6.5 An Example: The B.C.S. Model of Superconductivity
165
ω
s
T (y) =
1
2π
2π
0
ω
F
T (σ(exp(ιξ 3 ))(y)) d ι, 0 < T < T c .
(6.5.26)
Let us denote by ω
n
T the (extremal) KMS-state at β = T
−1 corresponding to the
values (6.5.24) of F ω = F. The states ω
n
T (T > 0) are interpreted as describing the
‘normal conducting phase’, and the states ω
s
T (T < T c ) represent the ‘superconducting phase’. The mixtures ω T := λω
s
T + (1 − λ)ω
n
T are also (τ
Q
, β = T
−1
)-KMS
states at 0 < T < T c , 0 ≤ λ ≤ 1. The equilibrium states of the considered system
can be defined as the thermodynamic limits of the (unique) Gibbs states of local systems (A
J
; τ
J
), |J | < ∞, where τ
J
t ∈
∗ - Aut A
J is generated by the local Hamiltonians Q
J defined in (6.1.1). According to [168], these thermodynamic limits coincide
with ω
n
T for T ≥ T c , whereas for 0 < T < T c the limit J → leads to the state ω
s
T .
6.5.6 The ground states of (A; τ
Q
):
Let us consider now an extremal τ
Q -ground state ω of our system, ω ∈ EK ∞ . Let
F ω be the corresponding classical stationary point in supp E g . The restriction ω
0 of
ω to the subalgebra A 0 is the unique ground state of the generator X (β
Q
F ω
), (6.5.15),
corresponding to its eigenvector χ(F ω ) ∈ C
2 with the minimal eigenvalue:
n(F) ·σ χ(F) = χ(F), F ∈ su(2)
∗
.
(6.5.27)
Due to the uniqueness of the ground state ω
0
∈ S(A 0 ) corresponding to a given F ω ∈
su(2)
∗ , any extremal τ
Q -ground state is an π(()-invariant product state. Conversely,
the π(()-invariant product state constructed from a vector χ(F) defined in (6.5.27)
will be a pure ground state of (A; τ
Q
) if the ‘consistency condition’ [(χ 1 , χ 2 ) is here
the scalar product in C
2 ]
(χ(F), X (ξ)χ(F)) = F(ξ), ξ ∈ su(2),
(6.5.28)
will be satisfied. This is a consequence of the considerations in Sect. 6.4. Let us solve
(6.5.28) for F. For ξ := ξ j ( j = 1, 2, 3) one has
(χ(F), X (ξ j )χ(F)) =
1
2
n j (F), j = 1, 2, 3,
(6.5.29)
where n j (F) is defined in (6.5.17). The obtained condition
n j (F) = 2F(ξ j ), j = 1, 2, 3,
(6.5.30)
leads to the following possibilities for F = F ω , ω ∈ EK ∞ :
165
ω
s
T (y) =
1
2π
2π
0
ω
F
T (σ(exp(ιξ 3 ))(y)) d ι, 0 < T < T c .
(6.5.26)
Let us denote by ω
n
T the (extremal) KMS-state at β = T
−1 corresponding to the
values (6.5.24) of F ω = F. The states ω
n
T (T > 0) are interpreted as describing the
‘normal conducting phase’, and the states ω
s
T (T < T c ) represent the ‘superconducting phase’. The mixtures ω T := λω
s
T + (1 − λ)ω
n
T are also (τ
Q
, β = T
−1
)-KMS
states at 0 < T < T c , 0 ≤ λ ≤ 1. The equilibrium states of the considered system
can be defined as the thermodynamic limits of the (unique) Gibbs states of local systems (A
J
; τ
J
), |J | < ∞, where τ
J
t ∈
∗ - Aut A
J is generated by the local Hamiltonians Q
J defined in (6.1.1). According to [168], these thermodynamic limits coincide
with ω
n
T for T ≥ T c , whereas for 0 < T < T c the limit J → leads to the state ω
s
T .
6.5.6 The ground states of (A; τ
Q
):
Let us consider now an extremal τ
Q -ground state ω of our system, ω ∈ EK ∞ . Let
F ω be the corresponding classical stationary point in supp E g . The restriction ω
0 of
ω to the subalgebra A 0 is the unique ground state of the generator X (β
Q
F ω
), (6.5.15),
corresponding to its eigenvector χ(F ω ) ∈ C
2 with the minimal eigenvalue:
n(F) ·σ χ(F) = χ(F), F ∈ su(2)
∗
.
(6.5.27)
Due to the uniqueness of the ground state ω
0
∈ S(A 0 ) corresponding to a given F ω ∈
su(2)
∗ , any extremal τ
Q -ground state is an π(()-invariant product state. Conversely,
the π(()-invariant product state constructed from a vector χ(F) defined in (6.5.27)
will be a pure ground state of (A; τ
Q
) if the ‘consistency condition’ [(χ 1 , χ 2 ) is here
the scalar product in C
2 ]
(χ(F), X (ξ)χ(F)) = F(ξ), ξ ∈ su(2),
(6.5.28)
will be satisfied. This is a consequence of the considerations in Sect. 6.4. Let us solve
(6.5.28) for F. For ξ := ξ j ( j = 1, 2, 3) one has
(χ(F), X (ξ j )χ(F)) =
1
2
n j (F), j = 1, 2, 3,
(6.5.29)
where n j (F) is defined in (6.5.17). The obtained condition
n j (F) = 2F(ξ j ), j = 1, 2, 3,
(6.5.30)
leads to the following possibilities for F = F ω , ω ∈ EK ∞ :