isospin to the vector field $v_{\mu}$ but how is the value of $v_{\mu}$ determined?
• In some papers they use $\frac{\mu_I}{2}$ in the coupling, whereas others just use $\mu_I$. This makes a difference in the fraction factor in the Lagrangian. For the first case we would get $\frac{f_{\pi}^2}{16}$ and for the other (most common) $\frac{f_{\pi}^2}{4}$
• @Alex You check the normalization of $f_\pi$ by gauging $SU(2)_L$ and computing the pion decay vertex $f_\pi q^\mu$. You check the normalization of $\mu_I$ by computing the shift in the energy of the pion (isopspin 1), $m_\pi\pm\mu_I$. Jan 7 '17 at 15:53