6 p-Wave Superconductivity and d-Vector Representation
181
quantization z-axis is taken along k. Such a |S z = 0 state is equivalent to the spinsinglet case as regards susceptibility, leading to a vanishing susceptibility. In fact, it
can be shown that for such an A 1u state, on average, the susceptibility is reduced to
two-third of the normal state susceptibility at T = 0, as if for a given field direction,
one-third of the spins were in the |S z = 0 state [1].
At the opposite, for an ESP state, where Cooper pairs are formed only with spins
of the same direction, we expect no change of the susceptibility for fields along the
quantization axis. However, this does not tell us what to expect in the perpendicular
directions.
Maybe the easiest way to understand if the spin susceptibility is reduced or not for
a given field direction, and whatever the order parameter, is to rewrite the d-vector
with the new quantization axis in this field direction, and check in this representation,
whether or not the z-component of the d-vector (corresponding to the amplitude of
the |S z = 0 state for that direction) is zero. Changing the quantization axis, and
rewriting the d-vector for this new quantization axis, amounts to rotate the reference
frame, or rotate in the opposite direction the d-vector in spin space (see Sect. 6.5.1). It
is easy to see [see (6.11) for the rotation of the d-vector] that the z-component of the
d-vector, when changing the quantization axis for the x- or y-axis, is, respectively,
−d x or d y . What it means is that, in the case of strong spin–orbit coupling, where
the d-vector cannot reorient depending on the field (H) direction:
• The |S z = 0 component of the order parameter, where z is the field direction, is
proportional to the d-vector projection along the field direction [which generalizes
(6.24), which had a physical meaning only for real d-vectors]: if (d·H) is non-zero,
there will be at least a partial suppression of the spin susceptibility and so, Pauli
depairing for the upper critical field, for fields applied in this direction.
• Whatever the d-vector, there is always at least one direction, where there will be
Pauli depairing (otherwise, d should be the null vector).
Coming back to the question of ESP states, if the phase between the
↑ and
↓ is
constant on the Fermi surface and it is a unitary state, on top of the quantization axis,
there is another direction (hence a whole plane) for which there is no change of the
spin susceptibility (and no Pauli depairing) and a perpendicular direction for which
the spin susceptibility is completely suppressed (see Sect. 6.11.5). For example, if d
is of the form (6.26)
d = ψ
⎛
⎝
1
2
(−
↑
+
↓
)
−
i
2
( (
↑
+
↓
)
0
⎞
⎠ ,
with
↑
=
↓ , then there is no Pauli depairing for fields in the x-z-plane and full
Pauli depairing (as in the singlet case) for field along the y-axis. If the ESP state is
non-unitary (|
↑
| = |
↓
| on some part of the Fermi surface), then for sure, there is
at least partial Pauli depairing in the two directions perpendicular to the quantization
axis and still no Pauli depairing for fields along the quantization axis.
181
quantization z-axis is taken along k. Such a |S z = 0 state is equivalent to the spinsinglet case as regards susceptibility, leading to a vanishing susceptibility. In fact, it
can be shown that for such an A 1u state, on average, the susceptibility is reduced to
two-third of the normal state susceptibility at T = 0, as if for a given field direction,
one-third of the spins were in the |S z = 0 state [1].
At the opposite, for an ESP state, where Cooper pairs are formed only with spins
of the same direction, we expect no change of the susceptibility for fields along the
quantization axis. However, this does not tell us what to expect in the perpendicular
directions.
Maybe the easiest way to understand if the spin susceptibility is reduced or not for
a given field direction, and whatever the order parameter, is to rewrite the d-vector
with the new quantization axis in this field direction, and check in this representation,
whether or not the z-component of the d-vector (corresponding to the amplitude of
the |S z = 0 state for that direction) is zero. Changing the quantization axis, and
rewriting the d-vector for this new quantization axis, amounts to rotate the reference
frame, or rotate in the opposite direction the d-vector in spin space (see Sect. 6.5.1). It
is easy to see [see (6.11) for the rotation of the d-vector] that the z-component of the
d-vector, when changing the quantization axis for the x- or y-axis, is, respectively,
−d x or d y . What it means is that, in the case of strong spin–orbit coupling, where
the d-vector cannot reorient depending on the field (H) direction:
• The |S z = 0 component of the order parameter, where z is the field direction, is
proportional to the d-vector projection along the field direction [which generalizes
(6.24), which had a physical meaning only for real d-vectors]: if (d·H) is non-zero,
there will be at least a partial suppression of the spin susceptibility and so, Pauli
depairing for the upper critical field, for fields applied in this direction.
• Whatever the d-vector, there is always at least one direction, where there will be
Pauli depairing (otherwise, d should be the null vector).
Coming back to the question of ESP states, if the phase between the
↑ and
↓ is
constant on the Fermi surface and it is a unitary state, on top of the quantization axis,
there is another direction (hence a whole plane) for which there is no change of the
spin susceptibility (and no Pauli depairing) and a perpendicular direction for which
the spin susceptibility is completely suppressed (see Sect. 6.11.5). For example, if d
is of the form (6.26)
d = ψ
⎛
⎝
1
2
(−
↑
+
↓
)
−
i
2
( (
↑
+
↓
)
0
⎞
⎠ ,
with
↑
=
↓ , then there is no Pauli depairing for fields in the x-z-plane and full
Pauli depairing (as in the singlet case) for field along the y-axis. If the ESP state is
non-unitary (|
↑
| = |
↓
| on some part of the Fermi surface), then for sure, there is
at least partial Pauli depairing in the two directions perpendicular to the quantization
axis and still no Pauli depairing for fields along the quantization axis.