9Superconductivity and Higgs mechanism

Continuous symmetry breaking without Goldstone modes.

9.1. Phenomena

For some metal, if one reduces the temperature, the resistivity suddenly drops to zero at some critical temperature TC.

† Zero resistivity

† A second order phase transition at Tc (discontinuity in the specific heat)

† Meissner effect (B = 0 in a superconductor).

† Magnetic field suppresses TC. If B is larger than some critical value BC, the superconducting phase disappears.

9.2. Superfluid and Bose–Einstein condensate

9.2.1. Phenomena

In some experiments, the viscosity of liquid helium-4 (bosons) vanishes below certain temperature (2.17 K).

Experiment 1: Connecting two containers with a tube and put some He4 liquid into both containers. Due to the viscosity and frictions, if wewant to push the liquid He from one container to the other, we need to apply a pressure and the pressure becomes bigger if we reduce diameterof the tube. However, below the critical temperature, the fluid can flow from one side to the other without applying a pressure. This fluid withzero viscosity are known as a superfluid, which has no fiction/viscosity.

Experiment 2: Put a torsion pendulum into liquid He, at high temperature, because of the viscosity, the amplitude of the rotation decreaseswith time. At low temperature (below Tc), the friction and viscosity remain none zero.

Puzzle: if we measure the viscosity using the first experiment, we find it is zero in the superfulid phase. However, the second experiment tellsus that the viscosity is none-zero. A Contradiction? Not really.

9.2.2. The two-fluid picture:

Below Tc, the He-4 liquid is a mixture of two types of fluids: some fraction of the He atoms forms a superfuild while other He4 atoms form anormal fluid. The density for the superfulid part is called rS and the density for the normal component is called rN . The density of the He liquidis the sum of them r = rS + rN . The superfulid part has no viscosity but the normal fluid has viscosity. Above TC, rS = 0 and r = rN . BelowTC, rS increases from 0 as the temperature is reduced. In experiment #1, when we push the fluids, the superflluid component can flow betweenthe two contains without any viscosity, so we always get some flow even if the pressure is very small (so we get viscosity=0). The normal fluidwill not flow when the pressure is small enough and thus doesn’t contribute to the flow. In other words, this experiment is blind to the normalcomponent and can only see the superfluid part. In experiment #2, both the normal and superfluid components touches the pendulum. Thenormal fluid part induces damping. The superfluid has no friction, but the pendulum cannot see it. The pendulum only sees the normal compo-nent, which gives it damping. So we find nonezero viscosity.

9.2.3. Bose-Einstein condensate

For simplicity, let’s ignore interactions and consider a free boson gas (in helium-4 the interactions are fairly strong and it is not a good approxi-

118 Phys620.nb

mation to ignore interactions. However, as far as the phenomenon of superfluidity is concerned, we can get the same physics even if we ignorethe interaction). The occupation number for the quantum state with momentum p is:

(9.1)np = 1

expep - m kB T - 1

Here ep = p2 2 m. Because the occupation number is non-negative np ¥ 0, m < ep for any p. For ep = p2 2 m, this means that m§0.

Total number of particles:

(9.2)N =k

1

expek - m kB T - 1=

V

2 p Ñd „d p

1

expep - m kB T - 1=

V

2 p Ñ34 p p2 „ p

1

expep - m kB T - 1

The density of particles

(9.3)r =N

V=

1

2 p2 Ñ3 p2 „ p1

exp p2

2 m- mkB T - 1

=2 m kB T32

2 p2 Ñ3 x2 „ x

1

expx2 - m kB T - 1

where x = p 2 m kB T .

The density r is a function of m and T : r = rm, T. If we plot n as a function of m and T , we get the following figures. The l.h.s. shows a 3Dplot of rm, T. The r.h.s. is the same function shown in a contour plot.

-10 -5 0

05

1015

20

0

50

100

-10 -8 -6 -4 -2 00

5

10

15

20

m

T

Fig. 1. Density as a function of m and T (in arbitrary units)

From the contour plot, we can see that if we fix the density, for each temperature, we can find the corresponding chemical potential. However, aconstant density curve stops at certain critical temperature Tc, at which the chemical potential reaches zero (remember that m cannot be largerthan 0). This means that below Tc, we cannot find a value of m to keep the density fixed. The value of Tc is determined by the density. At Tc,because m=0, we have:

(9.4)r =2 m kB Tc32

2 p2 Ñ3

0

¶

x2 „ x1

expx2 - 1=

2 m kB Tc32

2 p2 Ñ3

1

4p z 3

2

= m kB Tc 2 p Ñ232 z 3

2

Here z32 = 2.61238, with zs =n=1¶ n-s

(9.5)Tc =2 p Ñ2

m kB

r z3223

Q: What happens below Tc?

A: Bose–Einstein condensate

It turns out that the equation we have above is not entirely correct, because we used the integral „d p. For a quantum system (with finite size),

we know that all quantum states are discredited, so we should use sum instead of integral. When the system size becomes larger and larger, theseparation between different states becomes smaller and smaller, so that we can turn the sum into an integral. However, in order to turn a suminto an integral, we need to assume that the function is smooth enough and has no “spikes”, but this condition is violated at and below Tc. At Tc,we have m=0, so the occupation number is:

Phys620.nb 119

(9.6)np = 1

exp p2

2 m kB T - 1

This function has a singularity at p = 0. np = 0 = 1 0 = ¶. This means that there are large number of particles in the state p = 0 (the groundstate). However, when we compute the total number of particles, we use integral to substitute the sum.

(9.7)N =k

1

expek - m T - 1=

V

2 p Ñ34 p p2 „ p

1

exp p2

2 m- mT - 1

In this integral, the particles on the ground state p = 0 are not taken into consideration, because there is a p2 factor in the integral. At p = 0, thisprefactor p2 is zero. Therefore, using an integral is a bad approximation here, because it ignores the contribution from the ground state, which isthe most important state below TC and contains a very large number of particle (As will be shown below, above Tc it is safe to use the integralformula). So here, we should add the ground state contribution back into the total particle number

(9.8)N = N0 +4 p V

2 p Ñ3 p2 „ p1

exp p2

2 m- mT - 1

where N0 is the particle number in the ground state, and the integral gives us the particle number on all excited states. Similarly, for density, wehave

(9.9)r = r0 +4 p

2 p Ñ3 p2 „ p1

exp p2

2 m- mT - 1

where r0 is the density of the particles on the ground states and the integral gives the density for all particles on the excited states. Above Tc,

(9.10)r0 =np = 0

V=

1

V

1

exp-m T - 1

In the thermodynamic limit V ض, r0 = 0. So we get our old integral formula back:

(9.11)r =4 p

2 p Ñ3 p2 „ p1

exp p2

2 m- mT - 1

However, below Tc, r0 cannot be ignored because:

(9.12)r0 =np = 0, m = 0

V=¶

¶= ?

How can we determine r0? We know r = r0 + rexcited, so

(9.13)r0 = r - rexcited = r -4 p

2 p Ñ3 p2 „ p1

exp p2

2 mT - 1

= r - m kB T 2 p Ñ232 z 3

2

=m kB

2 p Ñ2

32z 3

2

TC32 - T32

We can plot r0 as a function of T .

0.5 1.0 1.5 2.0TTc

0.2

0.4

0.6

0.8

1.0

r0r

Fig. 2. r0 as a function of T.

9.2.4. Bose-Einstein condensate and two fluid picture

120 Phys620.nb

In BEC, we separate particles into two groups: particles in the ground state and particle in excited states. This is closely related with the two-fluid picture in superfluid liquid He. Here, r0 is the superfluid part and rexcited is the normal part. However, we need to keep in mind twoimportant differences between BEC and He4, which come from interaction effects.

† For free bosons, BEC is a third order transition (the third order derivative of the free energy shows some discontinuity). For interacting bosons, the transition turns into a second order one.

† For free bosons, the normal component vanishes at T = 0. For interacting particles, there will always be some normal component even at T = 0.

9.2.5. U(1) Phase symmetry breaking

r0 ∫ 0 is the necessary condition for super fluidity but not sufficient. We know that any phase transition need to be associated with a symmetrybreaking (unless it is a topological transition). So which symmetry is broken in the superfluid transition? To answer this question, let’s considera theory for the “normal” component. Here, we treat the particles in the ground state as the “background” (or say as our “vacuum”). Because weare not considering particle in the ground state, we can use the integral formalism safely (quantum field theory). However, there is a price oneneeds to pay for using this integral formulas. Because we ignored the ground states, the particle conservation law is violated. so br ∫ 0, andthis is our order parameter.

For operator br, its expectation in the normal phase must be zero, due to the particle conservation law. However, in the superfluid phase,because we ignored the ground states, which has a large number of particles, b ∫ 0 and the amplitude of b gives us the superfluid

density b ∂ r0 .

How do we know that b ∂ r0 ? The simplest way to show it is via dimension analysis. „d r b†r br = N is the total number of

particle. N is a dimensionless quantity, so b†r br must have the dimension of r-d (density). So b must have the dimension of the square

root density. So, b ∂ r0 . Therefore, we know that b = 0 above Tc and b ∫ 0 below Tc. So we can use b as our order parameter.

Define yr = br as our order parameter, we can write down an phenomenological theory (the Ginsburg-Landau free energy):

(9.14)Fy, y* = „ r gT“y* “y + aT y* y + bT y* y2 + ...

For stability reasons, we assume g>0 and b>0 and a = a0T - Tc. similar to what we learned from our thermodynamics class, we need tominimize the free energy. Because g>0, we’d better have “y = 0 (uniform y).

(9.15)Fy, y* = „ r a0T - Tc y 2 + bT y 4

For T > Tc, the minimum of F is reached at y=0. For T < TC, the minimum of F has y = a0Tc - T 2 b . However, the phase of y can

take any value. The phase is chosen “randomly”. This can be seen by plotting F as a function of the real and imaginary parts of y, we get thefamous Mexican hat free energy:

Fig. 3. F as a function of Re(y) and Im(y)

0.5 1.0 1.5 2.0TTc

0.2

0.4

0.6

0.8

1.0

y r

Fig. 4. y as a function of T.

Phys620.nb 121

For T > Tc, F has only one minimum with y=0. So we only have one ground state. However at T < Tc, the y=0 state becomes unstable and

there are infinite number of minimum of F on the ring of y = a0Tc - T 2 b . This means that the system have infinite number of

degenerate ground state. Which ground state the system falls into is determined by chance and they all have the same opportunity. Here theHamiltonian (as well as the GL free energy) has the U(1) phase symmetry. b Ø b ‰Â f. However, the ground state doesn’t have this U(1)symmetry. (b is not invariant under b Ø b ‰Â f, because bØ b ‰Â f). So the low temperature superfluid phase breaks the U(1) phasesymmetry.

How to understand this phase symmetry breaking?

For an quantum state, the absolute phase has no impact on any physics observables, because we can shift the phase of all quantum states bysome constant. However, in a BEC state, because there is a huge number of particles on the ground states, the phase of the ground state give usa reference point. Now, we can measure the phase of any excited states by comparing it with the phase of the particles on the ground state. Dueto the existence of this reference point, the system no longer has the U 1 symmetry. Here, particles on the ground state pick up a phase, whichis very similar to the ferromagnetic state, where the spin pick up a preferred direction.

9.2.6. Goldstone mode

Let’s consider the ordered phase and we assume that the order parameter takes some expectation value

(9.16)y = y0 = y0 ‰Â f0

For a < 0, by minimizing the GL free energy, we find that y0 = -a 2 b .

Now, let’s consider small fluctuations around y0

(9.17)yrØ = y0 +dyr

Ø exp  f0 + dfrØ = -a 2 b + dyr

Ø exp  f0 + dfrØ

Here, dy is a real function, which describe the fluctuations of the amplitude of the order parameter. df is also a real function, which describe thephase fluctuations of the order parameter.

(9.18)

Fy, y* = „ r gT“ y* “y + aT y* y + bT y* y2 + ... =

„ r g “dyrØ2 + 2 a dyr

Ø2 + y0 2 „ r g “dfrØ2 + …

Here, we find one massive mode (the amplitude model dy) and one massless model (the phase model df, i.e. the Goldstone mode).

9.3. Cooper Pair and the BCS theory

Superconductivity and superfluid are very similar (no resistivity vs no viscosity), which implies that the fundamental physics may be similar.But electrons are fermions, and we cannot have more than 1 electron on quantum state, so there is no way to have BEC for fermions (whichrequires many particles to stay on the ground state). How can fermions form a BEC state? Well, one fermion cannot condensate, but a pair offermions is a boson and these pairs can condensate. These fermion pairs are known as Cooper pairs, and this theory of superconductivity isknown as the BCS theory (by Bardeen, Cooper and Schrieffer).

This idea may sound crazy, because electrons the same charge. Therefore, they have repulsive interactions. How can a bunch of particles withrepulsive interactions form pairs?

9.3.1. Electron phonon interactions

A rigorous treatment on electron-phonon interactions requires quantum field theory. Due to time-limitation here are just describe the physicalpicture without rigorous proof. Key conclusions:

† Electron-phonon interactions are crucial for conventional superconductors.

† Electron-phonon interactions induce an effective attractive interactions between electrons.

The first conclusion are based on the isotopic effect in conventional superconductors. If superconductor are purely due to electrons and latticesplays no role, the transition temperature Tc should be independent on the mass of the nucleons. However, experiments shows that the transitiontemperature Tc depends strongly on the mass of the nucleons, if we use different isotopes of a given element type of atom (e.g. Hg 203, Hg 202,Hg 200, Hg 199, Hg 198). This shows a hint that superconductor are related with the lattice motion (phonon).

The second conclusion are based on this physics picture

† An electron carries negative charge, which will attract nearby nucleons. So locally, the spacing between the nucleons will be reduced.

122 Phys620.nb

† Electron mass is much smaller than nucleon mass. So electrons moves much faster.

† When electron flies away, the lattice distortion will not recover immediately (because lattice moves slower). So locally, there is a higher density of positive charges here, which will attract other electrons.

† Combining all the things mentioned above, we found that an electron will attract other electrons, through lattice distortions. And this is phonon mediated attraction between electrons.

Theoretically, the attraction are induced when two electrons exchange a virtue phonon, which can shown rigorously in quantum field theory.

9.3.2. Model Hamiltonian

The key conclusion in the BCS theory is that: in a Fermi liquid, as long as there are some attractive interactions (no matter how weak theinteractions are), the Fermi liquid state will become unstable below some temperature Tc. This instability will lead to the formation of Cooperpairs.

Now, let’s consider electrons with attractive interactions. The energy contains two parts, the kinetic energy and the potential energy.

(9.19)E =k,sek nsk +

1

2

r,r'

s,s'V r - r ' nsr ns'r '

If we translate this into the form of second quantization, we get the Hamiltonian

(9.20)H =k,sek cks

† cks +1

2

s,s'

r,r'V r - r ' crs

† crs cr's'† cr's'

We can transfer the interaction part into the k space.

(9.21)H =k,sek ck,s

† ck,s +1

2

q,k,k'

s,s'Vq ck+q2,s

† ck-q2,s ck'-q2,s'† ck'+q2,s'

Let’s simplify the interaction term by only looking at one scattering channels

(9.22)H =k,sek ck,s

† ck,s +k,k'

Vk,k' ck† c-k

† ck' c-k'

This interaction is only part of the electron interactions. Here we ignored all other type of interactions. This turns out to be a good approxima-tion, because this interaction is the key which induces the pairing while all other interactions have little effects.

For simplicity, we assume that Vk,k' = V < 0 (attractive).

(9.23)H =k,sek ck,s

† ck,s + V k,k'

ck† c-k

† ck' c-k'

9.3.3. the Mean field approximation:

In BEC, the order parameter is b+. Here, pairs of electrons forms a BEC state, so the order parameter is k ck† c-k

†. In other words, in the

condensed phase, c† c† ∫ 0. We can write:

(9.24)ck† c-k

† = ck† c-k

† + ck† c-k

† - ck† c-k

†(9.25)ck' c-k' = ck' c-k' + ck' c-k' - ck ' c-k'

The physical meaning for these two formulas are: the operator ck† c-k

† has average value ck† c-k

†. On top of the average value, the

operator fluctuates around this average value with fluctuations ck† c-k

† - ck† c-k

†. If we believe that the fluctuations of the operator is

small, the last term in the formula above is small. Here we can define Xk' = ck' c-k' , as a result, Xk* = ck

† c-k†. So we have

(9.26)ck† c-k

† = Xk* + ck

† c-k† - Xk

*(9.27)ck' c-k' = Xk' + ck' c-k' - Xk'

We rewrite the interactions in terms of the average values Xk' and Xk* and fluctuations, ck

† c-k† - Xk

* and ck' c-k' - Xk'

(9.28)

V

2

kck

† c-k† ck' c-k' = V

k,k 'Xk

* + ck† c-k

† - Xk*Xk' + ck' c-k' - Xk' =

V k,k'

Xk* Xk' + Xk

*ck' c-k' - Xk' + ck† c-k

† - Xk* Xk' + ck

† c-k† - Xk

* ck' c-k' - Xk'.

Phys620.nb 123

The last term is fluctuations times fluctuations. If fluctuations is small, the last term would be the smallest part. We will ignore this small term,and this approximation is known as the mean-field approximation (ignore the fluctuations*fluctuations part). Within this approximation, theHamiltonian becomes

(9.29)HMF =kek ck,

† ck, + ek ck,† ck, + V

kXk

* k'

ck' c-k' + V k

Xk k

ck† c-k

† - V kk'

Xk* Xk'

The mean-field Hamiltonian only have quadratic terms in c and c† (very similar to an free Fermi system).

Define V k Xk and V k Xk,

(9.30)HMF =kek ck,

† ck, + ek ck,† ck, + D

* ck c-k + D ck† c-k

† - D* D

V

If we work with fixed chemical potential, m, it make s more sense to consider H - m N , instead of H

(9.31)HMF - m N =kek - m ck,

† ck, + ek - m ck,† ck, + D

* ck c-k + D ck† c-k

† - D* D

V

We can write it in terms of a two-by-two matrix

(9.32)HMF - m N =k ck,

† c-k, e k - m DD* m - ek

ck,

c-k,†

-D* D

V

This is very similar to the Hamiltonian of a 2-band systems.

9.3.4. Bogoliubov transformation

Similar as in a 2-band systems, we should diagonalize the matrix using a unitary transformation, and here, this transformation is known as theBogoliubov transformation. Define:

(9.33)g1 k = u ck, + v c-k,†

(9.34)g2 k = -v* ck, + u* c-k,†

where u and v are complex numbers with u 2 + v 2 = 1. It is easy to check that the conjugate operators are

(9.35)g1 k† = u* ck,

† + v* c-k,

(9.36)g2 k† = -v ck,

† + u c-k,

The inverse transformation is:

(9.37)ck, = u* g1 k - v g2 k

(9.38)c-k,† = v* g1 k + u g2 k

and

(9.39)ck,† = u g1 k

† - v* g2 k†

(9.40)c-k, = v g1 k† + u* g2 k

†

It is also easy to check that the new operators g1 k, g2 k, g1 k† and g2 k

† satisfies the anti-commutation relations (so they are fermions). Here, Icheck on anti-commutation relation

(9.41)g1 k

†, g1 k =u ck,

† + v c-k, , u ck, + v c-k,† = u2 ck,

†, ck, + v2 c-k, , c-k,† + u v ck,

†, c-k,† + u v c-k, , c-k, = u2 + v2 = 1.

Therefore, these operators creates and annihilates fermions, and these fermions are known as Bogoliubov quasi-particles.

(9.42)HMF - m N =k g1 k

† g2 k† ÿ u v

-v* u* ÿ e k - m D

D* m - ek ÿ u* -vv* u

g1 k

g2 k - D* D

V

124 Phys620.nb

(9.43)

u v-v* u*

ÿ e k - m DD* m - ek ÿ u* -v

v* u = u v

-v* u* ÿ ek - m u* + D v* -ek - m v + D u

D* u* + m - ek v* -D* v + m - ek u=

ek - m u* u + D v* u + D* u* v + m - ek v* v -2 e k - m v u + D u2 - D* v2

-2 ek - m u* v* - D v*2 + D* u*2 ek - m v v* - D u v* - D* u* v + m - ek u* u

Since we want to diagonalize the matrix, -2 e k - m v u + D u2 - D* v2=0

(9.44)2 ek - m v u = D u2 - D* v2

Define:

(9.45)u = ‰Â fu cos c and v = ‰Â fv sin c and D = D ‰Â f

(9.46)2 ek - m cos c sin c ‰Â fu+ fv = ‰2  fu+ f D cos2 c - ‰2  fv- f D* sin2 c

So we have

(9.47)fu + fv = 2 fu + f = 2 fv - f

and

(9.48)2 ek - m cos c sin c = D cos2 c - sin2 cFor the phases, we get fu - fv = -f, while fu + fv can take any value. Without loss of generality, we can choose fu = -fv = -f 2. For c wehave,

(9.49)ek - m sin 2 c = D cos 2 c

So tan 2 c =D

ek-m .

In conclusion, if we choose u = ‰- f2 cos c and v = ‰Â f2 sin c with c = 1

2arctan

D

ek-m , the two-by-two matrix turns into a diagonal matrix:

(9.50) ek - m cos 2 c + D sin 2 c 00 -ek - m cos 2 c - D sin 2 c

= e k - m 2 + D 2 0

0 - e k - m 2 + D 2

(9.51)HMF - m N =k g1 k

† g2 k† ÿ Ek 0

0 -Ek g1 k

g2 k - D* D

V

with Ek = ek - m2 + D 2 .

For the top band, we have energy Ek ¥ D . for the bottom band, the dispersion relation -Ek § - D . So we have energy gap 2|D|, if |D|>0.

A superconductor is an insulator of Bogoliubov quasi-particles!

9.3.5. D=?

The value of D is determined by self-consistent condition.

(9.52)

D = V kc-k ck = V

kv g1 k

† + u* g2 k† u* g1 k - v g2 k =

V ku* v g1 k

† g1 k - u* v g2 k† g2 k = V

ku* v

1

exp Ek

kB T + 1

- u* v1

exp -Ek

kB T + 1

(9.53)D ‰Â f = V ‰Â f k

sin 2 c

2 1

exp Ek

kB T + 1

-1

exp -Ek

kB T + 1

(9 54)

D = V k

sin 2 c

2

exp- Ek

2 kB T

exp Ek

2 kB T + exp- Ek

2 kB T-

exp Ek

2 kB T

exp- Ek

2 kB T + exp Ek

2 kB T =

Phys620.nb 125

V k

sin 2 c

2

exp- Ek

2 kB T - exp Ek

2 kB T

exp Ek

2 kB T + exp- Ek

2 kB T= -V

k

D

2 e k - m 2 + D 2

tanhEk

2 kB T

(9.55)D = -V k

D

2 e k - m 2 + D 2

tanhEk

2 kB T

|D|=0 is always a solution for this equation. If |D|∫0,

(9.56)1 = -V

2

k

1

e k - m 2 + D 2

tanhe k - m 2 + D 2

2 kB T

For repulsive interactions, this equation has no solution, because the l.h.s.>0 while the r.h.s.<0. However, for V<0, we have

(9.57)1 =V

2

k

1

e k - m 2 + D 2

tanhe k - m 2 + D 2

2 kB T

The equations may have some solution. In a Fermi liquid, typically only fermions near the Fermi surface will contribute and near the Fermisurface, the sum of momentum k can be written as an integer of energy.

(9.58)1 =V

2 N0 „ e 1

e - m 2 + D 2

tanhe - m 2 + D 2

2 kB T

where N(0) is known as the density of the states, which can be measure directly in STM. The range of the interaction is determined by theDebye frequency, which is the typical energy scales of the phonons (remember that the attractive interaction comes from phonons).

(9.59)1 =V

2N0

m-eD

m+eD

„ e

tanh e k - m 2 + D 2 2 kB T

e k - m 2 + D 2

If we define x=e-m,

(9.60)1 =V

2N0

-eD

+eD

„ x

tanh x 2 + D 2 2 kB T

x 2 + D 2

= V N0 0

+eD

„ x

tanh x 2 + D 2 2 kB T

x 2 + D 2

For T > Tc, there is only one solution for the self-consistency equation, D=0.

For T < Tc, we have one solution D=0, which corresponds to the unstable solution for the G-L free energy. In addition, we also have a solutionwith |D|=D(T), where D(T) is a function of T.

Because |D|=0 at Tc, we know that

(9.61)1 = V N0 0

+eD

„ xtanhx 2 kB TC

x= V N0

0

eD2 kB TC

„ xtanhx

x

In the weakly coupling limit [|V|<<1/N(0)], kB Tc << eD, so the upper limit of the integral is very large eD 2 kB TC >> 1. For this limit,

(9.62)1 = V N0 0

eD2 kB TC

„ xtanhx

xº N0 V ln 1.13

eD

kB TC

So,

(9.63)Tc = 1.13eD

kB

exp- 1

N0 V

The critical temperature is proportional to the Debye frequency. In addition, it depends on the density of states N(0) and interaction strengths|V|. As the interaction becomes weaker and weaker, Tc goes down to zero.

126 Phys620.nb

-1.0 -0.8 -0.6 -0.4 -0.2 0.00.0

0.1

0.2

0.3

0.4

N0V

k BT

ceD

Fig. 5. Tc as a function of the interaction strength.

Order parameter at zero temperature can be found as:

(9.64)1 = V N0 0

+eD

„ x1

x 2 + D 2

= V N0 lneD + D0

2 +eD2

D0

In the weak coupling limit [V N0<< 1], we have D0 << eD

(9.65)1 = V N0 lneD + D0

2 +eD2

D0

º V N0 ln2 eD

D0

(9.66)D0 = 2 eD exp -1

N0 V

In particular, the ratio between D0 and Tc is a universal constant:

(9.67)D0

kB T=

2

1.13= 1.764

This is true for all weakly-correlated superconductors.

9.4. Effective gauge theory

9.4.1. the GL free energy

For super fluids, we learned above that the GL free energy is

(9.68)Fy, y* = „ r gT Ñ2 “y* “y + aT y* y + bT y* y2 + ...

Here, because Cooper pairs are charged particles (with charge 2 e), the order parameter is coupled with gauge fields (E and B fields). Using thestandard technique of minimal coupling, it is easy to show that

(9.69)Fy, y* = „ r gT-Â Ñ“-2 e AØ y* -Â Ñ“-2 e A

Ø y + aT y* y + bT y* y2 + ...

Here, we substitute pi = -Â Ñ ∑i into pi = -Â Ñ ∑i-2 e Ai, which is called minimal coupling. The factor 2 e is because a Cooper pair carriescharge 2 e.

9.4.2. Anderson-Higgs mechanism

Let’s consider the ordered phase and we assume that the order parameter takes some expectation value

(9.70)y = y0 = y0 ‰Â f0

Now, let’s consider small fluctuations around y0

Phys620.nb 127

(9.71)yrØ = y0 +dyr

Ø exp  f0 + dfrØ

Here, dy is a real function, which describe the fluctuations of the amplitude of the order parameter. df is also a real function, which describe thephase fluctuations of the order parameter.

(9.72)

Fy, y* = „ r gT-Â Ñ“-2 e AØ y* -Â Ñ“-2 e A

Ø y + aT y* y + bT y* y2 + ... =

„ r g -Â Ñ“-2 e AØ dyr

Ø*-Â Ñ“-2 e AØ dyr

Ø + 2 a dyrØ2 + y0 2 gT Ñ2 „ r “ f -

2 e

ÑAØ 2

+ …

Here, we find one massive mode (the amplitude model dy), same as the charge neutral case. However, for the phase mode df, we can make agauge transformation

(9.73)AØØ A

Ø+

Ñ

2 e“f

This transformation absorbs the f mode into the gauge field A. Thus, the free energy becomes

(9.74)Fy, y* = „ r g -Â Ñ“-2 e AØ dyr

Ø*-Â Ñ“-2 e AØ dyr

Ø + 2 a dyrØ2 + y0 2 gT Ñ2 „ r A2 + …

We find two things here

(1) The phase mode disappeared. (i.e. There is no Goldstone mode).

(2) The gauge field now has one additional term in the free energy ~A2. Remember that in a gauge theory, typically we only have terms like∑A2. In k-space, this means that the dispersion relation of a photon is w2 = k2. However, once we have Ò A2, the dispersion becomesw2 = k2 + Ò, and the coefficient of the A2 term is the square of the mass of photons, i.e. in a superconductor, photons become massive particles.This is known as the Anderson-Higgs mechanism.

This mechanism generates mass for a gauge field. In the standard model, gluons, which mediate strong interactions, obtain mass from thismechanism. Same is true for W and Z bosons, which mediate weak interactions.

9.4.3. Meissner effect and magnetic levitation

In the GL free energy,

(9.75)

Fy, y* =

„ r g -Â Ñ“-2 e AØ dyr

Ø*-Â Ñ“-2 e AØ dyr

Ø + 2 a dyrØ2 + y0 2 gT Ñ2 „ r “ f -

2 e

ÑAØ 2

+ …

to minimize the last term, “ f - 2 e

ÑAØ= 0. Therefore, the magnetic field

(9.76)B = “ μA =Ñ

2 e“ μ“f = 0

There is no magnetic field inside a superconductor. This is the Meissner effect.

9.4.4. Critical field Bc

The Meissner effect expels the B field, which costs energy B2 2 (times the size of the system). If this energy cost is larger than the energy gain

to form Cooper pairs D, the system will not want to form a superconducting state. This defines a critical field BC = 2D . If B field is abovethis value, the system turns into a normal metal (paramagnetic)

9.4.5. quantization of a magnetic flux.

If we have a superconducting ring, we can integrate the gauge field around the ring, which is just the magnetic flux.

(9.77) A ÿ„ r = B ÿ„ S

We know that ∑m f = Am, so

128 Phys620.nb

(9.78) A ÿ„ r = ∑m fÿ„ r = 2 n p

Here we used the fact that the order parameter is single-valued. Therefore, if we go around the ring, the phase of the order parameter can onlychange by 2 n p. If we put the unit back,

(9.79) A ÿ„ r = ∑m fÿ„ r = 2 n pÑ

2 e=

n p Ñ

e

Notice that the denominator here is 2e instead of e. This means that the fundamental particles in a superconductor carries electronic charge 2e,which is a cooper pair. The quantization of the magnetic flux (to 2pÑ/2e instead of 2p Ñ/e) is a direct experimental evidence for the formation ofCooper pairs.

This is also one of the best techniques to measure Ñ

This is also the direct experimental evidence for cooper pairs (fundamental charge is 2e instead of e).

9.4.6. Lagrangian.

If we write down the Lagrangian for a superconductor, we know that it must contain gauge fields and the phase of the order parameter. Theelectrons should not appear if we only care about low energy physics smaller than the superconducting gap (remember that superconductors areinsulators for Bogoliubov quasi-particles, so the fermions cannot move similar to an insulator. Similar to an insulator, we don’t need to considerthe motion of electrons). The amplitude of the order parameter is not important either, because their value is fixed by the GL free energy. Sohave can write down the theory as:

(9.80)L =1

2 „ r

Øe0 E2 -

B2

m0

+ LS∑m fr - 2 e AmrÑ

It is important to notice that the superconductor part LS cannot depends on f itself, because f is not gauge invariant. However, it can be afunction of ∑m fr - 2 e AmrÑ, which is gauge invariant. For simplicity, lets use the theorist units: 2e=Ñ=1.

(9.81)L =1

2 „ r

Øe0 E2 -

B2

m0

+ LS∑m fr - Amr

9.4.7.charge and currents

Currents and charge

(9.82)Ji =dLS

dAi

= -dLS

d ∑if

(9.83)r =dLS

dA0

= -dLS

d ∑0f

9.4.8. Zero resistivity

We know

(9.84)r =dLS

dA0

= -dLS

d ∑0f

We also know that the canonical momentum p is defined as

(9.85)p =dLS

d ∑0f

So, r = -p. Therefore, the Hamiltonian should be considered a function of -r and f, instead of ∑0f and f.

The equation of motion is:

(9.86)∑0f =dHS

dp= -

dHS

dr

Phys620.nb 129

On the other hand, we know that

(9.87)dHS

drr = V r

where V is the voltage. This is because E = rr V r „ r

Therefore,

(9.88)∑0fr = -V rFor a static system, the phase of the order parameter should be time-independent, ∑0f = 0. So we have V r=0. This means that if we have aconstant current in the system (the system is static), V = 0. No voltage but finite current. This is superconductivity.

9.4.9. The Josephson junction.

Josephson junction has two superconductors separated by a thing layer of insulator. The Lagrange of a Josephson junction LJ depends on thephase difference between the two superconductors, which is also gauge invariant

(9.89)L = LE and B + LS1 + LS2 + Ljunction

(9.90)Ljunction = A FDfwhere A is the area of the junction. FDf is a function of the phase difference between the two superconductors Df = fL - fR. It is easy tonotice that FDf must be a periodic function with periodicity 2 n p Ñ 2 e = n p Ñ e. Again the denominator 2 e is due to the fact that Cooperpairs have charge 2 e.

(9.91)FDf = FDf + n p Ñ eFor a charge neutral particle,

(9.92)Df = „ r ÿ “f

However, for charged particles, in the presence of a gauge field A, the gauge invariance Df shall take the form

(9.93)Df = „ r ÿ “f - A

This formula is gauge invariant.

Therefore the current cross the junction is

(9.94)J =d Ljunction

d A= A F ' Df d Df

d A= A F ' Df d A

d A= A F ' Df

Now we apply a voltage V across the junction. Because we know that

(9.95)∑0fr = -V r,it is easy to notice that

(9.96)Df = -V t + constant

Therefore:

(9.97)J = A F ' -V t + constantBy applying a fixed V , we found that the current is changing with time. Because F is a periodic function, F’ is also a periodic function. So J isaperiodic function of t, and the periodicity is p Ñ/e DV

9.5. High Temperature superconductivity

High Tc compounds are a family of materials: La2-x Srx CuO4, Bi2 Sr2 Can-1 CunO2 n+4+x (BSCCO), YBa2 Cu3 O7-x, etc

130 Phys620.nb

9.5.1. Lattice structure† Common feature: CuO2 planes separated by some insulating layers

† Undoped compounds are Mott insulators

† 1 hole per unit cell.

† Cu site has lower potential energy, so holes want to stay on Copper, instead of Os

† Strong Coulomb repulsion on the Cu site (it costs a lot of energy to put two holes on the same site). Basically, we can only have one hole per Cu site,

† Holes cannot move, because if we move a hole to a neighboring site, we will create double occupancy, which is prohibited due to strong repulsion.

† Each Cu site has one hole, and the hole carrier spin 1/2. So we have a lattice of spins. These spins form an anti-Ferromagnetic ordering.

9.5.2. Dope the system

Q: How to change electron density?

Option #1: Dope the system (replace some of the atom by another type of atom with different number of electrons, or remove some atoms fromthe systems).

† Dope electron or hole? Depending on the elements, we may be adding electron or removing electron from the system. Adding electron is know as “electron doping”. Add hole is known as “hole doping”.

Option #2, Gating

† too weak for high TC compounds

Q: Which technique is good?

A: If one can choose from these two options, gating is in principle better, because it doesn’t induce disorders. But in many cases gating onlychange the electron density by a small amount, so we have no choice by using the option #1.

Some high Tc compounds are hole doped and some are electron doped. But in most cases, “high Tc” means hole doped high Tc compounds,because hole doped case have a higher Tc.

9.5.3. Phase diagram

Fig. 6. Schematic Phase diagrams of hole-doped (right side) and electron doped (left side) high Tc superconductors (from wikipedia)

9.5.4. Tc

Phys620.nb 131

Fig. 7. Tc vs year

Iron based superconductor shows some similarity to electron doped high-Tc compounds (but they also have major differences). They havesimilar Tc. Both of them have strong e-e interaction, but the e-e interaction is weaker in comparison to hole-doped high Tc.

9.5.5. The order parameter (d-wave)

Order parameter is a function of momentum

(9.98)Dk = ck c-k

In conventional superconductors (BCS theory), we assume D is isotropic, Dk = ck c-k = constant. But in general it is a function of kØ

(9.99)Dk = DkØ = D k , qk

The dependence on |k| is not important, because we know only electrons near the Fermi surface matters, k~kF .

(9.100)Dk = DkØ = D k , qk º DkF , qk = Dqk

Let’s consider the case Dqk is a real function. For conventional SCs and high Tc SCs, we can always make D real by choosing a proper gauge,but there are cases where D is NOT real and cannot be made real.

For conventional superconductors, Dqk varies as a function of q, but the sign never change (which is called s-wave paring). For high Tc

superconductors, Dqk changes sign for four times as we go around from qk = 0 to 2 p. It has a d-wave symmetry and thus is called d-waveparing.

The superconducting phase of high TC materials are described very well by a d-wave pairing state, if one don’t ask why the electrons pair up atsuch a high temperature. The mystery of high Tc is in fact not about the superconducting phase, but about the normal phase. The normal phaseis a bad metal and we don’t really have a theoretical understanding about it. Since we don’t know the normal phase, we don’t know whyelectron turns into a superconductor.

9.5.6. Normal phase† Like a Fermi liquid at high doping.

† Not a Fermi liquid at low doping (known as nonFermi liquid or strange metal, or pseudo-gap phase)

Pseudo gap:

For the under-doped side, there is an temperature scale, T*, at which the behavior of many experiential measurable quantities changes. Differentexperiments give different T*. Below T*, the density of state suggests that there seems to be a insulating gap, but not a full insulating gap (apseudo-gap).

Q: Whether this is related with the superconducting gap or this is due to some other ordering tendency?

132 Phys620.nb

A: Unclear.

Q: Whether T* is a phase transition or a crossover?

A: Unclear. Because different experiments give different T*, it is probably a crossover (phase transition should have a well-define transitiontemperature Tc and all experiment should give the same Tc). However, in the presence of disorder, the transition temperature Tc many havesome distribution. So it is NOT impossible to think T* as a phase transition.

9.5.7. Mechanics for superconductivity: under debating

What we know:

† Not phonon driven

† too weak to get such a high TC.

† Isotope effect is weak.

† Phonon prefer s-wave pairing

† Very likely, it is due to e-e interaction and probably involves spin

† e-e interaction is very strong

† spin-spin coupling is very strong

† TC is too high to be explained by other interactions, which are typically too weak to give such a high TC.

Q: How can repulsion lead to pairing?

A: Unclear, but there are some hints.

Phys620.nb 133