(1)1 Exam TFY4230 - SPRING 2020 Solution 1 hX2i

(1)

1 Exam TFY4230 - SPRING 2020

Solution 1

hX²i=

* _N X

i=1 N

X

j=1

i∆_ij∆_j +

=

N

X

i=1 N

X

j=1

ijh∆_i∆_ji=

N

X

i=1 N

X

j=1

ijδ_ij =

N

X

i=1

i²

(2)

Solution 2

P(p_x) = Z ∞

0

P(p_x, x) dx

= Z ∞

0

Cδ

Kx−(E− p²_x 2m)

dx

= Z ∞

0

C δ

˜

x−(E− p²_x 2m)

1 Kd˜x

= C

K

(3)

Solution 3

S =k_Bln Γ_n=k_Bln (n²) =k_Bln E²

²

dS dE = 1

T ⇒ 2k_B E = 1

T ⇒ E = 2k_BT

(4)

Solution 4

Z = 1 h²

Z dp_x

Z dp_y

Z dx

Z

dye^−β

p2 x+p2

y 2m −βa

= 1

h² π2m

β Ae^−βa

= 2πm

h² k_BT Ae^−βa

(5)

Solution 5

Partition function:

Z = Z ∞

0

Ce^−E/E⁰e^−βEdE = C

1 E0 +β Mean energy:

hEi=− ∂

∂β lnZ = ∂

∂β ln 1

E₀ +β

= 1

1

E0 +β = E0kBT E₀+k_BT

(6)

Solution 6

Partition function:

Z = 1 h³

Z d³p

Z

d³re^−β^p

2 2m−βU

=

2πm βh²

3/2 V

2 +V 2e^−β

Mean energy:

hEi = − ∂

∂β lnZ = 3

2β +

e^β+ 1 = 3

2k_BT + e^β+ 1 The mean kinetic energy is hTi= ³₂k_BT, hence :

hUi=

e^β+ 1 ⇒ P = 1 e^β+ 1

(7)

Solution 7

Canonical distribution:

P(p_x, x) =Ce^−βH =Ce^−β^p

2x

2m−βk_BT_L^x

Which implies:

P(x) = ˜Ce⁻^L^x Normalization:

Z ∞ 0

dxP(x) = 1 ⇒ P(x) = 1 Le⁻^x^L Probability that x > L

Z ∞ L

P(x)dx= Z ∞

L

1

Le⁻^L^xdx= e⁻¹

(8)

Solution 8

The rectangle is deformed into a parallelogram with the same height and base as the rectangle. This implies area is conserved. At the same time the parallelogram is stretched so that the circumference goes to infinity.

(9)

Solution 9

Z = 1 h²

Z d²p

Z

d²re^−β

p2 x+p2

y

2m −βU

= 2πm h²β

Z

d²re^−βU

= 2πm h²β

π(R²−R²₀) +πR²₀e^β

(10)

Solution 10 Kinetic energy:

H =

N

X

i=1

~ p²_i 2m Mean energy:

Z = 1

h^2NN! Z

d²p₁· · · Z

d²p_N Z

d²r₁· · ·d²r_Ne^−βH

= 1

h^2NN! Z

d²p₁e^−β

~ p2

1 2m

NZ d²r₁

N

= 1

h^2NN!(2πmk_BT)^NA^N

(11)

Solution 11

The number of ways of placing N particles one two energy levels such that 2 of them are in in the ground state is ^N₂

= _(N_−2)!2!^N^! = ^N^(N−1)₂ .

Another way of seeing this is to choose one of the particles first, then we have N particles to choose between. When we choose the second particle we have N −1 to choose. Which gives N(N−1) possibilities. It does not matter which order we choose them, and we therefore divid by 2, i.e. N(N −1)/2.

(12)

Solution 12

The factor N! makes entropy an extensive property both in the canonical and micro-canonical ensemble.

(13)

Solution 13

The heat capacity in the low temperature limit comes mostly from long wavelength elastic defor- mations (phonons) in the solid.

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Solution 14

Easiest solution: This is the partition function of a system with only one energy level.

There can therefore be no fluctuations in energy, i.e. ∆E² = 0.

Direct calculation:

hEi=− ∂

∂β lnZ = ∂

∂ββN =N

∆E² =− ∂

∂βhEi=− ∂

∂βN = 0

(15)

Solution 15

Method 1: when T is large the average energy hEi goest to zero, since the spins have random orientations. Then number of states is 2³when the spins are random, i.e the entropy isS =k_Bln 2³ = 3k_Bln 2. Which gives the free energy F ≈ hEi −T S=−3 ln (2)k_BT at high temperature.

Method 2:

Z = X

s1,s2,s3

e^βJ(s¹^s²^+2s²^s³⁾

= X

s1,s2

e^{βJ s}¹^s² e^{2βJ s}² + e^{−2βJ s}²

= X

s1,s2

e^{βJ s}¹^s² e^2βJ+ e^−2βJ

= 2 e^βJ+ e^−βJ

e^2βJ+ e^−2βJ When T k_BT we have βJ 1, andZ ≈2³ Free energy:

F =−k_BT lnZ ≈ −k_BT ln (2³) =−k_BT3 ln (2)

(16)

Solution 16

The minimum energy is obtained when the spins s₁ and s₂ are parallel, and s₂ and s₃ are anti- parallel. I.e. the mininum energy isE =−2J, which is the mean energy at zero temperature. When T = 0 the spins are locked in the same position, there is no thermal fluctuations that can flip the spins (degree of freedom frozen out). I.e. we haveC = 0.

(17)

Solution 17

There are 2³ = 8 spin configurations. If all the spins point in the same direction, (s₁, s₂, s₃) = (1,1,1) or (s₁, s₂, s₃) = (−1,−1,−1), the energy is H = 3J. The only other configurations available is when two spins are parallel and one anti-parallel with the two others, for example (s₁, s₂, s₃) = (1,1,−1) or (s₁, s₂, s₃) = (−1,1,−1), in which case the energy is H = −J, the minimum energy in the system. Since there is more than one state with this energy, the entropy is non-zero. I.e atT = 0 we have hEi=−J and S >0.

(18)

Solution 18 Partition function:

Z = X

s1=±1

· · · X

s2N=±1

e^βh(s¹^+s²^+···+s^N^)−βh(s^N+1^+s^N+2^+···+s^2N⁾

= X

s1=±1

e^βhs¹

!N

 X

sN+1=±1

e^−βhs^N+1





N

= e^βh+ e^−βhN

e^−βh+ e^βhN

Free energy:

F =−1

β lnZ =−2N

β ln e^βh+ e^−βh

(19)

Solution 19

Z = X

φ1=±1

X

φ2=±1

e^{βJ φ}²¹^φ² = 2 e^βJ+ e^−βJ Mean energy:

hEi=− ∂

∂β lnZ =−J e^βJ−e^−βJ e^βJ+ e^−βJ hEi=−Jhφ²₁φ₂i ⇒ hφ²₁φ₂i= e^βJ−e^−βJ

e^βJ+ e^−βJ

(20)

Solution 20

When T = 0 the system is in a minimum energy state. The minimum of −hs₁ +s²₁ is when s1 = 1. The same applies of course to all the other spins. There is only one ground state, hence the entropy isS = 0.

(21)

Solution 21

The Hamiltonian is:

H =−hcos (θ1)−hcos (θ2)− · · · −hcos (θN)

In the low temperature limit θ_i is close to zero, and we can do a series expansion:

H ≈ −h

1− θ²₁ 2

−h

1−θ₂² 2

− · · · −h

1− θ²_N 2

The equipartition theorem then gives:

hHi ≈ −N h+N1 2k_BT

(22)

Solution 22

Chemical potential of the system:

µ_s = ∂F

∂N =−k_BTln (N)

In thermal equilibrium the chemical potential of the system and the reservoir must be equal:

µ_s =µ ⇒ −k_BT ln (N) = µ ⇒ N = exp

− µ k_BT

(23)

Solution 23

In thermal equilbrium we must have

µ=µ_s Which implies:

k_BTln (N λ²

A ) =−+k_BT ln (N_sλ² A_s ) Dividing by kBT and taking the exponential of both sides gives:

N λ²

A = exp (−

k_BT)N_sλ² A_s Which implies:

N_s=NA_s

A exp ( k_BT)

(24)

Solution 24

Average particle number:

hNi= ∂

β∂µln Θ = ∂ β∂µ

βµ+ (βµ)²

= 1 + 2βµ Average energy:

hEi = µhNi − ∂

∂β ln Θ

= µ(1 + 2βµ)−(µ+ 2µ²β)

= 0

(25)

Solution 25 Mean value :

hNi = ∂ β∂µln Θ

= ∂

β∂µ[M βµ−β]

= M

Fluctuations around mean:

∆N² = ∂

β∂µhNi= 0

(26)

Solution 26

Θ =

Nc

X

N=0

Z_N

=

Nc

X

N=0

e^βN

= e^β(N^c⁺¹⁾−1 e^β−1 Ifβ1 :

Θ≈e^βN^c

(27)

Solution 27 Free energy:

F = −1 βln (Z)

= −1

βln (K)− 1 β

N

3 ln (V) + N 5 ln (β) Pressure:

p=−∂F

∂V = N 3V k_BT Mean energy:

hEi=− ∂

∂β lnZ = N 5k_BT Heat capacity:

C = ∂

∂ThEi= N 5 k_B

(28)

Solution 28

The gas acts as a particle reservoir for adsorption on the two beads. In thermal equilibrium we must have

µ₁ =µ_g and

µ₂ =µ_g Which implies µ₁ =µ₂ and :

−₁+k_BT lncN₁ =−₂+k_BTlncN₂ Which implies:

N₁

N₂ = exp

₁−₂ k_BT

(29)

Solution 29

At low temperatures the spins in the Ising model are all parallel and locked in that configuration.

There is not enough thermal energy to flip a spin and the heat capacity is therefore zeroC0 = 0. At high temperatures the spin directions are random and the mean energy tends towards zerohEi= 0, the mean energy does not change if we increase the temperature, therefore the heat capacity is zero:

C∞ = 0.

(30)

Solution 30

Summing over all configurations results in the partition function:

Z = X

θ1,θ2,θ3

e^{β(cos (θ}¹^{)+cos (θ}²^{)+cos (θ}³⁾⁾

= X

θ1

e^β^{cos (θ}¹⁾X

θ2

e^β^{cos (θ}²⁾X

θ3

e^β^{cos (θ}³⁾

=

e^β+ e^β^{cos (α)}+ e^β^{cos (−α)}³

=

e^β+ 2e^β^{cos (α)}3

(31)

Solution 31 Mean energy:

hEi= Z ωD

0

g(ω)ωdω e^β¯^hω−1 Ifg(ω)∝ω we get

hEi ∝ Z ωD

0

ω²dω

e^β¯^hω−1 = 1 (β¯h)³

Z β¯hωD

0

x²dx e^x−1 Which implies hEi ∝1/β³ ∼T³ at low temperatures.

The heat capacity therefor scales as C ∼T².

(32)

Solution 32 Canonical partition function:

Z_N = X

s1,s2,···,sN

e^−βH^N

= X

s1

e^βhs¹X

s2

e^βhs²· · ·X

sN

e^βhs^N

= e^βh+ e^−βhN

Grand partition function:

Θ =

∞

X

N=0

e^βµNZ_N

=

∞

X

N=0

e^βµN e^βh+ e^−βh^N

This geometric series converges if e^βµ e^βh+ e^−βh

<1, in which case we get:

Θ = 1

1−e^βµ(e^βh+ e^−βh)

(33)

Solution 33

Mean number of particles:

hNi = ∂ β∂µln Θ

= ∂

β∂µ V λ³e^βµ

= V

λ³e^βµ At high temperatures βµ is small and we get

ρ= hNi V ≈ 1

λ³ Thermal de Broglie length:

λ= h

√2πmk_BT ∝ T^−1/2

When µ is kept constantρ therefore diverges at high temperatures:

ρ∝T^3/2

(34)

Solution 34 Partition function:

Z = 1 h

Z ∞

−∞

dp_xe^−β^p

2x 2m

Z L 0

dze^−βmgz

= 1 h

2πm β

1/2

1

mgβ 1−e^−βmgL Pressure:

p = − ∂

∂LF

= 1 β

∂

∂LlnZ

= 1 β

∂

∂Lln 1−e^−βmgL

= mge^−βmgL 1−e^−βmgL

= mg

e^βmgL−1

(35)

Solution 35

Using Stirlings formula the entropy can be written as:

S =k_Bln (Γ_n)≈k_Bln (N!)−k_Bnln (n) +k_Bn−k_B(N −n) ln (N −n) +k_Bln (N −n) Which implies:

dS dE = 1

dS dn = kB

ln

N −n n

Hence:

n

N −n = e⁻^{kB T}

(36)

Solution 36

The possible occupation numbers of energy levels and corresponding total energy:

(n₀, n₁, n₂) energy (1,1,0) (1,0,1) 2 (0,1,1) 3 (2,0,0) 0 (0,2,0) 2 (0,0,2) 4

Z = 1 + e^−β+ 2e^−2β+ e^−3β+ e^−4β

(37)

Solution 37 The possible occupation numbers of energy levels and corresponding total energy:

(n₀, n₁, n₂) energy (1,1,0) (1,0,1) 2 (0,1,1) 3

Z = e^−β+ e^−2β+ e^−3β

(38)

Solution 38

From the expression for the energy we get d= ¯hωdn which implies that the density of states is g(ω) = _¯_hω¹ .

The number of excited states is therefore hN_ex =i 1

¯ hω

Z ∞ 0

d e^β(µ−)−1

(39)

Solution 39

Total number of particles:

hNi = hn₀i+hn₁i+hn₂i

= 1

e^−βµ −1 + 1

e^β(−µ)−1 + 1 e^β(−µ)−1

= 1

e^−βµ −1 + 2 e^β(−µ)−1 Half of the particles are in the ground state:

hn₀i= N

2 ⇒ e^−βµ = 1 + 2 N

Half of the particles are in the excited states:

hn₁i+hn₁i= N

2 ⇒ e^β(−µ)= 1 + 4 N This implies :

e^β= 1 + _N⁴

1 + _N² ⇒ k_BT = ln (^N+4_N+2)

(40)

Solution 40

hEi=

p²_x+p²_y+p²_z 2m

= 3

2kBT = 3 2β

∆E² =− ∂

∂βhEi=− ∂

∂β 3 2β = 3

2(k_BT)²