Question

Consider two single-particle states, A and B, in a system of fermions, where ${ϵ}_A=\mu -x$and ${ϵ}_B=\mu +x;$ that is, level A lies below $\mu$ by the same amount that level B lies above $\mu$. Prove that the probability of level B being occupied is the same as the probability of level A being unoccupied. In other words, the Fermi-Dirac distribution is "symmetrical" about the point where $ϵ=\mu$.

Short answer

It is proved that the probability of level B being occupied is the same as the probability of level A being unoccupied.

Step-by-step solution

Step 1. Givne Information

We are given that the level A lies below $\mu$ by the same amount that level B lies above $\mu$.

We have to prove thatthe probability of level B being occupied is the same as the probability of level A being unoccupied.

Step 2. Fermi-Dirac distribution

Using the Fermi-Dirac distribution to calculate the probability of state B being occupied as follows:

$P\left(B\text{occupied}\right)=\frac{1}{e^{\frac{4\left({\epsilon }_0-\mu \right)}{UT}}+1}$

Here, ${\epsilon }_B$is the energy for the occupled state of $B,\mu$ is the chemical potential, k is the Boltzmann's constant, and T is the temperature.

The energy for the occupied state of B is,

${\epsilon }_B=\mu +x$

Substitute $\mu +x$for ${\epsilon }_B$ in the equation and simplifying, we get

$$\begin{gathered} P\left(B\text{occupied}\right)=\frac{1}{e^{\frac{t(\mu +x-\mu )}{kT}}+1} \\ =\frac{1}{e^{\frac{x}{kT}}+1} \end{gathered}$$

Step 3. Probability of state A

The probability of state A being unoccupied is as follows,

$P\left(A\text{Unoccupied}\right)=1-P\left(A\text{occupied}\right)$

Using the Fermi-Dirac distribution, the probability of state A being occupied is as follows,

$P\left(A\text{occupied}\right)=\frac{1}{e^{\frac{+\left({\epsilon }_A-\mu \right)}{NT}}+1}$

Where ${\epsilon }_A$ is the energy for the occupied state of A.

Substituting the values, we get

$P\left(A\text{Unoccupied}\right)=1-\frac{1}{e^{\frac{+\left({\epsilon }_A-\mu \right)}{kT}}}$

The energy for the occupied state of A is,

${\epsilon }_A=\mu -x$

Substituting the values, we get

$$\begin{gathered} P\left(A\text{Unoccupied}\right)=1-\frac{1}{e^{\frac{+(\mu -x-\mu )}{kT}}+1} \\ =1-\frac{1}{e^{\frac{-x}{kT}}+1} \\ =\frac{e^{\frac{-x}{kT}}}{e^{\frac{-x}{kT}}+1} \\ =\frac{1}{1+e^{\frac{x}{kT}}} \end{gathered}$$

Step 4. Proving the Probability

Now, substituting $P\left(A\text{unoccupied}\right)=\frac{1}{1+e^{\frac{x}{kT}}}$in the equation,

$P\left(B\text{occupied}\right)=\frac{1}{e^{\frac{x}{kT}}+1}$, we get

$P\left(B\text{occupied}\right)=P\left(A\text{unoccupied}\right)$

Hence, the probability of level B being occupied is same as the probability of level A being unoccupied or the Fermi-Dirac distribution being symmetrical about the point where $\epsilon =\mu$.