Mon., Nov. 8:
# Mentioned unusual statistics, the possibility of identical particles
whose state is changed
by a permutation. This may be how certain quasiparticles in systems
exhibiting the fractional
quantum hall effect behave. One reference on this is section 4, ``Non-abelian
and Projective
Statistics", in a paper by F. Wilczek, Some
Examples in the Realization of Symmetry
(Nucl.Phys.Proc.Suppl. 68 (1998) 367-373). Another reference is
A Chern-Simons
Effective Field Theory for the Pfaffian Quantum Hall State by E.
Fradkin et. al.
# Fermi sea is the filled states up to the Fermi energy
E_F = p_F^2/2m, where p_F is
the Fermi momentum equal to hbar times the Fermi wavevector
k_F. The top of the Fermi sea
is called the Fermi surface. For free electrons in a large box
it is a sphere. In a condensed
matter environment such as a metal, taking into account all interactions,
the notion of filling
up the lowest energy states still makes sense, and leads to a Fermi
surface in momentum
space that can have a distorted shape and even nontrivial topology.
# COOPER PAIRS
# References: handout pages from P.G. de Gennes, Superconductivity
of Metals and Alloys;
Chapter 8 in Baym; other specialized books, e.g., J.R. Schrieffer,
Theory of Superconductivity;
or M. Tinkham, Introduction to Superconductivity.
# Mechanism of pairing plays role in superfluidity and superconductivity
in condensed matter and in
neutron stars, and for nucleons outside closed shells in nuclei. (How
about in large molecules?)
# Historically, two pieces of experimental info. guided the theory
of superconductivity:
1. specific heat at low T goes like exp(-Delta/kT), indicating an energy
gap for low energy excitations.
2. critical temperature decreases as m^(-1/2) with increasing mass
of nuclear isotopes (isotope effect),
indicating that dyamics of ion cores plays a role.
There will be a gap if electrons form bound states. If we single out
two electrons, they
repel each other, but they are attracted to the ions, so they can attract
each other
via the intermediary of displaced ions. Even if this attraction is
too weak to form a bound state
by itself, we'll see that in the presence of the Fermi sea it can always
form a state with energy
less than that of two electrons at the Fermi surface, which is a bound
state. The Fermi sea is thus
unstable to the formation of pairs. More massive ions displace
less, hence the effective
electron-electron attraction is weaker , hence the binding energy is
smaller, hence the gap
is smaller, hence the critical temperature is lower, in accordance
with observations.
Wed., Nov. 10
# Cooper (Phys. Rev. 104, 1189 (1956)) found a
simple model that displays the existence of the pair gap.
Think of adding two electrons to the Fermi sea.
Neglect the interaction between the electrons in the Fermi sea, and
between the extra electrons and
the Fermi sea electrons, and adopt a potential <p1p2|V|p1'p2'>
between the extra electrons.
In the absence of impurities the total momentum P=p1+p2 is conserved,
so the potential has the form
<p1p2|V|p1'p2'> = delta_PP' V_kk', where k= (p1-p2)/2 is the relative
momentum, and similarly for k'.
The simple potential adopted by Cooper is either zero or -V/L^3 , being
non-zero only if the energies
corresponding to all four momenta are between the Fermi energy E_F
and E_F + E_D,
where E_D is hbar times the Debye frequency, the highest frequency
for lattice oscillations.
Typically E_D/E_F ~ 1/100. Note that since the potential depends only
on the magnitudes of
the momenta, it is nonzero only on (relative) s-wave wavefunctions.
These are symmetric under
pair interchange, hence the spin state of the two bound electrons must
be anti-symmetric,
i.e. the spin singlet state.
(The pairs are in an s-wave state for metallic superconductors,
a p-wave state for 3He, and (mostly?) d-wave for high-T_c superconductors.)
# For fixed total momentum P, the region in momentum space where the
potential is nonzero is largest
when P=0, and the size drops rapidly as P grows.
-------------------------------------------------------------------------------------------------
[[ I did not give a complete explanation why in class but let me give
the reason here.
You can visualize it in momentum space: let P = p1 + p2 and k = (p1-p2)/2.
Then p1 = P/2 + k and p2 = P/2 - k. In k-space draw a spherical shell
of finite thickness,
with with inner and outer radii p_F and p_max=sqrt[2mE_max], where
E_max=E_Fermi + E_Debye,
centered on P/2, and another such shell centered on -P/2.
The allowed k's are the ones for which |p1| = |P/2 + k| and |p2|
= |-P/2 + k| are between p_F and p_max,
and these are precisely those k's in the intersection of the two shells.
(Draw and label a picture
to see this.) When P=0, the shells coincide, so you get the whole volume
of the shell.
As |P| grows the volume of the intersection shrinks rapidly.]]
--------------------------------------------------------------------------------------------------
Hence the most significant binding happens for pairs with P=0.
Let us just consider these, so p1=k and p2=-k.
The pair wavefunction g(k) in momentum space vanishes unless k > k_F,
since the extra electrons
are excluded from the Fermi sea. The Schrodinger equation with
the reduced mass mu=m/2 reads
k^2/m g(k) + Sum_k' V_kk' g(k') = (2E_F + E)g(k),
which has the solution
g(k) = C/(k^2/m - 2E_F - E), where C = - Sum_k' V_kk' g(k').
Consistency then requires that
1 = - Sum_k' V_kk' /(k'^2/m - 2E_F - E), the gap equation.
Rewrite this in terms of an energy integral. Let x = k'2/2m - E_F.
Introducing the density of states N(x) per unit energy interval
per unit volume, we have
1 =V int_0^E_D dx N(x)/ (2x - E).
The density of states N(x) is fairly constant in the interval of integration,
so we get approximately
1 = (N(0)V/2) ln[(2E_D - E)/(-E)],
with the solution
E = -2E_D/(exp(2/N(0)V)-1).
If N(0)V is small we get approximately
E = - 2E_D exp(-2/N(0)V), the gap.
According to Baym, typically N(0)V ~ 1/4,
so we get exp(-8) ~ 3 10^-4, and 2E_D ~ 2E_F/100 ~ 0.1 eV,
E ~ - 3 10^-5 eV ~ -0.3 K. The critical temperature is some ten
times this...
# Remarks:
1. Without the Fermi sea a weak potential doesn't have a bound state.
The sum in the gap equation with E_F=0 is positive and finite at E=0
since N(x) ~ x^1/2 near zero energy.
If V is small enough, there is therefore no solution.
2. The gap energy is not analytic in V. The bound state cannot be found
by perturbing in V.
3. The length scale of the pair wavefunction works out to be
~(E_F/E)/k_F, i.e. huge: ~ 10 microns
(?? I think it's more like 1 micron.) (In high-T_C cuprate superconductors
it's only a few angstroms.)
4. The model shows that there is an instability for the ground state
to "condense" into Cooper pairs.
This involves all the electrons near the top of the Fermi sea, each
paired with every other, so to speak.
The pairing is strongest when the pair momentum is zero. If there is
a net momentum of the pairs,
i.e. a current, then the pairing is strongest when all pairs
have the same momentum.
(Since the gap is smaller when the pair momentum is not zero,
the critical temperature is lower in the presence of a supercurrent.)
The current persists, since individual electrons cannot be scattered
without breaking their
Cooper bond with all the other electrons. (At least I think
this is what Baym meant...)
Fri., Nov. 12
# Remarks about the Cooper pair problem that I didn't have a chance
to make Wednesday.
(They are included in Wednesday's minutes.)
# CHARGED PARTICLE IN A MAGNETIC FIELD
Classically, F = (q/c) v x B. The cross product tells the particle
which way to accelerate.
Quantum mechanically, the same can be true in the Heisenberg equations
of motion.
But in the Schrodinger picture, how does the wave know which way to
refract?
Answer: the vector potential enters the Schrodinger equation,
not the magnetic field...
just as the scalar potential enters, not the electric field.
# Classically, in the Hamiltonian formalism, the Hamiltonian is H =
(p-(e/c)A)^2/2m
for a charge e in a vector potential A. The magnetic field is B = curl
A.
p is the canonical momentum, with Poisson bracket {x_i, p^j}
= delta^i_j.
Since the Hamiltonian is the energy, which in this case is just the
kinetic energy (1/2)mv^2,
the kinetic (or "kinematic" or "mechanical") momentum
is mv = p - (e/c)A.
Hamilton's equations for the position and momentum can be written as
dx^i/dt = {x^i,H}, dp_i/dt = {p_i,H}.
# Canonical quantization is the rule of substituting operators
for x^i and p_i in the classical
Hamiltonian, and assuming canonical commutation relations
[x^i,p_j]=i hbar delta^i_j.
With this rule, Heisenberg's equation for an operator is dA/dt
= (1/i hbar)[A,H] is guaranteed
to reproduce the classical equation of motion (up to operator ordering
ambiguities, which produce
terms of linear or higher order in hbar) since the commutator obeys
the same algebraic rules as
the Poisson bracket:
(i) antisymmetry [A,B] = -[B,A],
(ii) bilinearity [A,B+C] = [A,B]+[A,C],
(iii) ``Leibniz" identity [A,BC] = [A,B]C + B[A,C].
Experiment shows that canonical quantization of the classical hamiltonian
for a particle in an external
magnetic field (with the operator order as written above) yields the
correct description of a quantum
particle. (This operator ordering is dictated by gauge-invariance.
See the minutes to come.
If we don't do something perverse, like writing p^2 = x p^2 x^-1, the
order in this case is determined
by Hermiticity, which chooses p.A+A.p over 2A.p, for example.)
# As a first example, let's look at a particle in a uniform magnetic
field in the z-direction. Classically, the motion in the z-direction is
free, while that in the xy plane is a circle of any radius with angular
frequency
w_c = |e|B/mc, the cyclotron frequency. The kinetic energy in
the xy plane is KE = (1/2)m w_c^2 r^2.
Classically, circular orbits of arbitrarily small radius and low energy
exist. Quantum mechanically, the
uncertainty relation prohibits the limit of zero velocity and definite
location, so there is a ground state
with non-zero energy, called the lowest Landau level. Above
the ground state are evenly spaced energy
levels, called Landau levels.
This can be seen roughly using the semiclassical Bohr-Sommerfeld quantization
condition:
\oint p_i dx^i = nh.
Here p_i = mv_i + (e/c) A_i is the canonical momentum.
Using Stokes' theorem, \oint A_i dx^i is the magnetic flux pi
r^2 B through the circular orbit.
The Bohr-Sommerfeld quantization then yields r^2 = 2 n hbar/m w_c,
hence E = n hbar w_c.
The correct quantum result is actually E = (n+1/2) hbar w_c, like a
harmonic oscillator.
In fact the particle in a magnetic field is formally identical to a
harmonic oscillator (see hw#10).
# As our second example, consider a particle confined to a one-dimensional
ring of radius b in the
x-y plane, with a magnetic flux through the ring but a magnetic field
which vanishes everywhere on
and near the ring. Classically, the particle would be uninfluenced
by the distant magnetic field.
But quantum mechanically, there is an influence.
Using Stokes' theorem, the vector potential along the loop can be taken
as A=Phi/(2pi b),
where Phi is the magnetic flux. Schrodinger's eqn thus reads
(1/2m)[-ihbar d/d(bs) - (e/c)Phi/2pi b]^2 f(s) = E Y(s),
with s the angle around the ring.
Continuity of the wave function imples f(2pi) = f(0).
Trying a simple periodic ansatz f(s)=exp(ins), for some integer n,
we find the energy spectrum
E = [n - (e/|e|)(Phi/Phi_0)]^2 (hbar^2/2mb^2),
where Phi_0 =hc/|e| is the flux quantum, Phi_0 =
4.135 10^-7 Gauss-cm^2
# This example illustrates the Aharonov-Bohm effect: a quantum
particle feels not the magnetic
field (which is zero in this case), but the vector potential...or rather
the loop integral (flux) associated
with the potential. Actually, the particle feels the flux only modulo
integer multiples of the flux quantum:
the spectrum of the hamiltonian is unchanged when the flux changes
by a multiple of the flux quantum.
As the flux grows from zero to Phi_0, the energy levels shift up for
one sign of n and down for the other, finally coinciding with the original
levels.
# In the free particle case, the Bohr-Sommerfeld quantization condition
is equivalent to the statement
that the flux through the orbit is an integer number of flux quanta.
For n=1 this yields r_0 = 3.63 10^-6 cm (B/1 Tesla)^-1/2.