In Fermi-Dirac statistics the probability of being in a certain energy state is

$$f(E) = \left[1 + \exp\left(\frac{E-E_F}{k T}\right)\right]^{-1}$$

In the area that I'm looking at the texts always assume the population's energy is much greater than the Fermi Energy and so approximate this as the Boltzmann Distribution.

However, I am interested in the probability and am wondering if there is a way I can express the difference $E - E_F$ as a function of the temperature, the work function or some other common parameters?

If anyone has any insight I would appreciate some guidance.

Thank you,


  • $\begingroup$ Only a simple thought but Fermi's Energy $E_{F}$ is a parameter, so what you actually want is how does the energy $E$ depends on temperature, $E\equiv E(T)$ $\endgroup$
    – J L
    Commented May 29, 2013 at 21:00
  • 5
    $\begingroup$ Sorry, but I think this is a non-sense. Here you have a probability $f(E)$ to be in a energy state E. So, if you take global parameters like internal energy $U$, these are mean quantities, for instance, the internal energy $U = <E> = \int dE E f(E)$. So you cannot have any relations between E and U, or with E and any other global parameter. $\endgroup$
    – Trimok
    Commented May 30, 2013 at 14:30
  • $\begingroup$ Did you understood Boltzmann distribution ? Because it's essentially the same thing as the Fermi function: the occupation probability depends on an energy scale and on temperature as well, isn't it ? The Fermi energy $E_{F}$ is a tabulated parameter for materials. Maybe you could find this answer physics.stackexchange.com/a/65624/16689 interesting as well. $\endgroup$
    – FraSchelle
    Commented Jun 28, 2013 at 22:17

3 Answers 3


Yes, you can find a relation between $E_F$ and other functions. However remember that this value is a parameter, not a function.

The average number of particles in state $k$ in Fermi-Dirac statistic is

$$<n_k> = \frac{1}{e^{\beta(E-\mu)}+1}$$

With $\beta = 1/kT$. At a very low temperatures, you find that $<n_k>$ can only have two values: 0, if $E_F > \mu$ or 1, if $E_F < \mu$, that is a state can be empty or occupied, so we can write $<n_k>=\theta(E_F-E)$. In this expression $\theta(x-a)$ is the Heaviside step function.

You can use this on the integrals to calculate density and pressure. If you remember, these integrals are:

$$\rho(\mu,T) = 2\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty} E^{1/2}<n_k> dE$$ $$P(\mu,T) = \frac{2}{3}\pi g { \left ( \frac{2m}{h^2} \right ) }^{3/2} \int _0 ^{+\infty}E^{3/2}<n_k> dE$$

Where $g$ is the spin degeneration. This integrals can be solved easily with $<n_k>=\theta(E_F-E)$. With that, you will find expressions for a gas strongly degenerated, at $T\simeq 0$. Using that you find that you can write $E_F$ in terms of the density:

$$E_F = {\left ( \frac{6\pi}{g} \right ) }^{2/3} \frac{h^2}{2m}\rho^{2/3} = \mu(T=0)$$

Here you can see that $E_F$ is a constant value, and it equals the value of $\mu$ at $T\simeq0$. You can also write, from the second integral, $P(T=0)=\frac{2}{3}\rho E_F$, and using the expression $P = \frac{2U}{3V}$, valid for an ideal, non-relativistic quantum gas, you can write:

$$U(T=0) = \frac{3}{5}\rho E_F$$

So now you have the Fermi energy related with $U$. Remember, this is for a quantum gas where the temperature is really low. Note that all the equations are constants, calculated for $T\rightarrow 0$. If you want another aproximation, for higher $T$, the integrals on density and pressure can be approximated as follows:

$$I = \int _0 ^{+\infty} \frac{f(E)}{<n_k>} dE \simeq F(\mu) + \frac{\pi^2}{6}(kT)^2{\left ( \frac{df(E)}{dE} \right ) }_{E=\mu}$$


$$F(E) = \int _0 ^E f(x)dx$$

I'm not going to demonstrate this expression here because it is very long, however, I think you can find it in some classic books like McQuarrie or Pathria. Using this, you can calculate another expression for pressure and density, also for low $T$, but not for $T\simeq0$. To properly work with that you'll have to use lot of approximations during the calculus. You can find now $\mu(T)$

$$\mu(T) \simeq E_F \left [ 1- \frac{\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

And use this find the dependence of some thermodynamics potentials with $T$:

$$U(T) = \frac{3}{5} N {E_F}\left [ 1+ \frac{5\pi^2}{12} {\left ( \frac{kT}{E_F} \right ) }^2 \right ]$$

$$S(T) = \frac{\pi^2Nk^2T}{2E_F}$$

I think this are some of the expressions you wanted. The first approximation works for $T\simeq 0$ and the second one for low $T$.


Although I'm not 100% sure about exactly what you are asking, I'm trying to answer what I understood.

Since, $E - E_F$ is an energy, it can be expressed in terms of energy scale set by temperature. In other words, there always exists a temperature $T$: $k_B T = E - E_F$, where $k_B$ is the Boltzmann constant. Of course I've assumed $E > E_F$. Therefore, $E - E_F$ can be converted to $T$ by dividing it by $k_B$.

So the general idea is to find out the expression of energy (eg. $k_B T$) in terms of the parameter of interest (eg. $T$). Equate the expressions (eg. $k_B T = E - E_F$) and find the energy scale in terms of this parameter of interest.

Does this answer your question?


$E$ is $E$, the independent variable, so I interpret your question to be "can $E_F$ be expressed in terms of some other common parameters?"

For real solids, not in any meaningful way. For most models of solids, not in any simple way. Perhaps yes for a free electron model of a solid, but that completely ignores the lattice, and so throws out a lot of important properties, for example thermal expansion.


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