Can Fermions Exhibit Negative Chemical Potential Like Bosons?

[Vrbic]

Is possible and what does it mean if a chemical potential is negative?
I mean that for boson it means that in environment is "needed" boson (photon) and is possible to create him. Is it true?
And what about for fermions? Could it mean that it is pleasent for environment to creat some fermion? Or is it nonsense to have negative chemical potential?

 

[BvU]

HI,
You must have googled and found e.g. this . So: yes, it's possible. If it's negative that means a new member particle is welcome because they all huddle together so nicely.

 

[vanhees71]

The answer to your question depends on the Hamiltonian used to describe the grand-canonical potential. The Statistical Operator is
$$\hat{\rho}=\frac{1}{Z} \exp[-\beta (\hat{H}-\mu \hat{N})],$$
where $\beta=1/(k_{\text{B}} T)$ is the inverse temperature (up to a trivial conversion factor due to the SI units, called Boltzmann's constant), $\hat{H}$ is the Hamiltonian of the system, and $\hat{N}$ is a conserved quantity (in non-relativistic physics of gases usually the particle number). The partition sum is defined as
$$\hat{Z}=\mathrm{Tr} \{\exp[-\beta (\hat{H}-\mu \hat{N})] \},$$
introduced to properly normalize the statistical operator.

I guess you deal with non-relativistic particles. Let's take the most imple example of an ideal gas, where you can neglect the interactions among the molecules in comparison to their kinetic energy, i.e., the Hamiltonian reads (for simplicity I assume spin-zero particles)
$$\hat{H}=\sum_{\vec{p}} \frac{\vec{p}^2}{2m} \hat{N}(\vec{p}), \quad \hat{N}=\sum_{\vec{p}} \hat{N}(\vec{p}).$$
Here $\hat{N}$ is the number operator, and I have assumed that the particles are trapped in a cubic box of length $L$. Imposing convenient periodic boundary conditions, then the momenta are $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3$.

The mean occupation number in the modes is given by the Bose-Einstein (Fermi-Dirac) distribution:
$$\langle N(\vec{p}) \rangle=f_{\text{B/F}}(\vec{p}),$$
with
$$f_{\text{B/F}}(\vec{p})=\frac{1}{\exp[\beta(\vec{p}^2/(2m)-\mu)]\mp 1}.$$
Now it's clear that you must have $\beta >0$ so that the total mean particle number and energy are finite. For fermions you can have $\mu \in \mathbb{R}$, but for bosons you must have $\mu < 0$, because otherwise you'd have a singularity in the Bose-Einstein distribution and it wouldn't be positive semidefinite as it must be as a proper single-particle phase-space distribution.

In our here considered case of a finite volume (which you always should consider first in many-body theory) this constraint $\mu < 0$ is no problem, because you can accommodate any positive average particle number at any given temperature by any choice of $\mu<0$, because in the limit $\beta \rightarrow \infty$, i.e., $T \rightarrow 0$, you simply have the ground state of the system, i.e., all bosons occupying the single-particle groundstate at $\vec{p}=0$. That's called Bose-Einstein condensation.

Only if you take the "thermodynamic limit", i.e., making the volume very large and instead of considering the discrete momenta of the finite-volume case making $\vec{p}$ continuous and substituting all sums (e.g., when calculating the partition sum or rather its logarithm, the grand-canonical potential) by integrals, you have to treat the zero-mode explicitly, because due to the integration measure $\mathrm{d}^3 p=p^2 \mathrm{d} p \mathrm{d}^2 \Omega$ the singularity at $\vec{p}=\mu=0$ is integrated over, and the integral only counts the particles in the excited states. Thus in the thermodynamic limit if you have Bose-Einstein condensation you must set $\mu=0$ and at a given temperature evaluate how many particles are in the excited states and then set the number of the particles in the ground state accordingly to get the total average number of particles right.

 

[Vrbic]

Very nice summary and I mostly understand. Thank you very much.
But I have few question:
1)

Imposing convenient periodic boundary conditions, then the momenta are ⃗p∈2πLZ3\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3.

Does it come from boundary condition on wave function? I suppose at the edge of box it is zero. Right?

2)

n our here considered case of a finite volume (which you always should consider first in many-body theory) this constraint μ<0\mu < 0 is no problem, because you can accommodate any positive average particle number at any given temperature by any choice of μ<0\muβ→∞\beta \rightarrow \infty, i.e., T→0T \rightarrow 0, you simply have the ground state of the system, i.e., all bosons occupying the single-particle groundstate at ⃗p=0\vec{p}=0. That's called Bose-Einstein condensation.

Only if you take the "thermodynamic limit", i.e., making the volume very large and instead of considering the discrete momenta of the finite-volume case making ⃗p\vec{p} continuous and substituting all sums (e.g., when calculating the partition sum or rather its logarithm, the grand-canonical potential) by integrals, you have to treat the zero-mode explicitly, because due to the integration measure d3p=p2dpd2Ω\mathrm{d}^3 p=p^2 \mathrm{d} p \mathrm{d}^2 \Omega the singularity at ⃗p=μ=0\vec{p}=\mu=0 is integrated over, and the integral only counts the particles in the excited states. Thus in the thermodynamic limit if you have Bose-Einstein condensation you must set μ=0\mu=0 and at a given temperature evaluate how many particles are in the excited states and then set the number of the particles in the ground state accordingly to get the total average number of particles right.

Here you speak only about bosnos. Right?

3) I plotted fermi statistic for constant $p$ and variable $\mu$. It is monotonically growing. Does it say that the bigger $\mu$ the bigger probability in frame of same momentum?

4) How I determine $\mu$? Is it purely experimental constant? I mean in practice. (I know in many textbooks are formulas for different quantities but if you deal such problem using numeric method it is useless)

 

[vanhees71]

ad 1) Of course, in the thermodynamic limit the specific boundary conditions don't play much of a role. Thus it's more convenient to consider periodic boundary conditions for the single-particle modes. Then you have exponential plane waves, $\exp(\mathrm{i} \vec{p} \cdot \vec{x}$. In order to fulfill the periodic boundary conditions $\psi(\vec{x}+L \vec{e}_j)$ (for $j \in \{1,2,3 \}$) you must of course have $p_j = 2 \pi/L n$ with $n \in \mathbb{Z}$. With rigid boundary conditions you get cosine and sine functions. It's a good excercise to check that all results in the chiral limit are the same (for the ideal gas of course).

ad 2) Yes, this is all about bosons. For fermions, there's no such problems as with Bose-Einstein condensation in the chiral limit for bosons. You can simply have $\mu \in \mathbb{R}$ and thus adjust for any given temperature the chemical potential to get the given average particle number.

ad 3) yes. Obviously the FD distribution is monotoneously rising with rising chemical potential.

ad 4) Look at the derivation of the grand canonical ensemble. It describes the case that you have a system of particles (e.g., our ideal gas) in contact with a heat bath (i.e., it can exchange energy with a large reservoir of particles at a given temperature), and you can exchange particles with the heat bath. Temperature and chemical potential are Lagrange parameters to find the state of maximum entropy, under the constraints that the mean energy (internal energy $U$) and the mean number of particles $\overline{N}$ is fixed. Thus temperature and chemical potential are used to get these given values of $\overline{E}=U$ and $\overline{N}$.

 

[Vrbic]

ad 1) Of course, in the thermodynamic limit the specific boundary conditions don't play much of a role. Thus it's more convenient to consider periodic boundary conditions for the single-particle modes. Then you have exponential plane waves, $\exp(\mathrm{i} \vec{p} \cdot \vec{x}$. In order to fulfill the periodic boundary conditions $\psi(\vec{x}+L \vec{e}_j)$ (for $j \in \{1,2,3 \}$) you must of course have $p_j = 2 \pi/L n$ with $n \in \mathbb{Z}$. With rigid boundary conditions you get cosine and sine functions. It's a good excercise to check that all results in the chiral limit are the same (for the ideal gas of course).

ad 2) Yes, this is all about bosons. For fermions, there's no such problems as with Bose-Einstein condensation in the chiral limit for bosons. You can simply have $\mu \in \mathbb{R}$ and thus adjust for any given temperature the chemical potential to get the given average particle number.

ad 3) yes. Obviously the FD distribution is monotoneously rising with rising chemical potential.

ad 4) Look at the derivation of the grand canonical ensemble. It describes the case that you have a system of particles (e.g., our ideal gas) in contact with a heat bath (i.e., it can exchange energy with a large reservoir of particles at a given temperature), and you can exchange particles with the heat bath. Temperature and chemical potential are Lagrange parameters to find the state of maximum entropy, under the constraints that the mean energy (internal energy $U$) and the mean number of particles $\overline{N}$ is fixed. Thus temperature and chemical potential are used to get these given values of $\overline{E}=U$ and $\overline{N}$.

Thank you very much. Now I understand very well.