Itinerant Spin Dynamics in Structures of ... - Jacobs University

Appendix B: Linear Response 133
in combination with 〈[J,P]〉 = Tr([P,ρ 0 ]J). We get
σ µν (ω) = V
∫ β
0
dx
∫ ∞
0
dtTr(ρ 0 J ν (0)J µ (t+ix))e i(ω+iη)t ,
using (1/V)Ṗ = J, with the system volume V, β = 1/k BT and the density matrix
ρ 0 = e−βH 0
Tr(e −βH 0 )
.
(B.5)
(B.6)
Extracting the time dependency of the current operator J µ D (t+ix) = ei(t+ix)H J µ e −i(t+ix)H
and using the eigenvector basis {|i〉} we can write the conductivity tensor in the following
way
σ µν (ω) = V
∫ β
0
dx
∫ ∞
0
dt ∑ 〈
∣
∣
∣∣m
〈m|J ν (0)|n〉 n∣e i(t+ix)En J µ e
〉e −i(t+ix)Em i(ω+iη)t·
m,n
(B.7)
·Tr(ρ 0 a † m a na † p a q).
(B.8)
We used H = ∑ m E ma † ma m . For the next step we need the identity
Tr(ρ 0 a † ma n a † pa q ) = δ mq δ np f(E m )(1−f(E n )),
(B.9)
where f(E) is the Fermi distribution function.
Proof. The factor δ mq δ np is due to momentum conservation. The second part is yield by
commutator relation
Tr(ρ 0 a † ma n a † pa q ) = Tr(ρ 0 a † ma n (δ nm −a m a † n))
(B.10)
= Tr(ρ 0 (a † m a nδ nm −a † m a na m a † n )) (B.11)
= Tr(ρ 0 (a † ma n δ nm −a † ma m a n a † n)) (B.12)
= Tr(ρ 0 (a † m a nδ nm −a † m a m(1−a † n a n)) (B.13)
= Tr(ρ 0 (a † ma n δ nm −n m (1−n n )). (B.14)
Applying Eq.(B.9) to Eq.(B.7) and evaluating the integral over x we get
σ µν (ω) = V
∫ ∞
0
dt ∑ f(E m )(1−f(E n )) 1−e−β(En−Em)
〈m|J ν |n〉〈n|J µ |m〉
(E
m,n
n −E m )
} {{ }
f(Em)−f(En)
En−Em
·e it(ω+iη+En−Em) . (B.15)