Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2015
Prof. Dr. A. Mirlin Blatt 07, 100 Punkte
Dr. U. Karahasanovic, Dr. I. Protopopov Besprechung, 12.06.2015
1. Thermodynamics of spin systems: arbitrary spin
(5 + 10 + 5 + 10 = 30 Punkte, m¨undlich) (a) The partition function reads
Z =
" S X
Sz=−S
eµβBSz
#N
=
"
sinhBµ(2S+1)2k
BT
sinh2kBµ
BT
#N
. (1)
(b) We compute free energy
F =−N kBTln
"
sinhBµ(2S+1)2k
BT
sinh2kBµ
BT
#
. (2)
Entropy reads S =N kBln
"
sinhBµ(2S+1)2k
BT
sinh2kBµBT
#
+ N Bµ 2kBT
coth Bµ
2kBT −(2S+ 1) cothBµ(2S+ 1) 2kBT
(3) The specific heat reads
CB =T ∂TS = B2M µ2 4kBT2
"
1 sinh2 2kBµBT
− (2S+ 1)2 sinh2Bµ(2S+1)2k
BT
#
. (4)
At low temperatures nothing dependence on spin.
CB≈ B2M µ2 4kBT2 exp
−Bµ kBT
. (5)
In this regime the heat capacity is determined by the gap in the spectrum.
At high temperatures the heat capacity also goes to zero due to the fact that the spectrum is limited from above
CB= B2M S(S+ 1)µ2
3kBT2 . (6)
(c) We find the magnetization M = N µ
2
(2S+ 1) cothBµ(2S+ 1)
2kBT −coth Bµ 2kBT
(7) At large fields all spins are in the ground state and the magnetization is simply µN S. At smallB the magnetization is linear,M ∝B and
χ= µ2S(S+ 1)
3kBT . (8)
This is the Curie law.
(d) Let us consider now the adiabatic reduction of magnetic field. The process is rever- sible and the entropy of the system is constant. The entropy is given by Eq. (3).
We observe that the condition S = constant is equivalent to T /Bconstant. So the temperature at the end of the process is αT.
2. The paramagnetic response of electron gas (Pauli paramagnetism) (5 + 10 + 10 + 10 = 35 Punkte, schriftlich) (a) The single-particle states in our system are labeled by the momentum p and the
spin projection sz =±1/2 and their energies are ǫ(p, sz) = p2
2m −2µsBsz (9)
The Ω potential reads (as usual for non-interacting fermions) Ω(T, V, µ, B) =−T V X
sz
Z d3p (2π~)3 lnh
1 +e−(ǫ(p,sz)−µ)/Ti
(10) Thus Ω(T, V, µ, B) is related to the Ω(T, V, µ)-potential at zero field by
Ω(T, V, µ, B) = 1
2Ω(T, V, µ+µsB) +1
2Ω(T, V, µ−µsB) (11) The potential Ω(T, V, µ, B) was computed in the lectures
Ω(T, V, µ) =−2T V
Z d3p (2π~)3 lnh
1 +e−(p2−2µm)/2mTi
=−2kBT V λ3T f5/2
eµ/T (12) f5/2(z) =
∞
X
n=1
(−1)n+1 zn n5/2 (13) (b) We can now compute the magnetic moment of the gas
M(T, V, µ) =−∂BΩ(T, V, µ, B) =−µs 2
∂Ω(T, V, µ+µsB)
∂µ − ∂Ω(T, V, µ−µsB)
∂µ
(14) We see that at small B magnetization is linear in B and the susceptibility (per particle) is
χT =−µ2s N
∂2Ω(T, V, µ)
∂µ2 = µ2s N(T, V, µ)
∂N(T, V, µ)
∂µ . (15)
Here N(T, V, µ) is the number of particles.
(c) We are now in position to study the high-temperature limit. At T ≫ EF via a dealing with a dilute gas of fermions which is described by Boltzmann statistics.
The chemical potential µis large end negative here. Correspondingly, Ω(T, V, µ) =−2kBT V
λ3T f5/2 eµ/T
≈ −2kBT V
λ3T eµ/kBT (16) N(T, V, µ)≈ 2V
λ3T eµ/kBT (17)
∂N(T, V, µ)
∂µ = 1
kBTN(T, V, µ) (18) Thus
χT = µ2s
kBT. (19)
This coincides with the result of exercise 1 (in the corresponding answer in exercise 1 we should put S= 1/2 andµ= 2µs).
(d) Let us consider the zero temperature limitT ≪EF. In this case N = 2V 4πp3F
3(2π~)3 = (2mµ)3/2
3π2~3 (20)
∂N
∂µ = 3 2
N
µ (21)
We find
χ= 3µ2s
2µ ≡ 3µ2s
2EF (22)
3. Spin systems and negative temperatures
(5 + 15 + 15 = 35 Punkte, m¨undlich) (a) Let us denote by n the number of spins pointing down. Obviously, n = 0, . . . , N
and the energy
E =−ǫ0N+ 2nǫ0. (23)
The number of states with given nand energy E =−ǫ0N + 2nǫ0 is
Γ(E) = N!
n!(N−n)!. (24)
(b) We now consider the microcanonical ensemble and find the entropy S(E) =kBln Γ(E) =kB[NlnN−nlnn−(N−n) ln(N −n)]
=−kBnln n
N −kB(N −n) ln 1− n
N
=−kB E
2ǫ0 +N 2
ln
1 2+ E
2ǫ0N
−kB
−E 2ǫ0 +N
2
ln 1
2 − E 2ǫ0N
. (25) The temperature of the system is defined by
1
T =kB∂ES(E) =−kB
ǫ0 arctanh E
N ǫ0 (26)
The temperature of the system becomes negative whenE >0. In this case we have more spins pointing up then those pointing down, The system is characterized by the population inversion. Let us imagine that we start atE =−N ǫ0 corresponding to T = 0. As the E grows so doesT. When E=−0 our system is characterized by T = +∞. Then, at E = +0,T =−∞. As E grows further towards E =N ǫ0 the temperature increases towardsT =−0. So, the natural ordering of temperatures is 0<+∞<−∞<−0. The energy as a function of temperature reads
E(T, N) =−N ǫ0tanh ǫ0
kBT (27)
(c) Let us now consider two systems ofN spins brought into the thermal contact. The final temperature can be deduced from the energy conservation
E(T,2N) =E(T1, N) +E(T2, N) (28) We find
1 T = kB
ǫ0 arctanh 1
2tanh ǫ0
kBT1 +1
2tanh ǫ0
kBT2
(29)
