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

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88 Chapter 5: <strong>Sp<strong>in</strong></strong> Hall Effect<br />

with c iα c † iα c iα|0〉 = c iα |0〉 and c iα c † jβ c jβ = c † jβ c jβc iα<br />

= 1 i<br />

v = 1 i<br />

∑<br />

ij<br />

αβ<br />

ψ ∗ n(i,α)[(r i −r j )H αβ<br />

ij ]ψ m(j,β), (5.15)<br />

∑<br />

(r i −r j )H αβ<br />

ij , (5.16)<br />

ij<br />

αβ<br />

where we used i,j and k as site <strong>in</strong>dices and the other for sp<strong>in</strong>. Us<strong>in</strong>g the def<strong>in</strong>ition <strong>of</strong> the<br />

sp<strong>in</strong> current, Eq.(5.6), we get accord<strong>in</strong>gly<br />

〈n|J z |m〉 = e ∑<br />

(r i −r j )ψ n (i,α){σ z ,H} αβ<br />

ij<br />

4i<br />

ψ n(j,β). (5.17)<br />

ij<br />

αβ<br />

Go<strong>in</strong>g <strong>in</strong>to momentum space we are left with<br />

v y = ∑ k y<br />

2ts<strong>in</strong>(k y )½+(α 2 σ x + ˜α 1 σ y )cos(k y ), (5.18)<br />

and for the sp<strong>in</strong> current<br />

J z x = ∑ k x<br />

ts<strong>in</strong>(k x )σ z . (5.19)<br />

The last Eq. follows from {σ z ,H D } = {σ z ,H R } = 0. To evaluate the rhs <strong>of</strong> Eq.(5.5) we<br />

need the matrix elements <strong>of</strong> both operators <strong>in</strong> the eigenvalue basis, which yields the pure<br />

imag<strong>in</strong>ary result<br />

〈∓|Jx|±〉〈±|v z y |∓〉 = ±it(α 2 2 − ˜α 2 1) s<strong>in</strong>(k x) 2 cos(k y )<br />

. (5.20)<br />

∆(k)<br />

Assum<strong>in</strong>gzerotemperaturetocalculateσ z xy ,Eq.(5.5)f<strong>in</strong>allysimplifiesto[NSSM05,MMF08]<br />

σ z xy ≡ σ SH = − ie<br />

V<br />

= 2 e V<br />

∑<br />

m,n<br />

∑<br />

f(E m )−f(E n )〈m|Jx z |n〉〈n|v y |m〉<br />

E m −E n (E m −E n )+iη , (5.21)<br />

I(〈m|Jx z |n〉〈n|v y |m〉)<br />

(E n −E m ) 2 +η 2 . (5.22)<br />

E m

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