A question on Fermi free gas model

In summary, the Fermi energy is calculated by considering the energy levels of fermions, which stack up higher due to the Pauli Exclusion Principle. However, this principle is always valid and states that different fermions cannot occupy the same eigenstate. This means that if fermions are in a superposition of different eigenstates, they cannot go to higher energy levels. Any wave function that violates this principle is not a valid solution to the Schrödinger equation for fermions. This includes wave functions with multiple particles that do not have a zero probability of being in the same state.
  • #1
funginator
1
0
When we are calculating the fermi energy, we say that each energy level is 2 fold degenerate and the fermions stack up into higher and higher energy level due to the Pauli Exclusion Principle. My question is: The Pauli Exclusion principle only says that we can't put different fermions into the same eigenstate, so if we have fermions which are in states of a superposition of different eigenstates with different quantum amplitudes, the fermions do not have to go to higher energy levels because it doesn't contradict the Pauli Exclusion principle anymore?
 
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  • #2
No, the exclusion principle is always valid. Say you have two particles which are both in a superposition of two eigenstates. You measure a particle in one state, then the other particle must be in the other state.

Or put it this way: A wave function which is a superposition of several eigenstates for several particles that does not give a zero probability of finding two particles in the same state, simply isn't a valid solution to the Schrödinger equation for fermions.
 
  • #3
alxm said:
Or put it this way: A wave function which is a superposition of several eigenstates for several particles that does not give a zero probability of finding two particles in the same state, simply isn't a valid solution to the Schrödinger equation for fermions.

I wonder if you mean that it's technically a valid solution of the differential equation but we have to throw it out because it violates the exclusion principle?

For example, if we take the ground state solution of the hydrogen atom and multiply it by any number greater than 1, then the resulting wave function is still a solution of the differential equation. We throw it out because it's not normalized, but it's still a valid solution of the differential equation.
 

1. What is the Fermi free gas model?

The Fermi free gas model is a theoretical model used in statistical mechanics to describe the behavior of a gas consisting of identical particles with no interaction between them.

2. How does the Fermi free gas model differ from the classical ideal gas model?

Unlike the classical ideal gas model, the Fermi free gas model takes into account the quantum nature of particles, specifically the Pauli exclusion principle. This means that particles cannot occupy the same quantum state, leading to different statistical behavior than the classical model.

3. What is the significance of the Fermi energy in the Fermi free gas model?

The Fermi energy is the highest occupied energy level at absolute zero temperature in the Fermi free gas model. It determines the maximum energy a particle can have while still obeying the Pauli exclusion principle. This energy level is also used to define the Fermi level, a key concept in condensed matter physics.

4. How is the Fermi free gas model used in research and applications?

The Fermi free gas model is used in a variety of fields, including materials science, solid state physics, and astrophysics. It is used to describe the behavior of electrons in metals, as well as in the study of neutron stars and white dwarfs. It also serves as a foundation for more complex statistical models in quantum mechanics.

5. What are the limitations of the Fermi free gas model?

The Fermi free gas model is limited in its ability to accurately describe systems with strong particle interactions, such as liquids and dense solids. It also does not take into account relativistic effects, which may be important for high-energy particles. Additionally, it assumes a uniform and non-interacting gas, which may not always reflect real-world systems.

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