Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2015
Prof. Dr. A. Mirlin Blatt 05, 100 Punkte
Dr. U. Karahasanovic, Dr. I. Protopopov Besprechung, 29.05.2015
1. Quantum description of rotations and vibrations in diatomic gas (5 + 5 + 10 + 10 + 10 = 40 Punkte, schriftlich) (a) The Hamiltonian of a single molecule reads
H= p21
2m1 + p22
2m2 +U(r1−r2) (1)
We introduce the center-of-mass coordinate R~ = (m1~r1+m2~r2)/(m+1+m2) and the relative coordinate ~r=~r1−~r2. The inverse transformation reads
~r1=R~ + m2~r m1+m2
, ~r2 =R~ − m1~r m1+m2
(2) The corresponding conjugate momenta are
P =p1+p2, p= m2
m1+m2
p1− m1
m1+m2
p2 (3)
The equation (3) can be proven in several different ways (cf. solution of the exercise 2 of exercise sheet 4). We illustrate here the quantum mechanical approach. We have
P~ =−i∂R~ =−i∂~r1
∂ ~R∂~r1 −i∂~r2
∂ ~R∂~r2 =−i∂~r1 −i∂~r2 =p1+p2. (4)
~
p=−i∂~r =−i∂~r1
∂~r ∂~r1−i∂~r2
∂~r∂~r2 =−i m2
m1+m2
∂~r1−i m1
m1+m2
∂~r2 = m2
m1+m2
p1+ m1
m1+m2
p2. (5)
The Hamiltonian is given by H= P2
2(m1+m2) + p2
2µ+U(r), µ= m1m2
m1+m2 (6) (b) The Schr¨odinger equation describing the internal motion of the molecule is
−~2
2µ∆ +U(r)
ψ=Eψ. (7)
Here ∆ is the Laplace operator. We write the Hamiltonian in spherical coordinates
~2
2µ −∆ˆr+Jˆ2 r2
!
ψ+U ψ=Eψ. (8)
Here
∆ˆr= 1 r2
∂
∂r
r2 ∂
∂r
(9) is the radial component of the Laplace operator and
Jˆ2 = 1 sinθ
∂
∂θ
sinθ ∂
∂θ
+ 1
sin2θ
∂2
∂φ2 (10)
is the angular momentum operator. The angular momentum is a conserved quantity.
The operator ˆJ2has eigenvalues ~2J(J+ 1). The eigenenergies of the molecule have degeneracies 2J + 1 due to the conservation of the z-projection of the angular momentum ˆJz. We can separate the variable in the Sch¨odinger equation via
ψ= φ(r)
r YJ,m(θ, φ) (11)
The purpose of 1/r in this formula is to bring the radial Laplace operator to the standard form.
(c) The Sch¨odinger equation for φreads
− ~2 2µ
d2φ
dr2 +~2J(J + 1)
2µr2 φ+U(r)φ=Eφ. (12)
The second term here is the centrifugal energy. The energy levelsǫJ,nof our molecule are enumerated by the angular momentum and the radial quantum numbern.
To understand the structure of the low-lying energy levels we consider first the case J = 0. Our potentialU has minimum atr =d. The wave functions corresponding to the energiesE ≪µω2d2are localized near this minimum and feel just the Harmonic potential. Thus,
ǫ0,n =~ω
n+1 2
. (13)
Let us now include the non-zero J. If we want our eigenenergy ǫJ,n to be small compared to mω2d2,J can not be too large. In this approximation the main effect of J 6= 0 is to change the value of the potential at the minimum (at d =r) from zero to ~2J(J+ 1)/2µd2 =~2J(J + 1)/2I. Thus
ǫJ,n= ~2J(J + 1) 2I +~ω
n+1
2
. (14)
(d) We are in position to consider thermodynamics. We have Frot =−kBTlnZrot, Zrot =
∞
X
J=0
(2J+ 1) exp
−~2J(J+ 1) 2IkBT
!N
(15) At low temperatures (T ≪~2/2IkB) the sum is dominated by the first terms
Frot ≈ −kBN T ln
1 + 3 exp
− ~2 IkBT
≈ −3kBN Texp
− ~2 IkBT
. (16) The entropy and the heat capacity now follow
Srot=−∂TFrot = 3N~2
IT e−~2/IkBT 1 +IkBT /~2
, (17)
Crot= 3N kB ~2
IkBT 2
e−~2/IkBT. (18) Note that the third low of thermodynamics is now satisfied. The heat capacity is exponentially small at low temperatures due to the presence of gap in the spectrum of rotations.
At high temperatures (T ≫~2/2IkB) the sum can be replaced by the integral over J. We get
Frot =−kBN T ln
2 Z ∞
0
dJJexp −~2J2/2IkBT
=−kBTln2IkBT
~2 (19)
This coincides with the result of the classical treatment of rotations in the exercise sheet 4. In particular, the rotational contribution to the heat capacity is given by
Crot =N kB. (20)
(e) The analysis of vibrations follows the same lines. We will give just the answers here.
Zvib = e−~ω/2kBT
∞
X
n=0
e−~ωn/kBT
!N
=
1 2 sinh~ω/2kBT
N
(21) Low temperatures:
Fvib =−N T kBe−~ω/kBT + ~N ω
2 , (22)
Cvib =kB ~ω
kBT 2
e−~ω/kBT. (23) High temperatures
Fvib=−N kBTlnkBT
~ω +N~ω
2 , (24)
Cvib =N kB. (25)
2. Ideal quantum gases out of equilibrium.
(15 + 15 + 10 = 40 Punkte, m¨undlich) (a) Let us consider thej-th group of states. We need to placeNj fermions intoνjstates.
Each state can be occupied by one fermion only. So the number of ways to do the placement is just the number of ways to selectNj states out of availableνj states
∆Γ(Nj) = νj!
Nj!(νj−Nj)!. (26) We now write down the entropy and apply the asymptotic expansion for
lnn!≈nlnn−n, n≫1. (27)
We get S=kBX
j
ln ∆Γ(NJ)
=kBX
j
[νjlnνj −NjlnNj−(νj−Nj) ln(νj −Nj)−νj+Nj+ (νj−Nj)] =
=kBX
j
[(νj−Nj +Nj) lnνj−NjlnNj−(νj−Nj) ln(νj −Nj)] =
=−kB
X
j
NjlnNj
νj + (νj −Nj) lnνj−Nj νj
=−kB
X
j
νj[njlnnj+ (1−nj) ln(1−nj)]. (28) This expression is identical in the form to the one derived in the lectures for the
entropy of a an ideal equilibrium Fermi gas. The only difference being that in Eq.
(28) the distribution of fermions is a generic non-equilibrium distribution.
In the limit nj ≪1 we get
S =−kBX
j
νjnjlnnj
e . (29)
This is actually the entropy of a non-equilibrium Boltzmann gas. Indeed, in the Boltzmann gas we are allowed to place particles into the states completely indepen- dently. Thus
∆Γ(Nj) = νjNj
Nj!. (30)
This leads immediately to (29).
In the limit 1−nj ≪1 we get Boltzmann gas of holes.
(b) Let us now consider Boson. The problem of computing ∆Γj can now be reformulated as follows. We have νj boxes (our states). They are distinguishable. We have Nj indistinguishable balls (our particle). We need to place the balls into boxes. This problem was solved in exercise 4a of exercise sheet 2. We get
∆Γ(Nj) = (Nj+νj−1)!
Nj!(νj−1)! (31)
We compute the entropy just as in the previous exercise. The result reads S=X
j
νj[(1 +nj) ln(1 +nj)−njlnnj]. (32) In the limit nj ≪1 this goes over into the entropy of a Boltzmann gas.
The limit of large occupation numbernj ≫1 gives S =kB
X
j
νjlnenj (33)
We will encounter applications of this formula in the discussion of phonon.
(c) Let us now derive the Fermi and Bose distributions from the maximal entropy principle. We start from fermions and the entropy
S=−kB
X
j
νj[njlnnj+ (1−nj) ln(1−nj)] (34) We fix the total number of particleP
jNj =νjnj and the total energyP
jEjνjnj in our system. We should maximize the entropy in the presence of these constraints.
We use the Lagrange multipliers to take constraints into account, i.e. we need to maximize
−kBX
j
νj[njlnnj+ (1−nj) ln(1−nj)]−λNX
j
νjnj−λEX
j
νjEjnj. (35) Differentiating with respect tonj we get
−lnnj+ ln(1−nj)−λN−EjλE = 0 (36) The solution
nj = 1
1 + exp (λN +EjλE) (37)
is just the Fermi distribution with temperatureT = 1/kBλE and chemical potential µ=−kBT λN.
The consideration of bosons is fully analogous.
(d) Ideal Bose gas near the condensation temperature.
(20 Punkte, m¨undlich) Let us first use the relation
g3/2(z)≈g3/2(1)−2p
π(1−z) (38)
and solve the equation for the chemical potential µ. We will prove Eq. (38) later.
We introduce notations
δT =T −Tc(n), δz= 1−z≈ − µ
kBTc(n), λc =
s 2π~2
mkBTc(n). (39) Our starting point is the equation
n= g3/2(z)
λ2T (40)
We perform expansion near T =Tc(n) n= 1
λ3cg3/2(1) + 1 λ3c
3δT
2Tcg3/2(1)− 1 λ3c2√
πδz (41)
and take into account the definition of critical temperature n= 1
λ3cg3/2(1). (42) We find
δz = 9
16πg23/2(1) δT
Tc 2
. (43)
Finally,
µ=−9kB 16π
(T −Tc)2
Tc g32/2(1). (44) We now prove Eq. (38). It is useful to derive first integral representation ofg3/2(z).
We have g3/2(z) =
∞
X
n=1
1 n3/2zn=
∞
X
n=1
1 nzn
Z dx
√2πe−nx2/2=− Z dx
√2πlnh
1−e−x2/2zi (45) Accordingly, for the difference g3/2(z)−g3/2(1) we have
g3/2(z)−g3/2(1) =− Z dx
√2π ln
1 + δz ex2/2−1
(46) Let us consider this integral in more detail. If we try to expand it inδz we immedia- tely get divergence at smallx. This tells us that the answer behaves as√
δz and this non-analytic behavior originates from small x. We can thus expand the expression under the integral at smallx and get
g3/2(z)−g3/2(1) =− Z dx
√2π ln
1 + 2δz x2
=−2√
πδz (47)
