8. Bosonic Systems 8.1. Bose Condensate

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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

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 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). E 1 O I = rad = ce(T ) Aτ 4 In the equilibrium the total force due to other ions = σT 4. affecting the ion R is ′ If the radiation arrives from a fixed direction (e.g. solar F = φ(R R ). − ′ ∇ − radiation) with intensity I, then the absorbed radiation XR power is naturally Since we are at a equilibrium this force must be zero. P = I A⊥ Thus the linear term in the series expansion of U vanishes. Up to the second order we are left with where A⊥ is the projected area perpendicular to the radiation. U = U eq + U harm, 8.3. Phonons where U eq is the potential energy of the equilibrium and

harm 1 ′ Classical harmonic lattice U = [u u ′ ]φ (R R ) 4 R,µ − R ,µ µν − We let the ions of a crystal to oscillate in the vicinity of ′ RRX their equilibrium position. We assume that µ,ν=x,y,z [u u ′ ] × R,ν − R ,ν 1. At the average equilibrium position the crystal is a ∂2φ(r) Bravais lattice. With every point R of the lattice we φµν (r) = . ∂rµ∂rν can thus associate an atom. The vector R, however, reperents only the average position of the ion. If we are not interested in the quantities related to the equilibrium of the crystal ( total energy, total volume, 2. Typical deviations from the equilibrium positions are total compessibility, . . .) we can forget the term U eq. The small as compared with the interatomic distances. harmonic potential is usually written more generally as

According to the hypothesis 1 the atoms of the crystal harm 1 ′ U = u D (R R )u ′ . can be identiﬁed with the Bravais lattice points R; e.g. 2 R,µ µν − R ,ν RR′ r(R) stands for the actual position of the ion associated Xµν

4 The former expression can be obtained by setting even at higher temperatures the measured heat • capacities do not approach the Dulong-Petit limit. ′ ′′ ′ Dµν (R R )= δRR′ φµν (R R ) φµν (R R ). − − − − R′′ X Einstein’s model Let us assume that every ion of the crystal moves in a The heat capacity of classical lattice similar harmonic potential well indpendently of each The classical ideal gas is gives a good description for a other. Then we can quantize the harmonic oscillators: real gases at high enough temperatures. However, the † 1 classical description for solid crystal heat capacity is not H = ¯hωE(a a + ), xs xs 2 xs well described by classical dynamics. X The volume element of the the 3N dimensional classical where the parameter phase space formed by the N ions of the lattice is kBTE 1 1 ωE dΓ= duR dPR = duR,µdPR,µ ≡ ¯h h h R R,µ Y Y is the Einstein frequency common for all 3N oscillators and the canonical partition sum is and TE the corresponding Einstein temperature. The partition function of one single harmonic oscillator is −βH Z = dΓ e . ∞ −βhω¯ (a†a+ 1 ) −βhω¯ (n+ 1 ) Z Zharm(ω) = Tr e 2 = e 2 n=0 The total energy E is then X − 1 βhω¯ − 1 βhω¯ −βhω¯ n e 2 1 ∂ = e 2 e = E = dΓ e−βH H = ln Z. 1 e−βhω¯ n Z −∂β X − Z 1 When we change variables, = 1 . 2 sinh( 2 β¯hω) −1/2 uR = β u¯R Since the number of modes is 3N the canonical partition −1/2 PR = β P¯ R, function is

∞ ∞ ∞ 3N 1 the partion function can be written as −βhω¯ E (nj + ) Z = e j=1 2 · · · 2 n1=0 n2=0 n3N =1 PR eq harm X X X P Z = dΓexp β + U + U 3N ∞ − 2M −βhω¯ (n+ 1 ) 3N Z E 2 X = e = Zharm(ωE) −βUeq −3N 1 = e β du¯ dP¯ j=1 n=0 h R R × Y X Z R −3N Y TE ¯ 2 = 2 sinh . PR 1 ′ 2T exp u¯R,µDµν (R R )¯uR′,ν "− 2M − 2 − # X X The heat capacity is Because all dependence on the temperature is outside of ∂E ∂ ∂ ∂ ∂ ln Z the integral the energy can be calculated easily C = = ln Z = k T 2 V ∂T −∂T ∂β ∂T B ∂T 2 2 ∂ −βUeq −3N ∂ (T /2T ) E = ln(e β const.) = k T T ln Z =3Nk E . −∂β × B 2 B 2 ∂T sinh (TE/2T ) eq = U +3NkBT.

C V The heat capacity is 3 N k B D ulong-Petit 1 experim ental ∂E C = =3Nk . E i n s t e i n v ∂T B

This expression for heat capacity, due to the lattice T vibrations, is known as Dulong-Petit’s law. Experimentally Normal modes of the harmonic crystal One-dimensional Bravais lattice at low temperatures the heat capacity is smaller than If the separation of the lattice points in the one • the one obtained from the Dulong-Petit law. When dimensional Bravais lattice is a the lattice points are na, we approach the temperature T = 0 the heat n an integer. Every lattice point na is associated with capacity tends to zero, due to quantum eﬀects. one atom.

5 We assume that in this one dimensional lattice only the where K describes the interaction of the ions na and nearest neighbours interact. Using the notation na + d, and G the interaction of na + d and (n + 1)a. The classical equations of motion are K = φ′′(x), ∂U harm the harmonic potential of the lattice is Mu¨1(na) = −∂u1(na) 1 = K[u1(na) u2(na)] U harm = K [u(na) u((n + 1)a)]2. − − 2 − G[u (na) u ((n 1)a)] n − 1 − 2 − X ∂U harm The classical equations of motion are Mu¨2(na) = −∂u2(na) ∂U harm = K[u (na) u (na)] Mu¨(na)= − 2 − 1 − ∂u(na) G[u (na) u ((n + 1)a)]. − 2 − 1 = K[2u(na) u((n 1)a) u((n + 1)a)]. − − − − Again we look for a solution of the form We assume that the N points of the lattice form a ring, u (na) = ǫ ei(kna−ωt) i.e. the deviations satisfy the boundary conditions 1 1 i(kna−ωt) u2(na) = ǫ2e . u((N + 1)a)= u(a); u(0) = u(Na). Substituting these into the equations of motion we end up We seek solutions of the form with the linear homogenous simultaneous equations 2 −ika u(na,t) ei(kna−ωt) [Mω (K + G)]ǫ1 + (K + Ge )ǫ2 = 0 ∝ − ika 2 (K + Ge )ǫ1 + [Mω (K + G)]ǫ2 = 0. To satisfy the boundary conditions we must have − This system has a non-trivial solution only if the ikNa e =1. coeﬃcient determinant vanishes. From this we obtain K + G 1 We see that the allowed values for k are ω2 = K2 + G2 +2KG cos ka. M ± M 2π n k = , n integer. p a N The ratio of the amplitudes is ika Substituting the exponential trial into the equation of ǫ1 K + Ge = ika . motion we see that the angular velocity ω must satisfy ǫ2 ∓ K + Ge | | 2K(1 cos ka) K 1 For every allowed wave vector k (counting N) we get two ω(k)= − =2 sin ka . solutions. Altogether the number of the normal modes is M M | 2 | r r now 2N. We consider couple of limiting cases. The solutions represent a wave advancing in the ring with Case 1. k π/a the phase velocity c = ω/k and with the group velocity The angular≪ velocities of the modes are now v = ∂ω/∂k. Clearly, ω(k) is invertible only if π

′ ′ ǫ = ǫ . Dµν (R R )= Dνµ(R R ), 1 ∓ 2 − − Case 3. K G so the matrix D is symmetric. 3. R D(R)=0 The dispersion≫ relations are now We move every ion R to R + d. This is equivalent with translating the whole lattice by theP amount d. The 2K G potential energies of the original and the translated ω = 1+ M O K lattices are equal; in particular at the equilibrium 0, i.e. r 2G 1 G 0 = d D (R R′)d ω = sin ka 1+ , µ µν − ν r M 2 O K RR′ Xµν and the amplitudes satisfy = Ndµdν Dµν (R) . ǫ ǫ . µν R ! 1 ≈∓ 2 X X The frequency of the optical branch is now independent Since the vector d is arbitrary we must have on the wave vector. Its magnitude corresponds to the vibration frequency of a molecule of two atoms with equal D(R)= . masses and coupled with the spring constant K. XR On the other hand, the acoustical branch is the same as The classical equations of motion in the case of the linear chain. harm Case 4. K = G ∂U ′ Now we have a Bravais lattice formed by single atoms Mu¨R,µ = = Dµν (R R )uR′,ν , − ∂uR,µ − ′ − with the primitive cell length a/2. XR ν or in the matrix notation Extra: Three dimensional Bravais lattice of single atoms ′ Using the matrix notation the harmonic potential can be Mu¨ = D(R R )u ′ R − − R written more compactly R′ X

harm 1 ′ form a system of 3N equations. Again we seek solutions U = u D(R R )u ′ . 2 R − R of the form RR′ i(k·R−ωt) X uR = ǫe . Independent on the interionic forces the matrix Here the polarisation vector ǫ tells us the direction of the ′ D(R R ) obeys certain symmetries: motion of the ions. Furthermore we require that for every − ′ ′ 1. Dµν (R R )= Dνµ(R R) a − − primitive vector i the solutions satisfy the Born-von This property can be verified by exchanging the order of Karman boundary conditions differentiations in the definitions of the elements of D: u R a u R 2 ( + Ni i)= ( ), ′ ∂ U Dµν (R R )= . − ∂u (R)∂u (R′) when the total number of primitive cells is N = N N N . µ ν u= 1 2 3 These conditions can be satisfied only if the wave vector

2. D(R)= D( R) k is of form − Let us consider a lattice where the displacements from the n n n equilibrium are u . In the corresponding reversal lattice k = 1 b + 2 b + 3 b . R N 1 N 2 N 3 the displacements are u . Since every Bravais lattice 1 2 3 − −R has the inversion symmetry the energies of both lattices Here bi are vectors in the reciprocal lattice and ni must be equal, no matter what the deviations uR are, i.e. integers. We see that we get diﬀerent solution only if k is restricted into the 1st Brillouin zone, i.e. there are harm 1 ′ U = u D(R R )u ′ exactly N allowed values for the wave vector. 2 R − R ′ We substitute the trial into the equations of motion and RRX 1 ′ end up with = ( u )D(R R )( u ′ ) 2 2 − −R − − −R Mω ǫ = D(k)ǫ, ( ) RR′ ∗ X where 1 ′ = u D(R R)u ′ , −ik·R 2 R − R D(k)= D(R)e RR′ X XR for an arbitrary uR. This can be valid only if is so called dynamical matrix. For every allowed k we have as the solution of ( ) three eigenvalues and vectors. D(R R′)= D(R′ R). The number of normal modes∗ is therefore 3N. − −

7 Employing symmetry properties of D(R) we can rewrite primitive cell. Every ion in the primitive cell adds one the dynamical matrix as degree of freedom so the total number of modes at a given wave vector k is 3p. The corresponding frequences 1 −ik·R ik·R i D(k) = D(R)[e + e 2] are ωs(k), where now s =1, 2, 3 and i =1, 2,...,p. The 2 − corresponding displacements are XR i = D(R)[cos(k R) 1]. i i i(k·R−ωs(k)t) · − us(R,t)= ǫs(k)e . XR The polarizations are no more orthogonal but satisfy Thus the dynamical matrix is p i ∗ i 2 1 ǫ (k) ǫ ′ (k)= δ ′ . D(k)= 2 D(R) sin ( k R). s · s ss − 2 · i=1 XR X Analogically with one dimensional lattice 3 of the modes We see that D(k) is a real and symmetric function of k. are now acoustical and the rest 3(p 1) modes optical. Since D(R) is symmetric D(k) is also symmetric. We − rewrite the equation ( ) as ∗ Extra: Quantum mechanical treatment Let us consider the harmonic Hamiltonian D(k)ǫs(k)= λs(k)ǫs(k). harm 1 2 H = PR As the eigenvalues of a real and symmetric matrix λs(k) 2M R are real and the eigenvectors ǫs(k) can be X 1 ′ orthonormalized, i.e. + uRD(R R )uR′ 2 ′ − ′ RR ǫ (k) ǫ ′ (k)= δ ′ , s,s =1, 2, 3. X s · s ss describing the lattice. Let ωs(k) and ǫs(k) be the The polarizations of three normal modes are ǫs(k) and frequences and polarizations in the corresponding the angular velocities correspondingly classical lattice. We deﬁne the operator aks so that

λ (k) 1 −ik·R ω (k)= s . aks = e ǫs(k) s M √N · r XR Let us assume now that the mutual interaction of the ions Mω (k) 1 s u P decreases rapidly with the increasing separation. Strictly R + i R . "r 2¯h s2¯hMωs(k) # speaking we assume that The Hermitean conjugate a† of the operator a is lim D(R)= (R−5). ks ks R→∞ O † 1 ik·R a = e ǫs(k) Then, at long wavelength, i.e. when k , we have ks √N · ≈ XR 1 1 sin2( k R) ( k R)2 Mωs(k) 1 2 · ≈ 2 · uR i PR . "r 2¯h − s2¯hMωs(k) # and k2 The operator a† is called the phonon creation operator D(k) (kˆ R)2D(R). ks ≈− 2 · and a the phonon annihilation operator. R ks X We employ the canonical commutation relations for the ˆ 2 Let cs(k) be the eigenvalues of the matrix position and momentum 1 (kˆ R)2D(R). [uR,µ, PR′,ν ] = i¯hδµν δRR′ −2M · R [uR,µ,uR′,ν ] = [PR,µ, PR′,ν ]=0, X We see that at small wave vectors the frequency is the identities

ˆ ik·R 0, k is not a reciprocal vector ωs(k)= cs(k)k. e = N, k is a reciprocal vector R Thus the dispersion of all three modes is a linear function X of k so all three modes are acoustical. In general cs(kˆ), and ˆ eik·R =0, R =0 together with ωs(k), depend also on the direction k of the 6 propagation in addition to the mode s. Xk together with the property of an orthogonal vector set Extra: Three dimensional lattice with base 3 We proceed exactly like in the case the one dimensional [ǫs(k)]µ[ǫs(k)]ν = δµν . lattice with base. We assume that there are p ions in the s=1 X 8 One can easily show that the creation and annihilation We cut the spectra at the Debye frequency operators obey the commutation relations • kBTD † ′ ωD , [aks,ak′s′ ] = δkk′ δss ≡ ¯h † † [a ,a ′ ′ ] = [a ,a ]=0. ks k s ks k′s′ where TD is the corresponding Debye temperature, With the help of the creation and annihilation operators and which is chosen so that the number of modes equals the number of modes within the ﬁrst Brillouin the operators uR and PR can be written as zone.

1 ¯h † uR = (aks + a−ks) In each mode j the density of states is √N 2Mωs(k) × ks s X 3 ik·R L V ǫs(k)e 2 2 dNj = 4πk dk = 2 3 ω dω. 2π 2π cj i ¯hMωs(k) † PR = − (aks a−ks) √N r 2 − × Thus the total density is Xks ik·R ǫs(k)e . V 2 1 dN = + ω2dω. 2π2 c3 c3 The Hamiltonian is now t l 1 H = ¯hω (k)(a† a + ). Because the total number of modes is s ks ks 2 ks ωD V 2 1 X 3N = dN = + ω3 This is simply the Hamiltonian of the system of 3N 6π2 c3 c3 D Zω=0 t l independent harmonic oscillators whose energies are correspondingly we get as the Debye frequency 1 N 2 1 −1 E = (nks + )¯hωs(k). ω3 = 18π2 + 2 D V c3 c3 Xks t l Here nks the eigenvalues of the occupation number † and correspondingly as the density of states operatorn ˆks = aksaks, i.e. nks =0, 1, 2,.... 9N 2 Debye’s model dN(ω)= 3 ω dω (ω<ωD). ωD It is always possible to diagonalize a harmonic Hamiltonian (D symmetric). This is most convenient in The canonical partition function is momentum space, where diﬀerent lattice momenta k ∞ ∞ ∞ 1 become indepedent. Harmonic potential gives rise to −β hω¯ s(k)(n + ) Z = e ks ks 2 harmonic oscillators, and in order to get the exact · · · n =0 n =0 n =0 solution we should evaluate the partition function X1 X2 3XN P − 1 βhω¯ (k) e 2 s † 1 −β hω¯ s(k)(a a + ) = , Z = Tr e ks ks ks 2 1 e−βhω¯ s(k) ∞ ks − 1 Y P−β hω¯ s(k)(n + ) ks ks 2 = e . from which we can derive as the free energy {n =0} Xks P 1 −βhω¯ s(k) The k-summation goes over all lattice momenta, F = ¯hωs(k) +kBT ln 1 e 3 2 − k n ks ks = 2π/(Na), where we assume lattice size V = (aN) X X h i and periodic boundary conditions. 0-point energy In the ﬁrst Brillouin zone there are 3N modes, n =0 . . . (N 1). or | {z } i − The solution requires the knowledge of the dispersion ∞ 9N 2 −βhω¯ relation ω (k). A reasonable realistic and analytical F = F0 + kBT dω ω ln 1 e . s ω3 − approximation is provided by the Debye’s model: D Z0 ∂F ∂S At low temperatures only the contribution of the low Because S = ∂T and CV = T ∂T , we have • 2 − energetic phonons is prominent, so we take into ∂ F CV = T ∂T 2 , and we obtain account only the acoustic modes: 2 tranverse and 1 − longitudinal polarisations. T C =3Nk f D . V B D T we assume that the small k dispersion relation holds: • ω (k) = c k Here l l 3 x y4ey f (x)= dy ωt(k) = ctk. D x3 (ey 1)2 Z0 −

9 is the so-called Debye function. Typical Debye 9. Fermion systems temperatures TD 9.1. Electron gas Au 170 The ideal Fermi gas is a good approximation for example Cu 315 for the conducting electrons in a metal. Fe 420 h¯2k2 When the single particle energies are ǫk = the Cr 460 2m density of 1-particle states is (gs: spin degeneracy) B 1250 C (diamond) 1860 g 2m 3/2 s √ ω1(ǫ) = V 2 2 ǫ Note: The higher TD the stiﬀer, harder crystal. 4π ¯h Behaviour of C : 3/2 V 2m T = V 2πg √ǫ. → ∞ s h2 Because x 3 2 fD(x) dy y =1, The density of particles is x→→0 x3 Z0 we have N¯ g 2m 3/2 ∞ √ǫ ρ = = s dǫ . CV 3NkB, V 4π2 ¯h2 eβ(ǫ−µ) +1 → Z0 and we end up with the Dulong-Petit heat capacity. The energy per particle will be T 0 → Because ∞ ǫ3/2 E 0 dǫ eβ(ǫ−µ) +1 ∞ 4 y ǫ¯ = = . 3 y e constant N¯ ∞ ǫ1/2 fD(x) dy = , R0 dǫ β(ǫ−µ) x→∞ 3 y 2 3 e +1 → x 0 (e 1) x Z − R we get Degenerate Fermi gas 4 3 Assume that kB T µ. Let us write the average 3 12π T ≪ C (T ) const. T = Nk . occupationn ¯ as a step function + correction: V → × 5 B T D 1 = θ(µ ǫ)+ h(ǫ µ), C V β(ǫ−µ) D e b y e e +1 − − where 1 a T h(x) = sign(x) . e l e c t r o n i c eβ|x| +1 contribution T a T 3 The function h(x) deviates from zero only at the narrow < Contribution from the electron gas is linear in T , but domain x kBT µ. | |∼ ≪ tiny, and usually negligible. h ( x )

x = e - m

Let’s evaluate the integral

∞ φ(ǫ) dǫ β(ǫ−µ) 0 e +1 Z ∞ = dǫ φ(ǫ) [θ(µ ǫ)+ h(ǫ µ)] 0 − − Z µ ∞ = dǫ φ(ǫ)+ dǫ h(ǫ) [φ(µ + ǫ) φ(µ ǫ)] 0 0 − − Z ∞ Z + dǫ h(ǫ)φ(µ ǫ). − Zµ The last term is of the order 1 h(µ)= e−µ/kB T eµ/kB T +1 ≈ and can be discarded.

10 If φ(ǫ) regular enough in the vicinity of ǫ µ we can The Fermi momentum is expand ≈ pF =¯hkF . 1 φ(µ + ǫ) φ(µ ǫ) 2φ′(µ)ǫ +2 φ′′′(µ)ǫ3 + . − − ≈ 3! · · · The density is

Now g 2m 3/2 µ ρ = s dǫ √ǫ ∞ φ(ǫ) 4π2 ¯h2 dǫ Z0 β(ǫ−µ) 3/2 0 e +1 gs 2mǫF Z µ = 6π2 ¯h2 dǫ φ(ǫ) ≈ 0 Z ∞ or ′ 2 z gs +2φ (µ)(kB T ) dz ρ = k3 . ez +1 2 F Z0 6π 1 ∞ z3 ′′′ 4 The spin degeneracy factor of electrons is +2φ (µ) (kB T ) dz z 3! 0 e +1 1 Z gs =2 2 +1=2, so + · · · · k3 and we end up with Sommerfeld’s expansion ρ = F . 3π2 ∞ φ(ǫ) µ π2 dǫ dǫ φ(ǫ)+ (k T )2φ′(µ) eβ(ǫ−µ) +1 ≈ 6 B For the average energy per particle we get Z0 Z0 4 1 7π 4 ′′′ dx x3/2 + (kB T ) φ (µ)+ . 0 2/5 360 ǫ¯ = ǫF = ǫF · · · 1 1/2 2/3 R0 dx x This was obtained again using Riemann ζ-function: 3 = ǫFR . ∞ 5 1 ∞ ux−1 1 ζ(x)= u du = x Γ(x) 0 e 1 k The total energy is Z − Xk=1 3 3 ¯h2 6π2ρ 2/3 (Check E = ǫ N = N . 5 F 5 2m g http://mathworld.wolfram.com/RiemannZetaFunction.html) s Now 1 1 2 Since = , 5/3 −2/3 eu +1 eu 1 − e2u 1 E = constant N V , − − × and thus we have ∂E 2 E ∞ x−1 p = = 1 u 1−x du = ζ(x) 2 ζ(x). − ∂V N −3 V Γ(x) eu +1 − Z0 or 2 Now pV = E, 3 ζ(1) = which is actually valid at all temperatures (homework). ∞ π2 Metallic electron gas ζ(2) = 6 When we write the density as ζ(3) = 1.2020 . . . N¯ 1 π4 ρ = = ζ(4) = V 4 πr3 90 3 i ζ(5) = 1.0369 . . . and deﬁne the dimensionles number π6 ζ(6) = ri 945 rs = , a0 T Temperature = 0 where a0 is the Bohr radius Now 2 n¯(ǫ)= θ(µ ǫ) 4πǫ0¯h − a0 = =0.529A˚, me2 and h(x)=0. The Fermi energy is we can see that

2 2 1.613 1030 1 ¯h kF ǫF = µ = . ρ = 3· 3 . 2m rs m

11 For metals we have In order to obtain the energy, we use

< < ∞ 3/2 1.9 rs 5.6. ǫ ∼ ∼ dǫ eβ(ǫ−µ) +1 The Fermi wave number will become 0 Z 2 2 5/2 π 2 µ + (kB T ) √µ + 1 3 9π 1.92 kF = = . ≈ 5 4 · · · a r 4 a r 2 0 s r 0 s 2 5 k T = ǫ5/2 1+ π2 B + . The Fermi velocity is 5 F 12 ǫ · · · " F # pF ¯hkF 1.92¯h vF = = = Now the energy/particle is m m ma0rs 4.2 106 m ∞ ǫ3/2 dǫ β(ǫ−µ) = · . ¯ǫ(T ) = 0 e +1 rs s ∞ ǫ1/2 R0 dǫ eβ(ǫ−µ) +1 For example in aluminium 2 R3 5 2 kBT km c = ǫF 1+ π v = 2029 = . 5 " 12 ǫF # F s 148 2 2 2 3 π kB T The Fermi temperature or the degeneracy temperarure T = ǫF + . F 5 4 ǫ is deﬁned so that F kB TF = ǫF . The heat capacity can now be written as

Now 2 2 2 ∂Nǫ¯ π kB π T 2 2 CV = = N T = NkB . ¯h 1.92 3.69 ∂T 2 ǫF 2 TF ǫF = 2 = 2 13.6eV. 2ma0 rs rs This is small when compared e.g. with the speciﬁc heat of binding the Maxwell-Boltzmann gas (C = Nk 3 ), or the Debye energy of V B 2 | {z } heat capacity, because T T . This is understandable hydrogen D F because the number of those≪ particles that can be excited Because with the thermal energy kB T is MB or BE gases 1eV = 11604k K, due to the Pauli exclusion∼ principle.≪ At T few K the B ∼ electronic CV dominates. we have el. Debye 3 1.92 2 Why CV T but CV T ? T = 13.6 11604K. This is due∝ to the Fermi surface:∝ consider the number of F r · s states in k-phase space around the T = 0 state, where For aluminium the Fermi temperature is TF = 136000K. ∆k kBT . For phonons the ﬂuctuations are around In general, the metals satisfy k =∼ 0, with number of states

T TF , ∆N (∆k)3 T 3. ≪ ∼ ∝ so the metallic electron gas is highly degenerate. For electrons the T = 0 state is a Fermi sphere, with Speciﬁc heat ∆N k2 ∆k T. Let now T > 0, but T TF . We can now solve µ = µ(T ) ∼ F ∝ N¯ ≪ by requiring that V = ρ remains constant when we vary the temperature. With the help of the Sommerfeld Pauli’s paramagnetism expansion we get The magnetic moment of the electron is

3/2 e 2 4π2 ¯h2 ∞ ǫ1/2 µ = s ǫ3/2 = ρ = dǫ −m 3 F g 2m eβ(ǫ−µ) +1 s Z0 or 2 2 3/2 π 2 1 µz = µBσz, = µ + (kB T ) + 3 12 √µ · · · − where e¯h eV so that we can write µ = =5.66 10−5 B 2m · T 2 π2 1 2 µ3/2 1+ (k T )2 + = ǫ3/2. and 3 8 B ǫ2 · · · 3 F 2 F σ = s = 1. z ¯h z ± From this we get for the chemical potential the expression In an external magnetic ﬁeld the energy of an electron is 2 2 π kB T 2 µ(T )= ǫF 1 + . p − 12 ǫ · · · ǫpσz = ǫp± = µzB = ǫp + µBBσz " F # 2m −

12 when the kinetic energy is From this we get

2 p mµB ǫp = . δkF = 2 B 2m −¯h kF We treat electrons as non-interacting so that the grand and canonical partition function is as before, provided that we 3 2 replace ǫp ǫp + µBBσz. The occupation numbers of kF kF → ρ± = δkF the states are now 6π2 ± 2π2 k3 k mµ 1 = F F B B. 2 2 2 n¯pσz =n ¯p± = . 6π ∓ 2π ¯h eβ(ǫp+µB Bσz −µ) +1 The relative polarization is

n p s ρ ρ 3mµ r + − = B B n p - − 2 2 ≡ ρ+ + ρ− − ¯h kF n 3µ p + = B B. − 2ǫF e p Because the metallic electron gas is strongly degenerate The magnetization per volume element is (T T ), we can restrict the temperature T = 0. ≪ F N e k M = µz = ρµB σz = ρµBr e k - e k + V h i − h i −

e F = m or 3 µ2 M = ρ B B. 2 ǫF k k F- k F + The susceptibility is, according to its deﬁnition, The Fermi wave vectors can be determined from the ∂M ∂M conditions χ = = µ . ∂H 0 ∂B ¯h2k2 F + + µ B = µ 2m B Pauli’s paramagnetic susceptibility is then ¯h2k2 F − 2 µBB = µ. 3 µB 2m − χ = µ0ρ 2 ǫF Because the number density is

g provided that T TF and µBB ǫF . ρ = s k3 , ≪ ≪ 6π2 F In aluminium the electron density is the spin population densities are ρ =1.82 1029m−3 · k3 ρ = F + and the Fermi energy + 6π2 3 2 kF − 1.92 ρ− = 2 . ǫ = 13.6eV = 11.7eV. 6π F 2.07 If the strength of the magnetic ﬁeld is The susceptibility ǫF B0 = , −5 2 µB 3 (5.66 10 ) χ = 4π 10−7 1.82 1029 · the magnetic energy is of the same order as the Fermi 2 · · · · · 11.7 5 Vs 1 (eV)2 energy. For metals ǫF 5eV, so B0 10 T. Thus the ≈ ≈ 3 2 realistic magnetic ﬁelds are B0 and we can work at the Am m T eV ≪ 2 small B limit. Denoting eVVs m2 = 9.4 1013 · Am4 Vs kF ± = kF δkF , ± = 9.4 1013 1.6 10−19 · · · so = 1.5 10−5 2 2 2 2 2 · ¯h kF ± ¯h kF ¯h kF µB B = δkF µB B is small because only the electrons very close to the Fermi 2m ± 2m ± m ± surface can be polarized magnetically. ¯h2k2 = µ = F . 2m

13 9.2. Two dimensional electron gas Confine the system in a rectangular cell The Hamiltonian for a free electron in the magnetic field z 6 B = A ∇× - y is given by B © 1 e 2 x = i¯h + A . 6 H0 2m∗ − ∇ c Lx In a magnetic field electrons move in circular (or more generally spiralling) motion, with angular frequency, - A cyclotron frequency © ω = eB/m∗c. c - Ly The size of the cycles is

1 2 ℓ0 = (¯hc/eB) , Using periodic boundary conditions we have the magnetic length.

Consider electrons 2πny ky = , ny =0, 1, 2,... Ly ± ± conﬁned to xy-plane. • and 2πny 2 subjected to a perpendicular magnetic ﬁeld B zˆ. X = ℓ0, 0 X

We can choose the gauge The number of allowed values of ny, i.e. the degeneracy of each Landau level, is A = (0, xB, 0) LxLy e Φ Ns = 2 = Φ = , and the single particle Hamiltonian is 2πℓ0 hc Φ0

2 where Φ0 = hc/e is the flux quantum. Each electron 2 1 2 eB within Landau level takes up area 2πℓ . Thus, on each 0 = p + py + x . 0 H 2m∗ x c Landau level there is exactly one state for each flux " # quantum and for each spin polarization. Making the ansatz φ = eiky yψ(x), we see that the When Ne is the number of electrons in an area and Ns equation for ψ becomes that of a harmonic oscillator: the number of flux quanta we define the filling fraction as

2 2 Ne n0 T 1 2 e B 2 ν = =4.136 15 −2 . p + (x X) ψ(x)= Eψ(x), Ns 10 m B 2m∗ x c2 − To treat the spin we note that where there should be the Zeeman coupling term ¯hc 2 • X = ky = kyℓ0. − eB − = µ B = gµ Bs HZeeman · − B z Thus, the eigenenergies are the discrete Landau levels in the Hamiltonian. Here g is the Lande factor and 1 µB the Bohr magneton. E = n + ¯hω , n =0, 1, 2,.... n 2 c in addition to the Zeeman term there are no spin • dependent terms in the Hamiltonian, not even in the ∗ with ωc = eB/m c. interacting many body system. The full eigenfunctions are the problem can be solved disregarding the spin. At • 2 2 later stages we can add the total Zeeman energy iky y −(x−X) /2ℓ x X φnX = e e 0 Hn − , ℓ0 EZeeman = gµBBSz. where X gives the center for the oscillatory motion.

14 Quantum Hall states Consider an experiment like

j x B

+ −

+ Ey - − + V − L

+ −

We observe that E x the Hall resistivity develops plateaus with • Ey h V ρxy = = 2 , n =1, 2, 3,.... H jx ne This quantization condition is obeyed with extreme accuracy. In fact, the current ISO standard for resistivity defines 25812.807 ρ = Ω. xy n at the same time the diagonal resistivity practically • The conductivity σ and the resistivity ρ are defined by vanishes. j = σE, E = ρj. Classically the diagonal and Hall For the moment we assume that the electrons are conductivities are polarized. If the current carrying electrons fill up exactly 2 n Landau levels, it can be shown that ρxy = h/ne and ρxx = 0. The plateaus can be explained by noting that in an ideal pure 2DEG the density of states is a • series of δ-peaks separated byhω ¯ c. In real systems there are impurities and the δ-peaks j n e2τ 1 x 0 • are spread and between the Landau levels there are σxx = 2 ≡ Ex m 1 + (ωcτ) localized states, which do not contribute to jx n0ec σxx conductivity. σxy = + , ≡ Ey − B ωcτ

h w c

h w c D(E) D(E)

E Filled Empty E E F where τ is the relaxation time. In particular ρ = B/n ec. Experimentally the resistivities behave When B is changed, Landau levels shift: when a xy − 0 • like (smeared) level crosses ǫF , number of delocalized

15 (Hall conducting) electrons changes rapidly: jump in the plateaus correspond to full Landau levels, conductivity. • the Landau levels are energetically far from each When ǫ is between the Landau levels, little change • other as compared to typical electron-electron • F in number of conducting electrons: Hall resistivity is interaction energies (at least when ν <5). constant. ∼ the mutual electronic interactions play practically no The Hall conductivity can be written in the form • role.

n0ec σxy = + ∆σxy, While this single particle picture is suﬃcient in the IQHE − B it cannot explain the FQHE where where, according to the Kubo formula, the contribution the Landau levels are only partially ﬁlled, so that from a localized state α to ∆σxy is • | i there is room for the Coulomb intra level interaction. f(E )ec • ∆σα = α . xy B It turns out that the correlations due to the electron interaction are essential in the FQHE. Here f(E) is the Fermi distribution function. When the number of electrons changes we observe (at T = 0) that 9.3. Relativistic electron gas The rest energy of an electron is as long as the Fermi level lies within the localized • mc2 =0.511keV states, σxy remains constant. if all states below the Fermi level are localized, the and the relativistic total energy • terms in σxy cancel exactly and σxy = 0. 2 2 2 ǫp = (mc ) + (cp) for QHE to exist there must be extended states in p2 • Landau levels. = mcp 2 + + . 2m · · ·

Fractional quantum Hall effect Denote by mc 12 −1 Increasing the magnetic field (i.e. reducing the electron kc = =2.59 10 m density) furthermore one finds resistivities to behave like ¯h · the Compton wave number of an electron and by

2π −12 λc = =2.43 10 m kc · its Compton wavelength. Since p =¯hk, we have

2 2 ǫk = c¯h k + kc .

Using again the periodic boundaryp conditions in volume V = L3, we obtain 2π k = (n ,n ,n ), L x y z

The plateaus in the Hall resistivity and the minima in the and we have the density at T =0 longitudinal resistivity correspond to filling fractions k3 ρ = F . p 2 ν = , 3π q 2 1/3 When kF = (3π ρ) is of the order kc, the relativistic where corrections must be taken into account. The corresponding density is p and q are small integers (<11). • ∼ k3 1 q is an odd integer. ρ = c =5.87 1035 • c 3π2 · m3 This behaviour is called as the Fractional Quantum Hall 106 density of metallic electron gas Effect (FQHE) as opposed to the previous Integer ≈ × Quantum Hall Effect (IQHE). Regarding the IQHE we We have an ultrarelativistic electron gas when kF kc or note that correspondingly ρ ρ . ≫ ≫ c

16 Let us consider cold relativistic material, i.e. let us If neutron pressure is not suﬃcient black hole. • → assume T TF . The total energy≪ is Typical properties of a white dwarf: 4 kF 2 2 2 the diameter of the star 2R 10 km. dk k c¯h k + kc • ≈ E = Nǫ¯ = N 0 , kF 57 dk k2 the total number of nuclei NN 10 . R 0 p • ≈ 30 where R the mass M 10 kg M⊙, where • ≈ 30 ≈ M⊙ =1.989 10 kg is the mass of the sun. kF /kc 2√ 2 · 2 0 dx x 1+ x ¯ǫ = mc 10 −3 6 kF /kc the mass density ρ 10 kg m is about 10 the dx x2 m R 0 • density of the sun or≈ of the earth. × kF /kc 2 √ 2 2 R 2 dx x 1+ x 1 = mc + mc 0 − the number density of electrons ρ 1036m−3. Then kF /kc 2 • ≈ R 0 dx x k k , so the electron gas is only moderately F ≈ c is the average electronic energy.R At the non relativistic relativistic. In inner parts the gas can be much limit we have denser and thus ultrarelativistic. 22 17 kF /kc dx x2[ 1 x2 + ] the pressure p 10 Pa 10 atm. ǫ¯ mc2 1+ 0 2 · · · • ≈ ≈ ≈ kF /kc 2 7 " R 0 dx x # the temperature in inner parts T 10 K T⊙. • ≈ 10≈ 2 Since the Fermi temperature is TF 10 K T we, 3 kRF = mc2 1+ + , however, have a cold electron gas. ≈ ≫ 10 k · · · " c # We can estimate the size of the star as follows: from which our earlier results can be derived, provided Let p(r) be the pressure at the distance r from the center that the rest energy of electrons is taken into account. of the star, g(r) the corresponding gravitational At the ultrarelativistic limit kF kc we get ≫ acceleration and ρm(r) the density. kF /kc 3 p A 2 0 dx x 3 ǫ¯ mc = c¯hkF . ≈ kF /kc 2 4 p + d p R0 dx x r + d r p r Thus the energy densityR is ( p + d p ) A r m A d r g E 3 = (3π2)1/3c¯hρ4/3 V 4 The condition for the balance of hydrostatic mechanical forces is and the pressure dp ∂E = g(r)ρ (r). p = dr − m − ∂V N Now at the ultrarelativistic limit (compare photon gas!) GM(r) g(r)= , r2 1 E 1 p = = (3π2)1/3c¯hρ4/3. 3 V 4 where M(r) is the mass inside of the radius r and Nm2 G =6.673 10−11 White dwarf · kg2 In a properly functioning star the pressure due to high temperature (caused by nuclear reactions, mainly 2H is the gravitational constant. We get the pair of equations He) and the gravitational force are in balance. When→ the dp(r) M(r)ρ (r) nuclear fuel is consumed the star shrinks and pressure = G m increases. If the mass of the star is large enough all dr − r2 material will become ionized. Depending on the mass of dM(r) 2 = 4πr ρm(r). the star the ﬁnal state can be dr white dwarf, if the Fermi pressure of the degenerate Because in nuclei there are roughly as many neutrons as • electron plasma prevents further compression. protons, and, on the other hand, there are as many protons as electrons, we have neutron star if the electron pressure is not enough to • compensate the gravitational force. The matter ρm(r) 2mpρ(r). compresses further, and electrons + protons ≈ neutrons. The degeneracy pressure of neutrons→ may Here stop further collapse. m =1.673 10−27kg p ·

17 is the proton mass and ρ(r) the number density of the we get the condition electrons. As a ﬁrst approximation the electron density of a star can M 2/3 R 2 M 2/3 =1 c , be taken to be a constant ρ. Then M − R M c c 8 M(r)= πm ρr3 where 3 p 1/2 3/2 9π ¯hc 57 and the total mass Mc = mp 0.52 10 mp 512 Gm2 ≈ · p 8 M = πm ρR3, ¯h 9π 1/3 M 1/3 3 p R = c 4700km. c mc 8 m ≈ p when R is the radius of the star. The pressure must now satisfy the diﬀerential equation For the radius of the star we get

1/3 1/3 dp 16 2 2 M M = πmpρ Gr R = R 1 . dr − 3 c M − M c " c # with the boundary condition that the pressure vanishes at We see that the white dwarf has the maximum mass the surface, i.e. M = M . A more careful calculation shows that the mass p(R)=0. c of a white dwarf cannot exceed Chandrasekhar’s limit, Integrating the diﬀerential equation we get for the about 1.4M⊙, without collapsing to a neutron star or a pressure at the center black hole. 8π p = Gm2ρ2R2. 9.4. Other Fermionic systems 3 p Nuclear matter Since the electron gas is not quite ultrarelativistic we The mass density of heavy nuclei is calculate more accurately than before. The average electronic energy is ρ 2.8 1017kg m−3. m ≈ · kF /kc dx x2√x2 +1 When we assume that the proton and neutron densities ǫ¯ = mc2 0 kF /kc 2 are equal the Fermi wave numbers of both gases are R 0 dx x kF /kc 3 1 1 15 −1 R dx x 1+ 2 + kF 1.36 10 m = mc2 0 2 x · · · ≈ · kF /kc 2 R 0 dx x and the Fermi energies 3 k 3 k = mc2 F +R c + . 4 k 4 k · · · ǫF 38MeV. c F ≈ 2 From this we can get for the pressure Since mnc = 938MeV, the nuclear matter is non relativistic. The attractive nucleon interactions cancel the ¯hc p = (k4 k2k2 + ) pressure due to the kinetic energy. 12π2 F − c F · · · Neutron star 1 m2c2 = (3π2)1/3¯hcρ4/3 1 + . When the mass of a star exceeds the Chandrasekar limit 4 − ¯h2(3π2ρ)2/3 · · · the Fermi pressure of the electrons is not enough to This the equation of state of the relativistic electron gas. cancel the gravitational force. The star continues its We require that the pressures obtained from the equation collapse. The star forms a giant nucleus where most of state and from the hydrodynamic balance conditions electrons and protons have transformed via the reaction are equal in the center, i.e. p+e− n+ ν → e 8π 2 2 Gmpρ R = to neutrons. The radius of the star is 3 1 m2c2 (3π2)1/3¯hcρ4/3 1 + . R 10km, 4 − ¯h2(3π2ρ)2/3 · · · ≈ the nucleon count When we substitute the electron density (as a function of 57 NN 10 the mass and radius) ≈ and the mass density 3M ρ = 18 −3 3 ρ 10 kg m . 8πmpR m ≈

18 The pressure acting against the gravitation is mostly due to the pressure of the degenerate Fermi gas and to the strong, at short distances very repulsive nuclear forces. (Crude 1st approximation: substitute m = me mn in white dwarf result.) → Quark matter When nuclear matter is compressed 2–10 times denser that in atomic nuclei. the nucleons start to ”overlap” and their constituent quarks form a quark plasma. Liquid 3He 1 The nucleus is p+p+n and the nuclear spin 2 . At low temperatures the nuclear spin determines the statistics, i.e. 3He atoms are Fermions. The Fermi temperature corresponding to the normal density is ǫF TF = 5K. kB ≈ Since the mutual interactions between 3He atoms are considerable the 3He matter forms an interacting Fermi liquid. The 3He liquid has two super phases (A and B). These are in balance with the normal phase at the critical point T T 2.7mK < F . c ≈ 1000