The Kubo formula of the electric conductivity

Transcription

1 The Kubo formula of the electric conductivity

2 We consider an electron under the influence of an electric field E(t). This system may be described by a Hamiltonian of the form Ĥ(t) = Ĥ0 e i E i (t)x i, (1) with Ĥ0 as reference Hamiltonian. For simplicity, we assume a spatially homogeneous field. We further stipulate that E i = E k δ ik and set ex k = ˆD. The time evolution of the system is given by the Liouville-von Neumann equation: ˆρ(t) = 1 [Ĥ(t), ˆρ(t)]. (2) t We split the density operator ˆρ into a equilibrium part, ˆρ 0, and a timedependent part ˆρ(t), according to (ˆρ = ˆρ 0 + ρ(t)). ˆ We then approximate by ˆρ t 1 ([Ĥ0, ˆρ(t)] E k (t)[ ˆD, ˆρ 0 ]. (3) where use has been made of the relation [ˆρ 0, Ĥ0] = 0. This equation is solved ˆρ(t) = 1 t [ ˆD(t t), ˆρ 0 ]E k (t )dt. (4) We operate here in the interaction picture, implying that ˆD(t) Û 0 (t) ˆDÛ0(t), where the propagator Û0(t) is defined as exp( Ĥ0t). i The change in an observable Ô as a function of time can then be found from O(t) T r{ ˆρÔ} = 1 t [ ˆD(t t), ˆρ 0 ]ÔE k(t )dt (5) We extend the upper integration limit to positive infinity by introducing the temporally non-local function γ: such that γô ˆD(t) = 1 Θ(t)T r{[ ˆD( t), ˆρ 0 ]Ô}, (6) O(t) = We factorize γ to separate the theta distribution: γ(t t )E k (t )dt (7) and observe [1] γô ˆD(t) = θ(t)αô ˆD(t) (8) αô ˆD(t) = 1 T r{[ˆρ 0, ˆD( t)]ô} = 1 T r{ˆρ 1 0[ ˆD( t), Ô]} = T r{ˆρ 0[ ˆD(0), Ô(t)]} (9) 1

3 In the following step, we identify the operator Ô with the current density operator ĵ l. Setting j l (t = 0) = 0, we obtain from Eqs. (7) - (9): and j l (t) = θ(t t )αô ˆD(t t )E k (t )dt (10) αô ˆD(t t ) = 1 T r{ˆρ 0[ ˆD(0), ĵ l (t t )]}. (11) We find an expression for the conductivity from the Fourier transform of 10, where the Fourier transform of f(t) is defined as f(ω) = f(t) exp( iωt)dt, and the back transform as f(t) = 1 f(ω) 2π exp(iωt)dω. Thus = j l (ω) = j l (t) exp( i(ω + iη)t)dt Θ(t t )αô ˆD(t t ) exp( i(ω + iη)(t t ))dt E k (t ) exp( i(ω + iη)t )dt = σ(ω)e k (t ) exp( i(ω + iη)t )dt. (12) We include here an infinitesimal increment iη to ensure that the interaction vanishes exponentially as t, t. Summarizing: with j l (ω) = σ lk (ω)ẽk(ω), (13) and Ẽ k (ω) = E k0 (t ) exp( i(ω + iη)t )dt, (14) σ lk (ω) = 1 0 T r{ˆρ 0 [ ˆD(0), ĵ l (t)]} exp( i(ω + iη)t)dt. (15) We bring the right-hand side of this equation into a more symmetric form, using the relation where αâ ˆB(ω) = i ω α Ĉ ˆB (ω), (16) Ĉ(t) = dâ (t). (17) dt 2

4 To justify this result, we operate with the complete bases m m and n n, and set Thus, we have αâ ˆB(t) = 1 ˆρ 0 = m p m m. (18) p m (A mn B nm e iωmnt B nm A mn e iωmnt ), (19) where ω mn (Em En). For the Fourier transform of αâ ˆB(t) we find αâ ˆB(ω) = 2π p m (A mn B nm δ(ω + ω mn ) B mn A nm δ(ω ω mn )). (20) If Ĉ = dâ dt = 1 [Â, Ĥ0], then αĉ ˆB(ω) = 2π i p m ( ω nm A mn B nm δ(ω + ω mn ) ω mn B mn A nm δ(ω ω mn )), (21) from which we obtain Eq.(16). We identify  with the dipole moment operator and derive the following relation between ˆD and ĵ k : ĵ k (t) = 1 d ˆD a d (t). (22) dt where a d is the d-dimensional real space volume per particle. As we use Eq. (17) to substitute for ˆD(0) in Eq. (15), we obtain: σ lk (ω) = ad ω 0 T r{ˆρ 0 [ĵ k (0), ĵ l (t)]} exp( i(ω + iη) t)dt. (23) This expression for σ lk (ω) is referred to as Kubo s formula of conductivity. Using the identity with ĵ l = e a d ˆv l, (24) ˆv l = i [Ĥ, x l] = i [Ĥ0, x l ] (25) we express the current density operators ĵ through velocity operators ˆv. Further, we consider the transverse conductance in the two-dimensional case,d = 2, as relevant for the quantum Hall effect in a graphene sheet. With k = x and l = y, this yields: 3

5 σ xy (ω) = e2 ωa 2 0 T r{ˆρ 0 [ˆv x (0), ˆv y (t)]} exp( i(ω + iη)t)dt. (26) In order to carry out the trace, we adopt the basis { m }, where We note that Ĥ 0 m = E m m. (27) ˆρ 0 m = F (E m ) m, (28) where F denotes the Fermi-Dirac distribution, and n ˆv y (t) m = n Û 0 (t)ˆv yû0(t) m = exp( i (E m E n )t) n ˆv y m. (29) With the help of the latter statements, and after inserting a complete basis n n = ˆ1 between the operators ˆv x and ˆv y, the integration in Eq. (26) can be performed directly. This results in [2] σ xy (ω) = e2 m ˆv x n n ˆv y m iωa d (F (E m ) F (E n )). (30) ω + η + E m E n Expanding 1 ω+η+e m E n σ xy (ω) = e2 a d up to first order with respect to ω, we obtain m ˆv x n n ˆv y m (F (E m ) F (E n )). (31) ω η + E m E n Note that the zeroth-order term of this expansion does not contribute, as the sum over m, n vanishes for this contribution. Considering the DC conductivity (ω = 0) at zero temperature, and setting η = 0, we find σ xy (ω) = e2 a d (Θ(E F E m ) Θ(E F E n )) m ˆv x n n ˆv y m (E m E n ) 2. (32) The factor following the summation sign is non-zero only if E m < E F < E n or E n < E F < E m. Summing over these two configurations results in σ xy = e2 a d E m<e F,E n>e F m ˆv x n n ˆv y m m ˆv y n n ˆv x m (E m E n ) 2. (33) In the following, we focus on electronic states in lattices, as described by Bloch functions (see (3.116)), assuming m, n to be located in the first Brillouin zone. From Eq. (25), we conclude 4

6 n ˆv i m = n x iĥ0 m n Ĥ0x i m where i = x, y. Likewise: = E m E n n x i m = E m E n n m, k i (34) m ˆv i n = E n E m m n. (35) k i The latter relation involves a shift of the derivative k i from the ket to the bra position of the scalar product. This is legitimate, as the position operator in k-space (=i k i ) is hermitian in the region considered here, namely a Brillouin zone, and thus a space with periodic boundary conditions. By Eqs. (34) and (35), it holds that σ xy = ie2 a d m n n m m n n m. (36) k x k y k y k x E m<e F,E n>e F The sum over the unoccupied states can be expressed through the sum over the occupied states by employing the completeness relation: n n = 1 n n. (37) E n>e F E n<e F Transforming Eq. (36) by use of the identity (37) we find [3] σ xy = ie2 a d m m m m. (38) k x k y k y k x E m<e F k The contribution to σ xy due to the second term on the right-hand side of Eq. (37) yields an expression that is odd with respect to the exchange of the indices m and n and thus vanishes upon summing over these indices. In what follows, we will consider the two-dimensional case which is of relevance for the planar graphene sheet. In order to take formula (38) into its ultimate form we make two further steps. First, we average the conductivity as given by Eq. (38) over the first Brillouin zone (BZ), involving integration with respect to k x, k y over the Brillouin zone while simultaneously dividing through its area, (2π)2 1 : a d σ BZ xy = ie2 h BZ dk x dk y 2π E m<e F Secondly, we introduce the vector field A k by defining m m m m. (39) k x k y k y k x 1 Each k-space area cell contributes dkxdky (2π) 2 text. to the electron density, see p.91 of the main 5

7 A k,x = i E m<e F m m, A k,y = i k x E m<e F m m. (40) k y Substituting Eq. (40) into Eq. (39), we find σ xy BZ = e2 dk x dk y h 2π [ A k,y A k,x ]. (41) k x k y BZ or, dropping the superscript BZ: σ xy = e2 dk x dk y h 2π ( k A k ) z. (42) BZ By use of Stoke s theorem, the right-hand side may be expressed as a line integral over the contour of the first Brillouin zone: σ xy = e2 dk A k. (43) 2πh 6

8 Bibliography [1] T.O.Stadelmann, Antidot Superlattices in InAsGaSb Double Heterostructures: Transport Studies, Ph.D. Thesis, Oxford [2] G.Tkachov, Topological Insulators, Pan Stanford Publishing, 2016 [3] D.J.Thouless, M.Kohmoto, M.P.Nightingale, M.de Nijs, Phys.Rev.Lett.49, 405 (1982) 7