Karlsruher Institut für Technologie
Institut für Theorie
der Kondensierten Materie
Theorie der Kondensierten Materie I
WS 2016/2017
Blatt 7
Lösungsvorschlag
Prof. Dr. A. Shnirman
PD Dr. B. Narozhny, M.Sc. T. Ludwig
1. Boltzmann equation in the presence of spin-orbit interaction:
(a) Derive the kinetic equation for a system with SO interaction
Consider first a signle spin. It can be described by a 2×2 density matrix ρ satisfying
the equation of motion
dρ
= i[ρ, H],
dt
where he Hamiltonian H is also a 2 × 2 matrix. The diagonal elements of ρ give the
probabilities to
nd the system in a state with sZ = ±1/2:
hsz i = Trρsz = (ρ11 − ρ22 )/2.
The off-diagonal elements of ρ take into account the possibility to find the spin in
a state which is a coherent superposition of the states with sz = ±1/2, e.g. a state
with a definite projection of sx .
Consider now the kinetic equations describing a particle with spin. Now, the density
matrix involves not only spin variables, but also phase space variables. Treating the
latter semicalssically, we find
dρ
= i[ρ, H] + I[ρ].
dt
As a result, the kinetic equation has the form
∂ρ
∂ρ
∂ρ
+v
+ ṗ
= i[ρ, H] + I[ρ].
∂t
∂r
∂p
Finally, using the explicit form of the Hamoltonian
H=
we find
p2
+ Ω(p)σ,
2m
∂ρ
∂ρ
+ i[Ω(p)σ, ρ] − eE
= I[ρ].
∂t
∂p
(b) Spin and charge distribution functions
Here we use the explicit form for a density matrix where the phase space part is
treated semiclassically
1
ρ = f (t, p) + S(t, p)σ,
2
where f (t, p) is the usual (charge) distribution function, S(t, p) is the spin distribution function, and the first term is proportional to the unit matrix (no texplicitly
written).
Substituting the above density matrix into the kinetic equation we find
∂S
1
∂f
∂S
1 ∂f
+
σ + i[Ω(p)σ, Sσ] − eEα
− eEα
σ = I[ρ].
2 ∂t
∂t
2
∂pα
∂pα
The commutator can be simplified using the properties of the Pauli matrices
[Ω(p)σ, Sσ] = Ωi (p)Sj [σi , σj ] = 2iΩi (p)Sj ijk σk = 2i[Ω(p) × S]σ.
This way we find the matrix equation
1 ∂f
∂S
1
∂f
∂S
+
σ − 2[Ω(p) × S]σ − eEα
− eEα
σ = I[ρ].
2 ∂t
∂t
2
∂pα
∂pα
The matrix equation can be reduced to a set of scalar equations by multiplying the
equation by the Pauli matrices and evaluating the trace. We find
∂f
∂f
− eEα
= TrI[ρ],
∂t
∂pα
∂Si
∂Si
1
− 2[Ω(p) × S]i − eEα
= Trσi I[ρ].
∂t
∂pα
2
Finally, let us assume the simplest τ -approximation for the collision integral
I[ρ] = −
δρ
δf
δSσ
=− −
.
τ
2τ
τ
Substituting this into the above kinetic equations we find
δf
∂f
∂f
− eEα
=− ,
∂t
∂pα
τ
∂Si
∂Si
δSi
− 2[Ω(p) × S]i − eEα
=−
.
∂t
∂pα
τ
(c) Rashba spin-orbit coupling
For the case of the Rashba coupling,
Ω = α(py , −px ),
the vector product in the kinetic equations has the form
Ω × S = α(−px Sz , −py Sz , px Sx + py Sy ) = −αpSz + α(S ⊥ · p)ez ,
where S ⊥ = (Sx , Sy ) and ez is the unit vector in the z-direction.
The equations for the spin distribution function now have the form
∂Sz
∂Sz
δSz
− 2αS ⊥ · p − eEα
=−
,
∂t
∂pα
τ
∂S ⊥
δS ⊥
∂S ⊥
+ 2αpSz − eEα
=−
.
∂t
∂pα
τ
(d) Equilibrium distribution function
The Hamiltonian
H=
p2
+ Ω(p)σ,
2m
has the following eigenvalues
± (p) =
p2
± αp.
2m
Without writing down the eigenvectors, we may relate the Hamiltonian in the diagonal form to the original basis using a unitary transformation
2
p
+
H=U
+ αpσz U,
2m
with
U + αpσz U = Ω(p)σ.
The equilibrium occupation numbers of the eigenstates are given by the Fermi functions nF [σ (p)]. Hence the equilibrium density matrix in the basis of the eigenstates
is given by
o
1n
nF [+ (p)] + nF [− (p)] + σz nF [+ (p)] − nF [− (p)] .
ρ̃0 =
2
The density matrix in the original basis can be found using the same unitary rotation
Ω(p)σ
1
+
nF [+ (p)] + nF [− (p)] + nF [+ (p)] − nF [− (p)]
.
ρ0 = U ρ̃0 U =
2
αp
Reading off the expressions for the charge and spin distributions we find (with
= p2 /(2m))
f0 = nF [+ (p)] + nF [− (p)] −→ 2nF (),
α→0
Ω(p)
S ⊥,0 = nF [+ (p)] − nF [− (p)]
−→ n0F ()Ω(p),
αp α→0
Sz,0 = 0.
(e) Linear response solution
Consider the equations for the spin distribution function in the steady state. Introducing small deviations from equilibrium,
S ⊥ = S ⊥,0 + δS ⊥ ,
Sz = δSz ,
we linearize the equations for the spin distribution function and find
: 0 + 2αδS p = δS /τ,
2α
S
⊥,0 p
⊥,0
z
∂S ⊥,0
= −S ⊥ /τ.
∂pα
Clearly, in the absence of the electric field the solution is zero, δS = 0, proving that
we have the correct equilibrium distribution.
Solving the above equations, we find
2αpδSz − eEα
δS ⊥ = eτ Eα
∂S ⊥,0
4eα2 τ 3
+
(E · S ⊥,0 )p.
∂pα
1 + 4α2 τ 2 p2
(f) Average spin polarization
The average spin is given by
Z
hsi =
d2 p
Trρs =
(2π)2
Z
d2 p
S.
(2π)2
Given that δSz is an odd function of the momentum components pα , we conclude
that
hsz i = 0.
The planar polarization we calculate choosing the x-direction along the in-plane
electric field,
E = (E, 0, 0).
In this case, only the y-component of the magnetization is nonzero. For weak spinorbit coupling (α → 0) we find
Z
p2y
d2 p
3 3
n0 (p2 /2m).
hsy i = 4eEα τ
(2π)2 1 + 4α2 τ 2 p2 F
At T = 0,
n0F (p2 /2m) = −
1
δ(p − pF ),
vF
and the integral can be easily evaluated.
The final result is given by
1
α3 τ 3 p2F
hsy i = − eEm
.
π
1 + 4α2 τ 2 p2F