How Does Coulomb Repulsion Affect a Fermionic Two-Level System?

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In summary, the problem statement discusses a fermionic system with two states and energy levels, where the number of particles in each state can only be 0 or 1. The Hamiltonian of the system includes a Coulomb-repulsion term when both states are occupied. The conversation then goes on to discuss the grand canonical partition function, <N> and <U> in different scenarios. To solve for the grand canonical partition function, it is necessary to consider all possible microstates and their corresponding energies and number of particles.
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Selveste
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Homework Statement
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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 don't know what to do as there is no [itex] U [/itex] is the expressions I have found.
 
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Selveste said:
[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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Related to How Does Coulomb Repulsion Affect a Fermionic Two-Level System?

What is a Fermionic two-level system?

A Fermionic two-level system is a quantum mechanical system that can exist in one of two states, known as the ground state and the excited state. This system follows the rules of Fermi-Dirac statistics, which govern the behavior of fermions, particles with half-integer spin.

What are some examples of Fermionic two-level systems?

Some examples of Fermionic two-level systems include electrons in an atom, protons and neutrons in a nucleus, and quarks within a hadron. These systems can also be artificially created in quantum computing and quantum information processing experiments.

How are Fermionic two-level systems different from Bosonic two-level systems?

Fermionic two-level systems differ from Bosonic two-level systems in that they follow different statistics. While fermions follow the Fermi-Dirac statistics, bosons, particles with integer spin, follow the Bose-Einstein statistics. This leads to different behaviors and properties for these two types of systems.

What is the significance of studying Fermionic two-level systems?

Fermionic two-level systems are of great interest to scientists because they can exhibit interesting quantum phenomena, such as superposition and entanglement. They also have important applications in fields such as quantum computing, quantum information processing, and condensed matter physics.

What challenges are associated with studying Fermionic two-level systems?

One of the main challenges in studying Fermionic two-level systems is their susceptibility to decoherence, or the loss of quantum coherence due to interactions with the environment. This can make it difficult to observe and control their quantum behavior. Additionally, the complexity of these systems makes it challenging to model and understand their behavior using classical methods.

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