Lecture 17 12.11.2019 Quantum ideal gas and Fermi-Dirac Distribution Degenerate stars

Fys2160, 2019 1 What we learn today...

• How to count microstates using quantum statistics • Derive the Fermi-Dirac (F-D) distribution for the average number of fermions occupying a given microstate • What is the difference between Fermi-Dirac and Boltzmann distributions • What is the Fermi energy? What is a degenerate gas? • Derive the thermodynamics of a degenerate fermi gas at 0K • Application: finding the equilibrium size of white darf stars

Fys2160, 2019 2 Quantum ideal gas at constant T and N

The Quantum Ideal Gas gives us a robust way of defining countable microstates as energy levels of quantum particles. We consider a gas free quantum particles in a box of length L and volume � = � 0 � �� • Energy levels of a free particle are � = ��, � = � , � , � , � = �, �, �, ⋯ � ���� � � � �,�,�

(�) A microstate for N-particles is defined by the state at which particle «1» is in energy level �� , particles «2» is in (�) energy level �� and so forth. Even though we deal with quantum particles, the counting of microstates is done � = 1 in a classical sense by labeling each particle (with an ordering number) and placing each particle in a given (�) � = 2 energy level �� . The partition function of N particles is the sum over all these microstates � = 3 �(�) �(�) � � � �, � = � � � � ⋯ � � = � �, � ,

Where � �, � = , � � = is the thermal wavelength ()

This labeling is then artificially removed by dividing the N-partition function �� � with the number of permutations �! for ordering N indistinguishable particles �� � � = � � �! Fys2160, 2019 3 Quantum ideal gas at constant T and N

The statistics that we obtain from this counting of microstates is given by the Boltzmann distribution

The probability that the gas is in a given microstate �: P s = ()

• It turns out that there are other ways in which we can do the counting over the microstates. This is taking into account the quantum nature of the particles: fermions and bosons • It is easier to derive the quantum statistics for fermions and bosons when we consider that the quantum gas is at equilibrium with a reservoir with which it can exchange both heat and particles.

Fys2160, 2019 4 Quantum Gas at constant T and � Suppose we have a gas that is in contact with a reservoir (R) and they can exchange both heat and particles. At equilibrium, the gas has the same temperature � and chemical potential � as the reservoir. What is a microstate for such a gas? What is the probability that the gas is in a given microstate? Let us consider that the gas system is the microstate � with energy �(�). The probability that the system is in a microstate (�) is proportional to the number of microstates that the reservoir can occupy when the system is the microstate �. Suppose we consider two available microstates � � , �(�).

� � Ω � � = = � � Ω � �

Where �(�) is the number of particles in the microstate s, E � is energy of that microstate. (Using that S = k ln Ω and T�� = �� − ��� = −�� + ���).

Probability that the system at constant � and � is in a given microstate is 1 � � = � �(�, �)

• The normalization condition ∑ �(�) = 1, defines the grand partition function �(�, �)

Fys2160, 2019 5 �(�, �) grand partition function The grand partition function depends not only on � but also on �, and is given by the sum over all accessible microstates weighted by the Gibbs factor

� �, � = � • This sum can be done in two fundamental ways:

• «classical» statistics: summing over all the energy levels of particle 1, particle 2, particle 3. If the particle are indistinguishable, we have to remember to remove the over-counting by dividing with �! • «quantum» statistics: summing over all particles that are in a given energy level (lumps of particles), and then summing over all energy levels. Particle are never labelled and we do not need to worry about �! Fun fact: the �! comes naturally when we take the classical limit of the quantum statistics

Fys2160, 2019 6 Bosons vs. Fermions • Bosons: particles with integer net spin can occupy the same quantum state • Bosons are social • Photons, helium-4 atoms, alkali atoms, etc

• Fermions: particles with half-integer net spin can NOT share the same quantum state (Pauli exclusion principle) • Fermions are anti-social • Electrons, protons, neutrons, helium-3 atoms etc.

• How do we include this information in the counting of microstates?

Fys2160, 2019 7 When bosons and fermions are important

• Classical limit is when � ≫ �, the number of available microstates of a single particle is much greater than the number of particles. For � free quantum particles this means that ≫ � (�), (1) the average distance between particles must be greater than the themal wavelength of the particle, � (�) = .

• This condition broken in very dense fermionic matter (neutron stars) or ultracold bose matter (liquid helium-4 and Bose-Einstein condensates).

• We will now focus on an ideal gas of fermions

Fys2160, 2019 8 Ideal gas of fermions at constant T and �

�� • Earch fermion has the same energy levels as that of a free particle in a box are � = �� � = 1 � ���� � = 2 • The probability of N fermions occupying a given microstate with energy level � is � = 3

1 � � = � , �(�, �, �) A microstate is defined as a quantum state when we know all the quantum numbers of the particles occuping a given energy level. Hence, due to the Pauli exclusion principle, a microstate can either be empty � = 0 or occupied with just one particle � = 1. • The grand partition function for that energy level � �, �, � = 1 + � • Eventually the total grand partition function is obtained from summing over the energy levels. So, the counting over all microstates is done in two-stages: clumping all the particles that are in a given energy level, and then summing over all energy levels.

Fys2160, 2019 9 Fermi-Dirac distribution The average occupation number of fermions in a given energy level � is � �(��) � = � � � = � � � + � �(��) ��,�

Fermi-Dirac distribution for the occupation number � � = � ��(��)�

Important notice!The chemical potential � �, � is dependent on temperature and the density of particles � = . When we choose to fix T and � independently of each other, it means that the density �(�, �) of the gas must follow as function of � and �. The equilibrium state is independent on what we choose to control. This means that the relationship �(�, �) is independent on how we do the experiment. However, how we do the counting of microstates depends on how we let the system interact with the reservoir.

Fys2160, 2019 10 Boltzmann distribution

The average occupation number for quantum particles in the classical regime is N times the probability that one particle occupies that energy level

�� = �� � ,

� �� � �� Where P � = � is the Boltzmann distribution. Hence, �� = � . For an ideal gas, we have also �� �� () computed the chemical potential and obtained that �(�) = −�� ln = −�� ln . Thus, () Boltzmann distribution for the occupation number

(�) �(��) �� = � Classical limit of F-D distribution as high-T limit: �(�) ≪ �

� �(��) = ≈ �� �� = �(�) � ��(��) + � �

Fys2160, 2019 11 Fermi ideal gas at low temperatures

• Remember that microscopically the classical limit applies when ≫ � � . • In metal is ca. (0.2 nm)3, while for an electron in room temperature � = (4.3 ��)3. Hence the classical limit does not applied. The temperature is too low for the electron gas to behave as a classical gas

• We model the electrons in a metal as an ideal gas of non-interacting fermions.

• Electrons are fermions – hence Fermi-Dirac statistics apply: average number of particles ocuppying a microstate with energy � is given by: � � = � ��(��) + �

Fys2160, 2019 12 Fermi Dirac distribution at T=0 K

� � = � ��(��) + �

1, � < � It approaches the step function �� = , �� = �(� = �, �) is the Fermi energy 0, � > �

• The chemical potential at zero absolute temperature is called the Fermi energy

• Degenerate gas: all states below the Fermi energy occupied; all states above are free.

• Fermi energy �� is not a property of the individual fermion, but of the entire gas. This means that it depends on the density of fermions in the gas �� (�). • How do we describe the thermodynamics of a Fermi gas at 0K? What is the Fermi energy, total energy, and the pressure of such an electron gas?

Fys2160, 2019 13 Fermi energy of electron gas Electrons are modelled as be free particles in a box of V=L3. (no interactions with crystal lattice, ions, etc.). • Energy levels of a free particle are �� � = �� + �� + �� , � = �, �, �, ⋯ � ���� � � � �,�,�

• Fermi energy is energy corresponding to the highest energy level that is occupied � (radius of the Fermi surface)

���� � = ��� � ���� • How many microstates are occupied? That is how many particles we have. But is also twice (due to spin 1/2) the volume bounded by the Fermi surface (for positive n numbers). 1 4 3� / � = 2× × �� → � = 2 3 �

� � �� �� � �� � � � = → � � = � � �� ��� � �� �

Fys2160, 2019 14 Total energy of Fermi gas at 0K

∑ ∑ ∑ • We need to sum over all the occupied energy states � = 2× �

• By switching to spherical coordinates and evaluating of the positive «quadrant» of the sphere (divide by 1/8:

1 3 � �, � = 2× ×4� �� � � = ��(�) 8 5

• So can use the Fermi energy as indicator of the applicability of Fermi-Dirac statistics:

Fermi temperature

kT ✏F T = ✏ /k ⌧ F F Fys2160, 2019 15 Non-zero pressure at 0K!

• From the thermodynamic identity for the internal energy, �� = ��� − ��� + ���, we have that �� � = − �� , • Using the expression of the internal energy obtained for the degenerate gas at 0K, obtain the degeneracy pressure 3 � 2 � � = − � � = � 5 �� , 5 � • Pressure increases with reducing volume. For a nonrelativistic degenerate gas: � = �� • Degeneracy pressure for the fermions keeps the matter from collapsing under gravitational pull. It folows from Pauli’s exclusion principle.

Fys2160, 2019 16 Fermi gas at T>0K At zero temperature, all electrons are confined to the energy level that they occupy. There is no thermal energy to allow for jumps between energy levels (and hence energy fluctuations) However, at non-zero and very small temperature, the electrons near the Fermi energy migh have sufficient thermal energy (��) to jump on the excited states above the Fermi level. Electrons buried deep into the Fermi sea do not have enough thermal energy to escape the Fermi sea. 1 • To the lowest order, the internal energy increases with increasing temperature with a n¯FD = (✏ µ)/kT factor proportional to ×�� e +1 3 ⇡2 (kT)2 U = N✏F + N 5 4 ✏F

• This allows us to calculate the heat capacity of metals at temperatures smaller than Fermi temperature and show that it has a linear dependence on temperature @U ⇡2Nk2T C = = V @T 2✏ ✓ ◆V F Fys2160, 2019 17 White dwarf star

• White dwarf is a burnt out star with a stellar core composed mostly of electron- degenerate matter.

• Very dense object: its mass is comparable to that of the Sun, while its volume is comparable to that of Earth.

• Faint luminosity comes from the emission of stored thermal energy.

• No fusion. Thermal energy can not counteract the gravitaional collapse – but the degenerate electron pressure can, when the average distance between electrons is comparable to the Broglie wavelength!

Sirius A and Sirius B • What is the size of such a star (e.g., Sirus B)?

Fys2160, 2019 18 Size of a a white dwarf • Total energy of the star: � = E + E • Potential binding energy is due to the gravitational field:

E = − , where � is the total mass, � is the radius and � is the gravitational constant

• The kinetic energy is dominated by that of the � electrons that are highly degenerate (in the lowest energy levels), and nonrelativistic: 3 3 ℎ 3� 3 � ℎ 9� 1 E = �� = � = , 5 5 8� �� 5 2� 8� 8�� � • Using that one electron corresponds to one proton + one neutron: � = . • Equilibrium size is reached when energy is minimized: = 0 2 h 1 R =0.029 5/3 1/3 Gmemp M Fys2160, 2019 19 White dwarf star • Dwarf star with a larger mass has a smaller equilibrium radius! Higher mass gives more gravitational attraction. h2 1 R =0.029 5/3 1/3 Gmemp M

• If we choose the mass of the Sun, the radius of the dwarf star would be 7200 km. A bit more than radius of the Earth. The density would be M/V, giving 1.3x 109 kg/m3 (1.3 million times the density of water).

• Effectively we can use T=0K, since Fermi temperature is � = 2.3 � 109 �. • If the mass of dwarf star is larger than 3 times mass of the sun, it can be relativistic and unstable – its radius will tend to go to zero, and Carbon-oxygen dwarf can re-ignite when its mass is larger than 1.4 mass of the sun and 1a supernova explosion can happen. It can end up into neutron star or black hole... Stable dwarf will turn into red and black dwarf – after radiating its energy. • Nowadays the critical mass is calculated as 1.4 mass of the Sun. Fys2160, 2019 20