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Chapter 9

Degenerate Fermi Gases

Key Concepts

• Examples.

• Fermi energy.

• Fermi temperature.

• Degeneracy condition (T ≪ TF ).

Reading

Schroeder, Section 7.3.Kittel & Kroemer, Chapter 7.Ashcroft & Mermin,Solid State Physics, Ch. 2.Kittel, Introduction to Solid State Physics, Ch. 6.

9.1 What is a degenerate Fermi gas?

A degenerate Fermi gas is a system of fermions where quantum statistics has a significant effecton the macroscopic properties of the system. Examples include:

• Conduction (free) electrons in a metal

• Liquid 3He

• Neutrons in a neutron star

100

• Electrons in a white dwarf star

• Protons and neutrons in an atomic nucleus

We assume the temperature is sufficiently low that

vQ =

(h√

2πmkBT

)3

≫ V

N.

This is the opposite condition to that required for the validity of Boltzmann statistics. Here, we areconsidering the case where the fermions are packed very closely together so that quantum statisticsare important.

As a first step, we takeT = 0.

9.2 Zero temperature

ε

I

0

n (ε)FD

µ = ε F

Figure 9.1: Fermi-Dirac distribution atT = 0.

All single particle states withǫ < µ are occupied, and all withǫ > µ are unoccupied – mathemat-ically, we obtain this by lettingT → 0 in the Fermi-Dirac distribution. Physically, it makes sensebecause at absolute zero we expect all particles to be in their lowest possible states. But becausewe are looking at fermions, we can’t have more than one fermion per state (two if we considerspin). Therefore the particles fill up all the lowest energy states in order.

TheFermi energy is defined byǫF ≡ µ(T = 0). (9.1)

N.B. The chemical potential varies with temperature — we will show this in a moment.

101

The Fermi energy gives the cut off energy for the distribution atT = 0. We can see that the abovedefinition makes sense because

µ =

(∂U

∂N

)

S,V

≈ ∆U

∆N. (9.2)

At T = 0, if we add one particle (∆N = 1) then it must go in with an energyǫF (because that isthe next available level) so the energy of the system goes up by ∆U = ǫF . Thus

µ(T = 0) =∆U

∆N= ǫF .

The Fermi energy is determined by the total number of particles in the system — the more particles,the more states are needed to place them all, and so the higherthe Fermi energy.

We now calculateǫF for a system ofN free electrons confined to a cube of volumeV = L3. Wefirst find the allowed energy levels.

For a one dimensional box, the allowed wavelengths and corresponding momenta are

λn =2L

n, pn =

h

λn

=hn

2L,

as we have already shown previously, and this extends to the three components of momentum inthree dimensions.

In three dimensions, the allowed energies are

ǫ(p) =p

2

2m,

=1

2m(p2

x + p2y + p2

z),

=h2

8mL2(n2

x + n2y + n2

z). (9.3)

In “n-space”, a constant energy surface is a sphere. The number ofstates with energy less thanǫF

is the volume of a sphere with corresponding radiusnmax given by

ǫF =h2

8mL2n2

max, (9.4)

Because all particles have energy less thanǫF by definition, this must beN , and so we can findnmax and henceǫF in terms ofN .

N = 2 × volume of1

8of a sphere,

= 2 · 1

8· 4

3πn3

max,

=π

3n3

max, (9.5)

where the factor of two takes spin into account, and the factor of 1/8 is because we only includenx, ny, nz > 0. The Fermi energy is then

ǫF =h2

8mL2n2

max =h2

8m

(3N

πV

)2/3

. (9.6)

102

(compare with the calculation of the Debye wavevector). Note thatǫF only depends on the densityN/V , meaning it isintensive— if we double the size of the system,ǫF does not change.

We now reconsider the requirement for degeneracyV/N ≪ vQ. This is equivalent to

kBT ≪ ǫF . (9.7)

We can also define theFermi temperature by

TF ≡ ǫF

kB

, (9.8)

so that the degeneracy condition becomes

T ≪ TF . (9.9)

In a typical elemental metal,n = N/V ≈ 1022 − 1023cm−3, with ǫF ≈ 1–10 eV. Hence,TF ≈104−105 K, and so most metals at room temperature are highly degenerate and it is often sufficientto consider them to be atT = 0!

Figure 9.2: Free electron Fermi gas parameters for metals atroom temperature. Taken from Kittel,Introduction to Solid State Physics.

103

9.3 Internal energy at zero temperature

We can now calculate the internal energy per particle of a Fermi gas. We would expect that itwould be approximatelyǫF /2, since the particles have all energies between 0 andǫF .

U = 2∑

nx,ny ,nz

ǫ(n) ≈ 2

∫d3nǫ(n),

= 2

∫ nmax

0

dnn2

∫ π2

0

sin θdθ

∫ π2

0

dφǫ(n),

= π

∫ nmax

0

ǫ(n)n2dn,

=πh2

8mL2

∫ nmax

0

n4dn,

=πh2

40mL2n5

max =3

5NǫF . (9.10)

So the average energy per particle is3ǫF /5.

9.4 Degeneracy pressure

The Fermi gas has pressure even atT = 0

P = −(

∂U

∂V

)

S,N

,

= − ∂

∂V

(3

5N

h2

8m

(3N

π

) 2

3

V − 2

3

),

=2NǫF

5V=

2U

3V. (9.11)

This last equality is the same as for a classical ideal gas andis required by Newton’s laws. We canalso write

PV =2

5NkBTF . (9.12)

This pressure is due to theexclusion principle, which prevents two fermions being in the samestate. When the particles are degenerate (or close to it) theyresist being pushed closer together.

This degeneracy pressure prevents the gravitational collapse of white dwarf and neutron stars (seetutorial problem).

9.5 Bulk modulus

Thebulk modulus is the change in pressure over the fractional change in volume — a high bulkmodulus means the material is difficult to compress, while aneasily compressible material has a

104

Metal Free electronB MeasuredB(1010 dynes cm−2) (1010 dynes cm−2)

Li 23.9 11.5Na 9.23 6.42K 3.19 2.81Rb 2.28 1.92Cs 1.54 1.43Cu 63.8 134.3Ag 34.5 99.9Al 228 76.0

Table 9.1: Bulk moduli for some typical metals. The free electron value is that for a free electrongas at the observed density of the metal.

smallB. Mathematically,

B = −V

(∂P

∂V

)

T

=1

κ, (9.13)

whereκ is the compressibility. For a Fermi gas,

P =2

5N

ǫF

V∼ V − 2

3

V= V − 5

3 .

So then,

B =5

3P =

2

3N

ǫF

V=

2

3

N

VkBTF . (9.14)

Compare this to an ideal gas whereB = P = NkBT/V (check this!). Electrons in a (roomtemperature) metal have a much higher bulk modulus (becauseusuallyTF ≫ T ) and are thereforeless compressible than an ideal gas of similar density. Metals are composed of ions and electrons,and while the crystal of ions is hard to compress, the compressibility of metals is dominated bythat due to the degeneracy pressure of the electrons.

105

Chapter 10

Degenerate Fermi gases at low temperatures

Key Concepts

• T ≪ TF , butT > 0.

• Density of states.

• Temperature dependence of chemical potentialµ(T ).

• Sommerfeld expansion.

• Heat capacityCV (T ) ∼ γT .

• Pauli paramagnetism.

• Heavy fermions.

Reading

Schroeder, Section 7.3.Kittel & Kroemer, Ch. 7.Ashcroft & Mermin,Solid State Physics, Ch. 2, p. 661.Kittel, Introduction to Solid State Physics, Ch. 6.

10.1 Small non-zero temperatures

We now consider the case0 < T ≪ ǫF /kB = TF . This applies to metals at room temperature.Recall that the Fermi-Dirac distribution function looks like Figure 10.1, also see Figure 10.2.

106

ε

I

0

n (ε)FD

T = 0

ε F

K TB

Figure 10.1: Fermi-Dirac distribution at low temperatures.

Now a fraction of the electrons, of orderNkBT/ǫF have kinetic energykBT greater thanǫF . Thuswe expect the internal energy at low temperatures to be

U(T ) ≈ U(T = 0) + aN

(kBT

ǫF

)kBT,

wherea is a number of order one. The heat capacity at low temperatures will then be

CV (T ) =

(∂U

∂T

)

V

≈ aNk2

B

ǫF

T,

which is consistent with the third law of thermodynamics.

We now derive this result in a more systematic manner.

10.2 Density of states

This is an important concept with very wide-ranging applications in quantum physics. We havealready come across the density of states in treating both phonons and black body radiation, withoutactually identifying it.

The density of states, denotedg(ǫ) is the number of single particle states per unit energy — thenumber of states “near” energyǫ.

If we let N(E) be the total number of single particle states with energy less thanE then

N(E) =

∫ E

−∞

dǫ · g(ǫ)

It follows that

g(E) =dN(E)

dE, (10.1)

107

Figure 10.2: Fermi-Dirac distribution at various temperature forTF = 50000 K. The results applyto an ideal gas in three dimensions with a constant total number of particles. You can read off thechemical potential for each of these distributions from thegraph — the energy at whichf = 0.5.Figure taken from Kittel,Introduction to Solid State Physics.

that is, the number of states betweenE andE + dE is g(E) dE.

If {α} denotes all the single-particle energy states, we can also write

g(ǫ) =∑

α

δ(ǫ − ǫα) (10.2)

whereδ(x) is theDirac delta function, defined such that

δ(x) =

{∞ for x = 0,0 otherwise.

but with the special property that the function is normalised to one∫ ∞

−∞

dx δ(x) = 1.

It is also the derivative of thestep functionΘ(x) which is defined to be

Θ(x) =

{1 for x ≥ 0,0 otherwise.

108

It follows that

N(E) =

∫ E

−∞

dǫ g(ǫ),

=

∫ E

−∞

dǫ∑

α

δ(ǫ − ǫα),

=∑

α

∫ E

−∞

dǫ δ(ǫ − ǫα),

=∑

α

Θ(E − ǫα). (10.3)

which is the total number of states with energy less thanE, consistent with the original definition.

For a system ofN fermions atT = 0, ǫF is defined by

N = no. of energy levels withǫα ≤ ǫF

=∑

α

Θ(ǫF − ǫα)

=

∫ ǫF

−∞

g(ǫ) dǫ (10.4)

The total energy atT = 0 is

U(T = 0) =∑

α

ǫαΘ(ǫF − ǫα) =

∫ ǫF

−∞

dǫ ǫ g(ǫ). (10.5)

The above expressions generalise naturally to finite temperature:

N(T ) =∑

α

nFD(ǫα)

=

∫ ∞

−∞

dǫ g(ǫ)nFD(ǫ), (10.6)

and

U(T ) =∑

α

ǫαnFD(ǫα)

=

∫ ∞

−∞

dǫ ǫ g(ǫ)nFD(ǫ). (10.7)

In words, to findU we sum over all energies, adding the number of states with each energy timesthe (average) number of particles in each of those states. The Fermi-Dirac distribution atT = 0is a backwards step function with the jump down atǫF , and so the general expressions above areconsistent with the zero temperature limit.

10.2.1 Density of states in a three dimensional cube — method 1

We now find the density of states for a Fermi gas confined to a three dimensional cube, subject tothe boundary condition that the wave functionvanisheson the surface of the cube.

109

Previously, we showed that the allowed values of the energy of a single state particle (fermionorboson) are

ǫ(p) =p2

2m=

h2

8mL2(n2

x + n2y + n2

z),

for nx, ny, nz = 1, 2, . . .. Then

N(E) = total no. of states with energy≤ E,

= volume of1

8of the sphere inn-space,

=1

8· 4

3πn3

0,

whereE = h2n20/8mL2. This is similar to what we did for theT = 0 case, except that we are

no longer saying that all these states are filled, i.e.N 6= N . (We are considering particles with nospin.)

We thus have

N(E) =π

6

(8mL2

h2E

)3/2

,

=4π

3L3

(2m

h2

)3/2

E3/2. (10.8)

And so

g(E) =dN(E)

dE= 2πL3

(2m

h2

)3/2 √E. (10.9)

For fermions with spin-half we multiply this by two, becausethere are twice as many states.

10.2.2 Density of states in a three dimensional cube — method 2

Let’s repeat the derivation using the definition

g(ǫ) =∑

α

δ(ǫ − ǫα),

where the sum in this case is over alln = (nx, ny, nz), and

ǫα ≡ ǫ(n) =h2

8mL2(n2

x + n2y + n2

z).

We convert the sum to an integral (in polar co-ordinates) andif the energy spacings are small, thisis a good approximation.

g(ǫ) =∑

α

δ

(ǫ − h2

8mL2|n|2)

, (10.10)

=

∫ ∞

0

dn n2

∫ π/2

0

sin θ dθ

∫ π/2

0

dφδ

(ǫ − h2

8mL2n2

), (10.11)

=π

2

∫ ∞

0

dn n2δ

(ǫ − h2

8mL2n2

). (10.12)

110

When working with this definition of the density of states, thenext step is to make a substitutionin the delta function. Here we set

x =h2

8mL2n2,

so

n = 2L

(2m

h2

)1/2 √x,

n2dn = 8L3

(2m

h2

)3/2

x · 1

2

dx√x.

Thus,

g(ǫ) = 4πL3

(2m

h2

)3/2 ∫ ∞

0

dx√

xδ(ǫ − x),

= 2πL3

(2m

h2

)3/2 √ǫ,

which agrees with the previous result. We also defineg(ǫ ≤ 0) = 0 — there are no states withenergy less than zero in this case. Such states can correspond to bound states such as occur inpotential wells, etc.

10.3 Temperature dependence of the chemical potential

For a general density of statesg(ǫ), we said before that

N =

∫ ∞

−∞

dǫ g(ǫ)nFD(ǫ),

=

∫ ∞

−∞

dǫ g(ǫ)1

e(ǫ−µ)/kBT + 1. (10.13)

If N is fixed, then Eq. (10.13) determinesµ as a function of temperature — asT increases,µ mustdecrease.

To see why, note thatnFD has some symmetry aboutǫ = µ, in the sense that the probability ofa state at energyµ + δǫ being occupied is the same as the probability of a state atµ − δǫ beingunoccupied.

The total number of particles in the system is determined by Eq. (10.6). Let’s consider first aconstant density of statesg(ǫ) ≡ g.

If the number of particles in the system is conserved, then atnon-zero temperatures, the area underthe graph should still beN . Because of the symmetry ofnFD aboutµ, the missing area belowµfrom theT = 0 distribution must be equal to theadditionalarea aboveµ, and so the total area isindeed stillN .

111

εε = µ (T = 0)F

g (E)

area = N

εε = µ (T = 0)F

g (E)

area = N

g(E)n (E)FD

Figure 10.3: Non-constant density of states

Now, however, consider the case where the density of states is not constant, and increases withǫ,as shown in Fig. 10.3.

Now, the area under the graph to the right ofµ will be scaled more than the “empty” area abovethe line to the left, due to the higher density of states on theright. Therefore, the area of the graphis greater thanN , and it would appear the number of particles has increased!

This problem is resolved if, as the temperature increases,µ decreases slightly, so that the peak ofnFD moves to the left, and isn’t scaled as much byg(ǫ), reducing the overall size.

Figure 10.4: Temperature dependence ofµ for an ideal Fermi gas in a box. Figure taken fromSchroeder.

112

~5k TB

εµ

-dn /dεFD

Figure 10.5: The derivative of the Fermi-Dirac distribution is negligible everywhere except withina fewkBT of µ.

10.4 The Sommerfeld expansion

We now explicitly findµ(T ) whenkBT ≪ ǫF = µ(T = 0). From

N =

∫ ∞

−∞

dǫ g(ǫ)nFD(ǫ), (10.14)

we first integrate by parts to get

N = N(ǫ)nFD(ǫ)∣∣∣∞

−∞−∫ ∞

−∞

dǫ N(ǫ)dnFD

dǫ(ǫ). (10.15)

As ǫ → −∞, N(ǫ) → 0 andnFD(ǫ) → 1. Also, asǫ → +∞, nFD(ǫ) → 0, so the boundary termis zero.

ForkBT ≪ ǫF , −∂nFD/∂ǫ is sharply peaked atǫ = µ. We can see this mathematically by

−dnFD

dǫ= − d

dǫ

1

e(ǫ−µ)/kBT + 1,

=1

kBT

ex

(ex + 1)2. (10.16)

wherex = (ǫ − µ)/kBT .

We now change our integration variables in Eq. (10.14) fromǫ to x

N =

∫ ∞

−∞

dx N(µ + xkBT )ex

(ex + 1)2. (10.17)

We perform a Taylor expansion ofN(ǫ) aboutǫ = µ. This is reasonable, because the total numberof states varies slowly on the energy scale ofkBT whenT is small.

N(µ + xkBT ) = N(µ) + N ′(µ)xkBT +1

2N ′′(µ)(xkBT )2 + . . .

113

However, by definition,

N ′(µ) =∂N(ǫ)

∂ǫ

∣∣∣∣∣ǫ=µ

= g(ǫ = µ),

so that

N = N(µ)I0 + kBTg(µ)I1 +1

2(kBT )2g′(µ)I2, (10.18)

where

In =

∫ ∞

−∞

dx xn ex

(ex + 1)2,

I0 =

∫ ∞

−∞

dx

(− d

dx

)(1

ex + 1

)= 1,

I1 = 0, since the integral is antisymmetric aboutx = 0

I2 =π2

3. (See Appendix B5 in Schroeder)

Thus, we find that

N = N(µ) +π2

6g′(µ)(kBT )2 + . . . (10.19)

Now, we letµ(T ) = ǫF − a(kBT )2 + . . . (10.20)

[remembering thatǫF ≡ µ(T = 0)] and substitute in this expression and work consistently tosecond order inT :

N = N(ǫF ) − a(kBT )2g(ǫF ) +π2

6g′(ǫF )(kBT )2. (10.21)

But by definition of the Fermi energy,N(ǫF ) = N — there must be enough states with energy lessthanǫF to contain all the particles. Thus, we must have

a =π2

6

g′(ǫF )

g(ǫF ), (10.22)

and so to second order inkBT/ǫF ,

µ(T ) = ǫF − π2

6

g′(ǫF )

g(ǫF )(kBT )2, (10.23)

For a three dimensional system we haveg(ǫ) ∼ √ǫ so

g′(ǫF )

g(ǫF )=

1

2ǫF

and so we do have an expansion on powers ofkBT/ǫF .

114

10.5 Heat capacity

We now evaluate the internal energy

U(T ) =

∫ ∞

−∞

dǫ g(ǫ)nFD(ǫ)ǫ, (10.24)

for kBT ≪ ǫF . The calculation is similar in spirit to that for findingµ(T ).

Let h(ǫ) be defined by

h(ǫ) =

∫ ǫ

−∞

dE [E g(E)], (10.25)

noting thath(ǫF ) = U(T = 0). IntegratingU(T ) by parts gives

U(T ) =

∫ ∞

−∞

dǫ [ǫg(ǫ)] nFD(ǫ),

=

∫ ∞

−∞

dǫd

dǫ

(∫ ǫ

−∞

dE [Eg(E)]

)nFD(ǫ),

= h(ǫ)nFD(ǫ)|∞−∞ −∫ ∞

−∞

dǫ h(ǫ)

(d

dǫnFD(ǫ)

). (10.26)

Again, becausenFD(ǫ) → 0 asǫ → ∞ andh(ǫ) → 0 asǫ → −∞, the boundary term is zero.Proceeding as before,

U(T ) =

∫ ∞

−∞

dx h(µ + xkBT )ex

(ex + 1)2. (10.27)

We can expandh as a Taylor series

h(µ + xkBT ) = h(µ) + xkBTh′(µ) +1

2(xkBT )2h′′(µ) + . . . (10.28)

and substituting this into the integral gives

U(T ) = h(µ) +π2

6(kBT )2h′′(µ) + . . . (10.29)

Then, we previously showed that

µ(T ) = ǫF − π2

6

g′(ǫF )

g(ǫF )(kBT )2, (10.30)

so we can Taylor expandh(µ) aroundǫF , giving

h(µ) = h(ǫF ) − h′(ǫF )π2

6

g′(ǫF )

g(ǫF )(kBT )2. (10.31)

From the definition ofh(ǫ),

h′(ǫF ) = ǫF g(ǫF ), (10.32)

h′′(ǫF ) = g(ǫF ) + ǫF g′(ǫF ). (10.33)

115

Collecting terms, we have

U(T ) = U(T = 0) − π2

6ǫF g′(ǫF )(kBT )2 +

π2

6(g(ǫF ) + ǫF g′(ǫF )) (kBT )2. (10.34)

The second and fourth terms cancel exactly, leaving

U(T ) − U(T = 0) =π2

6g(ǫF )(kBT )2 (10.35)

It then follows immediately that

CV (T ) =

(∂U

∂T

)

V

= γT, (10.36)

where

γ =π2

3g(ǫF )k2

B. (10.37)

For non-interacting electrons in three dimensions, the Fermi energy is

ǫF =h2

8m

(3N

πV

) 2

3

, (10.38)

and the density of statesper unit volumeis

g(ǫF ) =π

2h3(8m)

3

2

√ǫF , (10.39)

which can be rewritten in terms of the Fermi energy as

g(ǫF ) =3N/V

2ǫF

. (10.40)

Notice thatg(ǫF ) is completely determined by the conduction electron density N/V . We can alsowrite

g(ǫF ) =m

(~π)2kF , (10.41)

wherekF is the Fermi wave number.

Note that for fixed density,g(ǫF ) scales withm. If g(ǫF ) is larger than for the free (non-interacting)electron model, we say the electrons have aneffective massgiven by

m∗ = meγobs

γfree

. (10.42)

10.6 Low temperature heat capacity of metals

For the total heat capacity of metals, we have now calculatedthat at low temperatures we have

CV (T ) = γ T + βT 3.↑ ↑

electrons phonons

Thus, plottingCV (T )/T againstT 2 should be a straight line with gradientβ and intercept atT = 0of γ.

116

Figure 10.6: Low-temperature measurements of the heat capacities per mole of copper, silver andgold. Figure taken from Schroeder.

10.7 Magnetic susceptibility of metals — Pauli paramagnetism

In a magnetic fieldB, the energy of the electrons becomes

ǫ↑ = ǫ + µBB, (10.43)

ǫ↓ = ǫ − µBB, (10.44)

due toZeeman splitting, whereǫ is the energy in the absence of a magnetic field. (The arrowsubscript represents whether the electron spin is parallelor anti-parallel to the magnetic field.Remember the intrinsic magnetic moment of elecrons is in theopposite direction to its spin as it isnegatively charged.)

If we neglect the orbital motion of the electron, the densityof states is

g↑(ǫ) =1

2g(ǫ − µBB), (10.45)

g↓(ǫ) =1

2g(ǫ + µBB), (10.46)

i.e. the density of states of the spin up electrons (for example) is half the density of states in nomagnetic field at an energyµBB less thanǫ.

The total number density of electrons of spinσ = (↑, ↓) is

nσ =

∫dǫ gσ(ǫ)nFD(ǫ), (10.47)

117

Figure 10.7: Experimental values of electronic heat capacity. Taken from Kittel,Introduction toSolid State Physics.

where

nFD(ǫ) =1

eβ(ǫ−µ) + 1. (10.48)

The chemical potentialµ(T ) is determined by

n = n↑ + n↓,

where

nσ =

∫dǫ gσ(ǫ)nFD(ǫ). (10.49)

g(ǫ) varies on an energy scale of∼ ǫF so sinceµBB ≪ ǫF , to a good approximation we have

gσ(ǫ) =1

2g(ǫ ± µBB),

=1

2g(ǫ) ± 1

2µBB g′(ǫ). (10.50)

and therefore

g↑(ǫ) + g↓(ǫ) =

(1

2g(ǫ) − 1

2µBBg′(ǫ)

)+

(1

2g(ǫ) +

1

2µBBg′(ǫ)

),

= g(ǫ). (10.51)

118

Thus,

n = n↑ + n↓,

=

∫dǫ g↑(ǫ)nFD(ǫ) +

∫dǫ g↓(ǫ)nFD(ǫ),

=

∫dǫ g(ǫ)nFD(ǫ). (10.52)

which is the same as the zero-field equation.

Hence, we can neglect theB field dependence ofµ and from Eq. 10.23, we have

µ(T ) = ǫF

[1 + O

(kBT

ǫF

)2]

. (10.53)

The magnetisation is

M = −µB(n↑ − n↓),

= −µB

∫ ∞

−∞

dǫ nFD(ǫ) [g↑(ǫ) − g↓(ǫ)] ,

= µ2BB

∫ ∞

−∞

dǫ nFD(ǫ)g′(ǫ). (10.54)

Integrating by parts, and using the fact thatnFD(ǫ) → 0 asǫ → ∞ andg(ǫ) → 0 for ǫ ≤ 0

M = µ2BB

∫ ∞

−∞

dǫ g(ǫ)

(−∂nFD(ǫ)

∂ǫ

).

ForkBT ≪ ǫF∼= µ, the function∂nFD(ǫ)/∂ǫ is sharply peaked atǫ = ǫF and so to leading order

in (kBT/ǫF )2

∫ ∞

−∞

dǫ g(ǫ)

(−∂nFD(ǫ)

∂ǫ

)= g(ǫF )

∫ ∞

−∞

dǫ

(−∂nFD(ǫ)

∂ǫ

),

= g(ǫF ). (10.55)

The magnetic susceptibilityχ is given by

χ =M

B= µ2

Bg(ǫF ) =3N

2V

µ2B

kBTF

. (10.56)

Note thatχ > 0 i.e. this isparamagnetism.

Working to second order in(kBT/ǫF )2 we find

χ(T ) = µ2Bg(ǫF )

[1 − π2

12

(kBT

ǫF

)2]

. (10.57)

Note that thisweaktemperature dependence is in contrast to that for immobile spin-half particleswhich obey Curie’s law

χ(T ) ∼ 1

T.

119

Figure 10.8: Temperature dependence of the magnetic susceptibility of metals. Figure taken fromKittel, Introduction to Solid State Physics.

120

Metal rs/a0 Theory106χ Measured106χLi 3.25 0.80 2.0Na 3.93 0.66 1.1K 4.86 0.53 0.8Rb 5.20 0.50 0.8Cs 5.62 0.46 0.8

Table 10.1: Comparison of free electron and measured Pauli susceptibilities. The discrepanciesbetween theory and experiments are mainly due to electron-electron interactions.

Figure 10.9: Temperature dependence of magnetic susceptibility of liquid 3He. χ(T ) ∼ const forT ≪ TF ∼ 1 K. χ(T ) ∼ C/T for T > TF (Curie’s law). Taken from A. L. Thomson et al., Phys.Rev.128, 509 (1962).

121

Figure 10.10: Temperature dependence of the heat capacity of liquid 3He. CV (T ) ∼ γT forT ≪ TF ∼ 1 K. γ ∼ m∗ increases with pressure. Taken from D. S. Greywall, Phys. Rev. B 27,2747 (1983).

122

10.8 Heavy fermion metals

To conclude this chapter we briefly mention heavy fermion metals. These are systems for whichthe interaction between neighbouring electrons so strong that electrons cannot be considered inde-pendently of one another. They were discovered in the late 1970s, and some of them are super-conductors. Examples include UPt3, UBe13, CeCu2Si2, LiV 2O4, . . . . Some typical parameters forthese materials are

γ ∼ 1 J/mol K2,m∗

me

≈ 100–1000.

χe(T ) is also enhanced by a factor of 100–1000 compared to elemental metals.

123

Chapter 11

Degenerate Bose Gases

Key Concepts

• Macroscopic occupation of the ground state

• Bose-Einstein condensation (BEC)

• Critical temperature for BEC

• Superfluid4He

• Two fluid model of superfluidity

• Ultra-cold atomic gases

Reading

Schroeder, Section 7.6Kittel & Kroemer, Chapter 7.Anglin and Ketterle, Nature416, 211 (2002).Nobel prize article 2001, linked from course webpage.Bose-Einstein Condensation in Dilute Gases, Pethick and Smith, QC175.47.B65 P48 2002. (ad-vanced textbook on BEC.)

124

11.1 Macroscopic occupation of the ground state

Earlier in the course we learnt that there are two types of particles in nature — bosons and fermions.The difference between the two is that while Pauli exclusionprinciple applies to identical fermions,any number of identical bosons can co-exist in the same quantum state.

We have already derived the Bose-Einstein distribution forthe mean occupation number of bosonsoccupying a single quantum state of energyǫs

nBE(ǫs) =1

e(ǫs−µ)/kBT − 1.

The chemical potential is something of a mysterious quantity, but (in my opinion) the easiest wayto deal with it is to treat it as a normalisation constant for an isolated system.

Because we know that physically the mean occupation number of all quantum states in a systemmust be positive, this means that we must haveµ < ǫ0, the lowest energy state in the system.However,µ can bearbitrarily close toǫ0 — if we setµ = ǫ0 − ∆, then

nBE(ǫ0) =1

e∆/kBT − 1≈ 1

1 + ∆/kBT − 1=

kBT

∆.

Thus if ∆ is really small, then we can have a huge number of atoms in the lowest energy level.Obviously there must be some constraint on∆, because in a real system there are only a finitenumber of particles.

11.2 More on the density of states

In this chapter we will consider a system ofN atomic bosons of massm confined to a volumeV = L3 (an “infinite box” potential.) However, in real experimentstrapping potentials are usuallynot spatially homogeneous, and it is worthwhile to be able tocalculate the density of states forthese systems for more general treatments of BEC.

For systems where the trapping potential varies slowly on the length scale of the de Broglie wave-lengthλT , we can calculate the density of states for a three dimensional system as

g(ǫ) ≈ 1

h3

∫d3rd3p δ[ǫ − H(r,p)],

whereH = p2/2m + U(r). In your assignment you will show that

g(ǫ) = 2π

(2m

h2

)3/2 ∫

V

d3r√

ǫ − U(r).

For the infinite box, the spatial integral simply gives the volume, and leaves us with the same resultfrom earlier

g(ǫ) = 2πV

(2m

h2

)3/2 √ǫ.

125

11.3 Bose-Einstein condensation (BEC)

ConsiderN bosons of massm confined to a volumeV = L3. At zero temperature all of theNatoms will be in the ground state which has an energy of

ǫ0 =h2

8mL2

(12 + 12 + 12

)=

3h2

8mL2. (11.1)

At very low temperatures it is possible to find a macroscopic numberN0 of theN bosons still inthe ground state. The value ofN0 is given by the Bose-Einstein distribution function

N0 =1

e(ǫ0−µ)/kBT − 1. (11.2)

By taking the Taylor expansion of the denominator, we can findthat the chemical potential as afunction of temperature is given by

µ(T ) ∼= ǫ0 −kBT

N0

asT → 0. (11.3)

Hence for low temperaturesµ(T ) is very close to the energy of the ground state.

The chemical potential can be determined by calculating thetotal number of bosons

N =∑

s

nBE(ǫs) =∑

s

1

e(ǫs−µ)/kBT − 1. (11.4)

We can approximate this sum by an integralas long askBT is significantly larger than the energylevel spacing. However, if there are a large number of bosons in the ground state, there will bea spike in the functionnBE(ǫ). This is not accounted for correctly in the integral. In fact, thisapproximation may be bad forseveralof the lowest energy levels. However, only the ground statewill be wildly incorrect, and so we treatN0 separately from the integral. Thus we can write

N ≈ N0 +

∫ ∞

0

g(ǫ)dǫ

eǫ/kBT − 1(11.5)

≡ N0(T ) + Nex(T ), (11.6)

whereg(ǫ) is the density of states andNex(T ) is the number of bosons in all the excited states.

For now, we will assume thatµ(T ) = 0 when there is a large occupation of the ground state. Wecan do this because

• the energy of the ground state approaches zero as the volume becomes large;

• any change inµ will have negligible effect on the occupation of the excitedstates whenN0

is large.

We now have

Nex(T ) =2√π

(2πm

h2

)3/2

V

∫ ∞

0

√ǫdǫ

eǫ/kBT − 1

=2√π

(2πmkBT

h2

)3/2

V

∫ ∞

0

√xdx

ex − 1(11.7)

126

where we have made the substitutionx = ǫ/kBT . This integral can be tackled by expanding thefraction as an infinite series (see assignment question). The general result is

∫ ∞

0

dxxn−1

ex − 1= ζ(n)Γ(n), (11.8)

where the gamma functionΓ(n) interpolates the factorial function (i.e. for integern, Γ(n) =(n − 1)!) and the zeta functionζ(n) is defined as

ζ(n) =∞∑

m=1

1

mn.

For the 3D box, we haven = 3/2, soΓ(3/2) =√

π/2 andζ(3/2) = 2.612. Thus

Nex(T ) = 2.612 ·(

2πmkBT

h2

)3/2

V = 2.612 · V

vQ

, (11.9)

wherevQ = λ3T is the quantum volume.

We can see from the equation forNex that asT increases, the value ofNex also increases. Wedefine thecritical temperature Tc as the temperature at which the number of excited particlesNex

is equal to the total number of particles

Nex(Tc) = N. (11.10)

So, as the temperature approaches the critical temperaturethe number of particles in the groundstateN0(T ) approaches zero. Since the initial assumption thatµ(T ) = 0 was based on the factthat a large number of particles were in the ground state, fortemperatures greater than the criticaltemperature the assumption thatµ(T ) = 0 is no longer valid.

A Bose-Einstein condensate forms when the temperature is below the critical temperature, whichcan be written as

kBTc = 0.527

(h2

2πm

)(N

V

)2/3

(11.11)

The fraction of particles in the ground state of a system of bosons is given by

N0(T )

N=

{1 − (T/Tc)

3/2 T < Tc,0 T ≥ Tc.

(11.12)

Note that Bose-Einstein condensation occurs approximately when the quantum volumevQ is equalto or greater than the volume per particleV/N — this is a simple physical picture.

11.4 Why does it happen?

From the maths we have seen that BEC does occur, but this doesn’t give much understanding ofwhy it occurs. The answer to this question was hinted at in the tutorial on quantum statistics.

127

For a gas of noninteractingdistinguishableparticles, we can treat them all separately using Boltz-mann statistics. A single particle has a reasonable chance of being found in any single particlestate whose energy is of orderkBT . The available number of such single particle states is quitelarge usually, and is of order of the single particle partition functionZ1. The probability of beingin the ground state is thus1/Z1, and as this applies to each particle individually, only a minisculefraction will ever be found in the ground state away fromT = 0.

From the perspective of the system, each totalsystemstate has its own probability. The configura-tion with all particles in the ground state will have a Boltzman factor of 1, whereas the Boltzmannfactor of a typical system state at temperatureT with U ∼ NkBT will be e−NkBT/kBT = e−N .This seems rather small — however, we must remember to multiply it by the number of differentarrangements of theN particles. ForZ1 single particle states, the number of arrangements isZN

1 ,and this factor overwhelms thee−N .

Now think about the case of identical bosons. When there are more available states than particles(Z1 ≫ N ) then the situation is similar to that of distinguishable particles. However, when we haveZ1 < N , then there arefar fewer arrangements of identical bosons as compared to distinguishableparticles. Thus the probability of the system being in its ground stateis much higher than fordistinguishable particles.

The number of system statesNs is roughly the number of ways of arrangingN indistinguishableparticles amongZ1 single particle states — which is mathematically the same asdistributingNunits of energy amongZ1 oscillators. It is given by

Ns ∼(

N + Z1 − 1N

)∼{

(eZ1/N)N whenZ1 ≫ N,

(eN/Z1)Z1 whenZ1 ≪ N.

(11.13)

If Z1 ≫ N then(eZ1/N)N ≥ e−N , then the probability of the system being in an excited stateisvery close to one. However, ifZ1 ≪ N , then the total number of states is not very large, and thee−N term becomes more significant making the probability of the system being in an excited statevery low, thus allowing a Bose-Einstein condensate to form.

It is worth comparing the transition temperature with the energy of the ground state of the system

kBTc = 0.527

(h2

2πm

)(N

V

)2/3

= N2/3 0.527h2

2πmL2,

ǫ0 =3h2

8mL2.

Both are of orderh2/mL2, however the BEC temperature is enhanced by a factor ofN2/3.

11.5 Important points

11.5.1 Thermodynamic quantities

The behaviour of the condensate occupation, the chemical potential, and the heat capacity of theBose gas in the region ofTc is shown in Fig. 11.1 for the case of the infinite box potential. All

128

thermodynamic quantities can be calculated relatively easily, but we have to resort to numericalsolution and we will not do so in the course. However, the samegeneral principles as used earlierapply.

An important point to note is the peak in the heat capacity atTc. This is relevant in the case ofsuperfluid helium (see later). The condensate fraction of the Bose gas does not contribute to theheat capacity or the entropy of the system. The heat capacitybehaves asCV ∼ T 3/2 belowTc, andreduces to3NkB/2 well aboveTc.

11.5.2 BEC in other potentials

BEC does not occur in every Bose gas system, and it is worth looking at the maths to understandwhy. A BEC forms when we haveµ → 0 so that the ground state contains a macroscopic numberof particles. This will only happen if the integral for the number of excited particles Eq. (11.7)reaches a limit asµ → 0. From the result given in Eq. (11.8), if the density of statesis given byg(ǫ) ∝ ǫn−1, then we must haven > 1 so that the zeta function

ζ(n) =∞∑

m=1

1

mn,

will converge. Ifn ≤ 1, the result for the number of excited state particles diverges, and so ourapproximation thatµ ≈ 0 was incorrect. This means that the excited states can accomodate allparticles at any temperature, and there is no BEC.

For a 2D homogeneous system, we haveg(ǫ) ∼ constant, and so there is no BEC.

11.5.3 Massless bosons

We have already considered ideal gases of bosons — both photons and phonons. However, wedo not get BEC in these systems. The reason is that there is no restriction on the particle number— phonons and photons can be created and destroyed at will, and soµ = 0 always. It is therequirement that particles be conserved that results in theformation of a BEC.

11.5.4 Examples of BEC

The phenomenon of BEC and macroscopic quantum phenomena areintricately linked, and thereare three real-world examples of where Bose-Einstein statistics are important. The first is in super-conductors, where electrons can form boundCooper pairs below a transition temperature. Theycan be thought of as composite bosons forming a BEC, and can flowwithout resistance throughthe conductor.

The second two examples are superfluid helium and trapped atomic gases, and we will considerthese in more detail in the following sections.

129

0 0.5 1 1.5 2

T / Tc

NN

0N

ex

(a)

0 0.5 1 1.5 2−1

−0.8

−0.6

−0.4

−0.2

0

µ / k

B T

T / Tc

(b)

0 0.5 1 1.5 20

0.5

1

1.5

2

CV /

N k

B

T / Tc

(c)

Figure 11.1: Thermodynamics of a Bose gas in the region of thecritical temperature for Bose-Einstein condensation. (a) The behaviour of the ground state occupation. (b) The chemical poten-tial as a function of temperature. (c) The behaviour of the heat capacity. Note the peak atTc, andCV → 3/2NkB for T ≫ Tc.

130

Figure 11.2: The specific heats of (a)4He and (b)3He showing the anomalies associated with theirrespective phase transitions. Note that the temperature scales in (a) and (b) are quite different,being a factor of103 lower for 3He.

11.6 Superfluid helium

Helium-4 is a bosonic atom that liquifies at a temperature of 4.2 K at atmospheric pressure. How-ever, it undergoes another phase transition atTc = 2.17 K to a superfluid phase (called Helium-II)that has essentially zero viscosity. This transition occurs at the “lambda-point”, so called becausethe heat capacity diverges atTc, and a plot ofCV versusT is similar to the Greek letter — seeFig. 11.2.

The phenomenon was discovered by Kapitsa, Allen and Misenerin 1938. In 1962 Landau receiveda Nobel prize for his theory of superfluidity and in 1978 Kapitsa also received a Nobel prize forhis work in superfluidity. The superfluid state involves macroscopic quantum coherence, similarto that responsible for superconductivity.

The theory of BEC in an ideal gas of atoms was first published byEinstein in 1924 (before fermionswere known about), but was pretty much ignored for several years. In fact, it was even suggested

131

Figure 11.3: Phase diagrams of4He (left) and3He (right). Neither diagram is to scale, but quali-tative relations between the diagrams are shown correctly.Not shown are the three different solidphases (crystal structures), or the superfluid phases of3He below 3 mK.

by Uhlenbeck that the prediction of a phase transition was incorrect (and there is evidence thatEinstein accepted this criticism!) However, when superfluidity was discovered Fritz London sug-gested that Bose condensation could be at the heart of the physics. In fact, using the ideal gasformula for the ideal gas BEC transition temperature gives agood estimate for4He. This idea wasfurther strengthened when it was found there was no similiarsuperfluid transition in fermionic3He(see Fig. 11.3). The existence of a BEC in superfluid4He was argued about for many years, butnow it is accepted that this is the case.

[N.B. thereis a superfluid transition in3He, but at milliKelvin temperatures. Here the fermionsform Cooper pairs, similar to electrons in a superconductor.Nobel prizes were awarded for thiswork in 1996 (Lee, Osheroff, and Richardson, experiment) and 2003 (Leggett, theory).]

There is one important fact about superfluid4He — it is necessary to consider theinteractionsbetween the atoms to explain superfluidity, and so it is necessary to go beyond the ideal gas model.The interactions are very strong, and the system is difficultto treat theoretically. However, theweakly-interacting Bose gas turns out to be a reasonable qualitative model, and much work wasdone on this in the 1950s and 1960s in the context of superfluid4He.

11.6.1 Two fluid model of superfluidity

For Bose-Einstein condensates we have

N = The total number of bosons,

= N0(T ) + Ne(T ),

= Number in the ground state+ number in excited states.

It turns out that a very good model for superfluid helium is to treat it as a mixture of two com-ponents, superfluid and normal fluid, with the proportion of each component depending on the

132

Property Superfluid Normal fluidViscosity 0 non-zeroThermal conductivity ∞ finiteEntropy 0 nonzero

Table 11.1: Properties of superfluid and normal fluid components

temperature. In terms of density we write

ρ = ρs(T ) + ρn(T ). (11.14)

However, it isnot the case that the superfluid component directly correspondsto atoms in theground state of the system. In fact, by using neutron scattering methods it has been determinedthat even in the absence of a normal component only about 10% of the atoms are in the condensatestate. This is due to the strong interactions between the atoms — this results in largedepletionofthe condensate.

11.6.2 Physical properties

Superfluid4He does not boil, i.e. no bubbles form when it is heated. This is because thermalconduction within the4He occurs via the superfluid component, which hasinfinite thermal con-ductivity. This means that if one part of the fluid is heated, the heat is transferred throughout thefluid immediately. Thus it isimpossibleto set up a temperature gradient within the fluid (which isrequired to form bubbles.)

Heat transfer within a superfluid occurs not by conduction, but by the movement of the normalcomponent. If a heat source is placed somewhere within a superfluid, and a heat sink placedelsewhere within the superfluid, the normal fluid component moves from the source towards thesink, and the superfluid component moves in the opposite direction. This motion occurs under theconstraint that the total density remains constant throughout the superfluid.

This fact results in some remarkable properties. Table 11.1contains a comparison of some of theproperties of the superfluid and normal fluid components.

11.6.3 Superleaks

A superleak is a flow of the superfluid component through a highly non-porous material that thesuperfluid can flow through, but the normal fluid component cannot pass through due to viscosity.

If in the system shown in Fig. 11.4 we haveT1 > T2 andP1 = P2, then for anormalconnectionthermal equilibrium is reached by the conduction of heat. Thus the fluid carries entropy from 1 to2.

For a superleak this is NOT possible because only the superfluid can pass from 1 to 2 and thesuperfluid carries no entropy. SupposeT1 were to slowly increase. For thermal and diffusive

133

Superleak

T ,P

1 2powder

superfluid

1 1

T ,P2 2

Figure 11.4: Representation of a superleak. The superfluid can pass through the powder due to itsvanishing viscosity, however the normal component cannot.

equilibrium to hold it is required thatµ1 = µ2. Using the thermodynamic identities

G = µN = U − TS + PV,

and

TdS = dU + PdV − µdN,

we get the equation

Ndµ = −SdT + V dP (11.15)

From the conditions for equilibrium we know that

µ1 − µ2 = 0 = −S(T1 − T2) + V (P1 − P2), (11.16)

and therefore

P1 − P2 =S

V(T1 − T2). (11.17)

This leads to several interesting phenomena.

11.6.4 Thermomechanical effect

It is possible to establish a pressure difference between two superfluids connected by a superleakby creating a temperature difference.

The capillary in Fig. 11.5 contains a superleak. The heater in the main section raisesT1 aboveT2,and this causesP1 to become greater thanP2. The result is that the superfluid is forced through thecapillary.

134

Figure 11.5: A fountain based on the thermomechanical effect

11.6.5 Mechanocaloric Effect

Similarly, creating a pressure difference between two systems connected by a superleak will causea temperature difference to arise. This is in contrast to a normal fluid, for which a pressure differ-ence causes a mass flow.

The magnitude of this effect can be found as follows

∆T

∆P=

V

S.

If we let ∆P = ρg∆z, then this gives

∆T

∆z=

ρg

S, (11.18)

whereρ is the mass per mole andS is the entropy per mole. For superfluid4He at aroundT = 1.3K we have

∆T

∆z∼ 1 mK cm−1.

Thus a temperature difference of only 1 mK will produce a height difference of1 cm! The quanti-tative details of this effect have been experimentally verified.

135

11.6.6 Other phenomena

Superfluid helium displays a variety of other interesting phenomena. These include

• “Second sound” — temperature waves rather than pressure waves, caused by the normal andsuperfluid fractions moving out of phase.

• Vortices and quantised circulation.

11.6.7 Video

If we have time, part of the video “Superfluid helium” by J.F. Allen, University of St. Andrewswill be shown. Jack Allen’s obituary follows at the end of this chapter.

11.7 Ultra-cold Bose gases

In 1995 at the University of Colorado in Boulder, a team of physicists lead by Eric Cornell andCarl Wieman suceeded in creating the first ever Bose-Einsteincondensate in a dilute gas of87Rbatoms. The condensate they formed consisted of only 2000 atoms at a temperature of about 100nK. This temperature was achieved by first laser cooling in a vacuum to microKelvin temperatures,before transferring the atoms to a magnetic trap. The walls of the trap were then lowered, allowingevaporative cooling to occur. In this process the hottest atoms in the velocity distribution escape,and the remaining atoms cool via collisions and “rethermalisation” (the process by which the atomsreturn to equilibrium).

At these temperatures you would normally expect87Rb to be a solid! However, if they are keptsufficiently dilute only two particle collisions will ever occur, and these must be elastic. Threeparticles are required to collide such that two form a molecule, and the third takes away the excessenergy. Therefore the gas will bemetastable— however this requires it to be 100,000 times lessdense than air!

The important feature of BEC in atomic gases is that the interactions are much, much weaker thanin superfluid helium, and therefore the system is much closerto the ideal Bose gas that Einsteinfirst imagined. In fact, quantities such asTc are not much shifted from the values you will calculatein your assignment. This also means that it is feasible to treat these systems theoretically (althoughit is still not easy!)

A wide variety of interesting phenomena have been observed in gaseous Bose-Einstein conden-sates, including vortices, vortex lattices, ultra-slow light (17 m s−1), molecular BEC, four-wavemixing, entangled atomic beams, BECs in periodic lattices, and so on. One of the most talkedabout possibilities is to make atom lasers from BECs, consisting of a continuous stream of atomssharing the same quantum mechanical phase. The Nobel Prize was awarded to researchers in thisfield in 1997 (laser cooling) and 2001 (BEC), and it remains a very active area today. There is

136

Figure 11.6: Shadow images of a degenerate Bose gas of87Rb, taken at UQ in February 2004.From left to right: aboveTc, slightly belowTc, well belowTc. There are about 30,000 atoms in thepicture on the right.

a large theory team at UQ (the Centre of Excellence for Quantum-Atom Optics), as well as anexperimental BEC group.

11.8 Ultra-cold Fermi gases

Even more recently, interest has exploded in the field of degenerate Fermi gases. In a similarmanner to Bose gases, they can be cooled to very low temperatures. However, it is more difficultwith fermions, as the collisional process between identical fermions that is required for evaporativecooling turns off at very low temperatures. The solution is either to use a Bose gas “refrigerator”,or to cool and trap two different spin states simultaneously.

Currently people are working on trying to observe superfluidity in two-component degenerateFermi gases, in the hope that this will teach us about the nature of high temperature superconduc-tivity. In particular there is interest in what is known as the “BEC-BCS crossover” — the transitionfrom having a BCS-like state of weakly paired fermions forming the superfluid to having stronglybound fermions making up a bosonic molecule and BEC. This is also a topic of research at UQ. Itis predicted that in the next few years a Nobel prize is likelyto be awarded in this area as well.

137

news and views

436 NATURE | VOL 411 | 24 MAY 2001 | www.nature.com

In late 1937, two young physicists at theRoyal Society Mond Laboratory inCambridge, UK, found that liquid heliumcould flow through very small capillarieswith essentially zero viscosity below atemperature of 2.17 kelvin. While theywere preparing a note for publication, theyheard of another paper, just submitted byP. L. Kapitza in Moscow, reporting similarresults. Both papers appeared side by sideon 8 January 1938 in Nature. Kapitza andthe young physicists Jack Allen and A. D.Misener had discovered the mysteriousphenomenon of superfluidity in liquid4He. It was the start of a golden period inlow-temperature physics. Allen, the lastsurvivor of this heroic time, died of astroke on 22 April 2001, aged 92. Most of Allen’s greatest discoverieswere made at the outset of his career (in1938 he was essentially a postdoc, workingwithout a supervisor, which suited himfine). Between 1937 and 1939 he and hisassociates at Cambridge produced astream of papers on superfluidity in liquid4He, this output coming to an abrupt endwith the start of the Second World War. In 1946, with normal life resuming, Allenorganized the first international low-temperature meeting at the University of Cambridge, UK. This was the birth of the major triennial event in the low-temperature physics community; thetwenty-third meeting will be held inHiroshima next year. In 1947, Allen was appointed professorof natural philosophy and head of physicsat the University of St Andrews inScotland, and two years later he became aFellow of the Royal Society. But althoughhe stayed active in the world of low-temperature physics, revitalizing researchat St Andrews and helping to run manyconferences and workshops, his ownresearch was never again central. Heretired from St Andrews in 1978. Jack Allen was born in Winnipeg,Canada, in 1908 (the year 4He was firstliquefied in Leiden, in the Netherlands).He received his BA from the University ofManitoba, where his father was head ofthe physics department. In 1929, he wentto study at the University of Torontowhere James F. McLennan had built up astrong physics department. In 1923 thishad been only the second laboratory in theworld to liquefy helium, and when Allenarrived he immersed himself in the newworld of cryogenics. He received his PhDin 1933, having already written ten paperson superconductivity. Late in 1935 he went

to Cambridge to work with Kapitza, onlyto find that Kapitza was being detained inMoscow, and was in the process of settingup what was to become the famousInstitute of Physical Problems. InKapitza’s absence, Allen effectively tookover the low-temperature work, althoughofficially the director was J. D. Cockroft. In 1935, Don Misener, a graduatestudent at Toronto, had carried out thefirst experimental study of the viscosity of liquid 4He. By then it was known thatliquid helium underwent some sort ofphase transition at 2.17 K, as there wereabrupt changes in various thermodynamicproperties. Misener’s work suggested thatthe viscosity decreased substantially whenthe liquid passed through this transition.Misener joined Allen at Cambridge in1937 to do his PhD, and the two set out tostudy the phenomenon by examining flowin thin capillaries.Misener’s 1935 experimental work hadalso attracted the notice of Kapitza inMoscow. Both groups reported theirindependent discovery of superfluid flowin 1938, with Kapitza being the first tocoin the term ‘superfluid’. It is puzzlingand unfortunate that when Kapitza finallyreceived a well-deserved Nobel prize inphysics in 1978, the citation concerningsuperfluidity made no reference to thework of Allen and Misener. Allen quickly found other dramaticmanifestations of superfluidity, all ofwhich involved the counterflow of thenormal and superfluid components or the‘clamping’ of the normal component with

fine powder. But after 1945 the Moscowgroup under Kapitza (helped by L. D.Landau, who developed a complete theoryof the two-fluid behaviour of superfluidhelium in 1941) dominated furtherresearch on quantum liquids. The study of superfluid 4He increasingly involvedmicroscopic theories and newexperimental probes such as neutronscattering, none of which interested Allen. Allen was the last of a generation ofindependent-minded classical physicistswho delighted in explaining the visibleworld. He prized his own ability, and thatof glass-blowers and technical people, tobuild experimental apparatus. Allen wasas proud of his invention of the O-ringvacuum seal as anything else he did. Itshould be no surprise that his greatestwork on superfluidity in liquid 4Heinvolved phenomena that could be seen.Indeed, it is most fitting that Allendiscovered the famous ‘fountain effect’ in1938 with the help of a pocket flashlight. Over a ten-year period Allen made amovie of the various two-fluid phenomenaexhibited by liquid 4He. The photographyof these effects was a real challenge,because liquid 4He is essentiallytransparent. This unique colour movie(the fifth edition was completed in 1982) is one of Allen’s great legacies to physics.Allen had a commanding presence and a dry sense of humour. He stronglyidentified with the physics of an earlierday, and I imagine that he would haveenjoyed talking with classical physicistssuch as Lord Rayleigh, Michael Faradayand Daniel Bernoulli more than withWerner Heisenberg and ErwinSchrödinger. Superfluidity is a dramaticvisible manifestation of quantummechanics, being the result ofBose–Einstein condensation in which amacroscopic number of 4He atoms occupythe same, single-particle quantum state. It is paradoxical that the phenomenon wasfirst observed by Allen: a great physicistwho wasn’t much interested in atoms. Walking the old stone streets of StAndrews, one quickly notices the elegantmetal historical plaques on manybuildings, commemorating famous peoplewho have been associated with thishistoric town and its university over thecenturies. Almost every plaque connectsits subject to physics — not surprising,because Jack Allen was the motivatingforce behind the sign committee. I verymuch hope that the town sees fit to sohonour Allen himself. Allan Griffin Allan Griffin is in the Department of Physics,University of Toronto, Toronto, Ontario, M5S1A7, Canada.e-mail: griffin@physics.utoronto.ca

ObituaryJohn Frank (Jack) Allen (1908–2001)

Co-discoverer ofsuperfluidity

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