- #1

- 7

- 0

- Homework Statement:
- a

- Relevant Equations:
- a

**Problem statement:**

Consider a fermionic system with two states [itex] 1,2 [/itex] with energy levels [itex] \epsilon_i, i=1,2 [/itex]. Moreover, the number of particles in state [itex] i [/itex] is [itex] n_i = 0,1 [/itex]. Let the Hamiltonian of the system be

[tex] H = \sum_{i=1}^2 \epsilon_i n_i + \sum_{i \neq j} U n_i n_j [/tex]

Here, [itex] U > 0 [/itex] is a Coulomb-repulsion present in the system if both states [itex] i [/itex] and [itex] j [/itex] are occupied.

**a)**Compute the grand canonical partition function [itex] Z_g [/itex] of the system by direct summation.

**b)**Let [itex] \epsilon_i = \epsilon; (i=1,2) [/itex], and compute [itex] <N> [/itex].

**c)**Let [itex] \beta U \gg 1 [/itex] and find [itex] <U> [/itex] in this limit. Finally, for [itex] \beta U \gg 1 [/itex], set [itex] \mu = \epsilon [/itex] and compute [itex] <U> [/itex] in this case.

**Attempt at solution:**

Just to make it clear:

[itex] n_k [/itex] is the number of particles in the state with wave number [itex] k [/itex].

[itex] \mu [/itex] is the chemical potential - the energy required to remove one particle from the system.

**a)**

[tex] Z_g = \sum_{[n_k]} e^{-\beta \sum_{k}(\epsilon_k-\mu)n_k} = \prod_k \sum_{n_k}e^{-\beta (\epsilon_k-\mu)n_k} = \prod_k \sum_{n_k=0}^{1}e^{-\beta (\epsilon_k-\mu)n_k}[/tex]

[tex] =\prod_k \Big( 1 + e^{-\beta(\epsilon_k-\mu)}\Big) = \Big(1+e^{-\beta (\epsilon_1 -\mu)}\Big)\Big( 1+e^{-\beta (\epsilon_2 -\mu)} \Big) [/tex]

[tex] = 1 + e^{-\beta (\epsilon_1 - \mu)} + e^{-\beta (\epsilon_2 - \mu)} + e^{-\beta (\epsilon_1 + \epsilon_2 - 2\mu)} [/tex]

**b)**

[tex] <N> = \frac{\partial ln Z_g}{\partial (\beta \mu)} = \frac{2e^{-\beta (\epsilon - \mu)}+2e^{-2\beta(\epsilon - \mu)}}{1 +2e^{-\beta(\epsilon - \mu)}+e^{-2\beta(\epsilon - \mu)}} [/tex]

**c)**

Here I dont know what to do as there is no [itex] U [/itex] is the expressions I have found.