The chemical potential enters the grand canonical ensemble, in statistical physics, as the Lagrange multiplier ensuring the conservation of particle number.

In QFT and relativistic theories in general, the conservation number of particle is not always a requirement. E.g. for a scalar field with U(1) theory, the conserved charge is actually the number of particles minus the number of antiparticles.

So how do we treat the chemical potential (and, for that matter, Bose and Fermi distributions derived from it) in the relativistic limit?

  • $\begingroup$ I might add, what ensembles exist where the Lagrange multipliers ensure the conservation of the conserved charges $\endgroup$
    – lurscher
    Commented Oct 18, 2017 at 23:03
  • $\begingroup$ We indeed treat it as the one associated to conserved charges. Famous examples are cosmological applications. $\endgroup$
    – Name YYY
    Commented Oct 19, 2017 at 8:13
  • 1
    $\begingroup$ Is chemical potential used in cosmology? $\endgroup$ Commented Jan 18, 2018 at 0:04

1 Answer 1


In principle, the chemical potential in field theories does not differ from the ordinary chemical potential. However, due to the local character of field theories, chemical potentials in field theories have additional properties:

1) In field theories, charges appear as volume integrals of densities: $$ Q = \int d^3x \rho(x) = \int d^3x J_0(x) $$ (The second expression on the right hand side expresses the fact (by Noether's theorem) that a charge density commuting with the Hamiltonian is the zeroth component of conserved current $J^{\mu} $).

Chemical potentials can also appear as densities thus for a typical chemical potential term: $$ \mu Q \rightarrow \int d^3x \mu(x) \rho(x) = \int d^3x \mu(x) J^0(x) $$ 2) In field theory, currents minimally couple to gauge fields (i.e., through terms of the type $J^{\mu} A_{\mu}$ . Thus the chemical potential term describes a coupling to the zeroth component of a current. Thus it must the zeroth component of an external gauge field. To summarize this point:

Chemical potentials in field theory are the zeroth components of external gauge fields:

$$\mu = A_0$$ 3) As in ordinary thermodynamics, in field theory, chemical potentials can be coupled only to conserved charges: $[Q, H] = 0$. For example there exists a Baryon number chemical potential, strangeness chemical potential, etc.

Field theories contain also anomalous symmetries whose corresponding currents are not conserved. However, these currents have non-conservation laws of the type: $$\partial_{\mu} J^{\mu} = \partial_{\mu} K^{\mu}(A)$$ For example in the case of axial anomaly $$ K^\mu (x) = -\frac{1}{16\pi^2} \ \varepsilon^{\mu\alpha\beta\gamma}\ t r \ [\frac{1}{2}\ A_\alpha (x)\ \frac{\partial}{\partial x^\beta} \ A_\gamma (x) + \frac{1}{3}\ A_\alpha (x) \, A_\beta (x)\, A_\gamma (x)]$$ The topological current $K^{\mu}$ is not globally defined, however, locally we do have a conserved current: $$\partial_{\mu} (J^{\mu} - K^{\mu}) = 0$$ It turns out that when an anomalous current is coupled to a chemical potential, one of the most exciting recent discoveries takes place, namely the "Chiral magnetic effect", which consists of a macroscopic manifestation of the chiral anomaly.

This effect can be summarized as follows: In the presence of chiral charge imbalance (thus a nonvanishing chemical potential $\mu_5$ coupled to the axial current), an electric current proportional to the axial current chemical potential is formed in the direction of the external magnetic field.

$$\vec{J} = \frac{e^2}{2 \pi} \mu_5 \vec{B}$$

All the explanations given above (and much more) appear in the following lecture by Kharzeev and references therein.

  • $\begingroup$ "Chemical potentials can also appear as densities": why? how? $\endgroup$ Commented Jan 14, 2018 at 17:50
  • $\begingroup$ @SuperCiocia, In field theory, all observables appear as integrals of densities, for example, momenta, angular momenta etc. The fields themselves (electric, magnetic) are densities (The Hamiltonian is a space integral over polynomials of the fields). $\endgroup$ Commented Jan 15, 2018 at 7:39
  • $\begingroup$ But where do you get the $\mu Q$ from? $\endgroup$ Commented Jan 15, 2018 at 11:15
  • $\begingroup$ In the particular case where $\mu(x) = \mu = const.$ (i.e., the chemical potential density is uniform), you get $\int d^3x \mu(x) \rho(x) = \mu \int d^3x \rho(x) = \mu Q$. In field theory we allow $\mu(x)$ to vary over space, thus it is more general than the standard definition. $\endgroup$ Commented Jan 15, 2018 at 11:21
  • 2
    $\begingroup$ @Rudyard Please see the following question and answer. They precisely address your remark. physics.stackexchange.com/questions/294783/…. A gauge function should be single valued on the configuration manifold. In our case, since we are considering a finite temperature statistical field theory, the path integral is Euclidean and time is $\beta$ periodic, thus the proposed gauge function is not suitable. $\endgroup$ Commented Sep 26, 2019 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.