The Nature of the Cooper Pair - University of Liverpool

Statistical and Low Temperature Physics (PHYS393) 

7a. Superconductivity 

- The Nature of the Cooper Pair 

Kai Hock 

2010 - 2011 

University of Liverpool

To develop the idea ... 

1. Estimate the attraction between two electrons. 

2. Guess a wavefunction for the electron pair. 

3. Solve Schrodinger’s equation for the energy. 

4. Obtain the wavefunction of the Cooper pair. 

5. Explain why a superconductor has zero resistance. 

(Weisskopf(1979)http://cdsweb.cern.ch/record/880131/files/p1.pdf) 

Superconductivity 1

When an electron passes in between two positive ions in a 

metal, it attracts the ions. 

The ions move closer together, creating a region that has 

slightly higher positive charge. This creates an attractive 

potential for other electrons. 

Superconductivity 2

The potential energy of an electron due to a positive ion is 

U = − e2 

4πɛ0r . 

This is normally cancelled out by the free electrons. If an ion is 

displaced, the change in the potential is the attractive potential 

that now appears. This may be estimated by the small change 

approximation: 

δU ≈ dU 

dr δr. 

Superconductivity 3

Using the small change approximation, the attractive potential 

is 

V ≈ − e2 

4πɛ0d 2δ 

where r is roughly the distance d between ions, and δ is the 

displacement of the ion. 

To estimate δ, consider the ion next the electron path. 

Superconductivity 4

The force from this electron is likely to be active over a 

distance d near the ion. At larger distance, it tends to get 

screened off by a rearrangement of the other electrons. 

Assuming that the electron is roughly at the Fermi energy, its 

velocity ve is related to the Fermi energy E F by: 

EF = 1 

2 mev 2 e . 

The attraction is active for a time 

τ = d 

. 

Superconductivity 5 

ve

The attractive force is the Coulomb force 

F = − e2 

4πɛ0r 2 

where r is roughly the distance d between between ions. 

So the impulse of the passing electron in the ion is 

F τ = − e2 

4πɛ0d 2 

This is like a kick on the ion. For small displacements, the 

motion of the ion is approximately simple harmonic. 


d 

ve 

.

In a simple harmonic motion with amplitude A and frequency ω, 

the maximum velocity at the rest position is 

v0 = Aω. 

This maximum frequency would the velocity of the ion as it 

takes off immediately after the kick, so 

v0 = 

where M is the mass of the ion. 

The amplitude A corresponds to the displacement δ that we 

want to find. The frequency ω would be the Debye frequency. 


F τ 

M

Recall that the Debye frequency ω D is the maximum vibrational 

frequency in a lattice. This happens when the ion spacing is 

equal to a half wavelength: 

Compare this with the picture of the ions attracted by a 

passing electron: 

In both cases, the electron cloud in between would be deformed 

in the same way. We can expect the same restoring forces on 

the ions, and therefore the same frequency. 

Superconductivity 8

Combining these ideas, 

v0 = Aω 

gives 

F τ 

M = δωD. Therefore the displacement is 

δ = 

F τ 

Mω D 

= 

e2 

4πɛ0d 2 

d 

1 

ve MωD Recall the formula for the Debye frequency: 

ω D = 

� � 

6Nπ2v 3 1/3 

where v is the speed of sound in the metal. 

We can thus estimate the displacement δ. 


V 

, 

.

The displacement then gives us the attractive potential energy: 

V ≈ − e2 

4πɛ0d 2δ. 

However, the ion will return to the rest position after a time. 

This time is half of the period of the simple harmonic motion, 

the period being 

T = 2π 

. 

ωD During the time T/2, the electron travels a distance 

l = veT 

2 


πve 

= . 

ωD

The electron would therefore leave a trail of deflected ions 

behind it. This trail that extends for a distance l, and only 

exists along a narrow tube in between ions. 

This net positive charge can attraction another electron that 

happens to be travelling in the opposite direction through this 

same tube. 

So there is an attractive potential between two such electrons 

that extends over the distance l. 

Superconductivity 11

We shall approximately represent the attractive potential as a 

function that is constant within the range l: 

where 

V (r) = 

� 

V0 = 

−V0 for r < l 

0 for r > l 

e2 

4πɛ0d 2δ. 

We have seen that the potential only exists within a narrow 

tube, so it is not exactly a spherical potential. 

However, if two electrons happen to be travelling in opposite 

directions and pass very close to each other, then the potential 

has the same effect as a spherical potential, because the 

electrons do not experience the potential outside of the tube. 


We now want to find out if two electrons can attract in the 

way just described, and form a bound state - a Cooper pair. 

A bound state refers to a wavefunction that has a finite size, 

like the wavefunction of an electron in an atom. This means 

that the value of the wavefunction falls off to zero after some 

distance. 

In contrast, a free electron is not in a bound state. Its 

wavefunction may be e ikx , which has the same amplitude at 

any distance. 

Since we treat the potential as effectively spherical, we can use 

the same form of the Schrodinger equation as for the hydrogen 

atom. 


We have seen that two electrons must travel in opposite 

directions, and the paths must be very close, before they can 

experience the attractive potential. 

Electrons are at the Fermi energy have a wavelength of the 

order of d, the spacing between ions. We can see this by 

comparing the kinetic energy with E F : 

� 2 k 2 F 

2me 

= �2 

2me 

� � 

3π2 2/3 

N 

k F is the wavevector at 2π/λ F the Fermi energy, where λ F is the 

wavelength. 


V 

.

V/N would be the volume associated with one electron, or ion, 

and this is of the order d 3 . Substituting these, we find 

� 2π 

λ F 

� 2 

= 

Solving for the wavelength gives 

� � 

3π2 2/3 

d 3 

λ F ≈ 2d. 

Since the distance between the paths of the two electrons is 

smaller than λ F , because both must be roughly in between the 

same pair of ions, we may assume that the angular momentum 

is zero. 

To see this, ... 


.

... recall the quantisation condition for the angular momentum 

in the Bohr atom: 

The angular momentum is quantised according to the number 

of wavelengths in the orbit. If the orbit is smaller than one 

wavelength, then there is zero wavelength, and the angular 

momentum must be zero. 


So if the distance between the electrons is smaller than the 

wavelength of each electron, then the angular momentum 

would be zero. In a simplistic way, we may imagine the the 

electron pair move against each other along very close, parallel 

paths, and that they equally likely to do so in any direction. 

That is, the wavefunction should be spherically symmetric. 


The Schrodinger equation is then of the same form as that for 

the the ground state electron of the hydrogen atom: 

− �2 1 

2µ r2 d 

R dr 

� 

r 2dR 

dr 

� 

+ V (r)R = ER. 

except that the electron mass is replaced by µ, the reduced 

mass me/2. We need to use this reduced mass because both 

electrons are moving, unlike a hydrogen atom in which the 

nucleus is approximately stationary. 

http://en.wikipedia.org/wiki/Reduced_mass 

To show that a bound state is possible, we need to solve this 

equation to see if the wavefunction has a finite size. 


If V (r) is constant, the solution is of the form 

ψ(r) = sin(kr). 

However, V (r) is not constant. It becomes zero beyond r = l. 

We can try a sum of sine functions: 

Ψ(r) = � 

ak sin(kr), 

where a k are unknown constants that we must find. 

k 

We now come the the important part. The need for this sum 

means that energy states with many wavevectors k must be 

used to make up the wavefunction. However, we are not 

allowed to use k values below the Fermi energy, because those 

states are fully occupied. 

Therefore, we must impose the condition that k must be more 

than k F , the wavevector at the Fermi energy. 

The detailed steps to solve the Schrodinger equation are given 

in the appendix. 


Here, we just state the solution to the Schrodinger equation. 

The wavefunction is 

Ψ = 

�ω 

�D 

0 

aε sin kr = 

�ω 

�D 

0 

1 

ε + ∆ 

sin kr 

where k is in steps of π/R, R being roughly the radius of a 

sphere similar in size to the metal. (R corresponds to the length 

L of the cube that is commonly used for electrons in metal.) 

ε is the energy above the Fermi energy. k is related to ε by 

− EF . 

2me 

The energy of the electron pair is 

ε = �2 k 2 

E = E F − ∆. 

∆ is called the binding energy, and is given by 

∆ = �ωD exp(− 2πkF ). 

melV0 


Let us see what the wavefunction looks like. 

Ψ = 

�ω 

�D 

0 

1 

ε + ∆ 

sin kr 

It is a function of r only, so it is spherically symmetric. It is a 

sum of many sine functions. The coefficient decreases with 

energy. When the energy ε goes from 0 to ∆, the coefficient 

decreases from 1/∆ to 1/(2∆). 

So we may suppose that the terms are only significant in the 

energy range from 0 to ∆. As energy changes, k changes. 

When r is large enough, sin kr starts to go out of phase for 

different wavevectors k. 


They will start interfering destructively, and the sum Ψ will 

decrease. This gives us a way of estimating the size of the 

wavefunction. 

Consider the minimum and maximum k. These correspond to 

the energies ε = 0 and 1/∆, relative to E F . The absolute 

energies are E F and E F + ∆. 

Consider the radius r = ρ at which the sine functions for these 

two values of k first get out of phase by π radians. This gives 

the largest cancellation, and we may take this radius ρ as an 

estimate of the size of the wavefunction. 


Let the difference in k be ∆k. The phase at ρ for each sine 

function is kρ. Since this difference is is π between the 

maximum and minimum k, so 

∆k.ρ = π. 

To express in terms of E F , we use 

E = p2 

. 

2me 

The small change approximation gives 

∆E = p∆p 

. 

me 

At the Fermi energy, and for an energy change ∆E equal to ∆, 

we have 

∆ = p F ∆p 

me 


= pF .�∆k 

. 

me

Using 

∆k.ρ = π 

we can solve for the size of the wavefunction: 

ρ = hp F 

2me∆ . 

Putting in typical values, we find the the size of the 

wavefunction is of the order of 10 −4 cm. This is much larger 

than the size of an atom. 


So we have found a bound pair - a wavefunction of two 

electrons that has a finite size. This is in spite of the fact that 

the kinetic energies of the electrons, which are close to E F , are 

much larger than the binding energy ∆. 

Mathematically, we have seen that this solution is obtained 

after we impose the condition that the wavefunction sum 

Ψ = � α k sin kr 

must not contain any wavevector below that Fermi level. 

We can interpret this physically. If the bound pair of electrons 

separate, they become plane waves and would have to occupy 

two of the energy states below E F . However, we are talking 

about 0 K, and all states below E F are fully. There is no room 

for any more electron, because of the exclusion principle. That 

is why the pair must, with an energy just below E F (by ∆), 

must remain bound. 


The bound pair of electrons is the Cooper pair. They would 

have opposite spins, so the resultant spin is zero. The Cooper 

pair is a boson. 

Because the pair of electrons have opposite momenta, the 

resultant momentum is also zero. If many Cooper pairs are 

formed, they would all occupy the same state - of zero 

momentum. This is now possible because the pairs are bosons. 

The result is a condensate that is very similar to the 

Bose-Einstein condensate. 

Exactly how many Cooper pairs are formed? 

In order to esimate the number, assume that all Cooper pairs 

have the same wavefunction: 

Ψ = 

ε� 

0 

1 

ε + ∆ 

sin kr, 

that is, the component wavevectors k cannot be below the 

Fermi level. We shall justify the assumption afterwards. 


Then suppose that two electrons at the Fermi level E F start 

pairing up. When they pair up, their energy falls by ∆, to 

2E F − ∆. 

Next, imagine that more electrons pair up. As more and more 

electrons pair up, the Fermi sea gets depleted, and the ”sea” 

level falls. (This is not to be confused with the Fermi energy, 

which is still defined as E F .) The electron pair would always 

have the same energy of 2E F − ∆, because they have the same 

wavefunction Ψ. 

This goes on until the sea level has fallen by ∆/2, to 

E F − ∆/2. A pair of electrons at this level would have already 

have a total energy of 2E F − ∆. There is no advantage of 

forming a bound pair, which also have the same energy. So the 

electrons stop pairing up. 


The number of Cooper pairs is therefore equal to half of the 

number of electrons with ∆/2 of the Fermi energy. This is 

given by 1 2 × 2g(E F )∆/2. 

Recall from the notes on electrons in metal that 

g(EF ) = 3N 

. 

4EF So the number of Cooper pairs is 

N ′ = g(E F ) ∆ 

2 

3N∆ 

= . 

8EF We can estimate the distance between Cooper pairs by dividing 

the volume of the metal V by N ′ to get the volume occupied by 

one Cooper pair. Taking this to be a cube, the side of the cube 

would give an estimate of the distance D between adjacent 

pairs: 

D = 


� 

V 

N ′ 

�1/3

Putting in the numbers, we would find that the distance D 

between the Cooper pairs is about 30d, where d is the distance 

between the ions. 

Since the size of the Cooper pair wavefunction is around 10 −4 

cm, this means that there is a lot of overlap between Cooper 

pairs. However, there must be a distance below which two 

Cooper pair wavefunctions cannot come any closer, just as the 

wavefunctions between two atoms must start repelling if they 

come too close. 

According to Victor Weisskopf, the average distance D between 

Cooper pairs is just this distance. This means that the Cooper 

pairs are packed in a compact way, just like liquid helium-4. It 

behaves like a liquid instead of a gas. 

(Weisskopf(1979)http://cdsweb.cern.ch/record/880131/files/p1.pdf) 


So, as in superfluid helium-4, we may expect phonons and 

rotons, and that the critical velocity be determined by the 

minimum roton excitation. 

If we think of the rotation as a rotation of two Cooper pairs 

about each other, we can relate the roton excitation to the 

minimum rotational energy. The rotational energy is quantised 

and the minimum is given by 2� 2 /meD 2 . 

If we put in the numbers, we would find that this is much larger 

than ∆, the binding energy for the Cooper pair. The Cooper 

pair would break before any roton gets excited. 

As long as there is not enough energy to break the Cooper 

pairs, the Cooper pair condensate flows with zero resistance. 

This condesate is also called the BCS superfluid. This explains 

why a superconductor has zero resistance. 


At temperature T , we would expect thermal energies of the 

order of k BT , either in the form of phonons or excited 

electrons. If this energy is more than ∆, then there is enough 

energy to break the Cooper pairs. If this happens, the metal 

returns to being normal conducting, with resistance. 

The transition temperature Tc is therefore related to the 

binding energy by 

k BTc ≈ ∆. 

The complete BCS theory gives the more accurate relation of 

1.76k BTc = ∆ 

which agrees well with measured values of Tc and ∆. 


APPENDIX: Solving for the Cooper pair wavefunction 


The Schrodinger equation has the same form as that for the 

ground state electron of the hydrogen atom: 

− �2 1 

2µ r2 d 

R dr 

� 

r 2dR 

dr 

� 

+ V (r)R = ER. 

except that the electron mass is replaced by µ, the reduced 

mass me/2. 

The Schrodinger equation can be rewritten in the familiar 1D 

form by using a new function 

ψ(r) = rR(r). 

Note this this is r times the actual wavefunction R(r). The 

Schrodinger equation then becomes 

− �2 

2µ 


d2ψ + V (r)ψ = Eψ. 

dr2

The attractive potential is very small. It is known in quantum 

mechanics that a bound state is not possible if this is so. 

http://en.wikipedia.org/wiki/Finite_potential_well 

However, the electrons here are not in free space. Those at the 

Fermi energy are not allowed to have smaller energies, because 

all the states are occupied. 

In the 1950s, Leon Cooper showed that when this happens, a 

bound state becomes possible. 


To show that a bound state exists, we need to solve the 

Schrodinger’s equation. If V (r) is constant, the solution is 

ψ(r) = sin(kr). 

However, V (r) is not constant. We may still try a solution in 

the form of a sum of sine functions: 

Ψ(r) = � 

ak sin(kr), 

where a k are unknown constants that we must find. 

k 

When we study electrons in metal, we learn how k is quantised 

by solving Schrodinger’s equation for a cube. Since we now 

have spherical symmetry, we shall use a sphere of radius R 

instead (not to be confused with the wavefunction R(r)). 


Therefore 

and 

sin(kR) = 0, 

k = nπ 

R . 

We shall need to make use of this later. 

Rearrange the Schrodinger’s equation: 

Substitute 

and get 

� 

k ′ 

� 

�2k ′2 

ak ′ 

me 

− �2 

me 

d2Ψ − EΨ = −V (r)Ψ. 

dr2 Ψ(r) = � 

ak ′ sin(k ′ r) 

− E 


� 

k ′ 

sin(k ′ r) = −V (r) � 

ak ′ sin(k ′ r). 

k ′

We apply the relation 

Multiply 

� R 

0 sin(kr) sin(k′ r)dr = 

� 

k ′ 

� 

�2k ′2 

ak ′ 

me 

− E 

by sin(kr) and integrate, we get 

where 

R 

2 a k 

� 

� � 2 k 2 

me 

� 

R/2 for k = k ′ 

0 otherwise 

sin(k ′ r) = −V (r) � 

ak ′ sin(k ′ r) 

− E 

� l 

� 

k ′ 

= − � 

ak ′Vkk ′, 

Vkk ′ = V0 

0 sin(kr) sin(k′ r)dr. 

The upper limit is l instead of R because V (r) is zero for r > l. 


k ′

The product of the waves sin(kr) and sin(k ′ r) would be 

oscillatory. 

If k and k ′ are not equal, then it oscillates between plus and 

minus, and the integral may cancel to zero. 

If k and k ′ are nearly equal, then the product is nearly always 

positive. 

This is likely to happen if, within the distance l, the two waves 

differ by less than half wavelength. This means 

or 

kl − k ′ l = π, 

k − k ′ = π/l. 


Recall that 

So 

Returning to 

V kk ′ = V0 

l = πve/ω D. 

k − k ′ = ω D/ve. 

� l 

0 sin(kr) sin(k′ r)dr, 

we can say that if k − k ′ is more than ω D/ve, the integral is 

approximately zero. 

If not, we may roughly take it to be the value when k = k ′ , 

which is l/2. Therefore 

V kk ′ = 


� 

lV0/2 for k − k ′ < ω D/ve 

0 for k − k ′ > ω D/ve

Return to the Schrodinger’s equation 

R 

2 a k 

� � 2 k 2 

me 

− E 

� 

= − � 

ak ′Vkk ′. 

We shall now change from k to energy variable. For reasons 

that will become clear, we define the variable ε to be the 

energy above E F . So 

ε = �2 k 2 

me 

k ′ 

− E F . 

We define the amount of energy that E is below the Fermi 

energy by 

The equation is now 

∆ = E F − E. 

R 

2 ak(ε + ∆) = − � 


k ′ 

a k ′V kk ′.

Next, we approximate � a k ′V kk ′ to an integral. To do this, we 

need the density of states. We know that 

k = nπ/R. 

So there is one state for every interval of π/R in k, or the 

density of state is 

g(k) = R/π. 

We now come to an important point. We are going to impose 

the condition that k must be above the Fermi level. We shall 

intepret this condition after we have found the bound state. 

Recall 

ε = �2 k 2 

me 


− E F .

So in energy variable, the density of states is 

g(ε) = g(k) dk 

dε 

At the Fermi energy, 

= R 

π .me 

2k . 

g(0) = meR 

. 

2πkF Note that we are taking the Fermi energy at the origin, so zero 

energy here means Fermi energy. Combining these ideas, we 

obtain 

� 

k ′ 

a k ′V kk ′ ≈ 

� 

a ε ′V εε ′g(0)dε ′ 

The lower limit of the integral is 0, the Fermi level. To find the 

upper limit, we recall that V kk ′ is zero if k − k ′ > ω D/ve. 


To change to energy variable, rewrite as δk > ω D/ve. Using 

we find 

δε = 2�2 kδk 

me 

ε = �2 k 2 

me 

− E F , 

= 2�pδk 

me 

Multiplying δk > ω D/ve by �2ve, we obtain 

δε > 2�ω D. 

So if ε − ε ′ is more than 2�ω D, V εε ′ is zero. 

The Schrodinger equations is 

= 2�veδk. 

R 

2 aε(ε + ∆) = −g(0) aε ′Vεε ′dε ′ . 

If ε is zero, Vεε ′ goes to zero when ε ′ > 2�ωD. So we may take 

the upper limit to be 2�ωD. Superconductivity 43 

�

However, ε is 2�ω D, then the upper limit must be even higher. 

We assume that aε is significantly different from zero only for 

ε

The right hand side of the Schrodinger equation 

aε(ε + ∆) = −B 

is a constant. Let this be C. So 

or 

� �ωD 

0 

aε(ε + ∆) = C, 

aε = C 

ε + ∆ . 

a ε ′dε ′ . 

Substituting this back into the Schrodinger equation, we get 

Integrating, we find 

1 = −B 

� �ωD 

0 

dε ′ 

ε ′ + ∆ . 

∆ = �ωD exp(−1/B) = �ωD exp(− 2πkF ). 

melV0 


We have found both aε and ∆. This means that we have found 

both the wavefunction Ψ and the energy E. 

In other words, we have solved the Schrodinger equation! 

The wavefunction is 

Ψ = 

�ω 

�D 

where k is related to ε by 

The energy is 

0 

aε sin kr = 

ε = �2 k 2 

2me 

�ω 

�D 

0 

− E F . 

1 

ε + ∆ 

sin kr 

E = EF − ∆ = E − �ωD exp(− 2πkF ). 

melV0 


Putting in the numbers, we find that ∆ is about 10 −4 smaller 

than the Fermi energy. 

Compared to �ω D, ∆ is about 10 −2 smaller. Since 

we have 

and 

So 

a �ωD = 

aε = C 

ε + ∆ , 

a0 = C 

∆ 

C 

�ω D + ∆ ≈ 

a �ωD ≈ 10 −2 a0. 

C 

100∆ + ∆ . 

This justifies our assumption that aε is significantly different 

from zero only for ε

The Nature of the Cooper Pair - University of Liverpool

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