Peeter Joot [email protected] April 4, 2013

PHY452H1S Basic Statistical Mechanics. Problem Set 7: BEC and phonons 1.1

Disclaimer

This is an ungraded set of answers to the problems posed.

Exercise 1.1

Bose-Einstein condensation (BEC) in one and two dimensions

a. Obtain the density of states N(e) in one and two dimensions for a particle with an energymomentum relation Ek =

h¯2 k2 . 2m

(1.1)

b. Using this, show that for particles whose number is conserved the BEC transition temperature vanishes in these cases - so we can always pick a chemical potential µ < 0 which preserves a constant density at any temperature. Answer for Exercise 1.1 Part a.

We’d like to evaluate Nd (e) ≡

∑ δ(e − ek ) k

Ld ≈ (2π)d

Z

d

d kδ

We’ll use δ(g(x)) = ∑ x0

h¯2 k2 e− 2m

δ(x − x0 ) , | g0 (x0 )|

(1.2)

! ,

(1.3)

where the roots of g(x) are x0 . With g(k) = e −

1

h¯ 2 k2 , 2m

(1.4)

the roots k∗ of g(k) = 0 are r

∗

k =±

2me h¯2

.

(1.5)

The derivative of g(k) evaluated at these roots are h¯2 k∗ m√ h¯2 2me =∓ m√ h¯ h¯ 2me =∓ . m

g0 (k∗ ) = −

(1.6)

In 2D, we can evaluate over a shell in k space ∞ m A 2πkdk (δ (k − k∗ ) + δ (k + k∗ )) √ 2 (2π) 0 h¯ 2me A ∗ m = k 2π h¯2 k∗

Z

N2 (e) =

(1.7)

or N2 (e) =

2πAm . h2

(1.8)

In 1D we have ∞ m L dk (δ (k − k∗ ) + δ (k + k∗ )) √ 2π −∞ h¯ 2me 2L m √ = . 2π h¯ 2me

N1 (e) =

Z

(1.9)

Observe that this time for 1D, unlike in 2D when we used a radial shell in k space, we have contributions from both the delta function roots. Our end result is r 2L m (1.10) N1 (e) = . h 2e Part b. To consider the question of the BEC temperature, we’ll need to calculate the density. For the 2D case we have

2

N A Z 1 d2 k = A f (ek ) A (2π)2 Z 1 2π Am ∞ 1 = de −1 βe 2 A h z e −1 0 Z 2πm ∞ 1 dx −1 x = 2 h β 0 z e −1 2πmkB T ln(1 − z) =− h2 1 = − 2 ln(1 − z). λ

ρ=

(1.11)

Recall for the 3D case that we had an upper bound as z → 1. We don’t have that for this 2D density, so for any value of kB T > 0, a corresponding value of z can be found. That is 2

z = 1 − e−ρλ = 1 − e−ρh

4 /(2πmk

2 B T)

.

(1.12)

N L Z dk 1 L f (ek ) L r2π Z 1 2L m ∞ 1 1 de √ −1 βe L h 2 0 ez e −1 s Z ∞ 1 2m x 1/2−1 h β 0 z −1 e x − 1 s 1 2m Γ(1/2) f 1−/2 (z), h β

(1.13)

1 − f (z). λ 1/2

(1.14)

For the 1D case we have ρ= = = = = or ρ=

See fig. 1.1 for plots of f ν− (z) for ν ∈ {1/2, 1, 3/2}, the respective results for the 1D, 2D and 3D densities respectively.

3

10

f - 1 HzL 2

8

6

f - 1 HzL f - 3 HzL 2

4

2

0.0

0.2

0.4

0.6

0.8

1.0

Figure 1.1: Density integrals for 1D, 2D and 3D cases We’ve found that f 1−/2 (z) is also unbounded as z → 1, so while we cannot invert this easily as in the 2D case, we can at least say that there will be some z for any value of kB T > 0 that allows the density (and thus the number of particles) to remain fixed.

Exercise 1.2

Estimating the BEC transition temperature

a. Find data for the atomic mass of liquid 4 He and its density at ambient atmospheric pressure and hence estimate its BEC temperature assuming interactions are unimportant (even though this assumption is a very bad one!). b. For dilute atomic gases of the sort used in Professor Thywissen’s lab , one typically has a cloud of 106 atoms confined to an approximate cubic region with linear dimension 1 µ m. Find the density - it is pretty low, so interactions can be assumed to be extremely weak. Assuming these are 87 Rb atoms, estimate the BEC transition temperature. Answer for Exercise 1.2 Part a.

With an atomic weight of 4.0026, the mass in grams for one atom of Helium is 4.0026 amu ×

g = 6.64 × 10−24 g 6.022 × 1023 amu = 6.64 × 10−27 kg.

(1.15)

With the density of liquid He-4, at 5.2K (boiling point): 125 grams per liter, the number density is

4

mass 1 × volume mass of one He atom 125g 1 = −3 3 × 10 m 6.64 × 10−24 g 125g 1 = −3 3 × 10 m 6.64 × 10−24 g = 1.88 × 1028 m−3

ρ=

(1.16)

In class the TBEC was found to be TBEC =

1 kB



ρ ζ(3/2)

 2/3

2π h¯2 M

 2/3 2π(1.0545717310−34 m2 kg/s)2 1 ρ 1.3806488 × 10−23 m2 kg/s2 /K 2.61238 M 2 / 3 ρ = 2.66824 × 10−45 K. M =

(1.17)

So for liquid helium we have TBEC = 2.66824 × 10−45 1.88 × 1028


2/3

1 K 6.64 × 10−27

(1.18)

= 2.84K. Part b.

The number density for the gas in Thywissen’s lab is ρ=

106 = 1024 m−3 . (10−6 m)3

(1.19)

The mass of an atom of 87 Rb is 86.90 amu ×

10−3 kg = 1.443 × 10−25 kg, 6.022 × 1023 amu

(1.20)

which gives us   2/3 TBEC = 2.66824 × 10−45 1024 = 1.85 × 10

Exercise 1.3

−4

1 K 1.443 × 10−25

(1.21)

K.

Phonons in two dimensions

Consider phonons (quanta of lattice vibrations) which obey a dispersion relation Ek = hv ¯ |k|

5

(1.22)

for small momenta |k|, where v is the speed of sound. Assuming a two-dimensional crystal, phonons only propagate along the plane containing the atoms. Find the specific heat of this crystal due to phonons at low temperature. Recall that phonons are not conserved, so there is no chemical potential associated with maintaining a fixed phonon density. Answer for Exercise 1.3 The energy density of the system is d2 k e 2 βe (2π) e − 1 Z N(e) e = de . βe V e −1

E = V

Z

(1.23)

For the density of states we have d2 k δ(e − ek ) (2π)2 Z ∞ 1 = 2π kdkδ(e − hvk) ¯ (2π)2 0 Z ∞  1 e  1 = kdkδ k − 2π 0 hv ¯ hv ¯ 1 e . = 2π (hv) ¯ 2

N(e) = V

Z

(1.24)

Plugging back into the energy density we have ∞ E 2π e2 = de V (hv) ¯ 2 0 e βe − 1 π (kB T)3 = ζ(3), (hv) ¯ 2

Z

(1.25)

where ζ(3) ≈ 2.40411. Taking derivatives we have CV =

3πk3B T 2 dE ζ(3). =V dT (hv) ¯ 2

6

(1.26)