Carsten Timm: Theory of superconductivity

10.2 Isotope effect
How can one check that superconductivity is indeed governed by a phonon-mediated interaction? BCS theory
predicts
(
)
1
k B T c , ∆ 0 ∝ ω D exp −
. (10.49)
V 0 D(E F )
It would be ideal to compare k B T c or ∆ 0 for superconductors that only differ in the Debye frequency ω D , not in
V 0 of D(E F ). This is at least approximately possible by using samples containing different isotopes (or different
fractions of isotopes) of the same elements.
The eigenfrequency of a harmonic oscillator scales with the mass like
ω ∼ 1 √ m
. (10.50)
The entire phonon dispersion, and thus in particular the Debye frequency, also scales like
ω qλ ∼ 1 √
M
, ω D ∼ 1 √
M
(10.51)
with the atomic mass M for an elementary superconductor. The same scaling has been found above for the
jellium model, see Eq. (8.73). Consequently, for elementary BCS weak-coupling superconductors,
k B T c , ∆ 0 ∼ M −α with α = 1 2 . (10.52)
This is indeed found for simple superconductors. The exponent is found to be smaller or even negative for materials
that are not in the weak-coupling regime V D(E F ) ≪ 1 or that are not phonon-mediated superconductors. In
particular, if the relevant interaction has nothing to do with phonons, we expect α = 0. This is observed for
optimally doped (highest T c ) cuprate high-temperature superconductors.
10.3 Specific heat
We now discuss further predictions following from BCS theory. We start by revisiting the heat capacity or specific
heat. The BCS Hamiltonian
H BCS = E BCS + ∑ E k γ † kσ γ kσ (10.53)
kσ
with
leads to the internal energy
E k =
U = ⟨H BCS ⟩ = E BCS + 2 ∑ k
√
ξ 2 k + |∆ k| 2 (10.54)
E k n F (E k ). (10.55)
However, this is inconvenient for the calculation of the heat capacity C = dU/dT since the condensate energy
E BCS depends on temperature through ⟨c −k,↓ c k↑ ⟩. We better consider the entropy, which has no contribution
from the condensate. It reads
∑
S = −k B [(1 − n F ) ln(1 − n F ) + n F ln n F ], (10.56)
where n F ≡ n F (E k ). From the entropy, we obtain the heat capacity
kσ
C = T dS dS
= −β
dT dβ
= 2k B β ∑ d
dβ [(1 − n F ) ln(1 − n F ) + n F ln n F ] = 2k B β ∑ dn F
[− ln(1 − n F ) − 1 + ln n F + 1]
} {{ } dβ
k
k
= ln n F
1−n = ln e −βE k = −βE k F
= −2k B β 2 ∑ k
d n F (E k )
E k . (10.57)
dβ
95