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Lecture 11: The Grand Canonical Ensemble

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Lecture 11: The Grand Canonical Ensemble

Schroeder Ch. 7.1-7.3

Gould and Tobochnik 6.3-6.5, 6.8

- Derivation of the Gibbs distribution
- Grand partition function
- Bosons and fermions
- Degenerate Fermi gases
- White dwarfs and neutron stars
- Density of states
- Sommerfeld expansion
- Semiconductors

- We have already described the canonical ensemble, which is defined as a collection of closed systems.
- If we relax the condition that no matter is exchanged between the system and its reservoir, we obtain the grand canonical ensemble.
- In this ensemble, the systems are all in thermal equilibrium with a reservoir at some fixed temperature, but they are also able to exchange particles with this reservoir.

- What is the probability of finding a member of the ensemble in a given state with energy and containing particles?
- Consider a system in thermal and diffusive contact with a reservoir, , whose temperature and chemical potential are effectively constant.
- If has energy and particles, then the total number of states available to the combined system is

- Since the combined system belongs to a microcanonical ensemble, the probability of finding our system with energy and particles is given by
- Since the reservoir is very large compared with our system, then and

- Expanding in a Taylor series about gives
- Writing the above expression in terms of temperature and chemical potential gives

- Since the combined system is in the microcanonical ensemble, then the probability of finding in any one state of energy and particles is given by
- Each of the exponential factors is called a Gibbs factor
- If we define the constant as , then for any state , the probability is given by
- This is the Gibbs distribution and it characterizes the grand canonical ensemble

- The quantity is called the grand partition function.
- By requiring that the sum of the probabilities of all states to equal 1, it can be shown that

- Consider a system consisting of a single hydrogen atom, which has two possible states:
- Unoccupied (i.e. no electron present)
- Occupied (one electron present in the ground state)

- Q: What is the ratio of the probabilities of these two states?

- If we neglect the spin states of the electron and the excited states of the hydrogen atom, this system has just two states
- Unoccupied:
- Occupied:

- The ratio of the probabilities of these two states s given by

- If we treat the electrons as a monatomic ideal gas, then the chemical potential for electrons is given by
- Therefore, we have

- Taking electron spin into account, the hydrogen atom now has two occupied states, each with the same energy, so the ratio of unoccupied to occupied atoms is
- Now, a free electron has two degenerate states so the chemical potential of the electron gas is
- Therefore, we have

- It can be shown that all the thermodynamic functions can be expressed in terms of the grand partition function and its derivatives.
- The average internal energy is
- The average number of particles is
- The generalized forces are
- The entropy is given by

- Recall that in the canonical ensemble, there is a relationship between the Helmholtz free energy and the partition function: .
- Using an analogous argument, we can derive the grand potential:
- The grand potential is the maximum amount of energy available to do external work for a system in contact with both a heat and a particle reservoir.

- The most important application of Gibbs factors is to quantum statistics: the study of dense systems in which 2+ identical particles have a probability of occupying the same single-particle state.
- Particles that can share a state with another of the same species are called bosons.
- Examples include photons, helium-4 atoms, etc.

- Particles that cannot share a state with another of the same species are called fermions.
- Examples include protons, neutrons, neutrinos, etc.

- Because individual particles have wave-like properties, the state of a particle can be described by a wavefunction
- The indistinguishability of quantum-mechanical particles implies that
- Observations show that both signs are possible for quantum-mechanical particles

For bosons

For fermions

- Suppose that the two-particle wavefunction can be decomposed into a product of single-particle wavefunctions
- For bosons, this equation guarantees that will be symmetric under interchange of and .
- For fermions, this equation guarantees that .
- The rule that two identical fermions cannot occupy the same state is called the Pauli exclusion principle
- Pauli also demonstrated that all particles with integer spins are bosons, whereas all particles with half-integer spin are fermions.

- When , the chance of any two particles occupying the same state is negligible.
- For an ideal gas, the chance of any two particles occupying the same state is negligible only if .
- There are a number of systems that violate this condition
- Neutron star
- Liquid helium
- Electrons in metals
- Photons

- Consider a single-particle state of a system whose energy when occupied by a single particle is . The probability of the state being occupied by particles is
- If the particles in question are fermions, then can only be 0 or 1, which implies that
- The average number of particles in the state (also called the occupancy of the state) is given by
- This distribution is called the Fermi-Dirac distribution

- The Fermi-Dirac distribution goes to zero when and goes to 1 when .
- At very low temperature, fermions will distribute themselves in the energy levels below the chemical potential and all the levels above are empty.
- As the temperature rises, energy levels above the chemical potential begin to be occupied.

- As an application of the Fermi-Dirac distribution, let’s examine degenerate Fermi gases.
- Examples of degenerate Fermi gases are
- Conduction electrons in a metal
- Electrons in a white dwarf star
- Neutrons in a neutron star

- Let’s first consider the properties of an low-temperature electron gas.

- At , the Fermi-Dirac distribution becomes a step function in which all single-particle states with energy less than are occupied, while all states with energy greater than are unoccupied.
- In this context, is also called the Fermi energy
- When a gas of fermions is so cold that nearly all states below are occupied, it is said to be degenerate.
- It can be shown that the Fermi energy is given by

- Thus, the Fermi energy is the highest energy of all the electrons.
- To calculate the total energy of all the electrons, we can add up the energies of the electrons in all occupied states.
- It can be shown that the total energy is

- Therefore, the average energy of the electrons is 3/5 the Fermi energy.
- The temperature that a Fermi gas would have to have in order for is called the Fermi temperature.
- The pressure of a degenerate electron gas (also called the degeneracy pressure) is given by
- The degeneracy pressure is what keeps matter from collapsing under the electrostatic forces that attract electrons and protons.

- A star that has consumed all its nuclear fuel will undergo gravitational collapse, but may end up in a stable state as a white dwarf or a neutron star.
- This occurs when the mass that remains in the core after the outer layers are blown away doesnot exceed a particular limit, called the Chandrashekar limit.
- Stars that succeed in forming such stable remnants owe their existence to the high degeneracy pressure exerted by electrons (for white dwarfs) and neutrons (for neutron stars).

- A white dwarf star can be considered as a degenerate electron gas.
- The nuclei present within the white dwarf balances the charge and provides the gravitational attraction that holds the star together.
- The total kinetic energy of the degenerate electrons is given by the Fermi energy

- If we assume that the star contains one proton and one neutron for each electron, then and thus we have
- The Fermi energy and Fermi temperature for the white dwarf star is given by
- Here, is the equilibrium radius of a white dwarf star, which can be determined by finding the minimum of the total energy

- It can be shown that the gravitational potential energy of a white dwarf is given by
- The total energy of a white dwarf can be given by
- The equilibrium radius of a white dwarf is determined by the minimum of the total energy
- For a one solar mass white dwarf, and thus, the Fermi energy and the Fermi temperature for the white dwarf star is given by

- A neutron star is made entirely of neutrons and is supported against gravitational collapse by degenerate neutron pressure.
- The total kinetic energy of the degenerate electrons is given by the Fermi energy
- Thus, the kinetic energy comes from the neutrons, and the number of these is simply . Therefore, we have
- The Fermi energy and the Fermi temperature for the neutron star is given by

- The total energy of a neutron star can be given by
- The equilibrium radius of a white dwarf is determined by the minimum of the total energy
- For a one solar mass white dwarf, and thus, the Fermi energy and the Fermi temperature for the neutron star is given by

- One property of a Fermi gas that we cannot calculate using the approximation is the heat capacity.
- Therefore, we must examine the Fermi gas at small nonzero temperatures
- To better visualize – and quantify – the behavior of a Fermi gas at small temperatures, we will introduce a new concept called the density of states.

- Using a suitable change of variables, it can be shown that the energy integral for a Fermi gas at zero temperature becomes
- The quantity in square brackets is the number of single particle states per unit energy, also known as the density of states.
- Using the density of states, we can obtain the total number of electrons at zero temperature

- For nonzero temperature, the total number and energy of electrons can be determined as follows
- Note that the chemical potential is the point where the probability of being occupied is exactly ½.
- In the limit , we can use the Sommerfeld expansion to evaluate the above integrals.

- Performing integration by parts gives
- The boundary term vanishes at both limits and using a suitable change in variables, we have

- We can make two approximations
- Extend the lower limit to
- Expand in a Taylor series about the point

- With these approximations, we have
- This integral can be performed giving

- Solving for gives
- Performing the same expansion for the total energy gives
- From this result, we can easily calculate the heat capacity

- Atoms in a metal are closely packed, which causes their outer shell electrons to break away from the parent atoms and move freely through the solid.
- The set of electron energy levels for which they are more or less free to move in the solid is called the conduction band.
- Energy levels below the conduction band form the valence band and electrons at energies below that are strongly bound to the atoms.
- The work function, , is the energy that an electron must acquire to escape the metal.

- In a conductor, the Fermi energy lies within one of the bands, whereas in an insulator, the Fermi energy lies within a gap.
- Therefore, at , the band below the gap is completely occupied while the band above the gap is unoccupied.
- Because there are no empty states close in energy to those that are occupied, the electrons are “stuck in place” and the material does not conduct electricity.

- A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature.
- The figure below shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor.

- As an approximation, let’s model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas with an appropriate zero point.
- Let’s model the density of states near the top of the valence band as a mirror image of this model.
- In this approximation, the chemical potential must always lie precisely in the middle of the gap, regardless of the temperature.

- The number of electrons in the conduction band is given by
- If the width of the gap is much greater than , then we can approximate the above integral as
- After a suitable change in variables, this integral can be evaluated to give

- The above result indicates that a pure semiconductor will conduct much better at higher temperature because there are much more electrons in the conduction band.
- Moreover, in order to produce an insulator, the gap would have to become significantly wider.