Fermionic two-level system

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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.
 

Answers and Replies

  • #2
TSny
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[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]
It looks like you are writing ##Z_g## for a system of non-interacting particles, which is not the case for this system where you have the "Coulomb interaction energy" ##U##.


A basic way to express ##Z_g## is as a sum over all possible microstates, ##s##, $$Z_g = \sum_s e^{-\beta (E_s - \mu N_s)}$$ where ##E_s## is the total energy of the system for the microstate ##s## and ##N_s## is the total number of particles in the microstate ##s##. The sum over all microstates includes states for all possible values of the total number of particles.

In your problem, it is easy to explicitly enumerate all the possible microstates. There are only a few. So, you should list all the possible microstates and for each microstate, ##s##, specify the value of ##E_s## and ##N_s##. Then you can easily write out the above expression for ##Z_g##.
 
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