low temperature properties of the bose-einstein and fermi ...troy/sflood/oxford2015.pdfbose-einstein...

Low Temperature Properties of theBose-Einstein and Fermi-Dirac

Equations

W. C. Troy

Low Temperature Properties of the Bose-Einstein and Fermi-Dirac Equations – p.1/46

Contents

1. The Models.

2. Einstein Model For Cooling A solid.

3. Fermi-Dirac Two State Paramagnetism Model

• W. C. T., Quart. Jour. Appl. Math., AMS (2012)

• W. C. T., Physics Letters A, No. 45 (2012)


1. The Models

N =D

e(ǫ−µ)

kT − 1, ǫ > µ (BE)

N =D

e(ǫ−µ)

kT + 1(FD)

N = most probable number of particles with energy ǫ.

µ = chemical potential, ǫ = hν = quantum of energy,

ν = frequency, h= Planck’s constant,

T = temperature (K), k = Boltzman’s constant.


Homework Problems

N =D

e(ǫ−µ)

kT − 1, ǫ > µ (BE)

Prove : limT→0+

N = 0. (1)

N =D

e(ǫ−µ)

kT + 1(FD)

Prove : limT→0+

N =

0, ǫ > µ

1, ǫ < µ(2)


Textbooks

1.) D. Halliday and R. Resnick, Fundamentals of Physics

2.) P. Tipler and R. LLewellyn, Modern Physics

3.) D. Schroeder, Thermal Physics

4.) D. Griffiths, Quantum Mechanics


2. Einstein Model For A Solid

• R. Feynman - 1982 - proposed that development of aquantum computer is theoretically possible.

The Question. Can a solid be cooled to a temperature

T0 > 0 where all quanta of thermal energy are drained

off, leaving the object in the ground state? If ‘yes,’

quantum effects are expected (superposition of states).

This result may lead to quantum computing devices.

• D. Powell, Moved By Light. Science News, May 7, 2011


1982- 2006

Laser cooling - techniques based on radiation pressure

remove energy and reduce vibrations:

2006 - vibrations in glass ’doughnuts’ were reduced bycooling to T=11 mK

2006 - wobbles in mirrors were reduced bycooling to T=10 mK


2006-2009

• Further refinements of laser techniques were developed

to cool sticks, sails, drums and other multi-atom objectsto low temperatures where vibrations are quelled.

• Papers ’flowed in’ as researchers competed to removeevery quantum of energy from an object, leaving it in theground state where quantum effects are expected

• D. Powell, Moved By Light. Science News, May 7, 2011


2009-2010

• 63 quanta left - Park and Wang, Nature Physics 2009

• 37 quanta left - Schliesser et al, Nature Physics 2009

• 30 quanta left - Groblacher et al, Nature Physics 2009

• 4 quanta left - Rocheleau et al, Nature 2010


2010 - 2011

• O’Connell et al (Nature, 2010) reduced a quantumdrum (one trillion atoms) to its ground state at

T0 ≈ 20mK

- Science magazine 2010 breakthrough of the year.

• Teufel et al (Nature, 2011) reduced a drum ( 10−13 kg) toits ground state where it stayed for 100 microseconds, muchlonger than the 6 nano second result of O’Connell et al.


TheoryOur Goal. Answer the question theoretically: can allquanta of thermal energy be drained from a solid?

T0 ≡ Lowest Possible Temperature

Models.

• Einstein 1907 Model For A Solid.

• Stat. Mech. Microcanonical Mds.: T0 > 0.

Goal: show that T0 > 0.


Einstein Theory For A Solid

Einstein (1907) developed a formula for specific heat

Cv = 1n

∂U∂T in terms of T and the quantum ǫ = hν.

A1. Each atom is a 3D quantum oscillator, which

is attached to a preferred position by a spring.

A2. q > 0 quanta of energy have been added to the solid.

A3. Each quantum has energy ǫ = νh.


Petit-Dulong Law of Specific Heat

• P. L. Dulong (1785 - 1838) - French physicist and chemist.

• A. T. Petit (1791-1820) - French physicist.

• 1819 - Petit-Dulong Law of specific heat:

Cv =1

n

∂U

∂T= 5.94

(

cal

gm K

)

.

• A. T. Petit and P. L. Dulong, "Recherches sur quelquespoints importants de la Théorie de la Chaleur," Annales deChimie et de Physique 10 (1819)


Specific Heat:Cv =1n

∂U∂T

• Dulong and Petit (1819): for all solids,

Cv = 5.94

(

cal

gm K

)

.

• Weber (1875), Kopp (1904), Dewar (1904):

0 < Cv << 5.94

for many solids at low T.


Einstein Formula (1907):

Cv = 5.94( ǫ

kT

)2 exp( ǫkT )

(

exp( ǫkT ) − 1

)2 , 0 < T < ∞.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

1

2

3

4

5

6C

v

Figure 1. Cv vs. scaled temperature for diamond.


θ = temperature where Cv = 12 (5.94) ≈ 2.97

Lewis and Randall: ‘Thermodynamics And The FreeEnergy Of Chemical Substances‘ (1923)


Behavior of Cv asT → 0+

Cv = 5.94( ǫ

kT

)2 exp( ǫkT )

(

exp( ǫkT ) − 1

)2 , 0 < T < ∞.

It is easily verified that

limT→0+

Cv = 0. (3)

Can T → 0 as the number of quanta decreases to zero?

To answer this, study the derivation of Cv.


Micro. Canonical Derivation

N = no. of atoms, N ′ = 3N = no. of degrees of freedom.

W = no. of ways to distribute q quanta of energyover N ′ degrees of freedom:

W =(q + N ′ − 1)!

q!(N ′ − 1)!.

Entropy:

S = k ln(W ) = k ln

(

(q + N ′ − 1)!

q!(N ′ − 1)!

)

.

Simplification: de Moivre - Stirling approximation to n!


James Stirling 1692-1770

• Scottish mathematician.

• 1715 - Oxford - expelled from Baliol college for

correspondence supporting the 1708 Scottish

"Gathering of the Brig o’ Turk" uprising against the Stuarts.

• 1725 - Venice - feared assassination for discovering

trade secrets of Venice glassmkaers. Newton helped

him return to England.


Stirling’s Formula

• 1733 - de Moivre, "Miscellanes Analytica,"

N ! ∼ [constant]√

N

(

N

e

)N

as N → ∞.

• 1733 - Stirling, "Methodus Differentials,"

N ! ∼√

2πN

(

N

e

)N

as N → ∞.

• Statistical Mechanics and Physics textbooks:

ln(N !) = N ln(N) − N, N >> 1.


Derivation of Cv

Apply ln(M !) = M ln(M) − M when M >> 1.

S = k(

ln((q + N ′ − 1)!) − ln(q!) − ln((N ′ − 1)!))

becomes

S = k(

(q + N ′ − 1) ln(q + N ′ − 1) − q ln(q) − (N ′ − 1) ln(N ′ − 1))

Therefore,

dS

dq= k ln

(

1 +N ′ − 1

q

)

.


Internal Energy:

U = qǫ + N ′ǫ

2and

dU

dq= ǫ.

Temperature:

1

T=

∂S

∂U=

dSdq

dUdq

=k

ǫln

(

1 +N ′ − 1

q

)

.

Conclusion I:

T0 = limq→0+

T = 0.


Invert Temperature Equation:

q =N ′ − 1

eǫ

kT − 1, N ′ = 3nNA.

U = qǫ + N ′ǫ

2=

(N ′ − 1)ǫ

eǫ

kT − 1+ N ′

ǫ

2.

Specific Heat: Cv = 1n

∂U∂T .

Cv = 5.94( ǫ

kT

)2 exp( ǫkT )

(

exp( ǫkT ) − 1

)2 , 0 < T < ∞.

Conclusion II:

Cv exists for all T > 0 and limT→0+

Cv = 0.


Widely quoted:

limT→0+

Cv = 0. (4)

Property (4) is mathematically questionable becauseits derivation is based on

ln(q!) = q ln(q) − q,

which loses accuracy as q → 0+.


The Error In Stirling’s Approximation At Low q:

ln(q!) = q ln(q) − q

When q=10 Relative Error = 13 Percent.

When q=2 Relative Error = 188 Percent.

When q=2 Stirling’s Approximation Gives

.69 = −.61

When q=1 Stirling’s Approximation Gives

0 = −1


New Derivation

Replace Stirling ’s approximation

ln(M !) = M ln(M) − M

with the exact formula

ln(M !) = ln(Γ(M + 1)),

where the Gamma function Γ(z) is defined by

Γ(z) =

∫

∞

0tz−1e−tdt, Re(z) > 0.

Basic Property: M ! = Γ(M +1) when M is a positive integer.


Entropy S = k ln(

(q+N ′−1)!

q!(N ′−1)!

)

becomes

S = k(

ln(Γ(q + N ′)) − ln(Γ(q + 1)) − ln(Γ(N ′)))

.

U = qǫ + N ′ǫ

2.

1

T=

dSdq

dUdq

1

T=

k

ǫ

(

Γ′(q + N ′)

Γ(q + N ′)− Γ′(q + 1)

Γ(q + 1)

)

∀q ≥ 0.


Temperature:

T =ǫ

k

(

Γ′(q + N ′)

Γ(q + N ′)− Γ′(q + 1)

Γ(q + 1)

)

−1

> 0 ∀q ≥ 0.

Specific heat:

CV =1

n

dU

dT=

1

n

dUdq

dTdq

=

−k

n

(

Γ′(q + N ′)

Γ(q + N ′)− Γ′(q + 1)

Γ(q + 1)

)2 [d

dq

(

Γ′(q + N ′)

Γ(q + N ′)− Γ′(q + 1)

Γ(q + 1)

)]

−1

> 0 ∀q ≥ 0.


Lowest Temperature:

T0 = limq→0+

T =νh

k

(

Γ′(N ′)

Γ(N ′)+ γ

)

−1

> 0.

γ = Euler’s constant. T0 is large when ν is large!

Lowest Specific heat:

limq→0+

Cv =

−k

n

(

Γ′(N ′)

Γ(N ′)− Γ′(1)

Γ(1)

)2[

d

dq

(

Γ′(q + N ′)

Γ(q + N ′)− Γ′(q + 1)

Γ(q + 1)

∣

∣

∣

∣

q=0

)]

−1

> 0.


Derivative Free Method

Goal: Derive temperature formula without differentaition.

S = k ln

(

(q + N ′ − 1)!

q!(N ′ − 1)!

)

.

U = qνh + N ′νh

2.

1

T=

∆S

∆U=

S(q + 1) − S(q)

U(q + 1) − U(q)=

k

νhln

(

q + N ′

q + 1

)

.

T 0 = limq→0

T =νh

k ln(N ′)> T0 =

νh

k

(

Γ′(N ′)

Γ(N ′)+ γ

)

−1

> 0.


Comparison With Experiment

O’Connell et al (Nature, 2010): ν = 6 × 109Hz, one trillion atoms.

Ground state at T0 = 20mK

New Temperature Formula:

Ground state at q = 0, T0 = 9.8mK


Diamond

T0 = limq→0+

T = 23.205 (K) and limq→0+

Cv = .63 × 10−20

250 500 750 1000 12500

1

2

3

4

5

6C

v Diamond

T(Kelvin)

New Cv

23.1 23.205 23.3

0.3

0.63

0.9

x 10−20

Cv

Blowup

Einstein Cv

New Cv

C0

T0 T


When N ! = Γ(N + 1), the new derivation gives

limq→0+

T = T0 > 0 and limq→0+

U = N ′ǫ

2= Ground State.

Does this happen experimentally?

Le et al, Science (2011), show how two diamonds can

exhibit quantum entanglement at room temperature.


3. Fermi-Dirac Model

N =D

e(ǫ−µ)

kT + 1(FD)

limT→0+

N =

0, ǫ > µ

1, ǫ < µ(5)

Two-state paramagnetism model (Schroeder, ThermalPhysics, p. 98):

Paramagnetic materials ( e.g. magnesium, molybdenum),

are attrated to a magnetic field, B̄. The material does

not retain magnetic properties when B̄ is removed.


Two State Model

Assumptions: a paramagnetic system has N >> 1

electrons attracted to external field B̄ = B0k̂, B0 > 0.

N1 >> 1 electrons are in the up-state,

N2 >> 1 electrons are in the down-state

N = N1 + N2

Energies: U1 = −µBB0, U2 = µBB0

µB = eh4πm = 9.27 × 10−24 Joules

Tesla

Total Energy: U = µBB0 (N2 − N1) = µBB0 (N − 2N1)


Entropy: S = k ln(

N !N1!N2!

)

= k ln(

N !N1!(N−N1)!

)

N >> 1, N1 >> 1, N − N1 >> 1

S ≈ k (N ln(N) − N1 ln (N1) − (N − N1) ln (N − N1))

Total Energy: U = µBB0 (N − 2N1)

1T = ∂S

∂U = dS/dN1

dU/dN1

Solve for N1 : N1 = N

exp“

−2µBB0

kT

”

+1


N1 =N

exp(

−2µBB0

kT

)

+ 1

limT→0+

N1 = N

Solve for T: T = − 2µBB0

k ln“

NN1

−1”

limN1→N−

T = 0

These limits are both invalid because the Stirling

approximation requires N − N1 >> 1


Entropy: S = k ln(

N !N1!(N−N1)!

)

= k ln(

Γ(N+1)Γ(N1+1)Γ(N−N1+1)

)

Total Energy: U = µBB0 (N − 2N1) = heB0

4πm (N − 2N1)

1

T=

∂S

∂U=

dS/dN1

dU/dN1

T =heB0

2πmk(

Γ′(N1+1)Γ(N1+1) − Γ′(N−N1+1)

Γ(N−N1+1)

)

limN1→N−

T =heB0

2πmk(

Γ′(N+1)Γ(N+1) − Γ′(1)

Γ(1)

) > 0


Application

BPPH (2,2-diphenyl-1-picrylhydrazyl) - paramagnetic powder

(1) Grobet et al, J. Chem Phys. (1978), p. 5225

(2) Schroeder, Thermal Physics, pp.106-108

N = 2.3 × 1023, B0 = 2.06, k = 1.38 × 10−23, µB = 9.27 × 10−24

limN1→N−

T = T1 =heB0

2πmk(

Γ′(N+1)Γ(N+1) − Γ′(1)

) = .05K > 0

Experiment: > 99 percent magnetization at T ≈ 2K


Acknowledgement

• Stefanos Folias• S. J. Anderson• Richard Field• Toby Chapman• Stuart Hastings


Debye Formula

Einstein Specific Heat (1907):

Cv = 5.94( ǫ

kT

)2 exp( ǫkT )

(

exp( ǫkT ) − 1

)2 , 0 < T < ∞.

Debye Specific Heat (1913):

Cv = 9Nk

(

T

TD

)3 ∫ TDT

0

x4ex

(ex − 1)2dx, 0 < T < ∞.


Teufel et al (Nature, 2011) single atom analysis relieson predictions of the Bose-Einstein equation

< N > =1

exp( ǫkT ) − 1

, 0 < T < ∞.

P1: Thermal energy is present at every positive T > 0.

P2: T → 0 as < N > → 0. i.e. T0 = 0.

P2: Quantum effects become important when < N > < 1.

< N > < 1 when T <ǫ

k ln(2).


One Atom Model

Current theory (e.g. Teufel et al, Nature, 2011) - theBose-Einstein equation

< N >=1

exp( ǫkT ) − 1

, 0 < T < ∞,

for a single atom represents the entire solid:

< N >= ave. no. of quanta in 1 particle state with energy ǫ.

ǫ = hν = quantum of energy,

ν = frequency, h= Planck’s constant,

T = temperature (K), k = Boltzman’s constant.


Teufel et al (Nature, 2011) single atom analysis relieson predictions of the Bose-Einstein equation

< N > =1

exp( ǫkT ) − 1

, 0 < T < ∞.

P1: Thermal energy is present at every positive T > 0.

P2: T → 0 as < N > → 0. i.e. T0 = 0.

P2: Quantum effects become important when < N > < 1.

< N > < 1 when T <ǫ

k ln(2).


Comparison With Experiment

O’Connell et al experiment: ground state achieved when

< N > = .07, ν = 6GHz and T = 20mK

The one atom model

< N > =< N/D >=1

exp( ǫkT ) − 1

predicts

Ground State At T = 106mK


low temperature properties of the bose-einstein and fermi ...troy/sflood/oxford2015.pdfbose-einstein...

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