I'm studying the QDC phase diagram at finite densities and I'm having some trouble understanding the role of the chemical potential.

It's clear how, if I increase the density of particles at $T=0$, I will eventually reach a deconfined phase. It's also clear that if $T$ grows, it will become easier to reach the said deconfined phase, and the density needed to do that will be lower. If $T$ is really high, the deconfined phase will exist at any density level. Then the phase diagram has a pretty defined face:
enter image description here
It's also clear how, since we want to use the relativistic QCD and not the semiclassical nuclear physics, the number of particles (and consequently their density) isn't a good choice for a parameter. Up to here, all is good.
We then choose to use the chemical potential, the conjugate variable to the (net) particle number in the Grand Canonical Ensemble, and the definition I'm given is "the energy needed to change the particle number", which I do not understand. Why, if this "energy cost" increases, I should have a phase transition? Why, if the temperature increases, the energy cost will be lower? Why does the phase diagram look exactly the same as the one with particle density?
enter image description here
The reason for my question(s) may as well reside in a lack of knowledge of the QCD phase structure, but I think that my misunderstanding regards the chemical potential itself.

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    $\begingroup$ It s probably more precise to say that, in this case, the chemical potential is an "energy per baryon" since the total baryon number is conserved. Each nucleoon has baryon number 1 and each quark 1/3 (-1/3 for antiquarks). Cool question :) $\endgroup$
    – Quillo
    May 3, 2022 at 7:28
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    $\begingroup$ @Quillo this may be crucial. If the chemical potential measures only the energy needed for a baryon then the phase transition becomes obvious: if you need energy higher than the baryon mass, the system doesn't want baryons, and deconfines. Does this make sense? $\endgroup$ May 3, 2022 at 10:47
  • $\begingroup$ I think that it is intuitively sound but still it s not obvious (at least to me) to see why the deconfinement line has that shape. At T=0 the deconfinement is for chem. pot. close to 1.1 GeV (comparable with but higher than the nucleon mass). At T>0 deconfinament is way before.. ok, if you re hot you re more likely to be deconfined, it makes sense but to me it's not intuitive that che chem pot at T=0 is higher than the nucleon mass. Maybe in a dense environment the mass of nucleons gets higher? $\endgroup$
    – Quillo
    May 3, 2022 at 16:33
  • $\begingroup$ Related: physics.stackexchange.com/q/27033/226902 , physics.stackexchange.com/q/299911/226902 , physics.stackexchange.com/q/7470/226902 . Status of QCD phase diagram, 2023: arxiv.org/abs/2301.04382 $\endgroup$
    – Quillo
    Mar 23 at 8:32

1 Answer 1


The phase diagram of QCD is not known/understood completely: neither from first principles nor experimentally. We have limited knowledge from collider experiment, neutron star mergers and theory. Both diagrams shown in the question are sketches/model computations and I fail to see how they look "look exactly the same" other than the fact that both include a phase transition(s) (between a confined and de-confined phase assuming I and II in the first figure refer to confinement?).

Before going into further details a few remarks about chemical potential: what we are dealing with here is baryon $\mu_B$ (or quark $\mu$) chemical potential, which is NOT related to a particle number but rather to baryon number $B$ (or quark number $N$). Which are both strictly conserved in QCD and measure the difference between quark and anti quark number $B=\frac{1}{3}N=\frac{1}{3}(N_q-N_\bar{q})$. In this sense baryon (or quark) chemical potential is the energy that can be absorbed or released due to a change of baryon number $B$ (or quark number $N$) -- through an imbalance between quarks and anti-quarks. Increasing the chemical potential leads to an increased imbalance between quarks and anti-quarks (higher $\mu$ leads to higher $N=N_q-N_\bar{q}$ meaning we have more quarks than anti-quarks).

The confinement-deconfinement phase transition for QCD is especially at non-zero $\mu$ not completely understood. Later I will discuss a simplified model but when it comes to QCD: closely related to the confinement-deconfinement transition is chiral symmetry breaking and restoration. Actually those phase diagram sketches at non-zero $\mu$ usually include information from model calculations of low-energy effective models for QCD. Those usually study the chiral condensate $\langle\bar{\psi}\psi\rangle$ which at low $T$ and $\mu$ acquires a nonzero expectation value $\langle\bar{\psi}\psi\rangle>0$ due to dynamic symmetry breaking: this is due to quantum fluctuations. Increasing the temperature "melts" this expectation value -- thermal fluctuations decease $\langle\bar{\psi}\psi\rangle$ and at $\mu=0$ we find a smooth crossover at $T_c$ to an approximately restored phase $\langle\bar{\psi}\psi\rangle\approx 0$. Chiral symmetry gets only approximately restored due to finite quark masses. When increasing $\mu$ -- the imbalance between quarks and anti-quarks -- the interactions between quarks (mediated by gluons) as well as bosonic quantum fluctuations (of light mesons and glouns) drive the approximate restoration of chiral symmetry at $\mu>0$. At higher chemical potentials (and low temperatures) we expect the diquark condensation $\langle\psi \mathcal{O} \psi\rangle>0$ and color-super conductivity. In thermodynamic equilibrium the physical ground state is the one with the lowest energy and depending on $\mu$ and $T$ this ground state corresponds to a state realizing different expectation values $\langle\bar{\psi}\psi\rangle$, $\langle\psi \mathcal{O} \psi\rangle$, $\ldots$. In the quark-gloun plasma (QGP) $\langle\bar{\psi}\psi\rangle\approx 0$, in the hadronic phase including vacuum $\langle\bar{\psi}\psi\rangle>0$ and for Color-superconductors $\langle\bar{\psi}\psi\rangle\approx0$ and $\langle\psi \mathcal{O} \psi\rangle>0$.

So far this discussion was not very specific/intuitive but when it comes to QCD most things are not really intuitive. But maybe the study of a very crude model with a confinement-deconfinement phase transition might help to illuminate on some points raised when it comes to translating axes in those phase diagrams. I will use the MIT bag model as a simple model to describe deconfined quark matter together with a pion-gas as a very simple model for the hadronic phase. This will lead to a very similar phase diagram to the one showed in the first plot in the question.

  • The MIT bag model: For the purpose of this discussion the MIT bag is a simple model for quark matter. In the MIT bag model hadrons (confined/colorless regions) are described as bags filled with free moving quarks and gluons. The space outside the bags is free of quarks and gluons and it is assumed that it costs energy to maintain these bags. The required energy per unit volume is refereed to as $B$ -- the bag constant. Inside a bag of volume $V$ we therefore have an energy of $B\,V$ plus the energy associated with the kinetic motions of quarks and gluons (for small volumes/bags there are additional contributions due to boundary conditions for the bag). Considering only the two light quark flavors ($u$ and $d$) and neglecting their mass the pressure in a large bag (in the deconfinement region/QGP) is given by the pressure of free, massless non-interacting quarks (fermions) plus the pressure of a gas of massless gluons (bosons) minus the bag constant (in this context bag pressure) $B$: $$ p_\mathrm{QGP}=-B+\frac{8\pi^2}{45}T^4+\sum_f\left(\frac{7}{60}\pi^2T^4+\frac{1}{2}T^2\mu_f^2+\frac{1}{4\pi^2}\mu_f^4\right) \tag{1}. $$ The quark number density $n_f$ per flavor $f$, the baryon number density $\rho$ and the entropy density $s$ are given by $$ \begin{align} n_f&=\frac{N_f}{V}=\frac{\partial p}{\partial \mu_f}=-\frac{\partial \Omega}{\partial \mu_f},\\ \rho&=\frac{B}{V}=\frac{1}{3}\sum_f n_f,\\ s&=\frac{\partial p}{\partial T}=-\frac{\partial \Omega}{\partial T}, \end{align} $$ with the grand canonical potential density/per unit volume $\Omega=-p$. Assuming iso-spin symmetric matter $\mu_d=\mu_u=\mu=\frac{1}{3}\mu_B$, where $\mu$ is the quark chemical potential and $\mu_B$ is the baryon chemical potential we can rewrite the expression for the pressure (1) and compute $n=n_u+n_d$, $\rho$ and $s$ explicitly for the MIT bag model $$ \begin{align} p_\mathrm{QGP}&=-B+\frac{8\pi^2}{45}T^4+\left(\frac{7}{30}\pi^2T^4+T^2\mu^2+\frac{1}{2\pi^2}\mu^4\right), \tag{2.1}\\ n_\mathrm{QGP}&=\frac{2 \mu ^3}{\pi ^2}+2 \mu T^2,\tag{2.2}\\ \rho_\mathrm{QGP}&=\frac{1}{3}n_\mathrm{QGP} = \frac{2 \mu ^3}{3 \pi ^2}+\frac{2 \mu T^2}{3},\tag{2.3}\\ s_\mathrm{QGP}&= \frac{74 \pi ^2 T^3}{45}+2 \mu ^2 T.\tag{2.4} \end{align} $$ In Eq. (2.2) we see the relation between quark chemical potential and baryon number (note we use the natural units typical in high-energy physics $c=\hbar=k_\mathrm{B}=1$ and for conversions to SI remember $\hbar c = 197.326\, \mathrm{MeV}\,\mathrm{fm}$).

  • Pion-gas: for the hadronic/confined phase we use a simple relativistic gas of $N_f^2-1=3$ pions (neglecting their mass) which has the following equation of state $$ \begin{align} p_\pi&=\frac{\pi ^2 T^4}{30}, \tag{3.1}\\ n_\pi&=0,\tag{3.2}\\ \rho_\pi&=\frac{1}{3}n_\pi = 0,\tag{3.3}\\ s_\pi&= \frac{2 \pi ^2 T^3}{15},\tag{3.4} \end{align} $$ note that $n_\pi=\rho_\pi=0$ as the pions do not carry baryon number ($n_\pi$ is the contribution of the pions to the quark number density and not to a non-existing density of pions). As such this model for the hadronic phase is at best incomplete since we do not consider baryons (like protons and neutrons) but it is sufficient for illustrative purposes.

Assuming a first-order phase transition (at constant pressure) between our simplified QGP and the hadronic phase we can compute a phase transition line: $$ \begin{align} &p_\mathrm{QGP}(\mu,T)\stackrel{!}{=}p_\pi(\mu,T)\\ &\Rightarrow T_c(\mu)=\frac{\sqrt{\frac{3}{34}} \sqrt{\sqrt{340 \pi ^2 B+55 \mu ^4}-15 \mu ^2}}{\pi }.\tag{4} \end{align} $$ Using Eq. (4) we could actually fit the bag constant to the known critical temperature of QCD $T_c(\mu=0)$ or we use a literature value. For a typical literature value $B=(220\,\mathrm{MeV})^4$ we find $T_c(\mu=0)=158.\,\mathrm{MeV}$ and the following phase diagram: mu-T-Diagram

which we can translate into the $n$-$T$-plane using Eq. (2.2) with Eq. (4):


When plotting quark number density, entropy density and energy density ($\epsilon=\mu n +T s -P$) it becomes apparent that we are indeed dealing with a first order phase transition:

mu-T-n mu-T-n mu-T-epsilon

It goes without saying that this is one of the most simplistic model of a QCD like confinement-deconfinement phase transition. Concerning the phase transition line in the $\mu$-$T$ and $n$-$T$ plane: it hits all the axes at a right angle and for small $T$ we find $$ (\mu_c(T),n_c(T))=(\mu_c(0),n_c(0))*\left(1-\frac{\pi T^2}{2 \sqrt{2} \sqrt{B}}+\ldots\right), $$ with $\mu_c(T=0)=463.\,\mathrm{MeV}$ and $n_c(T=0)=2.810\,\mathrm{fm}^{-3}$ which explains the extreme similarity between both phase diagrams (up to a rescaling of the respective horizontal axis).

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    $\begingroup$ Your lengthy answer is appreciated, but my question concerns your second paragraph. Why is that increasing $\mu$ increases $N$ (or $B$)? Isn't $\mu$ the cost to increase $N$, shouldn't a higher value of $\mu$ discourage new particles? How does the definition "the energy that can be absorbed or released due to a change of the particle number" leads to "higher $\mu$ leads to higher $N$"? $\endgroup$ May 3, 2022 at 14:39
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    $\begingroup$ In general a chemical potential is defined as the derivative of internal energy w.r.t. particle number. For QCD matter the internal energy grows no linearly with baryon number and thus $\mu$ increases with $n$ as indicated by Eq. (2.2). A higher $\mu$ does not discourage a higher $n$ it necessitates it. In fact for a ultra relativistic cold fermi gas $\mu$ is equal to the fermi momentum and an increase leads to more occupied states => more particles per volume. If you want to think of $\mu$ as the cost to increase $n$ fine: once you pay this cost you get a higher number density. $\endgroup$
    – N0va
    May 5, 2022 at 0:05

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