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COMMONWEALTH OF AUSTRALIA

Copyright Regulations 1969

WARNING

This material has been reproduced and communicated to you by or on

behalf of the University of Queensland pursuant to Part VB of the

Copyright Act1968 (the Act).

The material in this communication may be subject to copyright

under the Act. Any further reproduction or communication of this

material by you may be the subject of copyright protection under the

Act.

Do not remove this notice.

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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 effect

on the macroscopic properties of the system. Examples include:

Conduction (free) electrons in a metal Liquid 3He

Neutrons in a neutron star

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Electrons in a white dwarf star Protons and neutrons in an atomic nucleus

We assume the temperature is sufficiently low that

vQ =

h

2mkBT

3 V

N.

This is the opposite condition to that required for the validity of Boltzmann statistics. Here, we are

considering the case where the fermions are packed very closely together so that quantum statistics

are important.

As a first step, we take T = 0.

9.2 Zero temperature

I

0

n ()FD

= F

Figure 9.1: Fermi-Dirac distribution at T = 0.

All single particle states with < are occupied, and all with > are unoccupied mathemat-ically, we obtain this by letting T 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 because

we are looking at fermions, we cant have more than one fermion per state (two if we consider

spin). Therefore the particles fill up all the lowest energy states in order.

The Fermi energy is defined by

F (T = 0). (9.1)N.B. The chemical potential varies with temperature we will show this in a moment.

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

=

U

N

S,V

UN

. (9.2)

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

(T = 0) =U

N= F.

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 higher the Fermi energy.

We now calculate F for a system of N free electrons confined to a cube of volume V = L3. We

first find the allowed energy levels.

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

n =2L

n, pn =

h

n=

hn

2L,

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

three dimensions.

In three dimensions, the allowed energies are

(p) =p2

2m,

= 12m

(p2x +p2y +p

2z),

=h2

8mL2(n2x + n

2y + n

2z). (9.3)

In n-space, a constant energy surface is a sphere. The number of states with energy less than Fis the volume of a sphere with corresponding radius nmax given by

F =h2

8mL2n2max, (9.4)

Because all particles have energy less than F by definition, this must be N, and so we can find

nmax and hence F in terms ofN.

N = 2 volume of 18

of a sphere,

= 2 18

43

n3max,

=

3n3max, (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

F =h2

8mL2 n2max =h2

8m 3NV2/3

. (9.6)

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(compare with the calculation of the Debye wavevector). Note that F only depends on the densityN/V, meaning it is intensive if we double the size of the system, F does not change.

We now reconsider the requirement for degeneracy V /N vQ. This is equivalent to

kBT F. (9.7)We can also define the Fermi temperature by

TF FkB

, (9.8)

so that the degeneracy condition becomes

T TF. (9.9)

In a typical elemental metal, n = N/V

1022

1023cm3, with F

110 eV. Hence, TF

104105 K, and so most metals at room temperature are highly degenerate and it is often sufficientto consider them to be at T = 0!

Figure 9.2: Free electron Fermi gas parameters for metals at room temperature. Taken from Kittel,

Introduction to Solid State Physics.

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9.3 Internal energy at zero temperature

We can now calculate the internal energy per particle of a Fermi gas. We would expect that it

would be approximately F/2, since the particles have all energies between 0 and F.

U = 2 nx,ny,nz

(n) 2d3n(n),= 2

nmax0

dn n2

2

0

sin d

2

0

d(n),

=

nmax0

(n)n2dn,

=h2

8mL2

nmax0

n4dn,

=

h2

40mL2 n5max =

3

5 NF. (9.10)

So the average energy per particle is 3F/5.

9.4 Degeneracy pressure

The Fermi gas has pressure even at T = 0

P =

U

VS,N ,=

V

3

5N

h2

8m

3N

23

V2

3

,

=2N F

5V=

2U

3V. (9.11)

This last equality is the same as for a classical ideal gas and is required by Newtons laws. We can

also write

P V =2

5N kBTF. (9.12)

This pressure is due to the exclusion principle, which prevents two fermions being in the same

state. When the particles are degenerate (or close to it) they resist being pushed closer together.

This degeneracy pressure prevents the gravitational collapse of white dwarf and neutron stars (see

tutorial problem).

9.5 Bulk modulus

The bulk modulus is the change in pressure over the fractional change in volume a high bulk

modulus means the material is difficult to compress, while an easily compressible material has a

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Metal Free electron B Measured B(1010 dynes cm2) (1010 dynes cm2)

Li 23.9 11.5

Na 9.23 6.42

K 3.19 2.81

Rb 2.28 1.92Cs 1.54 1.43

Cu 63.8 134.3

Ag 34.5 99.9

Al 228 76.0

Table 9.1: Bulk moduli for some typical metals. The free electron value is that for a free electron

gas at the observed density of the metal.

small B. Mathematically,B = V

PV

T

=1

, (9.13)

where is the compressibility. For a Fermi gas,

P =2

5N

FV

V2

3

V= V

5

3 .

So then,

B =5

3P =

2

3N

FV

=2

3

N

VkBTF. (9.14)

Compare this to an ideal gas where B = P = N kBT /V (check this!). Electrons in a (roomtemperature) metal have a much higher bulk modulus (because usually TF 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 by

that due to the degeneracy pressure of the electrons.

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Chapter 10

Degenerate Fermi gases at low temperatures

Key Concepts

T TF, but T > 0. Density of states. Temperature dependence of chemical potential (T). Sommerfeld expansion.

Heat capacity CV(T)

T.

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 case 0 < T F/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.

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I

0

n ()FD

T = 0

F

K TB

Figure 10.1: Fermi-Dirac distribution at low temperatures.

Now a fraction of the electrons, of order NkBT /F have kinetic energy kBT greater than F. Thuswe expect the internal energy at low temperatures to be

U(T) U(T = 0) + aN

kBT

F

kBT,

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

CV(T) =

U

T

V

a Nk2B

FT,

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 have

already come across the density of states in treating both phonons and black body radiation, without

actually identifying it.

The density of states, denoted g() is the number of single particle states per unit energy thenumber of states near energy .

If we let N(E) be the total number of single particle states with energy less than E thenN(E) = E

d g()

It follows that

g(E) = d N(E)dE , (10.1)107

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Figure 10.2: Fermi-Dirac distribution at various temperature for TF = 50000 K. The results applyto an ideal gas in three dimensions with a constant total number of particles. You can read off the

chemical potential for each of these distributions from the graph the energy at which f = 0.5.Figure taken from Kittel, Introduction to Solid State Physics.

that is, the number of states between E and E+ dE is g(E) dE.

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

g() =

( ) (10.2)

where (x) is the Dirac delta function, defined such that

(x) =

for x = 0,0 otherwise.

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

dx (x) = 1.

It is also the derivative of the step function (x) which is defined to be

(x) =

1 for x 0,0 otherwise.

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It follows that

N(E) = E

d g(),

= E

d ( ),=

E

d ( ),

=

(E ). (10.3)

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

For a system ofN fermions at T = 0, F is defined by

N =no. of energy levels with

F=

(F )

=

F

g() d (10.4)

The total energy at T = 0 is

U(T = 0) =

(F ) =F

d g(). (10.5)

The above expressions generalise naturally to finite temperature:

N(T) =

nFD()

=

d g()nFD(), (10.6)

and

U(T) =

nFD()

=

d g()nFD(). (10.7)

In words, to find U 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 at T = 0is a backwards step function with the jump down at F, 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 to

the boundary condition that the wave function vanishes on the surface of the cube.

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Previously, we showed that the allowed values of the energy of a single state particle (fermion or

boson) are

(p) =p2

2m=

h2

8mL2(n2x + n

2y + n

2z),

for nx, ny, nz = 1, 2, . . .. ThenN(E) = total no. of states with energy E,= volume of

1

8of the sphere in n-space,

=1

8 4

3n30,

where E = h2n20/8mL2. This is similar to what we did for the T = 0 case, except that we are

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

We thus have

N(E) = 6

8mL2

h2E

3/2,

=4

3L3

2m

h2

3/2E3/2. (10.8)

And so

g(E) =d N(E)

dE= 2L3

2m

h2

3/2 E. (10.9)

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

10.2.2 Density of states in a three dimensional cube method 2

Lets repeat the derivation using the definition

g() =

( ),

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

(n) = h28mL2

(n2x + n2y + n

2z).

We convert the sum to an integral (in polar co-ordinates) and if the energy spacings are small, this

is a good approximation.

g() =

h

2

8mL2|n|2

, (10.10)

=

0

dn n2/20

sin d

/20

d

h

2

8mL2n2

, (10.11)

=

2 0 dn n2 h2

8mL2 n2 . (10.12)110

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When working with this definition of the density of states, the next step is to make a substitution

in the delta function. Here we set

x =h2

8mL2n2,

so

n = 2L2m

h2

1/2 x,

n2dn = 8L3

2m

h2

3/2x 1

2

dxx

.

Thus,

g() = 4L3

2m

h2

3/2 0

dx

x( x),

= 2L32mh23/2 ,

which agrees with the previous result. We also define g( 0) = 0 there are no states withenergy less than zero in this case. Such states can correspond to bound states such as occur in

potential wells, etc.

10.3 Temperature dependence of the chemical potential

For a general density of states g(), we said before that

N =

d g()nFD(),

=

d g()1

e()/kBT + 1. (10.13)

IfN is fixed, then Eq. (10.13) determines as a function of temperature as T increases, mustdecrease.

To see why, note that nFD has some symmetry about = , in the sense that the probability ofa state at energy + being occupied is the same as the probability of a state at beingunoccupied.

The total number of particles in the system is determined by Eq. (10.6). Lets consider first a

constant density of states g() g.

If the number of particles in the system is conserved, then at non-zero temperatures, the area under

the graph should still be N. Because of the symmetry of nFD about , the missing area below from the T = 0 distribution must be equal to the additional area above , and so the total area isindeed still N.

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= (T = 0)F

g (E)

area = N

= (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 with ,as shown in Fig. 10.3.

Now, the area under the graph to the right of will be scaled more than the empty area abovethe line to the left, due to the higher density of states on the right. Therefore, the area of the graph

is greater than N, and it would appear the number of particles has increased!

This problem is resolved if, as the temperature increases, decreases slightly, so that the peak ofnFD moves to the left, and isnt scaled as much by g(), reducing the overall size.

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

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~5k TB

-dn /dFD

Figure 10.5: The derivative of the Fermi-Dirac distribution is negligible everywhere except within

a few kBT of.

10.4 The Sommerfeld expansion

We now explicitly find (T) when kBT F = (T = 0). From

N =

d g()nFD(), (10.14)

we first integrate by parts to get

N = N()nFD()

d N()dnFDd (). (10.15)As , N() 0 and nFD() 1. Also, as +, nFD() 0, so the boundary termis zero.

For kBT F, nFD/ is sharply peaked at = . We can see this mathematically by

dnFDd

= dd

1

e()/kBT + 1,

=1

kBT

ex

(ex

+ 1)2

. (10.16)

where x = ( )/kBT.

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

N =

dx N( + xkBT) ex(ex + 1)2

. (10.17)

We perform a Taylor expansion of N() about = . This is reasonable, because the total numberof states varies slowly on the energy scale of kBT when T is small.

N( + xkBT) = N() + N()xkBT + 12 N()(xkBT)2 + . . .113

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However, by definition, N() = N()

=

= g( = ),

so that

N = N()I0 + kBT g()I1 + 12

(kBT)2g()I2, (10.18)

where

In =

dx xnex

(ex + 1)2,

I0 =

dx

d

dx

1

ex + 1

= 1,

I1 = 0, since the integral is antisymmetric about x = 0

I2 = 2

3. (See Appendix B5 in Schroeder)

Thus, we find that

N = N() + 26

g()(kBT)2 + . . . (10.19)

Now, we let

(T) = F a(kBT)2 + . . . (10.20)[remembering that F (T = 0)] and substitute in this expression and work consistently tosecond order in T:

N = N(F) a(kBT)2g(F) + 26

g(F)(kBT)2. (10.21)

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

a =2

6

g(F)

g(F), (10.22)

and so to second order in kBT /F,

(T) = F 2

6g(F)g(F)

(kBT)2, (10.23)

For a three dimensional system we have g() sog(F)

g(F)=

1

2F

and so we do have an expansion on powers of kBT /F.

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10.5 Heat capacity

We now evaluate the internal energy

U(T) =

d g()nFD(), (10.24)

for kBT F. The calculation is similar in spirit to that for finding (T).

Let h() be defined by

h() =

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

noting that h(F) = U(T = 0). Integrating U(T) by parts gives

U(T) =

d [g()] nFD(),

=

dd

d

dE[Eg(E)]

nFD(),

= h()nFD()|

d h()

d

dnFD()

. (10.26)

Again, because nFD() 0 as and h() 0 as , the boundary term is zero.Proceeding as before,

U(T) =

dx h( + xkBT)ex

(ex + 1)2. (10.27)

We can expand h as a Taylor series

h( + xkBT) = h() + xkBT h() +

1

2(xkBT)

2h() + . . . (10.28)

and substituting this into the integral gives

U(T) = h() +2

6(kBT)

2h() + . . . (10.29)

Then, we previously showed that

(T) = F 2

6

g(F)

g(F)(kBT)

2, (10.30)

so we can Taylor expand h() around F, giving

h() = h(F) h(F) 2

6

g(F)

g(F)(kBT)

2. (10.31)

From the definition ofh(),

h(F) = F g(F), (10.32)

h(F) = g(F) + F g(F). (10.33)

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Collecting terms, we have

U(T) = U(T = 0) 2

6Fg

(F)(kBT)2 +

2

6(g(F) + Fg

(F)) (kBT)2. (10.34)

The second and fourth terms cancel exactly, leaving

U(T) U(T = 0) = 26

g(F)(kBT)2 (10.35)

It then follows immediately that

CV(T) =

U

T

V

= T , (10.36)

where

=2

3g(F)k

2B. (10.37)

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

F =h2

8m

3NV

23, (10.38)

and the density of states per unit volume is

g(F) =

2h3(8m)

3

2

F, (10.39)

which can be rewritten in terms of the Fermi energy as

g(F) =3N/V

2F. (10.40)

Notice that g(F) is completely determined by the conduction electron density N/V. We can alsowrite

g(F) =m

()2kF, (10.41)

where kF is the Fermi wave number.

Note that for fixed density, g(F) scales with m. Ifg(F) is larger than for the free (non-interacting)electron model, we say the electrons have an effective mass given by

m = meobsfree

. (10.42)

10.6 Low temperature heat capacity of metals

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

CV(T) = T + T3.

electrons phonons

Thus, plotting CV(T)/T against T2 should be a straight line with gradient and intercept at T = 0

of.

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Figure 10.6: Low-temperature measurements of the heat capacities per mole of copper, silver and

gold. Figure taken from Schroeder.

10.7 Magnetic susceptibility of metals Pauli paramagnetism

In a magnetic field B, the energy of the electrons becomes

= + BB, (10.43)

= BB, (10.44)due to Zeeman splitting, where is the energy in the absence of a magnetic field. (The arrowsubscript represents whether the electron spin is parallel or anti-parallel to the magnetic field.

Remember the intrinsic magnetic moment of elecrons is in the opposite direction to its spin as it is

negatively charged.)

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

g() =1

2g( BB), (10.45)

g() =1

2g( + BB), (10.46)

i.e. the density of states of the spin up electrons (for example) is half the density of states in no

magnetic field at an energy BB less than .

The total number density of electrons of spin = (, ) is

n = d g()nFD(), (10.47)117

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Figure 10.7: Experimental values of electronic heat capacity. Taken from Kittel, Introduction to

Solid State Physics.

where

nFD() =1

e() + 1 . (10.48)

The chemical potential (T) is determined by

n = n + n,

where

n =

d g()nFD(). (10.49)

g() varies on an energy scale of F so since BB F, to a good approximation we haveg() =

1

2g( BB),

=1

2g() 1

2BB g

(). (10.50)

and therefore

g() + g() =

1

2g() 1

2BBg

()

+

1

2g() +

1

2BBg

()

,

= g(). (10.51)

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Thus,

n = n + n,

=

d g()nFD() +

d g()nFD(),

= d g()nFD(). (10.52)which is the same as the zero-field equation.

Hence, we can neglect the B field dependence of and from Eq. 10.23, we have

(T) = F

1 + O

kBT

F

2. (10.53)

The magnetisation is

M = B(n n),= B

d nFD() [g() g()] ,

= 2BB

d nFD()g(). (10.54)

Integrating by parts, and using the fact that nFD() 0 as and g() 0 for 0

M = 2BB

d g()

nFD()

.

For kBT F = , the function nFD()/ is sharply peaked at = F and so to leading orderin (kBT /F)

2

d g()

nFD()

= g(F)

d

nFD()

,

= g(F). (10.55)

The magnetic susceptibility is given by

=M

B= 2Bg(F) =

3N

2V

2BkBTF

. (10.56)

Note that > 0 i.e. this is paramagnetism.

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

(T) = 2Bg(F)

1

2

12

kBT

F

2. (10.57)

Note that this weak temperature dependence is in contrast to that for immobile spin-half particles

which obey Curies law

(T)

1

T.

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Figure 10.8: Temperature dependence of the magnetic susceptibility of metals. Figure taken from

Kittel, Introduction to Solid State Physics.

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Metal rs/a0 Theory 106 Measured 106

Li 3.25 0.80 2.0

Na 3.93 0.66 1.1

K 4.86 0.53 0.8Rb 5.20 0.50 0.8

Cs 5.62 0.46 0.8

Table 10.1: Comparison of free electron and measured Pauli susceptibilities. The discrepancies

between theory and experiments are mainly due to electron-electron interactions.

Figure 10.9: Temperature dependence of magnetic susceptibility of liquid 3He. (T) const forT TF 1 K. (T) C/T for T > TF (Curies law). Taken from A. L. Thomson et al., Phys.Rev. 128, 509 (1962).

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

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10.8 Heavy fermion metals

To conclude this chapter we briefly mention heavy fermion metals. These are systems for which

the 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, LiV2O4, . . . . Some typical parameters for

these materials are

1 J/mol K2, m

me 1001000.

e(T) is also enhanced by a factor of 1001000 compared to elemental metals.

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Chapter 11

Degenerate Bose Gases

Key Concepts

Macroscopic occupation of the ground state Bose-Einstein condensation (BEC) Critical temperature for BEC Superfluid 4He

Two fluid model of superfluidity

Ultra-cold atomic gases

Reading

Schroeder, Section 7.6

Kittel & Kroemer, Chapter 7.

Anglin and Ketterle, Nature 416, 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.)

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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 exclusion principle 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 for the mean occupation number of bosons

occupying a single quantum state of energy s

nBE(s) =1

e(s)/kBT 1 .

The chemical potential is something of a mysterious quantity, but (in my opinion) the easiest way

to 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 system

must be positive, this means that we must have < 0, the lowest energy state in the system.However, can be arbitrarily close to 0 if we set = 0 , then

nBE(0) =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 of N atomic bosons of mass m confined to a volumeV = L3 (an infinite box potential.) However, in real experiments trapping potentials are usuallynot spatially homogeneous, and it is worthwhile to be able to calculate the density of states for

these systems for more general treatments of BEC.

For systems where the trapping potential varies slowly on the length scale of the de Broglie wave-

length T, we can calculate the density of states for a three dimensional system as

g() 1h3

d3rd3p [ H(r,p)],

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

g() = 2

2m

h2

3/2 V

d3r

U(r).

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

from earlier

g() = 2V2mh2 3/2

.

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11.3 Bose-Einstein condensation (BEC)

Consider N bosons of mass m confined to a volume V = L3. At zero temperature all of the Natoms will be in the ground state which has an energy of

0 =h2

8mL2

12 + 12 + 12

=3h2

8mL2. (11.1)

At very low temperatures it is possible to find a macroscopic number N0 of the N bosons still inthe ground state. The value of N0 is given by the Bose-Einstein distribution function

N0 =1

e(0)/kBT 1 . (11.2)By taking the Taylor expansion of the denominator, we can find that the chemical potential as a

function of temperature is given by

(T) = 0 kBT

N0 as T 0. (11.3)Hence for low temperatures (T) is very close to the energy of the ground state.

The chemical potential can be determined by calculating the total number of bosons

N =s

nBE(s) =s

1

e(s)/kBT 1 . (11.4)

We can approximate this sum by an integral as long as kBT is significantly larger than the energylevel spacing. However, if there are a large number of bosons in the ground s tate, there will be

a spike in the function nBE(). This is not accounted for correctly in the integral. In fact, this

approximation may be bad for several of the lowest energy levels. However, only the ground statewill be wildly incorrect, and so we treat N0 separately from the integral. Thus we can write

N N0 +0

g()d

e/kBT 1 (11.5) N0(T) + Nex(T), (11.6)

where g() is the density of states and Nex(T) is the number of bosons in all the excited states.

For now, we will assume that (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 in will have negligible effect on the occupation of the excited states when N0

is large.

We now have

Nex(T) =2

2m

h2

3/2V

0

d

e/kBT 1

=2

2mkBTh2 3/2

V 0

xdx

ex 1 (11.7)

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

0

dxxn1

ex

1

= (n)(n), (11.8)

where the gamma function (n) interpolates the factorial function (i.e. for integer n, (n) =(n 1)!) and the zeta function (n) is defined as

(n) =m=1

1

mn.

For the 3D box, we have n = 3/2, so (3/2) =

/2 and (3/2) = 2.612. Thus

Nex(T) = 2.612

2mkBT

h2 3/2

V = 2.612 VvQ

, (11.9)

where vQ = 3T is the quantum volume.

We can see from the equation for Nex that as T increases, the value of Nex also increases. Wedefine the critical temperature Tc as the temperature at which the number of excited particles Nexis equal to the total number of particles

Nex(Tc) = N. (11.10)

So, as the temperature approaches the critical temperature the number of particles in the ground

state N0(T) approaches zero. Since the initial assumption that (T) = 0 was based on the fact

that a large number of particles were in the ground state, for temperatures greater than the criticaltemperature the assumption that (T) = 0 is no longer valid.

A Bose-Einstein condensate forms when the temperature is below the critical temperature, which

can be written as

kBTc = 0.527

h2

2m

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 volume vQ is equalto or greater than the volume per particle V /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 doesnt give much understanding of

why it occurs. The answer to this question was hinted at in the tutorial on quantum statistics.

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For a gas of noninteracting distinguishable particles, we can treat them all separately using Boltz-

mann statistics. A single particle has a reasonable chance of being found in any single particle

state whose energy is of order kBT. The available number of such single particle states is quitelarge usually, and is of order of the single particle partition function Z1. The probability of beingin the ground state is thus 1/Z1, and as this applies to each particle individually, only a miniscule

fraction will ever be found in the ground state away from T = 0.

From the perspective of the system, each total system state has its own probability. The configura-

tion with all particles in the ground state will have a Boltzman factor of 1, whereas the Boltzmann

factor of a typical system state at temperature T with U N kBT will be eNkBT/kBT = eN.This seems rather small however, we must remember to multiply it by the number of different

arrangements of the N particles. For Z1 single particle states, the number of arrangements is ZN1 ,

and this factor overwhelms the eN.

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 are far fewer arrangements of identical bosons as compared to distinguishableparticles. Thus the probability of the system being in its ground state is much higher than for

distinguishable particles.

The number of system states Ns is roughly the number of ways of arranging N indistinguishableparticles among Z1 single particle states which is mathematically the same as distributing Nunits of energy among Z1 oscillators. It is given by

Ns

N + Z1 1N

(eZ1/N)N

when Z1 N,(eN/Z1)

Z1 when Z1 N. (11.13)

IfZ1 N then (eZ1/N)N eN, then the probability of the system being in an excited state isvery close to one. However, if Z1 N, then the total number of states is not very large, and theeN 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

2m

N

V

2/3= N2/3

0.527h2

2mL2,

0 =3h2

8mL

2.

Both are of order h2/mL2, however the BEC temperature is enhanced by a factor of N2/3.

11.5 Important points

11.5.1 Thermodynamic quantities

The behaviour of the condensate occupation, the chemical potential, and the heat capacity of the

Bose gas in the region of Tc is shown in Fig. 11.1 for the case of the infinite box potential. All

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thermodynamic quantities can be calculated relatively easily, but we have to resort to numerical

solution and we will not do so in the course. However, the same general principles as used earlier

apply.

An important point to note is the peak in the heat capacity at Tc. This is relevant in the case of

superfluid helium (see later). The condensate fraction of the Bose gas does not contribute to theheat capacity or the entropy of the system. The heat capacity behaves as CV T3/2 below Tc, andreduces to 3N kB/2 well above Tc.

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 understand

why. A BEC forms when we have 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 as 0. From the result given in Eq. (11.8), if the density of states is given byg() n1, then we must have n > 1 so that the zeta function

(n) =m=1

1

mn,

will converge. If n 1, the result for the number of excited state particles diverges, and so ourapproximation that 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 have g() 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, we

do 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 so = 0 always. It is therequirement that particles be conserved that results in the formation of a BEC.

11.5.4 Examples of BEC

The phenomenon of BEC and macroscopic quantum phenomena are intricately linked, and there

are three real-world examples of where Bose-Einstein statistics are important. The first is in super-

conductors, where electrons can form bound Cooper pairs below a transition temperature. They

can be thought of as composite bosons forming a BEC, and can flow without resistance through

the conductor.

The second two examples are superfluid helium and trapped atomic gases, and we will consider

these in more detail in the following sections.

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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

/

kB

T

T / Tc

(b)

0 0.5 1 1.5 20

0.5

1

1.5

2

C

V

/NkB

T / Tc

(c)

Figure 11.1: Thermodynamics of a Bose gas in the region of the critical 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 at Tc, andCV 3/2N kB for T Tc.

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Figure 11.2: The specific heats of (a) 4He and (b) 3He showing the anomalies associated with their

respective 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 at Tc = 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 at Tc, and a plot of CV versus T is similar to the Greek letter seeFig. 11.2.

The phenomenon was discovered by Kapitsa, Allen and Misener in 1938. In 1962 Landau received

a Nobel prize for his theory of superfluidity and in 1978 Kapitsa also received a Nobel prize for

his work in superfluidity. The superfluid state involves macroscopic quantum coherence, similar

to that responsible for superconductivity.

The theory of BEC in an ideal gas of atoms was first published by Einstein in 1924 (before fermions

were known about), but was pretty much ignored for several years. In fact, it was even suggested

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Figure 11.3: Phase diagrams of 4He (left) and 3He (right). Neither diagram is to scale, but quali-

tative relations between the diagrams are shown correctly. Not shown are the three different solid

phases (crystal structures), or the superfluid phases of 3He below 3 mK.

by Uhlenbeck that the prediction of a phase transition was incorrect (and there is evidence that

Einstein 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 gas

formula for the ideal gas BEC transition temperature gives a good estimate for 4He. This idea was

further strengthened when it was found there was no similiar superfluid transition in fermionic 3He

(see Fig. 11.3). The existence of a BEC in superfluid 4He was argued about for many years, but

now it is accepted that this is the case.

[N.B. there is a superfluid transition in 3He, but at milliKelvin temperatures. Here the fermions

form Cooper pairs, similar to electrons in a superconductor. Nobel prizes were awarded for this

work in 1996 (Lee, Osheroff, and Richardson, experiment) and 2003 (Leggett, theory).]

There is one important fact about superfluid 4He it is necessary to consider the interactions

between 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 difficult to treat theoretically. However, the

weakly-interacting Bose gas turns out to be a reasonable qualitative model, and much work was

done on this in the 1950s and 1960s in the context of superfluid 4He.

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

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Property Superfluid Normal fluid

Viscosity 0 non-zero

Thermal conductivity finiteEntropy 0 nonzero

Table 11.1: Properties of superfluid and normal fluid components

temperature. In terms of density we write

= s(T) + n(T). (11.14)

However, it is not the case that the superfluid component directly corresponds to atoms in the

ground state of the system. In fact, by using neutron scattering methods it has been determined

that even in the absence of a normal component only about 10% of the atoms are in the condensate

state. This is due to the strong interactions between the atoms this results in large depletion of

the condensate.

11.6.2 Physical properties

Superfluid 4He does not boil, i.e. no bubbles form when it is heated. This is because thermal

conduction within the 4He occurs via the superfluid component, which has infinite thermal con-

ductivity. This means that if one part of the fluid is heated, the heat is transferred throughout the

fluid immediately. Thus it is impossible to set up a temperature gradient within the fluid (which is

required to form bubbles.)

Heat transfer within a superfluid occurs not by conduction, but by the movement of the normal

component. If a heat source is placed somewhere within a superfluid, and a heat sink placed

elsewhere within the superfluid, the normal fluid component moves from the source towards the

sink, and the superfluid component moves in the opposite direction. This motion occurs under the

constraint that the total density remains constant throughout the superfluid.

This fact results in some remarkable properties. Table 11.1 contains a comparison of some of the

properties 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 the

superfluid can flow through, but the normal fluid component cannot pass through due to viscosity.

If in the system shown in Fig. 11.4 we have T1 > T2 and P1 = P2, then for a normal connectionthermal equilibrium is reached by the conduction of heat. Thus the fluid carries entropy from 1 to

2.

For a superleak this is NOT possible because only the superfluid can pass from 1 to 2 and the

superfluid carries no entropy. Suppose T1 were to slowly increase. For thermal and diffusive

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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 that 1 = 2. Using the thermodynamic identities

G = N = U T S+ P V,and

T dS = dU + P dV dN,we get the equation

N d = SdT + V dP (11.15)From the conditions for equilibrium we know that

1 2 = 0 = S(T1 T2) + V(P1 P2), (11.16)and therefore

P1 P2 = SV

(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 superleak

by creating a temperature difference.

The capillary in Fig. 11.5 contains a superleak. The heater in the main section raises T1 above T2,and this causes P1 to become greater than P2. The result is that the superfluid is forced through thecapillary.

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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 cause

a 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 = gz, then this gives

T

z=

g

S, (11.18)

where is the mass per mole and S is the entropy per mole. For superfluid 4He at around T = 1.3K we have

T

z 1 mK cm1.

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

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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. Andrews

will be shown. Jack Allens 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 and

Carl Wieman suceeded in creating the first ever Bose-Einstein condensate in a dilute gas of 87Rb

atoms. The condensate they formed consisted of only 2000 atoms at a temperature of about 100

nK. 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, allowing

evaporative 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 atoms

return to equilibrium).

At these temperatures you would normally expect 87Rb to be a solid! However, if they are kept

sufficiently dilute only two particle collisions will ever occur, and these must be elastic. Three

particles are required to collide such that two form a molecule, and the third takes away the excess

energy. Therefore the gas will be metastable however this requires it to be 100,000 times less

dense 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 closer to the ideal Bose gas that Einstein

first imagined. In fact, quantities such as Tc are not much shifted from the values you will calculatein your assignment. This also means that it is feasible to treat these systems theoretically (although

it 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 s1), molecular BEC, four-wave

mixing, entangled atomic beams, BECs in periodic lattices, and so on. One of the most talked

about possibilities is to make atom lasers from BECs, consisting of a continuous stream of atoms

sharing the same quantum mechanical phase. The Nobel Prize was awarded to researchers in this

field in 1997 (laser cooling) and 2001 (BEC), and it remains a very active area today. There is

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Figure 11.6: Shadow images of a degenerate Bose gas of 87Rb, taken at UQ in February 2004.

From left to right: above Tc, slightly below Tc, well below Tc. There are about 30,000 atoms in the

picture on the right.

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

experimental 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 difficult

with fermions, as the collisional process between identical fermions that is required for evaporative

cooling 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 degenerate

Fermi 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 transition

from having a BCS-like state of weakly paired fermions forming the superfluid to having strongly

bound 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 likely to be awarded in this area as well.

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newsandviews

Inlate1937,twoyoungphysicistsattheRoyalSocietyMondLaboratoryinCambridge,UK,foundthatliquidheliumcouldflowthroughverysmallcapillarieswithessentiallyzeroviscositybelowatemperatureof2.17kelvin.Whiletheywerepreparinganoteforpublication,theyheardofanotherpaper,justsubmittedbyP.L.KapitzainMoscow,reportingsimilarresults.Bothpapersappearedsidebysideon8January1938inNature.KapitzaandtheyoungphysicistsJackAllenandA.D.Misenerhaddiscoveredthemysteriousphenomenonofsuperfluidityinliquid4He.Itwasthestartofagoldenperiodin

low-

temperaturephysics.Allen,thelastsurvivorofthisheroictime,diedofastrokeon22April2001,aged92.MostofAllensgreatestdiscoveries

weremadeattheoutsetofhiscareer(in1938hewasessentiallyapostdoc,workingwithoutasupervisor,whichsuitedhimfine).Between1937and1939heandhisassociatesatCambridgeproducedastreamofpapersonsuperfluidityinliquid4He,thisoutputcomingtoanabruptendwiththestartoftheSecondWorldWar.In1946,withnormalliferesuming,Allenorganizedthefirstinternationallow-tem

peraturemeetin

gattheUniversit

y

ofCambridge,UK.Thiswasthebirthofthemajortriennialeventinthelow-temperaturephysicscommunity;thetwenty-thirdmeetingwillbeheldinHiroshimanextyear.

In1947,AllenwasappointedprofessorofnaturalphilosophyandheadofphysicsattheUniversityofStAndrewsinScotland,andtwoyearslaterhebecameaFellowoftheRoyalSociety.Butalthoughhestayedactiveintheworldoflow-temperaturephysics,revitalizingresearchatStAndrewsandhelpingtorunmanyconferencesandworkshops,hisownresearchwasneveragaincentral.HeretiredfromStAndrewsin1978.

JackAllenwasborninWinnipeg,Canada,in1908(theyear4HewasfirstliquefiedinLeiden,intheNetherlands).HereceivedhisBAfromtheUniversityofManitoba,wherehisfatherwasheadofthephysicsdepartment.In1929,hewenttostudyattheUniversityofTorontowhereJamesF.McLennanhadbuiltupastrongphysicsdepartment.In1923thishadbeenonlythesecondlaboratoryintheworldtoliquefyhelium,andwhenAllenarrived he immersed himself in the new

toCambridgetoworkwithKapitza,onlytofindthatKapitzawasbeingdetainedinMoscow,andwasintheprocessofsettingupwhatwastobecomethefamousInstituteofPhysicalProblems.InKapitzasabsence,Alleneffectivelytookoverthelow-tem

peraturework,althou

gh

officiallythedirectorwasJ.D.Cockroft.In1935,DonMisener,agraduate

studentatToronto,hadcarriedoutthefirstexperimentalstudyoftheviscosityofliquid4He.Bythenitwasknownthatliquidheliumunderwentsomesortofphasetransitionat2.17K,astherewereabruptchangesinvariousthermodynamicproperties.Misenersworksuggestedthattheviscositydecreasedsubstantiallywhentheliquidpassedthroughthistransition.MisenerjoinedAllenatCambridgein1937todohisPhD,andthetwosetouttostudythephenomenonbyexaminingflowinthincapillaries.Miseners1935experimentalworkhad

alsoattractedthenoticeofKapitzainMoscow.Bothgroupsreportedtheirindependentdiscoveryofsuperfluidflowin1938,withKapitzabeingthefirsttocointhetermsuperfluid.ItispuzzlingandunfortunatethatwhenKapitzafinallyreceivedawell-deservedNobelprizeinphysicsin1978,thecitationconcerningsuperfluiditymadenoreferencetotheworkofAllenandMisener.Allenquicklyfoundotherdramatic

manifestations of superfluidity all of

finepowder.Butafter1945theMoscowgroupunderKapitza(helpedbyL.D.Landau,whodevelopedacompletetheoryofthetwo-fluidbehaviourofsuperfluidheliumin1941)dominatedfurtherresearchonquantumliquids.Thestudyofsuperfluid4Heincreasinglyinvolvedmicroscopictheoriesandnewexperimentalprobessuchasneutronscattering,noneofwhichinterestedAllen.Allenwasthelastofagenerationof

independent-mindedclassicalphysicistswhodelightedinexplainingthevisibleworld.Heprizedhisownability,andthatofglass-blowersandtechnicalpeople,tobuildexperimentalapparatus.AllenwasasproudofhisinventionoftheO-ringvacuumsealasanythingelsehedid.Itshouldbenosurprisethathisgreatestworkonsuperfluidityinliquid

4

Heinvolvedphenomenathatcouldbeseen.Indeed,itismostfittingthatAllendiscoveredthefamousfountaineffectin1938withthehelpofapocketflashlight.Overaten-yearperiodAllenmadea

movieofthevarioustwo-fluidphenomenaexhibitedbyliquid4He.Thephotographyoftheseeffectswasarealchallenge,becauseliquid4Heisessentiallytransparent.Thisuniquecolourmovie(thefiftheditionwascompletedin1982)isoneofAllensgreatlegaciestophysics.Allenhadacommandingpresence

andadry

senseofhumour.Hestrongly

identifiedwiththephysicsofanearlierday,andIimaginethathewouldhaveenjoyedtalkingwithclassicalphysicistssuchasLordRayleigh,MichaelFaradayandDanielBernoullimorethanwithWernerHeisenbergandErwinSchrdinger.Superfluidityisadramaticvisiblemanifestationofquantummechanics,beingtheresultofBoseEinsteincondensationinwhichamacroscopicnumberof4Heatomsoccupythesame,single-particlequantumstate.ItisparadoxicalthatthephenomenonwasfirstobservedbyAllen:agreatphysicistwhowasntmuchinterestedinatoms.WalkingtheoldstonestreetsofSt

Andrews,onequicklynoticestheelegantmetalhistoricalplaquesonmanybuildings,commemoratingfamouspeoplewhohavebeenassociatedwiththishistorictownanditsuniversityoverthecenturies.Almosteveryplaqueconnectsitssubjecttophysicsnotsurprising,becauseJackAllenwasthemotivatingforcebehindthesigncommittee.IverymuchhopethatthetownseesfittosohonourAllenhimself. AllanGriffinAllan Griffin is in the Department of Physics

Obituary

JohnFrank(Jack)Allen(19082001)

Co-discovererofsuperfluidity

UNIV

.

OF

STAND

REWS