8. Bosonic systems When the temperature is high or the density low, the n¯0 g3/2(z) term V is negligible as compared with λ3 , i.e. 8.1. Bose condensate T ¯ Number of particles N 3 λT = g3/2(z). The average number of particles is V Now g (z) is a positive monotonically increasing ∂Ω 3/2 N¯ = N = function and h i − ∂µ T ¯ 1 N 3 = λT = g3/2(0) = 0 eβ(ǫl−µ) 1 V l z=0 X − ¯ N 3 or λT = g3/2(1) = ζ(3/2)=2.612. V z=1 N¯ = n¯l. l N 3 X V l T We denote βµ z = e = fugacity, 2 . 6 1 2 so 1 n¯l = . z eβǫl z−1 1 1 − N¯ Let us consider free non-interacting gas when density is Let us choose the density ρ = V ja T so that large. Then ¯ ¯h2k2 p2 N 3 ǫ = l = l , λT =2.612, l 2m 2m V and and z = 1. If we still increase the density or decrease the ¯ βǫl N 3 1 e < . temperature the increase of the term V λT must originate ≤ ∞ from n¯0 λ3 , since z 1, i.e. Because stability requires ∂N/∂µ¯ > 0, the fugacity is V T ≤ restricted to lie between N¯ 3 N¯ 3 V λT = g3/2(z), when V λT < 2.612 0 z < 1 ¯ ¯ ≤ N λ3 = n¯0 λ3 + g (1), when N λ3 2.612. V T V T 3/2 V T ≥ or µ< 0. When We treat the state p = 0 separately, since the V corresponding occupation numbern ¯ can become λ3 2.612 , 0 T ≥ N¯ macroscopic: z the state p = 0 will be occupied macroscopically forming n¯ = . 0 1 z →z→1 ∞ the Bose-Einstein condensate. The formation starts when − We write the grand potential as the temperature is less than the critical temperature βµ 2 2/3 ΩBE = kBT ln 1 e 2π¯h ρ Tc = − 2 2 βµ −β h¯ k mkB 2.612 +kBT ln 1 e e 2m . − kX6=0 h i or the density greater than the critical density Let us define function g (z) so that α 3/2 mkBT ∞ ρc =2.612 . zn 2π¯h2 g (z)= . α nα n=1 X When T <Tc, the relative fraction of the condensate is Then (homework!) n¯ 2.612 V T 3/2 0 =1 =1 . V k T N¯ − λ3 N¯ − T B T c ΩBE = kBT ln(1 z) 3 g5/2(z). − − λT n 0 N 1 2 where λT = 2π¯h /(mkBT ). For the number of particles we get q ¯ V N =n ¯0 + 3 g3/2(z). T λT T C 1 Pressure 8.2. Black body radiation (photon gas) With the help of the grand potential the pressure is The photon is a relativistic massles boson, whose spin is S =1, so g =2S + 1 = 3. In the vacuum only transverse ∂Ω k T p = = B g (z), polarisation exists, so that g = 2. − ∂V λ3 5/2 T,N T The energy of a photon is 2 2 2 so ǫ(p) = (m0c ) + (pc) = pc =¯hkc. kB T above the p 3 g5/2(z) λT critical point p = Using frequency f or angular velocity ω, the energy becomes kB T kB T below the 3 g5/2(1) = 1.342 3 λT λT critical point. ǫ =¯hω =¯h2πf = hf. Since the wavelength λ is p transition border T 1 > T 2 > T 3 2π T 1 λ = , k T 2 we have T 3 c V f = This is a 2nd order phase transition (d2Ω/dV 2 λ discontinuous) ω = ck. C V Density of states Employing the periodic boundary conditions the wave T vector is T 2π C k = (n ,n ,n ), L x y z 4He liquid so the number of states in the vicinity of k is A second order phase transition to a super liquid state at the temperature Tc =2.17K. The expression given above, L 3 dNk = g dk 2 2π 2π¯h ρ 2/3 Tc = , V mkB 2.612 = g 4πk2dk. (2π)3 results in Tc =3.13K, which is close to the experimental value. With the help of the angular velocity this is C V V ω2 dω H e I I dNk = dNω = g 4π H e I (2π)3 c2 c Vω2dω = g . 2π2c3 T T C We denote now This is called a λ-transition. dNω = f(ω) dω, Two-liquid model so When T <T , we assume that 4He is a “mixture” of two ω2 dω c f(ω) dω = V . components: superfluid and normal fluid. Then π2c3 The sum over quantum states can be replaced by the ρ = ρs + ρn integration like j = j + j s n ∞ . = = dω f(ω) . · · · · · · 0 · · · Xl Xk,λ Z When T 0, then ρs 1, but n¯0 0.1. → ρ → N¯ →∼ Here k is the wave vector and This is due to the fact that 4He is not an ideal liquid: 4 L, left between He atoms there is λ = R, right a strong repulsion at short distances, • is the polarization. an attraction at longer distances. Photons obey the Bose-Einstein statistics • 2 In field theory the photon modes are equivalent to where independent harmonic oscillators, and the Hamiltonian of ∞ ∞ x3 the system can be written as dx = dx x3 e−nx ex 1 Z0 − n=1 † 1 X Hˆ = (¯hck) a a + . 1 1 π4 kλ kλ 2 = Γ(4)(1+ + . .)=6 ζ(4) = Xk,λ 24 34 × 15 Here the eigenvalues ofn ˆ = a†a are 0,1,. ; thus, the Thus the energy density obeys the Stefan-Boltzmann law energy of individual mode is 4 e(T )= σT 4, 1 c E = (n + )¯hω = n¯hω + 0-point motion. 2 where σ is the Stefan-Boltzmann constant Because the number of photons in a photon gas is not an 2 4 independent degree of freedom (depends on E), µ does π kB −8 W σ = 3 =5.67 10 4 . not appear as a Lagrange multiplier. Thus, according to 60¯h c2 · m2K the Bose-Einstein distribution the occupation of the energy state ǫ(ω) (Planck distribution) is Thermodynamics Now 1 n¯(ω)= . Ω= F µN = F, eβhω¯ 1 − − because µ = 0. Thus the free energy is The total energy is ∞ −βhω¯ ∞ F = k T dω f(ω) ln 1 e ¯hω B − E = dω f(ω) . Z0 eβhω¯ 1 ∞ Z0 − V 2 −βhω¯ = 2 3 kB T dω ω ln 1 e energy density π c 0 − The will be Z V k T ∞ E ¯h ∞ ω3 = B dx x2 ln 1 e−x = e(T )= dω π2c3(β¯h)3 − V π2c3 eβhω¯ 1 Z0 Z0 − k4 T 4 π4 ∞ = V B , = dω e(ω,T ), − π2c3¯h3 45 0 Z or where the energy density at the given angular velocity 4 σ 1 F = VT 4 = E. obeys Planck’s law of radiation −3 c −3 Here ¯hω3 e(ω,T )= . E = e(T )V π2c3(eβhω¯ 1) − is the total energy. The entropy is e ( w ,T) ∂F S = w − ∂T m a x or T 3 16 σ 3 T 2 < S = VT . T 1 < 3 c w The pressure is ∂F We can see that the maximum of the intensity follows the p = Wien displacement law −∂V or 4 σ ωmax = constant T. p = T 4. × 3 c At the long wavelength limit, λ hc or ω kB T , the kB T h¯ We see that the photon gas satisfies the relation energy density obeys the Rayleigh-Jeans≫ law ≪ 1 e(ω,T ) = const. ω2T. p = e. × 3 At a given temperature the energy density will be Radiation of a black surface ∞ ¯hω3 e(T ) = dω π2c3(eβhω¯ 1) Z0 − ¯h 1 ∞ x3 = dx π2c3 (β¯h)4 ex 1 Z0 − k4 T 4 π4 = B , π2c3¯h3 15 3 We can think that the emitting surface is a hole on a with the lattice point R. If u(R) is the deviation of the hollow container filled with isotropic black body ion R from its equilibrium then radiation. The radiation power can be determined by counting the number of photons escaping through the rR = R + uR. hole per time interval. Let φ(r) be the potential energy of two ions separated by q the distance r. The energy of the whole lattice is then 1 U = φ(rR rR′ ) 2 ′ − l ( q ) RRX 1 ′ = φ(R R + uR uR′ ). 2 ′ − − In the time interval τ the photons escaping to the RRX direction θ originate from the region whose depth is When we use the notation PR for the momentum of the ion R the total Hamiltonian is ℓ(θ)= cτ cos θ. P2 H = R + U. The total energy of photons going into the spherical angle 2m R element dΩ is X dΩ e(T )Acτ cos θ . 4π Harmonic approximation Thus, the total energy of the radiation escaping in time τ Because u interatomic distances, we can approximate is R U as a Taylor≪ series: π/2 dΩ N Erad = e(T )Acτ cos θ U = φ(R) θ=0 4π 2 Z R 1 π/2 X = e(T )Acτ dθ sin θ cos θ 1 ′ 2 + (uR uR′ ) φ(R R ) Z0 2 − · ∇ − 1 RR′ = Ae(T )cτ. X 1 2 ′ 4 + [(uR uR′ ) ] φ(R R ) 4 ′ − · ∇ − The radiation power per unit area (intensity) is RRX + (u3).