Precessing magnetic impurity on sc

1. Dynamical Shiba states Yi-Hua Lai 04/30/2017 1 / 4
Abastract
The goal of this note is to discuss the dynamical Shiba states induced from the single precessing
magnetic impurity in an s-wave superconductor. The reference is from arXiv: 1611.09722, 29
Nov., 2016.
Precessing Magnetic Impurity
Let's consider the spin of the magnetic impurity S(t) = S0 (sin θ sin(2ωt), sin θ cos(2ωt), cos θ),
precessing with the frequency 2ω, the magnitude S0, and the angle between z axis and itself θ.
Note that we don't use the conventional spherical coordinate such that
S(t) = S0 (sin θ cos(2ωt), sin θ sin(2ωt), cos θ) because S(t) = S0 (sin θ sin(2ωt), sin θ cos(2ωt), cos θ)
will be more convenient to be unitarily transformed to be time-independent later. Also. it does
not aect the actual physics.
The coupling term between the magnetic impurity and the spin is
S(t) · σ = s0 sin θ sin(2ωt)
0 1
1 0
+ sin θ cos(2ωt)
0 −i
i 0
+ cos
1 0
0 −1
= S0
cos θ −i sin θei(2ωt)
i sin θe−i(2ωt)
− cos θ
According to the note Shiba state from BdG, the eective Hamiltonian with precessing
magnetic impurity in the four-component Nambu basis Ψ = ψ↑, ψ↓, ψ†
↓, ψ†
↑
T
is
H(t) = ξp (1spin ⊗ τz) + ∆ (1spin ⊗ τx) − JS(t) · (σ ⊗ 1e−h) δ(r)
=




ξp − JS0 cos θδ(r) iJS0 sin θei(2ωt)
δ(r) ∆ 0
−iJS0 sin θe−i(2ωt)
δ(r) ξp + JS0 cos θδ(r) 0 ∆
∆ 0 −ξp − JS0 cos θδ(r) iJS0 sin θei(2ωt)
δ(r)
0 ∆ −iJS0 sin θe−i(2ωt)
δ(r) −ξp + JS0 cos θδ(r)




where ξp = p2
/2m−µ, ∆ is the superconducting gap, and J denotes the strength of the exchange
coupling between the magetic impurity with spin S and the electrons in the superconductor.
σi are the Pauli matrices operating in spin space, while τi are the Pauli matricess operating in
electron-hole space.
Rotating wave transformation
We will transform to the rotating frame of the precessing magnetic impurity, in which case the
eectibe Hamiltonian becomes static. Thus, we can perform the calculations similarly to the
static Shiba impurity.
The original time-dependent Schrodinger equation:
H(t)|ψ(t) = i
∂
∂t
|ψ(t)

2. Dynamical Shiba states Yi-Hua Lai 04/30/2017 2 / 4
Multiply the unitary operator U†
(t) on the left of both sides, and insert 1 = U(t)U†
(t) to the
equation,
U†
(t)H(t) U(t)U†
(t) |ψ(t) = i U†
(t)
∂
∂t
U(t)U†
(t) |ψ(t)
= i U†
(t)
∂
∂t
U(t) U†
(t)|ψ(t) + U(t)
∂
∂t
U†
(t)|ψ(t)
⇒ U†
(t)H(t)U(t) − i U†
(t)
∂
∂t
U(t) U†
(t)|ψ(t) = i
∂
∂t
U†
(t)|ψ(t)
Dene the new eective Hamiltonian ˜H(t) = U†
(t)H(t)U(t) − i U†
(t) ∂
∂t
U(t) and new wave
function | ˜ψ(t) = U†
(t)|ψ(t) , then we obtain the new transformed Schrodinger equation:
˜H(t)| ˜ψ(t) = i
∂
∂t
| ˜ψ(t)
If we choose
U(t) = eiσzωt
=




eiωt
0 0 0
0 e−iωt
0 0
0 0 eiωt
0
0 0 0 e−iωt




, then the new eective Hamiltonian becomes time-independent:
˜H(t) = U†
(t)H(t)U(t) + ω (σz ⊗ 1e−h)
=




ξp − JS0 cos θδ(r) iJS0 sin θδ(r) ∆ 0
−iJS0 sin θδ(r) ξp + JS0 cos θδ(r) 0 ∆
∆ 0 −ξp − JS0 cos θδ(r) iJS0 sin θδ(r)
0 ∆ −iJS0 sin θδ(r) −ξp + JS0 cos θδ(r)




+




ω 0 0 0
0 − ω 0 0
0 0 ω 0
0 0 0 − ω




= ξp (1spin ⊗ τz) + ∆ (1spin ⊗ τx) − JS(0) · (σ ⊗ 1e−h) δ(r) + ω (σz ⊗ 1e−h)
≡ ˜H
where the unitary transformation changes S(t) to the initial S(0) = S0 (sin θ sin Φ, sin θ cos Φ, cos θ).
Thus, we get the time-independent Schrodinger equation, which is the BdG equation:
˜Hψ(r) = Eψ(r)
⇒ [E − ξpτz − ∆τx − ωσz] ψ(r) = −JS(0) · σδ(r)ψ(0)
After the Fourier transform ψ(r) = [d3
p/(2π)3
] eip·r
ψp, we get
[E − ξpτz − ∆τx − ωσz] ψp = −JS(0) · σψ(0)
⇒ ψp = −J [E − ξpτz − ∆τx − ωσz]−1
S(0) · σψ(0) (1)
Note that
[E − ξpτz − ∆τx − ωσz]
=




E − ξp − ω 0 ∆ 0
0 E − ξp + ω 0 ∆
∆ 0 E − ξp − ω 0
0 ∆ 0 E − ξp + ω





3. Dynamical Shiba states Yi-Hua Lai 04/30/2017 3 / 4
and
[E − ξpτz − ∆τx − ωσz]−1
=






E+ξp− ω
( ω−E)2−(∆2+ξ2
p)
0 ∆
( ω−E)2−(∆2+ξ2
p)
0
0 E+ξp+ ω
( ω+E)2−(∆2+ξ2
p)
0 ∆
( ω+E)2−(∆2+ξ2
p)
∆
( ω−E)2−(∆2+ξ2
p)
0 E−ξp− ω
( ω−E)2−(∆2+ξ2
p)
0
0 ∆
( ω+E)2−(∆2+ξ2
p)
0 E−ξp+ ω
( ω+E)2−(∆2+ξ2
p)






Since only the matrix [E − ξpτz − ∆τx − ωσz]−1
depends on p, (1) can be inverse Fourier trans-
formed back to the real space in this form:
ψ(0) = −J [E − ξpτz − ∆τx − ωσz]−1 d3
p
(2π)3
(S(0) · σ) ψ(0) (2)
Cancelling ψ(0) on both sides, we get
14×4 + J [E − ξpτz − ∆τx − ωσz]−1 d3
p
(2π)3
(S(0) · σ) = 0 (3)
(a)
[E − ξpτz − ∆τx − ωσz]−1 d3
p
(2π)3
= − πν








(E− ω)
√
∆2−(E− ω)2
0 ∆√
∆2−(E− ω)2
0
0 (E+ ω)
√
∆2−(E+ ω)2
0 ∆√
∆2−(E+ ω)2
∆√
∆2−(E− ω)2
0 (E− ω)
√
∆2−(E− ω)2
0
0 ∆√
∆2−(E+ ω)2
0 (E+ ω)
√
∆2−(E+ ω)2








(b)
S(0) · σ = S0




cos θ −i sin θeiΦ
0 0
i sin θe−iΦ
− cos θ 0 0
0 0 cos θ −i sin θeiΦ
0 0 i sin θe−iΦ
− cos θ




J [E − ξpτz − ∆τx − ωσz]−1 d3
p
(2π)3
(S(0) · σ)
= − α








(E− ω) cos θ
√
∆2−(E− ω)2
−i(E− ω)eiΦ sin θ
√
∆2−(E− ω)2
∆ cos θ√
∆2−(E− ω)2
−i∆eiΦ sin θ√
∆2−(E− ω)2
i(E+ ω)e−iΦ sin θ
√
∆2−(E+ ω)2
− (E+ ω) cos θ
√
∆2−(E+ ω)2
i∆e−iΦ sin θ√
∆2−(E+ ω)2
− ∆ cos θ√
∆2−(E+ ω)2
∆ cos θ√
∆2−(E− ω)2
−i∆eiΦ sin θ√
∆2−(E− ω)2
(E− ω) cos θ
√
∆2−(E− ω)2
−i(E− ω)eiΦ sin θ
√
∆2−(E− ω)2
i∆e−iΦ sin θ√
∆2−(E+ ω)2
− ∆ cos θ√
∆2−(E+ ω)2
i(E+ ω)e−iΦ sin θ
√
∆2−(E+ ω)2
− (E+ ω) cos θ
√
∆2−(E+ ω)2








≡ − αB(E, θ, Φ)
Thus, (3) turns out to be
14×4 − αB(E, θ, Φ) = 0 (4)
To solve the eigen-energies, one can calculate the determinant of (4), i.e.
det (14×4 − αB(E, θ, Φ)) = 0 (5)

4. Dynamical Shiba states Yi-Hua Lai 04/30/2017 4 / 4
4 Special Cases of Shiba energy
Generally, the eigen-energies, which depend on θ and α, are too complicated. Hence, we here
only calculate 4 special cases.
(a) θ = 0,
E = ±
1 − α2
1 + α2
∆ + ω
which is the same result as the one of the static magnetic impurity, except that the energy shift
by ω. Indeed, we expect to reduce our case to the static one because θ = 0 means there is no
precession at all.
(b) θ = π
2
,
E = ±
(1 + α4
) − (1 − α4)2δ2 + 4α4
1 − α4
∆
where δ ≡ ( ω)/∆. This result highlights the non-linear behavior of the Shiba energy as a
function of the driving ω. Furthermore, when α → 1,
E ≈ ±
(1 + α4
) −
√
4α4
1 − α4
∆ = ±
1 − α2
1 + α2
∆
which is independent of the driving frequency.
(c) Full θ dependence, with α = 1,
E = ±
2δ2 − tan2
θ + tan2
θ
√
1 − 4δ2 cos2 θ
2
∆
where δ ≡ ( ω)/∆.
(d) Full θ dependence, with δ 1 and α = 1,
E = ±
2δ2 − tan2
θ + tan2
θ
√
1 − 4δ2 cos2 θ
2
∆
≈ ±
2δ2 − tan2
θ + tan2
θ(1 − 2δ2 cos2 θ)
2
∆
= δ 1 − sin2
θ∆
=
ω
∆
cos θ∆ = ω cos θ
s(α = 1) =
1 − α2
1 + α2
∆|α=1 ≈ 0
⇒ E ∼= s + ω cos θ