Bosons en fermions Indistinguishability We already came across a form of indistinguishably in the canonical partition function: for distinguishable particles Q = Λ 3N βe p r, r 2,..., r N ))dτ dτ 2... dτ N ) for indistinguishable particles Q = Λ 3N N! βe p r, r 2,..., r N ))dτ dτ 2... dτ N. 2) For ideal gases this becomes Q = V N Λ 3N N! = qn T /N!. We saw another type of indistinguishably for the rotational partition function of molecules in an ideal gas, which lead to an additional factor of σ, where σ is the symmetry number of the molecule. The last form on indistinguishably is a result of the spin character of the atomic nuclei. In terms of spin we can distinguish two groups of particles: fermions and bosons: Fermions have half-integer spins σ = 2, 3 2, 5 2,. Examples are electrons, protons, neutrons all with σ = 2 ), nuclei with N n + N p = 2n + odd) like H, T = 3 H), and 3 He. Bosons have integer spins σ = 0,, 2,. Examples are photons σ = ) and nuclei with N n + N p = 2n even) like D and 4 He. The Pauli exclusion principle holds for fermions. It states that each energy level can only be occupied by a limited number of fermions. Each fermion should have a unique set of quantum numbers, including spin. This means that a maximum of two electrons or H-nuclei can occupy one energy level: one with spin σ = + 2 and one with spin σ = 2. For bosons this limitation does not hold and at T = 0 K the ground state can be occupied by all N bosons with a total energy of E = Nɛ 0. The Boltzmann distribution that we have worked with so far, does not take this difference between fermions and bosons in consideration. For not too low temperatures, this is not a problem. The number of thermally accessible energy levels is then so large that on average the occupancy per energy level is very low n i ) and the Pauli exclusion principle hardly plays a role. Especially the translational energy levels help in this sense, for instance, in the case of electrons in a magnetic field which already came across several times. The magnetic field influences predominately two energy levels. The translational energy levels, that are hardly affected by the magnetic field, however, are still available and provide enough unique quantum numbers to not worry about the Pauli principle. Only for a very large number of particles in a limited volume, i.e., high densities, or for very low temperatures the Pauli exclusion principle starts to affect the value of n i N. An example of a high fermion density is the case of conduction electrons in metals. Here we shall derive an ression for the number of occupied energy levels in an ideal gas of fermions and bosons in an open system. For an open system the number of particles can vary and we cannot use the canonical ensemble where N, V, and T are constant. We should apply the grand canonical ensemble which has V, T, and the chemical potential µ constant. The latter we already used when deriving the ression for chemical equilibria where the number and nature
of the particles changes as a consequence of the reaction. In this ensemble the internal energy is given by du = T ds P dv + µ dm 3) where labels the type of particles. We use m as the number of particles of species and reserve n i for the number of particles in energy level i. The Gibbs energy G = U T S + P V now becomes dg = du T ds SdT + P dv + V dp dg = T ds P dv + µ dm T ds SdT + P dv + V dp dg = dg = µ dm SdT + V dp ) G dp + P T,m i ) G dt + T P,m i ) G m P,T,m i dm, 4) and the chemical potential for species of type is ) G µ =. 5) m P,T,m i 2 The grand canonical partition function For a canonical ensemble, N and V are constant and E varies between the systems of the ensemble. The canonical partition function that belongs to this ensemble is Q = QN, V, T ) = i [ βe i. The sum runs over i, the different realizations of the system. The canonical partition function is directly related to the Helmholtz free energy A through β ln Q = A. In the grand canonical ensemble, µ, V and T are constant and E and N vary. The grand canonical partition function that belongs to this ensemble is denoted by Ξ = Ξµ, V, T ). In the case we only have one type of particles this partition function becomes Ξ = i [ βe i + βµn i = N [βµnqβ, N, V ), 6) where i labels the different realizations of the system. In this case, not only E can vary but also the number of particles and for each realization E i and N i should be specified. The grand canonical partition function is directly related to P V The probability of finding a certain configuration is P V = ln Ξ 7) β P i = [ βe i + βµn i, 8) Ξ analogous to the canonical equivalent. This ression does not account for the nature of the particles, bosons or fermions, which as lained before is not a problem for most situations. Here, however, we use the grand canonical partition function to derive a distribution function over energy levels for both ideal gases of bosons and fermions. 2
For an ideal gas, we can use the single molecule levels with ɛ E i = ɛ n i) en N i = n i). 9) with n i) the occupancy of the single-particle state ɛ for the ith realization of the system in the ensemble. We will limit ourselves to an ideal gas containing only one type of particles. The grand canonical partition function is in that case Ξ = [ βe i +βµn i = [ β ɛ n i) µni) ) = [ βɛ µ)n i). i i i 0) The average occupancy of single-particle state with energy ɛ is given by n = i i n i) P i = ni) [ βe i µn i ). ) Ξ Here, we sum over all realizations of the ensemble i and for each configuration i we multiply the number of particles in single-particle state, n i), by the probability of configuration i to occur. If we make use of E i = ɛ ni), we find n = i ni) [ β ɛ ni) + βµn i Ξ = Ξ Ξ βɛ ) = ln Ξ βɛ ). 2) Here we use a different label, since the already labels the specific level which occupancy we are interested in. The probability of a configuration to occur, however, depends on all particles in that configurations and not only in the particles in single-particle state ; hence the sum over all states labeled by. For an ideal gas, the particles are independent and we can change the order of the sum and the product Ξ = i [ βɛ µ)n i) = n max n i) =0 [ βɛ µ)n i). 3) The sum now runs over the different single-particles states and the product over the possible occupancies of these states. 3 Ideal gas of bosons The Pauli exclusion principle does not hold for an ideal gas of bosons. The maximum occupancy of a single-particle state is therefore unlimited. If we consider again only one type of particles, we get for the partition function Ξ = n i) =0 [ βɛ µ)n i) = [. 4) [ βɛ µ) 3
Here we used to the same geometric progression that we have also used to derive an ression for the harmonic partition function. Substitution of this ression in 2) gives [ [ ln [ βɛ µ) n = = ln [ [ βɛ µ) = βɛ ) βɛ ) or ln [ [ βɛ µ) βɛ ) [ βɛ µ) [ βɛ µ), n = = ln [ [ βɛ µ) [ [ βɛ µ) = [ [ βɛ µ) βɛ ) [βɛ µ) This is the Bose-Einstein distribution function for an ideal gas of bosons. 4 Ideal gas of fermions The Pauli principle limits the number of particles in single-particle state to n = 0 or n =. The number of spin states is usually included in the density of states that we will introduce later. We again start from the ression of the grand canonical partition function for an ideal gas 3) but we will now include restriction due to the Pauli principle Ξ = i [ βɛ µ)n i) = n i) =0 [ βɛ µ)n i) = 5) [ + [ βɛ µ). 6) If we apply ression to equation 2), we obtain the average occupancy of the single-particle state or n = ln Ξ βɛ ) = ln [ + [ βɛ µ) = βɛ ) ln [ + [ βɛ µ) = ln [ + [ βɛ µ) = βɛ ) βɛ ) ln [ + [ βɛ µ) [ + [ βɛ µ) = [ βɛ µ) [ + [ βɛ µ) βɛ ) + [ βɛ µ), n = [βɛ µ) + This is the Fermi-Dirac distribution function for an ideal gas of fermions. 5 Meaning and use of the distribution functions We now have derived the distribution functions for ideal gases of fermions +) and bosons -), respectively, which are of the form n = 7) [βɛ µ) ±. 8) Here I would like to point out again that this derivation was made for an ideal gas and that the distribution function n describes the average number of particles in the single-particle 4
3 2.5 Bose-Einstein Maxwell-Boltzmann Fermi-Dirac 2 n.5 0.5 0-4 -3-2 - 0 2 3 4 βɛ µ) Figure : Comparison of the Fermi-Dirac, Bose-Einstein and Maxwell-Boltzmann statistics state at a temperature T. From the previous classes, we have third distribution function: the Maxwell-Boltzmann distribution function n = N [ βɛ q 9) with q the microcanonical N, V, E) or molecular partition function. Figure plots the three distributions as a function of βɛ µ). Here ɛ can take values between 0 for the ground state level and, in principle, infinity. Both β and µ depend on temperature and therefore does not simply go with /β. Depending on the sign and magnitude of µ, βɛ µ) can have negative values. For the Fermi-Dirac and the Maxwell-Boltzmann distribution functions, this leads to a physical result. The Bose-Einstein distribution function results in negative values of n for βɛ µ) < 0 and hence µ 0 to give physically relevant answer without singularity for all possible single-particle states ɛ 0. 5. The chemical potential for a boson gas The boundary condition N = n = N can be applied to determine the value for the chemical potential of the system. We can apply this to determine the value of µ for bosons to see that this is indeed µ 0. We start by first taking simple the limit for T = 0 K. At T = 0 K all particles are in the ground state since bosons do not follow the Pauli principle { 0 for > 0 n, T = 0) B-E = N for = 0. Since at T = 0 K all particles are in the ground state, we obtain N = n 0. Using the Bose-Einstein distribution function we get N = n 0 = [βɛ 0 µ) = [ µβ 5
or µ = β ln N + ). If the temperature goes to zero or β to infinity µ = lim ) β β ln N + lim β βn = 0 = µ 0 using ln + x) x. We see that indeed the sign of µ is negative. For an ideal gas of atomic bosons we can also derive an ression for T > 0 K. This is a bit more involved, but starts from the same boundary condition N = n = [βɛ µ) = [ βɛ µ) [ βɛ µ) = [ βɛ µ) {[ βɛ µ)} i = {[ βɛ µ)} i+ = i=0 i=0 {[ βɛ µ)} i = βµi) βɛ i) 20) i= i= In the case of an atomic boson gas, we only have translational levels with ɛn x, n y, n z ) = ɛn x ) + ɛn y ) + ɛn y ) and ɛn) = h2 n 2 8mL 2. 2) The latter describe the energy levels of a particle in a one-dimensional box and we have used this ression before to derive the translational partition function. Remember that the translational energy levels are quite close together and we will approximate the summation of in Eq. 20) by integrals over n x, n y, and n z N = i= βµi) β h2 n 2 ) ) x 8mL 2 i β h2 n 2 y 8mL 2 i β h2 n 2 ) z 8mL 2 i dn x dn y dn z. Apart from the factors i in the onents we have come across these integrals before in the derivation of the translational partition function. We will make the substitutions n a = n a i and dn a = idn a for a = x, y, and z N = i= = V Λ 3 βµi) i 3/2 i= βµi) i 3/2 β h2 n x 2 8mL 2 ) β h2 n y 2 8mL 2 ) β h2 n z 2 8mL 2 ) 22) dn xdn ydn z with Λ the de Broglie wavelength. This ression links the chemical potential to the number of particles at a certain temperature. Figure 2 plots the chemical potential as function of temperature. Indeed µ 0 and for T the chemical potential goes to minus infinity. At some critical temperature T c, µ becomes 0 and below this value there is no solution for µ. For T < T c, one can show that N = n 0 + V Λ 3 i 3/2 n 0 + 2.62 V Λ 3, 23) i= which means that only a limited amount of atoms can go into an excited state. All other atoms should condense into the lowest energy state. This is a Bose-Einstein condensate and is due to Bose-Einstein statistics and not to interactions between the atoms, since this is derived within the context of an ideal gas. 6
µ 0. -0. 0-0.2-0.3-0.4-0.5-0.6-0.7-0.8-0.9-0 0.5.5 2 2.5 T/T c Figure 2: The chemical potential of an ideal gas of atomic bosons. 5.2 The chemical potential for a fermion gas We will now look that the value of the chemical potential for a fermion gas in the limit of T = 0 K. We will later apply this to electrons in a conductor. At T = 0 K the energy levels are occupied with one particle from the ground state level until f, the Fermi level, because of the Pauli exclusion principle { for 0 < f n) F-D =. 0 for > f Using the Fermi-Dirac distribution function we see that which gives n) F-D = [βɛ µ) + { if lim [βɛ ɛ < µ µ) = β if ɛ > µ n) F-D = or µt = 0) = ɛ f, which is positive. 5.3 The high temperature limit { for 0 < ɛ µ 0 = ɛ f 0 for ɛ > µ 0 = ɛ f, For high temperature, a relatively large number of states will become thermally accessible, which means that the total number of particles N will be spread over many different states and n. In this case, the nature of the particle boson or fermion does not really play a role in the distribution of the particles over the single-particle states and all three distribution functions should coincide and follow the Maxwell-Boltzmann statistics. If n, then [βɛ µ)±, which gives [βɛ µ) and we can therefore approximate the occupancy n for both bosons and fermions by n [ βɛ µ) = [βµ [ βɛ. 24) To convert this into the Boltzmann distribution, we need to have an ression for µ which we again can obtain from N = n [βµ [ βɛ. 25), 7
Combing both ressions results in [ βɛ n N [ βɛ = N [ βɛ, 26) q in which we can recognize the occupancy according the Boltzmann distribution in the microcanonical ensemble. We can conclude that the distinction between bosons and fermions disappears at a certain temperature and the distribution follows that Boltzmann statistics that we have used so far. This situation changes for high density, where µ becomes very large. An example is for conduction electrons in metals which we will discuss next. Electrons in a conductor As an example of an ideal fermion gas, we consider the conduction electrons in a metal, an atomic crystal, for instance sodium Na). These conduction electrons can be described reasonably well by free electrons, even in the high densities that can be found in crystals. In this approximation, the single-particle energies are see lecture series Condensed Matter) ɛ = h2 k 2 2m with k = ˆxn,x + ŷn,y + ẑn,z ) π L, 27) where L 3 = V is the volume of the crystal. Since we consider electrons which are fermions, the maximum number of electrons per state is two. Again we use the boundary condition of N = n to find an estimate for the chemical potential. This is very similar to what we have done for the ideal boson gas, but we now need to consider the Fermi-Dirac function, which is not as easily converted into another ression. Again we replace the sum by integrals, since a crystal of appreciable size V the energy levels are close. We further need to consider a factor of 2 for the two spin states. The total number particles does becomes N = n = 2 0 π ) 3 2 L 8 0 0 dn x dn y dn z nn x, n y, n z ) F-D = dk x dk y dk z nk x, k y, k z ) F-D = 2V 2π) 3 d 3 knk) F-D. The Fermi-Dirac distribution function results in complicated integrals and we will restrict ourselves to the limit at T = 0 K. At this temperature, the Fermi-Dirac distribution becomes { for k kf nk, T = 0) = 28) 0 voor k > k f where k f is the k-value that corresponds to ɛ f, the Fermi energy. This is determined by the total number of fermions. If we consider k-space, which contains all possible k-vectors according to k = ˆxn,x + ŷn,y + ẑn,z ) π L, then all values k k f are within a sphere with radius k f. At T = 0 K, only these values contribute to the distribution function N = 2V 2π) 3 kf 0 θ φ k 2 sin θdkdθdφ = The value of k f and hence the value of ɛ f depends on the electron density: k f = 3π 2 N ) 3 V 2V 4π 2π) 3 3 k3 f = V 3π 2 k3 f. 29) with ɛ f = h2 k 2 f 2m = µ 0. 30) Taking Na with a lattice constant of 4.23 Å as an example, we find N V = 4.23 0 0 ) 3 m 3 which corresponds to a Fermi wave vector of k f = 7.3 0 9 m. The temperature that corresponds 8
to the Fermi energy is T f = ɛ f k B = h2 kf 2 2mk B 24000 K if we use the rest mass of the electron for m. This high temperature is a direct consequence of the high electron density and it means that at room temperature, the Fermi-Dirac statistics dominates the thermodynamic properties of the system. nε ) T=0 µ 0 2k B T T=0 ε 6 Density of states So far we have used the Fermi-Dirac and Bose-Einstein distribution functions n = [βɛ µ) ± 3) by taking the sum or the integral over the energy levels. For main cases, it is more convenient to work with nɛ) = 32) [βɛ µ) ± which is the average number of particles in a single-particle state with energy ɛ at a temperature T. In this case the ectation value of property A becomes A = n A dn)a) = dɛnɛ)ρɛ)aɛ), 33) where ρɛ) is the density of states, given by ρɛ) = Nɛ). 34) ɛ Please notice that Nɛ) is the number of states at energy ɛ and is therefore different from nɛ), the distribution function. In equation 33) nɛ) is the distribution function that describes the probability that a particle is in the state with energy ɛ, while in equation 34) ρɛ) is the number of states that can be occupied at this energy. The number of spin states should be considered here, since in the derivation of the Fermi-Dirac distribution function this was not considered and only a maximum of one particle could occupy any energy level. Hence for electrons we need to correct this number with a factor of two. 9