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Category: PHY 831

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1. Show that in the grand canonical ensemble, for a three-dimensional gas of

spin S particles with single-particle energies e = |~p|

2/2m, the pressure can be

written as

P = (2S + 1)

Z

d

3 p

(2πh¯)

3

|~v||~p|

3

f∓(e(~p)), (1)

where ~v = ~p/m.

2. Find the density of single-particle states for particles trapped in three-dimensional

parabolic potential. Assume the single particle energy levels are e = h¯ ω(mx +

my + mz) where mi = 0, 1, 2, …, ∞, i.e. neglect the zero point energy of the

quantum harmonic oscillator. How does this differ from the density states of

a gas with energy momentum relation e = |~p|c trapped in a box with side

lengths L, where c is a constant?

3. Consider a gas of N non-relativistic electrons (spin = 1/2) confined to a twodimensional area A with mass m in contact with a reservoir with temperature

T and chemical potential µ.

(a) Find the Fermi energy, eF of the system.

(b) Calculate the two-dimensional “pressure” (i.e. −

∂F

∂V

T,N

) of the system

when T = 0.

(c) What is the heat capacity of the electrons at fixed N (

∂E

∂T

N,V

) when T

eF, to first order in T?

(d) What is the heat capacity of the electrons at fixed µ (

∂E

∂T

µ,V

) when T

eF, to first order in T?

4. (From last week:) Assume there are N random variables labeled by i = {1, …, N}

that each obey the arbitrary normalized probability distribution g(x), so that

they have averages hx

n

i

i =

R

dxxng(xi). Assume that hxii = 0 and hx

2

i

i = σ

2

for all i. Show that the distribution of the average of these random variables,

x¯ = 1

N ∑i xi

, in the large N limit is given by

P(x¯) = 1

√

2πσ2/N

e

− Nx¯

2

2σ

2

,

which is essentially the central limit theorem, which says that the probability

distribution of the sum of a large number of random variables tends to a Gaussian (or normal) distribution. Therefore, it is maybe not so surprising that this

1

distribution shows up quite often in statistical mechanics. This also shows the

standard deviation of x¯ is ∝ 1/√

N, since we have R ∞

−∞

dxx2

exp(−x

2/2σ

2

) =

√

2πσ3/2, and the distribution of x¯ goes to a delta function in the large-N limit

[Hints: Use δ(x) = 1

2π

R ∞

−∞

dyeixy and R ∞

−∞

exp(iay − by2

) = √

π/b exp(−a

2/4b).]

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