PHZ7427: Spring 2014
BCS Theory
D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. NPB2114
I.
NOTATIONS
Second-quantized operators, a, b, as well as operators composed of them, do NOT
have ”hats”, i.e., a ≡ a† , etc.
kB = 1.
II.
HAMILTONIAN
The second-quantized version of the most general Hamiltonian describing the
electron-electron interaction reads
X
X
1
H=
εp a†pσ apσ +
Up,k;p0 ,k0 a†p0 σ a†k0 σ0 akσ0 apσ
(1)
2V
0 0
0
0
p,σ
pkp k ;σσ ;k6=k
where V is the system volume and σ =↑, ↓ is the spin projection. Notice that we
excluded processes with zero momentum transfer. This is the same constraint as we
encountered in the Hartree-Fock theory: processes with zero momentum transfer
correspond to the interaction of homogeneously distributed electrons but such an
interaction is cancelled by the interaction of electrons with ions and the interaction
among ions. Notice also that momentum conservation implies that p + k = k0 + p0 .
It is convenient to introduce the effective “statistical”
Hamiltonian H 0 = H − µN ,
P
where µ is the chemical potential and N = pσ a†pσ apσ is the number operator.
Then
X
X
1
H0 =
ξp a†pσ apσ +
Up,k;p0 ,k0 a†p0 σ a†k0 σ0 akσ0 apσ
(2)
2V
0 0
0
p,σ
pkp k ;σσ ;
where ξp = εp − µ. As in the Cooper problem, we assume that pairing involves only
electrons with opposite momenta and opposite spins, which means that k = −p,
k0 = −p0 , and σ 0 = −σ with − ↑=↓. The pairing interaction is assumed to be equal
to a negative constant within an interval −Λ..Λ around the Fermi energy. In the
phonon model, Λ = ~ωD . With these assumptions, the Hamiltonian is re-written as
X
λ X † †
0
†
H =
ξp apσ apσ −
ap0 ↑ a−p0 ↓ a−p↓ ap↑ .
(3)
V
0
0
p,σ
pp ;p6=p
2
III.
BOGOLIUBOV TRANSFORMATION
The idea due to Bogoliuibov is to introduce new fermions which are linear superpositions of the p ↑ and −p ↓ states:
bp↑ = up ap↑ − vp a†−p↓
bp↓ = up ap↓ + vp a†−p↑
(4)
The coefficients up and vp (“coherence factors”) can depend only on the magnitude
of p in an isotropic system. The admixture of the opposite momentum, opposite
spin state (measured by vp ) is non-zero due to the pairing interaction. You will have
to show in Homework # 4 that b and b† satisfy canonical commutation relations for
fermions,
bpσ b†pσ + b†pσ bpσ = 1,
(5)
provided that |up |2 + |vp |2 = 1. One can always choose these coefficients to be real,
so that
u2p + vp2 = 1.
(6)
Substituting Eq. (4) into Eq. (3) one obtains with the help of Eqs. (5) and (6)
i
X h
†
† †
†
2
2
2
0
ξp 2vp + up − vp bp↑ bp↑ + bp↓ bp↓ + 2up vp bp↑ b−p↓ + b−p↓ bp↑
H =
p
λX †
−
Bp0 Bp
V
0
(7)
pp
where
Bp =
u2p b−p↓ bp↑
−
vp2 b†p↑ b†−p↓
+ vp up
b−p↓ b†−p↓
−
b†p↑ bp↑
.
(8)
(Again, you will have to fill the gap in the derivation in Homework #4). Now, we
require the total energy of the system to be at its minimum at fixed values of the
occupation numbers npσ = hb†pσ bpσ i. (The last condition means that the entropy of
the system, which is defined by the occupation numbers, is fixed; thus we minimize
the energy at fixed entropy.) In equilibrium only the two combinations of b and b†
has non-zero expectation values: hb†pσ bpσ i = npσ and hbpσ b†pσ i = 1 − npσ . In the
absence of the magnetic field, np↑ = np↓ ≡ np , and also np = n−p in equilibrium.
The expectation value of the first line in Eq. (7) for H 0 is the given by
X
hH 0 i1st line =
2ξp vp2 + u2p − vp2 np
(9)
p
3
To average the second line, we recall that due to the constraint p 6= p0 the factors
Bp† 0 and Bp are to be averaged independently
#2
"
X
X
X
λ
hBp† 0 ihBp i = −
up vp (1 − 2np ) .
(10)
hBp† 0 Bp i =
V
0
0
0
0
p
pp ;p6=p
pp ;p6=p
Notice that only the last term in Eq. (8) contributes to the result. Collecting the
two contributions, we obtain for the total energy
#2
"
X
X
2
λ
2ξp vp + u2p − vp2 np −
E=
up vp (1 − 2np ) .
(11)
V
p
p
So far, the coefficients up and vp are arbitrary, except for the normalization condition
δE
(6). Now we choose up and vp in such a way that E is minimized, i.e., δu
= 0 (it
p
does not matter if we vary up or vp ). Using the constraint (6) and relabeling p → p1 ,
re-write the first term in Eq. (11) as
X
X
E1 =
2ξp1 vp21 + u2p1 − vp21 np1 =
2ξp1 1 − u2p1 + 2u2p1 − 1 np1
(12)
p1
p1
hence
δE1
= −4ξp up (1 − 2np ).
δup
The variation of the second term is
"
#2
X
δE2
λ δ
= −
up1 vp1 (1 − 2np1 )
δup
V δup p
1
X
λ
dvp
vp + up
(1 − 2np )
up0 vp0 (1 − 2np0 ).
= −2
V
dup
0
(13)
(14)
p
From Eq. (6), we find dvp /dup = −up /vp so that
X
δE2
λ vp2 − u2p
= −2
(1 − 2np )
up0 vp0 (1 − 2np0 ).
δup
V vp
0
(15)
p
Combining E1 and E2 together, we obtain the minimization condition


X
δE
2
λ
= − (1 − 2np ) 2ξp up vp − (u2p − vp2 )
up0 vp0 (1 − 2np0 ) = 0.
δup
vp
V
0
p
(16)
4
or
2ξp up vp = ∆(u2p − vp2 ),
(17)
where
∆≡
λX
up0 vp0 (1 − 2np0 )
V 0
(18)
p
Substituting vp =
q
1 − u2p into Eq. (17) and squaring the whole equation, we obtain
a quadratic equation for u2p
u4p
−
u2p
∆2
+
=0
4(∆2 + ξp2 )
(19)
with the solutions

u2p =

1
|ξp |

1 ± q
.
2
2
2
∆ + ξp
(20)
Hence

vp2 = 1 − u2p =

|ξp |
1

1 ∓ q
.
2
∆2 + ξp2
(21)
We can choose any sign here: indeed, what matters for the calculation of ∆ is the
product
v
v
u
u
1
∆
|ξp |
|ξp |
1u
u
= q
(22)
up vp = t1 ± q
t1 ∓ q
2
2
2
2
2
2
2
2
∆ + ξp
∆ + ξp
∆ + ξp
for both cases. Substituting Eq. (22) into Eq. (17) and replacing the sum over p by
an integral, we obtain a self-consistent equation for ∆
Z
λ
d3 p 1 − 2np
q
=1
(23)
2
(2π~)3 ∆2 + ξ 2
p
which is reminiscent of the Cooper equation for the bound state energy. The energy
∆ plays the role of a gap in the energy spectrum. Notice that Fermi function in the
equation above contains a yet unknown energy of excitations in this system
1
np =
,
(24)
Ep
exp T + 1
5
where Ep is yet to be found. However, if we define Ep as the excitation energy
above the ground state, then Ep > 0 and thus np = 0 at T = 0. In this case, the
equation for ∆(T = 0) ≡ ∆0 becomes
Z
d3 p
1
λ
q
=1
(25)
3
2
(2π~)
2
2
∆0 + ξp
Converting the integral over the p to that over ξp , assuming that the density of
states is nearly independent of the energy, and cutting the energy integration at
|ξ| = ~ωD , we obtain
Z
Z
dξ
λgF ~ωD
λgF ~ωD
λgF ~ωD dξ
p
=
ln
= 1 → ∆0 = ~ωD exp(−2/λgF ),
≈
2 0
2 ∆0 ξ
2
∆0
∆20 + ξ 2
(26)
where gF is the density of states at the Fermi energy. Note that this result is twice
smaller than the Cooper’s result for the bound state energy.