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Physics 181 Problem Set 10

Ye Zhao: yezhao@college.harvard.edu

April 13, 2012

1 Bosons in two dimensions

The particles in two dimensional surface of area A have a constant density of state given by

g() =4mA

h2

Hence the number of particles is given by

N =

0nBE()g() d =

4mA

h2

0

1

e() 1d

=4mA

h2

(kT ln

(1 e

kT

))Since the total number of particles have to remain the same, we can relate the chemical potentialand temperature by

Nh2

4mAkT= ln

(1 e

kT

) = kT ln

1 exp

Nh2

4mAkT

A plot of the chemical potential versus temperature is shown in the plot below.

0 2 4 6 8 10

20

15

10

5

0

T

Figure 1: A plot showing the chemical potential of the 2-D gas at different temperatures.

Hence we see that the chemical potential remains way below 0 for finite temperatures and it is only

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equal to 0 when T is 0. Hence there is a smooth transition from the gas phase to the liquid phasewithout going through a strange and rapid condensation as the BEC.

In order to be an abrupt BE C at a finite temperature. We need that g() as 0.

2 Bosons in a Harmonic Potential

2.1 Density of State

For n 1, the degeneracy of level n is approximately n2

2 . The spacing between energy levels is hf.Hence, the density of state which is the number of states per unit of energy is given by

g()hfn2

2=

(/hf)2

2 g() =

2

2(hf)3

2.2 Condensation Temperature

Using the same approach as we have done in class, we can write

N =

0nBE()g() d =

1

2(hf)3

0

2

e() 1d

Using the substitution x = kT which gives = kT x and d = kTdx, we have

N =

0nBE()g() d =

1

2

kT

hf

3 0

x2

ex 1d

Again we see that the prefactor decreases when temperature decreases, hence has to increasetowards zero in order for the number to remain constant. Hence at the critical temperature, wehave = 0. Hence

N =1

2

kTchf

3 0

x2

ex 1d = (3)

kTchf

3

Tc = 1.063hf

kN1/3

2.3 Comparison with Particles in a Box

The spring potential energy of the particle is given by 12ksL2 where the spring constant is related

to the frequency according to

ksm = = 2f

ks

mf

2

At temperature Tc, the average particle energy is of the order kTc. Hence we have

kTc mf2V2/3

1

V1/3

kTcm

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Substitute this into the temperature that we calculated in the previous part, we have

Tc h

kN1/3

1

V1/3

kTcm

(kTc)2

h2

m(kTc)

N

V

2/3

kTc h2

m

NV

2/3

Hence we see that the expression is exactly what we have derived in class up to a factor.

3 Solid-Vapor Equilibrium

3.1 Chemical Potential

The energy of the particle in the in the energy level n = (nx, ny, nz) is given by

En = (nx + ny + nz)h0 0

Hence the partition function of a single particle is

Z1 =

nx,ny,nz

e0eh0(nx+ny+nz) =e0

(1 eh0)3

For solids, the particles are located at indistinguishable lattice sites. Hence we have ZN = ZN1 . To

calculate the chemical potential, we first need to find out the free energy which is given by

F = kT ln ZN = N kT ln Z1

Hence we can calculate the chemical potential by taking the derivative with respect to N, ie

=F

N

= kT ln Z1 = kt ln e0

(1 eh0)3 = 0 + 3kT ln(1 eh0)

3.2 Vapor Pressure on Coexistence Line

The chemical potential of an ideal gas has been calculated before and is given by

g = kT ln

[V

N

2mkT

h2

3/2]

Equating this with the chemical potential calculated above for the solid, we have

0 + 3kT ln(1 eh0

)= kT ln

[V

N2mkT

h2 3/2]

(1 eh0

)3 [VN

2mkT

h2

3/2]= e

0kT

Using the ideal gas law P V = N kT to replace the ratio VN withkTP we have

P = kT e0/kT(

1 eh0/kT)32mkT

h2

3/2

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3.3 Low Temperature Limit

In the low temperature limit where kT h0 we have eh0/kT 0. Hence we have

P = kT e0/kT

2mkT

h2

3/2

4 BEC Questions

4.1 Box with two non-interacting boson gases

b)

The temperature may be different for the two gases as the critical temperature depends on themass of the gases as well.

4.2 Two Degenerate Ground States

c)

Since we are integrating over all energy levels, the fraction in the ground state is very small. Hencehaving a degeneracy of 2 in the ground state will not contribute significantly to the difference inthe temperature. Hence the condensation temperature is about the same.

4.3 Direction of Particle Flow

a)

The gas particles will flow from a higher chemical potential to a lower chemical potential which isfrom the BEC condensate state ( = 0) to the classical gas state ( < 0).

4.4 Necessary Condition for Bose Condensation at Low Temperature

b), d)

Both these conditions have to be satisfied in order to ensure that the integral in calculating thetotal number of particles has an analytical solution.

4.5 Ideal Gas of Photons in the Condensed Phase

DisagreeFor blackbody radiation, the number of particles is not conserved as the photons can disappearinto the walls of the blackbody.However, there have been some new research suggesting the possibility of finding photons in thecondensate state in a form of superlight.

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