8.044 Lecture Notes Chapter 9: Quantum Ideal Gases

1 Lecture Notes Chapter 9: Quantum Ideal Gases Lecturer: McGreevy Range of validity of classical Ideal gas .. 9-2. Quantum systems with many indistinguishable particles .. 9-4. Statistical mechanics of N non-interacting (indistinguishable) bosons or fermions 9-9. Classical Ideal gas limit of Quantum Ideal gas .. 9-13. Ultra-high temperature (kB T mc2 and kB T ) limit .. 9-15. Fermions at low temperatures .. 9-18. Bosons at low temperatures .. 9-33. Reading: Baierlein, Chapters 8 and 9. 9-1. Range of validity of classical Ideal gas For a classical Ideal gas, we derived the partition function 3/2. Z1N.. V 2 mkB T. Z= , Z1 = 3 = V , N! th h2. h where the length scale th 2 mkB T. is determined by the particle mass and the temperature. When does this break down? 1. If idealness' fails, if interactions become important. This is interesting and impor- tant but out of bounds for We'll still assume non-interacting particles. 2. If classicalness' fails. Even with no interactions, at low enough temperatures, or high enough densities, Quantum effects modify Z.

2 An argument that classicalness must fail comes by thinking harder about th , the thermal de Broglie wavelength . Why is it called that? Recall from that the de Broglie wavelength for a particle with momentum p is dB = hp . For a classical Ideal gas, we know the RMS. momentum from equipartition p~2 3 p p h i = kB T = pRMS h~p2 i = 3mkB T . 2m 2. We could have defined h h th = ;. pRMS 3mkB T. there's no significance to the numerical prefactor of 1 instead of 1 . 2 3. Now recall the significance of dB in QM: h dB = minimum size of a wavepacket with momentum p . p So we can infer that th minimum size of Quantum wavepackets describing atoms in a Quantum Ideal gas . The classical picture of atoms as billiard balls with well-defined trajectories only makes sense if 1/3. V. th typical spacing between particles =. N. V 1/3. And this inequality th 1T N.. is violated at low T or high density. The above inequality is the condition for validity of the classical treatment. 9-2.

3 V 1/3.. System T N. / th conclusion . air 300K 180 classical Liquid 4 He 4K not classical Conduction e in Cu 300 K not classical A second derivation of the same criterion, this time in momentum space: Consider solutions to the Schro dinger equation in an L L L box. The energy eigenstates of one particle are . m ~ (~. x) = sin (kx x) sin (ky y) sin (kz z) with k = m , and m = 1, 2, x L x x y y y z z z with energy eigenvalues ~2. kx2 + ky2 + kz2.. =. 2m Let's ask under what circumstances we can ignore the QM. The mean energy per particle in a classical Ideal gas is 23 kB T . How many 1-particle states with energies like this are there to put our N particles into? The number of states with 23 kB T is . 1 4 2m 23 kB T 3/2. 1/2. 8 3 ~2 volume of octant of sphere in ~k-space with radius 2mE. ~2. N ( ) = =. ( /L)3 volume in ~k-space per allowed grid point r !3 . mkB T V 3 2. = V 4 3 = 3 . h th . So: when N V3 , the number of particles is big compared to the number of possible one- th particle states for the particles to sit in.

4 So we need to worry about things like the Pauli Exclusion principle, which prevents us from putting more than one particle in each state. For N V3 , the classical analysis is fine we needn't worry about the particles needing to th occupy the same 1-particle state. 9-3. Quantum systems with many indistinguishable particles [This section is about Quantum mechanics . You've already encountered some of these ideas in , and will discuss this further in We'll come back in subsection and think about when this business reduces to classical mechanics .]. Consider two particles. Their state can be described by a wavefunction (x1 , x2 ). (We won't worry right now about how many coordinates of each particle ( how many dimensions). we have to specify.) If the particles are indistinguishable, then | (x1 , x2 )|2 = Prob finding a particle at x1 and a particle at x2.. Note that we make no specification of which particle is where. Indistinguishability requires: | (x1 , x2 )|2 = | (x2 , x1 )|2.

5 Swapping the arguments twice should give back the same , not just the same | |2 : (x1 , x2 ) = (x2 , x1 ). 9-4. Two choices: Bosons particles for which (x1 , x2 ) = + (x2 , x1 ). , the wavefunction is symmetric. It is a fact (observed experimentally, understood via Quantum field theory) that they have integer spin. : hydrogen atoms, 4 He, photons, phonons, magnons, gluons, Higgs bosons (?). The stat mech of a gas of them was developed by Bose and Einstein, so in the context of stat mech, these are called Bose-Einstein statistics . Fermions particles for which (x1 , x2 ) = (x2 , x1 ). , the wavefunction is antisymmetric. It is a fact (observed experimentally, understood via Quantum field theory) that they have half-integer spin (1/2, 3 ). : electron, proton, neutron, 3 He, 7 Li. The stat mech of a gas of them was developed by Fermi and Dirac, hence Fermi-Dirac statistics . For non-interacting Bose or Fermi particles (we will always assume this): energy eigenstates are always (symmetric or antisymmetric) linear combinations of products of single-particle energy eigenstates.

6 [Recall: there is more to QM than energy eigenstates, but they are enough to construct the partition function.]. 9-5. Wavefunctions of several bosons or fermions Consider for example two indistinguishable Quantum particles in a box. Label the possible states of one Quantum particle in the box by a fancy label which is a shorthand for all its Quantum numbers, , the wavenumbers (mx , my , mz ) of its wavefunction. If they are bosons, the wavefunction must be of the form (x1 , x2 ) = (x1 ) (x2 ) +. |{z} (x1 ) (x2 ). | {z } | {z }. an energy eigenstate another energy eigenstate makes it symmetric of one particle for one particle OK state of bosons | {z }. not sym or antisym hence not allowed For fermions: (x1 , x2 ) = (x1 ) (x2 ) . |{z} (x1 ) (x2 ). makes it antisymmetric OK state of fermions Working our way up to 3 indistinguishable particles: (x1 , x2 , x3 ) = (x1 ) (x2 ) (x3 ) (x1 ) (x2 ) (x3 ). (x1 ) (x2 ) (x3 ) (x1 ) (x2 ) (x3 ). (x1 ) (x2 ) (x3 ) (x1 ) (x2 ) (x3 ).

7 With + for bosons and for fermions. This time we count 1, 2, 3, many: 9-6. N indistinguishable particles: 1. Pick N single particle states , , .. 2. There is exactly one symmetric combination. This is an energy eigenstate for N bosons. 3. IF , , .. are ALL DIFFERENT then there is exactly one antisymmetric combi- nation. This is an energy eigenstate for N fermions. This fact that Fermions must all be in different single particle eigenstates is the Pauli Exclusion Principle. Many bosons can be in the same single-particle energy eigenstate. If the particles were distinguishable, there would be N ! states for each choice of , , .. If the particles are indistinguishable, there is (at most) one such state. 9-7. Occupation number representation of the many-particle state For either bosons or fermions, the state (x1 , x2 ..xN ) is fully specified by indicating 1. Which 1-particle states are occupied? 2. If bosons, how many particles are in each 1-particle state? (For fermions, this number can only be 0 or 1.)

8 Label the 1-particle states ( mx , my , mz for Ideal gas, or n, `, m for Hydrogen atoms). So: the state is specified by a set of integers called OCCUPATION NUMBERS: n ( ) # of particles in 1-particle state . when the many-particle state is . Fermions: n {0, 1} Bosons: n {0, 1, 2, }. These numbers also specify N, E, .., as follows. X. total # of particles in state : N= n ( ).. The sum here is over all possible states in which one of the particles could be; many of them will be unoccupied. This is not a sum over particles. Similarly, if = the energy eigenvalue of the 1-particle energy eigenstate . then the total energy in the many-particle state is X. total energy in state : E( ) = n ( ) .. 9-8. Statistical mechanics of N non-interacting (indistinguishable). bosons or fermions We'll use the canonical ensemble: an ensemble of copies of the system, all with the same N, T (hence, E varies amongst the copies in the ensemble), in contact with a heat bath at temperature T.

9 X. hEi = Ei p( ). |{z} | {z i}. i ensemble total energy prob of finding of N particles the N -particle state i in the ensemble ! X X. = n ( i ) p( i ). i ! X X. = n ( i )p( i ). i | {z }. hn i mean value of the occupation number of the 1-particle state . in the ensemble of N -particle states P. So: if we know hn i i n ( i )p( i ), then we know X. hEi = hn i .. A check on this formalism: ! X XX X X X. hn i = n ( i )p( i ) = n ( i ) p( i ) = N p( i ) = N. i i i | {z }. =N. So: how do we calculate the average occupation numbers hn i? 9-9. hn i for fermions X. hn i = n ( i )p( i ) p( ) = Z(N ) 1 e E( ). all N -particle states i 1 X. = n ( i )e Ei Z(N ) i 1 1 1. 1 X X X. = .. n e (n1 1 +n2 2 +..). Z(N ) n1 =0 n2 =0 n3 =0. | {z }. subject to the constraint n1 + n2 + .. = N. We just sum over all occupation numbers, consistent with the fact that there are N particles altogether. One of these sums is special, so let's write the two terms n = 0 and n = 1. explicitly: 1 1 1.

10 1 X X X. hn i = |{z}. 0 + e .. e| (n1 1{z +n2 2 +..). Z(N ) n =0 n =0 n =0. }. n =0 excluding n . |1 2. {z 3 }. doesn't include n . which is 1. so the rest are subject to the constraint n1 + n2 + .. = N 1. Repackage in a sneaky way which takes advantage of n = 0 or 1: 1 1 1. 1 X X X. hn i = e .. (1 n ) e| (n1 1{z +n2 2 +..). Z(N ) n =0 n2 =0 n =0. }. includes n . |1 {z 3 }. includes n . subject to the constraint n1 + n2 + .. = N 1. Notice that the n = 1 term in the summation gives zero; this means it doesn't matter if we include it in the Boltzmann factor. The n = 0 term in the summation gives what we had before. 1. hn i = e (Z(N 1) hn ifor N 1 particles Z(N 1)). Z(N ). Z(N 1) . = e (1 hn ifor N 1 particles ) (1). Z(N ). 9-10. The Fermi-Dirac distribution We want to rewrite our expression for hn ifermions as a function of T, . To do this, recall that F (T, V, N ) F (T, V, N 1) = kB T (ln Z(N ) ln Z(N 1)). Z(N 1). = = e /kB T . Z(N ). Now, in a box with three particles, hn iN =3 6= hn iN 1=2.