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Problem statement:
Consider a fermionic system with two states 1,2 with energy levels \epsilon_i, i=1,2. Moreover, the number of particles in state i is n_i = 0,1. Let the Hamiltonian of the system be
H = \sum_{i=1}^2 \epsilon_i n_i + \sum_{i \neq j} U n_i n_j
Here, U > 0 is a Coulomb-repulsion present in the system if both states i and j are occupied.
a) Compute the grand canonical partition function Z_g of the system by direct summation.
b) Let \epsilon_i = \epsilon; (i=1,2), and compute <N>.
c) Let \beta U \gg 1 and find <U> in this limit. Finally, for \beta U \gg 1, set \mu = \epsilon and compute <U> in this case.
Attempt at solution:
Just to make it clear:
n_k is the number of particles in the state with wave number k.
\mu is the chemical potential - the energy required to remove one particle from the system.
a)
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}
=\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)
= 1 + e^{-\beta (\epsilon_1 - \mu)} + e^{-\beta (\epsilon_2 - \mu)} + e^{-\beta (\epsilon_1 + \epsilon_2 - 2\mu)}
b)
<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)}}
c)
Here I don't know what to do as there is no U is the expressions I have found.
Consider a fermionic system with two states 1,2 with energy levels \epsilon_i, i=1,2. Moreover, the number of particles in state i is n_i = 0,1. Let the Hamiltonian of the system be
H = \sum_{i=1}^2 \epsilon_i n_i + \sum_{i \neq j} U n_i n_j
Here, U > 0 is a Coulomb-repulsion present in the system if both states i and j are occupied.
a) Compute the grand canonical partition function Z_g of the system by direct summation.
b) Let \epsilon_i = \epsilon; (i=1,2), and compute <N>.
c) Let \beta U \gg 1 and find <U> in this limit. Finally, for \beta U \gg 1, set \mu = \epsilon and compute <U> in this case.
Attempt at solution:
Just to make it clear:
n_k is the number of particles in the state with wave number k.
\mu is the chemical potential - the energy required to remove one particle from the system.
a)
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}
=\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)
= 1 + e^{-\beta (\epsilon_1 - \mu)} + e^{-\beta (\epsilon_2 - \mu)} + e^{-\beta (\epsilon_1 + \epsilon_2 - 2\mu)}
b)
<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)}}
c)
Here I don't know what to do as there is no U is the expressions I have found.