PHY303 � Statisti al Thermodynami s � Fall 2012 � Prof. D. Garanin����������������������������������������-Assignment 5, with solutions1 Statisti al thermodynami s of free lassi al parti lesPartition fun tion of lassi al parti les in 3d is de�ned asZclass =

ˆ

d3p

ˆ

d3r exp [−βE(p, r)] , (1)where E(p, r) is parti le's energy. Note that this expression has the unit of (momentum × distan e)3, unlike the quantumpartition fun tion that is dimensionless. De�ne the density of states of a free lassi al parti le in a box of volume V . By omparing it with the density of states for a quantum parti le in a rigid box, �nd the missing fa tor in Eq. (1) that would makethe lassi al partition fun tion mat h the quantum one. This will de�ne a quantum-me hani al � ell� in the phase spa e of a lassi al parti le. Show that this quantum-me hani al aspe t does not ontribute into the internal energy and heat apa ity ofthe lassi al parti les.Solution: The energy of the parti le onsists of the kineti and potential energy,E(p, r) =

p2

2m+ U(r), (2)so that the lassi al partition fun tion fa torizes

Zclass =

ˆ

d3pe−βp2/(2m)

ˆ

d3re−βU(r). (3)For free parti les there is no potential energy, and Zclass for parti les in a rigid box of volume V be omesZclass = V

ˆ

d3pe−βp2/(2m). (4)Choosing the kineti energy ε = p2/(2m) as the integration variable, one an rewrite this in the formZclass =

∞ˆ

0

dερclass(ε)e−βε, ρclass(ε) = 2πV (2m)3/2

√ε.Quantum-me hani al partition fun tion for this problem has the same form with

ρ(ε) =V

(2π)2

(

2m

~2

)3/2 √ε. (5)The two densities of states are related as

ρ(ε) =ρclass(ε)

(2π~)3that de�nes the missing fa tor in the above de�nition of the lassi al partition fun tion. Corre ting Zclass asZclass =

ˆ

d3pd3r

(2π~)3exp [−βE(p, r)] , (6)one obtains the dimensionless quantity that oin ides with the lassi al limit of the quantum partition fun tion Z. This formula an be used in the presen e of the potential energy as well.The interpretation of the above is the following. The (x, px) proje tion of the phase spa e of the parti le is dis retized into ells ∆x∆px = 2π~ = h, and the similar for other dire tion y and z. The ells have quantum origin and are related to theHeisenberg's un ertainty prin iple, stating that the produ t of un ertainties of measuring x and px of a quantum parti le is oforder h. The number of quantum ells in a limited region of x and pxis limited and it de�nes the number of di�erent states inthis region. It is impossible to have more di�erent states be ause there is no way to distinguish states that are too lose both in

x and in pxby measurement.Similarly quantum ells an be introdu ed in many-parti le problems and in problems with rotational degrees of freedom.Quantum ells is an external element in lassi al statisti al physi s. Statisti al averages of most physi al quantities (ex ept forthe entropy and related fun tions) are insensitive to the quantization of the phase spa e of the system be ause the orre tionfa tor introdu ed above an els. 1

2 Classi al parti les with gravityUsing the distribution fun tionf(p, r) =

1

Zclassexp [−βE(p, r)]for lassi al parti les with gravity, �nd the dependen e of parti le's on entration n and pressure P as the fun tion of the height.Set the minimal height (the earth level) to zero. Cal ulate the heat apa ity of this system and ompare it with the one for freeparti les.Solution: The energy of the parti le has the form

E =p2

2m+mgz.The lassi al partition fun tion of the parti le fa torizes:

Zclass =

ˆ

d3p

ˆ

d3r exp [−βE(p, r)] = ZkineticZpotential,whereZkinetic =

ˆ

d3pe−βp2/(2m), Zpotential =

ˆ

d3re−βmgz.Here we do not use the phase-spa e quantization fa tor from the pre eding problem be ause it will an el out in the �nalresults.The kineti part of Z an be al ulated asZkinetic =

∞ˆ

−∞

dpxe−βp2

x/(2m)

3

= (2πmkBT )3/2

.Assuming that the parti les are ontained to a verti al ylinder of the ross-se tion S (that is non-essential), for the potentialpartition fun tion one obtainsZpotential = S

∞ˆ

0

dze−βmgz =SkBT

mg.Thus

Zclass =S

g(2π)

3/2m1/2 (kBT )

5/2.Suppose there are N parti les in the system. Then the number of parti les in the element of the phase spa e is

dN = Nfdpxdpydpzdxdydz,where f is the distribution fun tion introdu ed above. The on entration of parti les is de�ned asn =

ˆ

d3pdN

dxdydz= N

ˆ

d3pf.Sin e f and Zclass fa torize, the integrals over momentum an el out and one obtainsn =

N

Zpotentiale−βmgz.One an he k that integrating this over the volume yields the identity N = N . Using the result for Zpotential, one gets the �nalresult

n =mgN

SkBTe−βmgzthat exponentially de reases with the height. For the pressure one does not have anything better than the ideal-gas formula

P = nkBT =mgN

Se−βmgz.The average internal energy is given by

U = −N∂ lnZclass

∂β= −N

∂ lnβ−5/2

∂β=

5

2

N

β=

5

2NkBT.The heat apa ity is

C =∂U

∂T=

5

2NkB.2

This result might be unexpe ted. There is the heat apa ity (3/2)NkB from three translational degrees of freedom. Additionally,there is a potential energy for the motion in the verti al dire tion. Its ontribution is NkB instead of the expe ted (1/2)NkB, aswere the ase for a vibrational degree of freedom. The reason for a di�erent result is that the theorem of the equidistribution ofthe energy over degrees of freedom is valid in the ases when the energy is a quadrati fun tion in momenta and deviations fromthe equilibrium positions (see the next problem). In our ase, the potential energy is a linear rather than a quadrati fun tionof z.3 Classi al harmoni os illatorsConsider lassi al parti les with the potential energyV (r) =

kr2

2in 3d. Cal ulate the partition fun tion, internal energy and heat apa ity.Solution: We start, as usual, with al ulating the lassi al partition fun tionZclass =

ˆ

d3pe−βp2/(2m)

ˆ

d3re−βkr2/2 = ZkineticZpotential. (7)One obtainsZkinetic =

∞ˆ

−∞

dpxe−βp2

x/(2m)

3

= (2πmkBT )3/2

.Similarly,Zpotential =

∞ˆ

−∞

dxe−βkx2/2

3

= (2πkBT/k)3/2

.ThusZclass ∝ T 3.The internal energy and heat apa ity are given by

U = −N∂ lnZclass

∂β= 3NkBTand

C =∂U

∂T= 3NkB.The fa tor 3 here is due to the three translational degrees of freedom of our system. Per ea h degree of freedom, there is NkB/2due to the kineti energy and the same amount due to the potential energy. This problem illustrates the equidistribution ofenergy over degrees of freedom in lassi al statisti al physi s.4 Phonons in 1d and 2dCal ulate the internal energy and heat apa ity of the system of harmoni phonons in one and two dimensions at low temperatures.Solution: Instead of al ulating the partition fun tion, it is more onvenient to use the dire t formula for the internal energy

U = U0 +

ˆ

∞

0

dωρ (ω)~ω

exp (β~ω)− 1, (8)where the upper limit of the ontegration has been set to in�nity at low temperatures. In 2d the phonon density of states is

ρ(ω) =Sω

2πv2,where S is the system's area. Dis arding the zero-point energy U0, one obtains

U =S~

2πv2

ˆ

∞

0

dωω2

exp (β~ω)− 1=

S~

2πv2

(

kBT

~

)3 ˆ ∞

0

dxx2

ex − 1=

ζ(3)S (kBT )3

πv2~2.The heat apa ity be omes

C =∂U

∂T= 3kB

ζ(3)S (kBT )2

πv2~2.3

5 Two intera ting Ising spinsConsider the model of two oupled spins with the HamiltonianH = −gµBB (S1,z + S2,z)− JS1,zS2,z.Here B is the external magneti �eld and J is the so- alled ex hange intera tion, ferromagneti for J > 0 and antiferromagneti for J < 0. The model above in whi h only z omponents of the spins are oupled is alled Ising model. The energy levels of thissystem are given byεm1m2

= −gµBB (m1 +m2)− Jm1m2,where the quantum numbers take the values −S ≤ m1,m2 ≤ S. Write down the expression for the partition fun tion of thesystem. Can it be al ulated analyti ally for a general S? If not, perform the al ulation for S = 1/2 only. Cal ulate theinternal energy, heat apa ity, magnetization indu ed by the magneti �eld, and the magneti sus eptibility. Analyze ferro- andantiferromagneti ases.Solution: The partition fun tion of the system is given byZ =

S∑

m1,m2=−S

e−βεm1m2 .For a general spin S one an perform analyti ally only one summation. One an use the results for a single spin in a magneti �eld and with h ≡ gµBB writeZS =

S∑

m1=−S

eβhm1

S∑

m2=−S

eβ(h+Jm1)m2 =S∑

m=−S

eβhmsinh [(S + 1/2)β (h+ Jm)]

sinh [β (h+ Jm) /2].The remaining sum most probably annot be al ulated analyti ally.For S = 1/2 the expression above simpli�es to

Z1/2 =

1/2∑

m=−1/2

eβhm2 cosh

[

β (h+ Jm)

2

]that is,Z1/2 = 2

{

eβh/2 cosh

[

β (h+ J/2)

2

]

+ e−βh/2 cosh

[

β (h− J/2)

2

]}

.This expression an be simpli�ed toZ1/2 = 2

[

eβJ/4 cosh (βh) + e−βJ/4]

. (9)Finally, in zero �eld the result simpli�es toZ1/2 = 4 cosh (βJ/4) .Let us al ulate the internal energy and heat apa ity in zero �eld. One obtains

U = −N∂ lnZ

∂β= −NJ

4tanh

(

βJ

4

)

,where N is the number of two-spin systems. In the limit of low temperatures the hyperboli tangent tends to 1 and one obtainsthe anti ipated result U = −NJ/4 (the two oupled spins are parallel for J > 0). In the ase of antiferromagneti oupling,J < 0, one has tanh (βJ/4) → −1 for T → 0, so that for both ferro- and antiferromagneti oupling one obtains U = −N |J |/4at zero temperature.The average spin value per two-spin system is given by

〈Sz〉 = 〈m1 +m2〉 =1

Z

∂Z

∂ (βh). (10)Using Eq. (9), one obtains

〈Sz〉 =eβJ/4 sinh (βh)

eβJ/4 cosh (βh) + e−βJ/4=

sinh (βh)

cosh (βh) + e−βJ/2.The sus eptibility an be obtained by di�erentiating this expression with respe t to the magneti �eld. In parti ular, at zero�eld

χ ∝ 1

1 + e−βJ/2.In the ferromagneti ase J > 0 the exponential is very small at low temperatures, βJ ≫ 1, so that the sus eptibility has aregular value, omparable with that of an isolated spin. To the ontrary, in the antiferromagneti ase J < 0 the exponentialis large and thus the sus eptibility is very small. Try to explain this in physi al terms and draw the dependen e 〈Sz〉 on themagneti �eld for the antiferromagneti oupling at low temperatures.4