## Physics Friday 86

Let us consider a model quantum system with three permitted spacial ‘orbital’ states; a particle in state n=1,2, or 3 has energies ε1, ε2, and ε3, respectively. Now suppose are particles are spin-½ fermions. Then the spin projection has two possible states, ms=½ and ms=-½; these are designated “up” and “down”. Thus, there are six possible states (n,ms), with n=1,2,3, and ms=±½. Since the particles are fermions, no more than one particle may be found in each of these states at a given time.
Now, let us consider this system in contact with a thermal reservoir with a fixed temperature T, and a reservoir of our fermions with Gibbs potential μ; this is to say, in the grand canonical ensemble. Compute the grand canonical partition sum. To compute it, we note that the grand canonical partition sum factors across non-interacting states, like the canonical partition sum does. Thus,
$\mathcal{Z}=z_{1,-\frac{1}{2}}z_{1,\frac{1}{2}}z_{2,-\frac{1}{2}}z_{2,\frac{1}{2}}z_{3,-\frac{1}{2}}z_{3,\frac{1}{2}}$. Now, each orbital state partition sum has only two terms: one for the empty state, with E=0 and N=0; and one for the occupied state, with E=εn and N=1. Thus, (assuming no magnetic field, so energy is independent of the spin direction):
$z_{n,m_s}=e^{-\beta(0-\mu\cdot0)}+e^{-\beta(\epsilon_n-\mu\cdot1)}=1+e^{-\beta(\epsilon_n-\mu)}$.
Now, we could also pair the states with the same spatial orbit n (but differing spin alignments):
$z_{n}=z_{n,-\frac{1}{2}}z_{n,\frac{1}{2}}=\left(1+e^{-\beta(\epsilon_n-\mu)}\right)^2=1+2e^{-\beta(\epsilon_n-\mu)}+e^{-2\beta(\epsilon_n-\mu)}$.
We can interpret this expanded product in terms of the four possible states of n: the empty state, two singly occupied states (of opposite spin), and one doubly-occupied state.
The probability that the state (n,ms) is empty is thus $\frac{1}{z_{n,m_s}}$, and the probability that it is occupied is $p_{n,m}=\frac{e^{-\beta(\epsilon_n-\mu)}}{z_{n,m_s}}=\frac{1}{e^{\beta(\epsilon_n-\mu)}+1}$.

Now, our grand canonical partition sum is
$\mathcal{Z}=\left[1+e^{-\beta(\epsilon_1-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_2-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_3-\mu)}\right]^2$. Thus, the grand potential is
$\Psi=-\frac{\ln\mathcal{Z}}{\beta}=-\frac{2}{\beta}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]$.
Now, the mean number of particles ⟨N⟩ can be found from Ψ:
$\begin{array}{rcl}\langle{N}\rangle&=&-\frac{\partial\Psi}{\partial\mu}\\&=&\frac{2}{\beta}\frac{\partial}{\partial\mu}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_1-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_2-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\beta{e^{-\beta(\epsilon_1-\mu)}}}{1+e^{-\beta(\epsilon_1-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_2-\mu)}}}{1+e^{-\beta(\epsilon_2-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_3-\mu)}}}{1+e^{-\beta(\epsilon_3-\mu)}}\right]\\&=&\frac{2}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_3-\mu)}+1}\end{array}$
Note that this is the same result we get were we to sum the probability of occupation over all six states, as we should expect.
Similarly, we could find the energy via $U=-\frac{\partial}{\partial\beta}(\ln\mathcal{Z})+\mu\langle{N}\rangle$, or we can use the occupation probabilities to get:
$U=\sum_{n,m}E_{n,m}p_{n,m}=\frac{2\epsilon_1}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2\epsilon_2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2\epsilon_3}{e^{\beta(\epsilon_3-\mu)}+1}$.
We can also find entropy via $S=k\left(\ln\mathcal{Z}+\beta{U}-\beta\mu\langle{N}\rangle\right)$ (see here), getting:
$S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right]$.

### 3 Responses to “Physics Friday 86”

1. Physics Friday 87 « Twisted One 151’s Weblog Says:

[…] Friday 87 By twistedone151 Continuing from last week with our simple fermion system, we now consider the system enclosed in walls which are impermiable […]

2. Physics Friday 88 « Twisted One 151’s Weblog Says:

[…] Friday 88 By twistedone151 Previously, I used the grand canonical formalism to derive the fundamental thermodynamic relation for a system […]

3. Physics Friday 91 « Twisted One 151’s Weblog Says:

[…] Friday 91 By twistedone151 When we modeled the ideal Fermi fluid (previous posts here, here, here, here, and here), we used the fact that the grand canonical partition function can be […]