# Fermi-Dirac distribution

1. Nov 24, 2015

### akk

How to find the normalization constant of Fermi-Dirac distribution function.

2. Nov 24, 2015

### Chandra Prayaga

I am not sure what you are referring to, but the F-D function gives the average occupation of an orbital:

<n> = 1 / [exp β(e - μ) + 1]. There are no normalization constants. I gave standard symbols. β is 1/kBT, where kB is Boltzmann's constant, and T is the absolute temperature, μ is the chemical potential.

3. Nov 24, 2015

### akk

, here A is the normalization constant for Mazwellian distribution. The normalization constnt for Fermi-Dirac is 2/{h}^{3},,where h is planck's constant. but i don't know how to normalize..don't know formula

4. Nov 24, 2015

### Chandra Prayaga

OK. The formula you pictured, is the Boltzmann probability distribution function. The normalization constant A in that case, is called the partition function, usually denoted by the letter Z. It is given by:

Z = ∑ exp[-βEi]. The summation is over all possible states. Ei is the energy of the state i. For each system, you need to know energies of all the states in order to calculate Z. If, as is the case for a classical gas, the energies are continuously distributed, the summation becomes an integral:

Z = ∫g(E) exp[-βE]dE where g(E) is the density of states function. This is the partition function for a single particle.

5. Nov 24, 2015

### Chandra Prayaga

Sorry. Just to add: The Fermi-Dirac distribution function is not a probability distribution. It gives the average occupation of an energy level, or orbital, and does not have a normalization constant to be calculated.

6. Nov 24, 2015

### akk

Thanks,, i may be wrong, but i found in many books that its normalization constant is
2\{h}^{3},,here 2 is due to two possible vales of the fermion spin and h is the volume of a quantum state in the phase space.

7. Nov 24, 2015

### Chandra Prayaga

The 2/h3 is part of the density of states function that I gave in my earlier post. If you could mention one or two of the books that you saw this in, perhaps I can clear it up for you.

8. Nov 24, 2015

### akk

ASTROPHYSICS PROCESSES by
Massachusetts Institute of Technology,

9. Nov 24, 2015

### Chandra Prayaga

The partition function Z which I gave in post #5 above can be evaluated for an ideal gas. It is given by:

Z = (V/ħ3) (mkBT/2π)3/2. So the constants that you mentioned above are indeed part of the partition function, which is the normalization factor for the probability distribution function.

10. Nov 24, 2015

### akk

thanks

11. Nov 29, 2015

### vanhees71

It depends a bit on the conventions used by your textbook. Of course, the physics is unique. Suppose we have a (spin-saturated) Fermi gas consisting of particles with spin $s$ (which is necessarily a number $s \in \{1/2,3/2,/\ldots \}$ because of the spin-statistics theorem). Then the average number of particles per unit phase space is given by
$$\frac{\mathrm{d} N}{\mathrm{d}^3 \vec{x} \mathrm{d}^3 \vec{p}} =f(\vec{p}) = \frac{g}{(2 \pi \hbar)^3} \frac{1}{1+\exp[\beta (E(\vec{p})-\mu)]},$$
where $\mu$ is a chemical potential for some conserved particle number (non-relativistic physics) or charge (relativistic physics), and $E(\vec{p})$ is the energy of a single particle as a function of its momentum. In relativistic physics one usually includes the rest energy in $E$, so that you have $E(\vec{p})=\sqrt{m^2 c^4+\vec{p}^2 c^2}$, while in non-relativistic physics you usually use the simple kinetic energy $E(\vec{p})=\vec{p}^2/(2m)$. In both cases $m$ is the same invariant mass of the particle. Further $\beta=1/k_{\text{B}} T$, where $T$ is the absolute temperature of the gas, and $g=2s+1$ is the spin degeneracy factor of the particles.

You don't need any arbitrary normalization constant here, because this just counts the number of quantum states per unit phase space, and this is determined from quantum theory uniquely. The equilibrium distribution then follows from maximizing the von Neumann-Shannon-Jaynes entropy of the corresponding statistical operator (here I work with the grand-canonical operator).

The semiclassical limit, leads to the Boltzmann equation, which is valid when the average particle occupation numbers are all very small compared to $1$. Then you can neglect the 1 in the denominator of the Fermi distribution, and you get the Maxwell-Boltzmann distribution
$$f(\vec{p})=\frac{g}{(2 \pi \hbar)^3} \exp[-\beta(E(\vec{p})-\mu)].$$
It is remarkable that there is the Planck action appearing in this classical formula. The reason is that in classical statistics we do not have a natural measure for phase-space cells, and the normalization is indeed arbitrary, but that on the other hand almost always does not matter, because here you can simply take an arbitrary normalization and use the fugacity factor $z=A \exp(\beta \mu)$. The natural choice for $A$ can only be inferred from quantum theory and is given by $g/[(2 \pi \hbar)^3]$.

12. Nov 29, 2015

### akk

Thank you very much . So basically the factor 1/{h^3} is not a normalization constant. It the volume element of a quantum state in phase space. and we do not get it by integration.
The Fermi-Dirac distribution function is

f(E)=\frac{1}{e^\frac{{E-E_{F}}}{k_{B}T}+1},

%\label{}
~~\mbox{when}~~~~~~ T\rightarrow0

~~~~f(E)=
\Big\{
\begin{array}{rcl}
&&1~~~~~~~ E<E_{F}\\
&&0~~~~~~~ E>E_{F}.
\end{array}

If i write it in terms of momentum and normalize it i get
\label{}
n=\int^{\infty}_{-\infty}f(p)d^{3}p.

\label{}
n=4\pi\int^{p_{F}}_{0}p^{2}dp.

it gives me expression for the fermi momentum instead of $(\frac{2}{2\pi \hbar})^{3}$.
\label{}
n_{0}=\frac{p_{F}^{3}}{3\pi^{2}\hbar^{3}}.

and that was mine confusion.

13. Nov 29, 2015

### vanhees71

The particle density is given by
$$n=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi \hbar)^3} f_{\text{F}}(\vec{p}),$$
where
$$f_{\text{F}}(\vec{p})=\frac{g}{1+\exp[\beta (E(\vec{p})-\mu)]}.$$
For $T \rightarrow 0$, i.e., $\beta \rightarrow \infty$ you get
$$f_{\text{F}}(\vec{p})=\Theta[\mu-E(\vec{p})].$$
For the non-relativistic particle this implies
$$n=\frac{g}{(2 \pi \hbar)^3} 4 \pi \int_0^{p_{\text{F}}} \mathrm{d} p p^2=\frac{g}{2 \pi^2 \hbar^3} \frac{p_{\text{F}}^3}{3}.$$
The Fermi momentum is determined by
$$\frac{p_{\text{F}}^2}{2m}=\mu \; \Rightarrow \; p_{\text{F}}=\sqrt{2 m \mu}.$$