Optical activity of noncentrosymmetric metals

0.18 MB·English

by  V. P. Mineev

Optical activity of noncentrosymmetric metals
V. P. Mineev1 and Yu Yoshioka2
1 Commissariat a` l’Energie Atomique, INAC/SPSMS, 38054 Grenoble, France
2 Osaka University, Graduate School of Engineering Science, Toyonaka, Osaka 560-8531, Japan
(Dated: January 13, 2010)
We describe the phenomenon of optical activity of noncentrosymmetric metals in their normal
and superconducting states. The found conductivity tensor contains the linear in wave vector off
diagonal part responsible for the natural optical activity. Its value is expressed through the ratio
0
of light frequency to the band splitting due to the spin-orbit interaction. The Kerr rotation of
1
polarization of light reflected from themetal surface is calculated.
0
2
PACSnumbers: 78.20.Ek,74.25.Nf,74.20.Fg

n
a
J I. INTRODUCTION Here, the gyrotropy tensor λ (ω) is an odd function of
ijl
3 frequency.
1 The gyrotropictensorhas the mostsimple formin the
The metals without inversion symmetry have re-

] cently become a subject of considerable interest mostly metals with cubic symmetry. In this case, the Drude
n due to the discovery of superconductivity in CePt Si.1 part of the conductivity tensor is isotropic σij(ω,0) =
3
o σ(ω)δ andthe gyrotropicconductivitytensorλ (ω)=
Now the list of noncentrosymmetric superconductors ij ikl
c has grown to include UIr2, CeRhSi 3, CeIrSi 4, Y C 5, λ(ω)eikl is determined by the single complex function
r- Li (Pd ,Pt ) B6, KOs O 7, and3 other co3mpo2und3s. λ(ω) = λ′(ω)+iλ′′(ω) such that a normal state density
p 2 1−x x 3 2 6 of current is
Thespin-orbitcouplingofelectronsinnoncentrosymmet-
u
riccrystalliftsthespindegeneracyoftheelectronenergy
s ǫ ∂E
. bandcausinganoticeablebandsplitting. TheFermisur- j= +σE+λrotE. (3)
t 4π ∂t
a facesplittingcanbeobservedbythedeHaas-vanAlphen
m
effect discussed theoretically in the paper Ref. 8. The In time representation λ is an operator being an odd
- band splitting reveals itself in the large residual value of function of operation of time derivative ∂/∂t.
d
the spin susceptibility of noncentrosymmetric supercon- In the superconducting state besides λrotE the gy-
n
o ductors at zero temperature.9 It also makes possible the rotropic part of current density contains also an addi-
c existence ofnonuniform superconducting states that can tional term proportionalto the magnetic induction B
[ be traced to the Lifshitz invariants in the free energy.10

Anothersignificantmanifestationofthebandsplitting is j =λrotE+νB. (4)
1 g
the natural optical activity.
v
3 The natural optical activity or natural gyrotropy is Here ν = ν(T) is a constant coefficient being equal to
1 wellknownphenomenontypicalforthe bodieshavingno zero in the normal state.14,15
1 centre of symmetry11. The optical properties of a nat- In this paper we present the derivation of current re-
2 urally active body resemble those of the magneto-active sponsetothe electro-magneticfieldwithfinite frequency
1. mediahavingnotimereversalsymmetry. Itexhibitsdou- and wave vector valid for the normal and the supercon-
0 ble circular refraction, the Faraday and the Kerr effects. ducting state of noncentrosymmetric metals with cubic
0 In this case, the tensor of dielectric permeability has lin- symmetry. We find that the gyrotropy conductivity λ is
1 ear terms in the expansion in powers of wave vector directly proportional to the ratio of the light frequency
:
v to the energy of the band splitting. Then making use
i ε (ω,q)=ε (ω,0)+iγ (ω)q , (1) theMaxwellequationsandcorrespondingboundarycon-
X ij ij ijl l
ditions at the surface of noncentrosymmetric metal de-
r whereγ (ω)isanantisymmetricthirdranktensorcalled rived in the Section V we calculate the Kerr rotation of
a ikl
the tensor of gyrotropy. polarization of light reflected from the surface of metal.
The spacial dispersion term in permeability has been
derivedbyArfiandGor’kov12inframeofmodelwhere”a
conductorlackingacenterofinversionissimulatedbyan II. CURRENT RESPONSE TO
ordered arrangement of impurities whose scattering po- ELECTRO-MAGNETIC FIELD
tentialisasymmetric”(seealso13). Weshallbeinterested
in gyrotropy properties of a clean noncentrosymmetric The current response of a clean metal to the electro-
metal. In metals, it is more natural to describe optical magnetic field at finite q and ω can be written following
properties in terms of spacial dispersion of conductivity thetextbookprocedure16. Inapplicationtooursituation
tensor having the following form: one has to remember that due to spin-orbital coupling
determined by the dot product of the Pauli matrix vec-
σ (ω,q)=σ (ω,0)−iλ (ω)q . (2) tor σ and pseudovector γ(k), which is odd in respect to
ij ij ijl l
2
momentum γ(−k) = −γ(k) and specific for each non- where
centrosymmetriccrystalstructure9,17,allthevaluessuch
i¯hω +ξ
as single electron energy G (ω ,k)=− n λ ,
λ n ¯h2ω2 +ξ2+|∆˜ (k)|2
n λ λ
ξαβ(k)=(ε(k)−µ)δαβ +γ(k)σαβ, (5) t (k)∆˜ (k)
F (ω ,k)= λ λ , (11)
velocity vαβ(k) = ∂ξαβ(k)/∂k, the inverse effective λ n ¯h2ωn2 +ξλ2 +|∆˜λ(k)|2
mass (m−1) = ∂2ξ (k)/∂k ∂k , the Green func-
ij αβ αβ i j and
tions Gαβ(τ1,k;τ2,k′) = −hTτakα(τ1)a†k′β(τ2)i and
Fαβ(τ1,k;τ2,k′) = hTτakα(τ1)a−k′β(τ2)i are matrices in t (k)=−λ γx(k)−iγy(k) .
λ
the spin space. Taking this in mind, we obtain γ2(k)+γ2(k)
x y
e2 q
ji(ωn,q)=− c Tr mˆ−ij1nˆe + Thefunctions∆˜λ(k)arethegapsintheλ-bandquasipar-
ticle spectrum in superconducting state. In the simplest
(2dπ3¯hk)3T ∞ {vˆi(k)Gˆ(0)(cid:2)(K+)vˆj(k)Gˆ(0)(K−) gmaopdefulnwcittihonBsCaSrepatihriengsainmteerainctiboonthvg(bka,nkd′s):=∆˜−V(gk,)th=e
Z m=−∞ +
X ∆˜ (k)=∆ and we deal with pure singlet pairing18.
+vˆ(k)Fˆ(0)(K )vˆt(−k)Fˆ+(0)(K )} A (ω ,q). (6) −
i + j − j n Transforming the Green functions using eqn. (9) into
i the band representation19, we obtain
The transposed matrix of velocity is determined as
vˆt(−k) = ∂ξˆt(−k)/∂k. The arguments of the Tr{vˆ(k)Gˆ(0)(K )vˆ (k)Gˆ(0)(K )
i + j −
zero field Green functions are denoted as K =
± +vˆ(k)Fˆ(0)(K )vˆt(−k)Fˆ+(0)(K )}=
(Ω ±ω /2,k±q/2). The Matsubara frequencies take i + j −
m n
the values Ωm =π(2m+1−n)T and ωn =2πnT. v++,i(k)G+v++,j(k)G++v++,i(k)F+v++,j(−k)F+† +
Onecanpassfromthespintothebandrepresentation,
v (k)G v (k)G +v (k)F v (−k)F† +
where the one-particle Hamiltonian −−,i − −−,j − −−,i − −+,j +
v (k)G v (k)G +v (k)F v (−k)F† +
+−,i − −+,j + +−,i − −+,j +
H0 = ξαβ(k)a†kαakβ = ξλ(k)c†kλckλ (7) v (k)G v (k)G +v (k)F v (−k)F†(12)
k k,λ=± −+,i + +−,j − −+,i + +−,j −
X X
as the trace of the matrices in eqn. (6). For the brevity,
has diagonal form. Here, the band energies are
weomitheretheargumentsoftheGreenfunctions. They
ξ (k)=ε(k)−µ+λ|γ(k)|, (8) are the same as in the upper two lines. The matrix ve-
λ
locity in the band representation is
suchthattwoFermisurfacesaredeterminedbyequations
ξλ(k) = 0. The difference of the band energies 2|γ(kF)| vλλ′(±k)=u†λα(±k)vαβ(±k)uβλ′(±k)=
characterizes the intensity of the spin-orbital coupling. ∂ξ (±k) ∂γ (±k)
The Fermi momentum with γ = 0 is determined by the 0∂k δλλ′ + l∂k τl,λλ′(±k), (13)
equation ε(k )=ε .
F F
The diagonalizationis made by the followingtransfor- where τˆ(k) = uˆ†(k)σˆuˆ(k) are hermitian matrices. We
mation have neglected20 the difference between uˆ(k) and uˆ(k±
q/2).
akα = uαλ(k)ckλ, (9) The explicit expressions for the τˆ(k) matrices are
λ=±
X
γˆ −γxγˆz+iγy
with the coefficients τˆ = x γ⊥ ,
x −γxγˆγz⊥−iγy −γˆx !
uu↑↓λλ((kk))==sλ |γ|2+γ|γxλ|+γzi,γy , τˆy = −γˆ γyγγˆˆγγzy⊥⊥+iγx −γy−γˆγγˆz⊥−yiγx !, (14)
2|γ|(|γ|+λγz) τˆz = γ⊥z −γγˆ ,
forming a unitary matrixpuˆ(k). (cid:18) γ z (cid:19)
The zerofield Greenfunctions in the bandrepresenta- where γˆ = γ/|γ|, γ = γ2+γ2. All the diagonal ele-
tion are diagonal and have the following form:9 ⊥ x y
mentsofthesematricesaqreoddfunctionsofthe momen-
G(λ0λ)′(ωn,k)=δλλ′Gλ(ωn,k), tduiamgocnoaml pcoomnepnotns.entCsoorrfevseplooncditiyglmy,atthreicepsro(1d3u)ctasreofevthene
Fλ(0λ)′(ωn,k)=δλλ′Fλ(ωn,k), (10) functionsofmomentum. Hence,thesetermsineqn. (12)
3
can produce only the terms proportional to even powers III. GYROTROPY CONDUCTIVITY IN THE
oftheproductkqbeingnotresponsibleforthegyrotropy NORMAL STATE
properties.
The off diagonalelements of τˆ(k) havethe mixed par- Substituting the Green function in the eqn. (20) and
ity. So, the dispersive terms proportional to the odd performing summation over the Matsubara frequencies,
powers of the product kq can arise only from the part we obtain
of eqn. (12) containing the off diagonalelements of τˆ(k)
matrices. Hence, for the calculation of gyrotropyof con- I = d3k kˆ f(ξ−(k−))−f(ξ+(k+)) −
ductivity, only the last two lines in eqn. (12) consisting l (2π¯h)3 l i¯hω +ξ (k )−ξ (k )
Z (cid:26) n − − + +
of interband terms are important. They are equal to f(ξ (k ))−f(ξ (k ))
+ − − +
− . (21)
∂∂kγl ∂∂γkm{τl,+−τm∗,+−[G−G+−F−F+†]+ i¯hωn+ξ+(k−)−ξ−(k+)(cid:27)
i j
Here f(ξ (k )) is the Fermi distribution function and
τ τ∗ [G G −F F†]}. (15) ± ±
l,−+ m,−+ + − + − k =k±q/2. By changing the sign of momentum k→
±
Using the identity −k in the first term under integral and making use that
ξ (k) is even function of k, we come to
τ τ∗ =τ∗ τ =δ −γˆγˆ +ie γˆ , λ
l,+− m,+− l,−+ m,−+ lm l m lmn n
where γˆl = γl/|γ|, one can rewrite the above expression I =2 d3k kˆ [ξ+(k+)−ξ−(k−)][f(ξ+(k+))−f(ξ−(k−))].
as l (2π¯h)3 l (ξ (k )−ξ (k ))2−(i¯hω )2
Z + + − − n
∂γ ∂γ (22)
l m{(δ −γˆγˆ )[G G +G G −F F† (16)
∂k ∂k lm l m − + + − − + Analytical continuation of this expression from the dis-
i j
crete set of Matsubara frequencies into entire half-plane
−F F†]+ie γˆ [G G −G G −F F† +F F†]}.
+ − lmn n − + + − − + + − ω > 0 is performed by the usual substitution iωn →
Starting this point we need the explicit form of spin- ω+i/τ.
orbit coupling vector γ(k). Its momentum dependence We shall work at frequencies smaller when the band
is determined by the crystal symmetry.9,17 For the cu- splitting ¯hω <γ0kF far from the resonance region ¯hω ≈
bic group G = O, which describes the point symmetry γ0kF but still in the collisionless limit ωτ >1 where one
of Li (Pd ,Pt ) B, the simplest form compatible with can decompose the integrand in powers of ω2:
2 1−x x 3
the symmetry requirements is
d3k
γ(k)=γ0k, (17) Il =2 (2π¯h)3kˆl[f(ξ+(k+))−f(ξ−(k−))]
Z
where γ0 is a constant. For the tetragonal group G = 1 (h¯ω)2
C ,whichisrelevantforCePt Si,CeRhSi andCeIrSi , × + , (23)
4v 3 3 3 ξ (k )−ξ (k ) (ξ (k )−ξ (k ))3
the spin-orbit coupling is given by (cid:26) + + − − + + − − (cid:27)
γ(k)=γ (k xˆ−k yˆ)+γ k k k (k2−k2)zˆ. (18) The frequency independent term in eqn. (23) corre-
⊥ y x k x y z x y
spondsto the currentdensity νB introducedineqn. (4).
The gyrotropy current j , which is linear with respect
g We are interested in linear in q part of density of cur-
tothewavevectorq,originatesfromthelastterminthe
rent. Expanding the integrand up to the first order in
eqn. (16). One can show that for the tetragonal crystal ∂ξ±q one can prove by direct calculation that this term
with the symmetry group G = C , for the electric field ∂k
4v vanishes. Thus, in the normal state ν = 0 as it should
lying in the basal plane the linear in the component of
be in gauge invariant theory (see Section V and14). The
wave vector q part of conductivity is absent. In that
frequency dependent term determines the current
follows we continue calculation for the metal with cubic
symmetry where γˆ = kˆ signγ0. We put γˆ = kˆ taking e2γ2
γ0 as a positive constant. Thus, we obtain for gyrotropy jig(ω,q)=ieijl c0¯hqm(h¯ω)2IlmAj(ω,q), (24)
current21
e2γ2 where
j (ω ,q)=ie 0I A (ω ,q), (19)
gi n ijl l j n
c d3k f(ξ )−f(ξ ) ∂ξ ∂ξ
I = kˆ −3 + − + + − +
I = d3k kˆ lm Z (2π¯h)3 l(cid:20) (ξ+−ξ−)4 (cid:18)∂km ∂km(cid:19)
l (2π¯h)3 l 1 ∂f(ξ+) ∂ξ+ ∂f(ξ−) ∂ξ−
Z + + . (25)
∞ (ξ −ξ )3 ∂ξ ∂k ∂ξ ∂k
×T [G (K )G (K )−F (K )F†(K )− + − (cid:18) + m − m(cid:19)(cid:21)
+ + − − + + − −
AftersubstitutionoftheFouriercomponentofthevec-
m=−∞
X
−G (K )G (K )+F (K )F†(K )].(20) torpotentialbytheFouriercomponentofanelectricfield
− + + − − + + − A=cE/iω, we obtain
Letus find firstthe gyrotropyconductivity inthe nor-
mal state. jg(ω,q)=e e2¯h3γ2ωq I E (ω,q). (26)
i ijl 0 m lm j
4
The integral I given by eqn. (25) consists of two Transforming the summation into an equivalent con-
lm
different contributions. The first partof it is determined tour integration16, eqn. (30) can be written as
bythedifferenceoftheFermidistributionfunctionforthe
quasiparticles in two bands, another one originates from ¯h ω′
thederivativesofthesefunctions. Performingintegration J+−(k,ω)= dω′tanh
4πi 2T
overmomentumspaceforthesphericalFermisurfacesin I
× [GR(ω′,k )−GA(ω′,k )]GA(ω′−ω,k )
the limit γ0kF ≪εF, we obtain + + + + − −
1 1 e2ω (cid:8)+[GR−(ω′,k−)−GA−(ω′,k−)]GR+(ω′+ω,k+)
jig(ω,q)=eijl 8 − 24 π2γ k qlEj(ω,q)= −[F+R(ω′,k+)−F+A(ω′,k+)]F−A(ω′−ω,k−)
(cid:18) (cid:19) e02ωF −[F−R(ω′,k−)−F−A(ω′,k−)]F+R(ω′+ω,k+) .
=e qE (ω,q). (27)
ijl12π2γ0kF l j (cid:9) (32)
Here,inthefirstline,wemarkoutthecontributionsfrom
Here, the Green functions are
the two parts of the integral (25). The corresponding
gyrotropy conductivity is
u2(k) v2(k)
GR,A(ω,k)= λ + λ , (33)
e2ω λ ¯hω−ǫ (k)±iδ ¯hω+ǫ (k)±iδ
λ λ
λ=i . (28) and
12π2γ k
0 F
t (k)∆ 1 1
FR,A(ω,k)= λ − ,
IV. GYROTROPY CONDUCTIVITY IN THE λ 2ǫ (k) ¯hω+ǫ (k)±iδ ¯hω−ǫ (k)±iδ
λ (cid:20) λ λ (cid:21)
SUPERCONDUCTING STATE (34)
where
To find the gyrotropy conductivity in the supercon-
ducting phase with the cubic symmetry, one needs to u2(k) 1 ξ (k)
λ = 1± λ , (35)
perform summation and integration in the eqn. (20) us- v2(k) 2 ǫ (k)
ing G and F in the superconducting phase. The integral λ (cid:27) (cid:18) λ (cid:19)
in eqn. (20) consists of two terms
d3k ǫ2(k)=ξ2(k)+∆2, (36)
I = kˆ[J (k,ω)−J (k,ω)], (29) λ λ
l (2π¯h)3 l +− −+
Z
and k =k±q/2. Taking into account
±
where
∞ GR−GA =−2πi[u2δ(ω−ǫ )+v2δ(ω+ǫ )], (37)
J (k,ω)=T [G (K )G (K )−F (K )F†(K )], λ λ λ λ λ λ
+− + + − − + + − −
m=−∞
X
(30)
and FR−FA = πitλ∆[δ(ω−ǫ )−δ(ω+ǫ )], (38)
λ λ ǫ λ λ
∞ λ
J (k,ω)=T [G (K )G (K )−F (K )F†(K )].
−+ − + + − − + + −
after integration with respect to ω′, we can rewrite eqn.
m=−∞
X
(31) (32) as:
1 ǫ (k ) ǫ (k ) u2(k )u2(k ) v2(k )v2(k )
J (k,ω)=− tanh + + −tanh − − + + − − + + + − −
+− 2 2T 2T ǫ (k )−ǫ (k )−ω ǫ (k )−ǫ (k )+ω
(cid:20)(cid:18) (cid:19)(cid:18) + + − − + + − − (cid:19)
ǫ (k ) ǫ (k ) u2(k )v2(k ) v2(k )u2(k )
+ tanh + + +tanh − − + + − − + + + − −
2T 2T ǫ (k )+ǫ (k )−ω ǫ (k )+ǫ (k )+ω
(cid:18) (cid:19)(cid:18) + + − − + + − − (cid:19)(cid:21)
1 ∆2 ǫ (k ) ǫ (k ) 1 1
+ + − −
− − tanh −tanh +
24ǫ (k )ǫ (k ) 2T 2T ǫ (k )−ǫ (k )−ω ǫ (k )−ǫ (k )+ω
+ + − − (cid:20) (cid:18) (cid:19)(cid:18) + + − − + + − − (cid:19)
ǫ (k ) ǫ (k ) 1 1
+ + − −
+ tanh +tanh + . (39)
2T 2T ǫ (k )+ǫ (k )−ω ǫ (k )+ǫ (k )+ω
(cid:18) (cid:19)(cid:18) + + − − + + − − (cid:19)(cid:21)
5
Here we have ignored the shifts in the arguments of the As in the normal state, the gyrotropic current consists
phase factors: t (k±q/2) ≈ t (k) leading to the small of two different contributions. The first part of it is de-
λ λ
corrections of the order of γ k /ε to the main terms. terminedbythe quasiparticlesfilling upthe statesinbe-
0 F F
Forthe secondtermunderintegralintheeqn. (29)we tween two Fermi surfaces. Another part originates from
have the derivatives of electron distribution functions in two
bands. The first contribution is not changed in the su-
J (k,ω)=J (−k,−ω). (40)
−+ +− perconducting state, at ∆ << γ k . But the second
0 F
contribution in the superconducting state is suppressed
Hence, the integral (29) can be rewritten as
incomparisonwithitsnormalvalueduetothegapinthe
d3k quasiparticlespectrum. Neglectingthetermsoftheorder
I = kˆ [J (k,ω)+J (k,−ω)]. (41)
l (2π¯h)3 l +− +− of∆2/γ02kF2 onecansubstituteunderintegralsthederiva-
Z
tivesoftheparticledistributionfunctions∂n(ξ )/∂ξ by
± ±
It means that we should work with the doubled even
thederivativesofthequasiparticlesdistributionfunction
part of eqn. (39). Expanding the integrand in powers of ∂f(ǫ )/∂ǫ , where f(ǫ) = (1−tanh ǫ )/2. This leads
ω after long but straightforwardcalculations we come to ± ± 2T
to the temperature dependence of gyrotropy coefficient
the following formula λ(T) determined by the integral I3
lm
d3k
I =2 kˆ[n(ξ (k ))−n(ξ (k ))] e2ω 1
l (2π¯h)3 l + + − − λ(T)=i (1− Y(T)). (48)
Z 8π2γ0kF 3
1 (h¯ω)2
×(cid:26)ξ+(k+)−ξ−(k−) + (ξ+(k+)−ξ−(k−))3(cid:27) Here Y(T) = 41T coshd2ξǫ/2T is the Yosida function. It
is useful to comparethis formulawith the corresponding
d3k (h¯ω)2 R
−2∆2 kˆ normal state equation (27).
(2π¯h)3 l(ξ (k )+ξ (k ))(ξ (k )−ξ (k ))3
Z + + − − + + − − Besidesofthistermthereisanotherpartofthecurrent
tanhǫ+(k+) tanhǫ−(k−) originating of the integral I1 . It yields the coefficient
× 2T − 2T , (42) lm
ǫ (k ) ǫ (k )
+ + − − ! e2γ k
0 F
ν(T)= (1−Y(T)) (49)
where 6π2¯h2c
1 ξ ǫ such that the total gyrotropy coefficient Λ(T) in the su-
n(ξ)= 1− tanh (43)
perconducting state is
2 ǫ 2T
(cid:18) (cid:19)
is the distribution function of electrons over energies. e2 ¯hω 1 2γ0kF
Λ(T)=i 1− Y(T) − (1−Y(T))
At ∆/γ k ≪ 1 the second integral is obviously much 4π2¯h 2γ k 3 3h¯ω
0 F (cid:20) 0 F (cid:18) (cid:19) (cid:21)
smallerthanthe firstone. So,wecometo the expression (50)
To find a relationship of the gyrotropy conductivity
d3k
I ∼=2 kˆ[n(ξ (k ))−n(ξ (k ))] with observable optical properties one has to develop
l (2π¯h)3 l + + − − electrodynamic theory of noncentrosymmetric metals.
Z
1 (h¯ω)2
× + (44)
ξ (k )−ξ (k ) (ξ (k )−ξ (k ))3
(cid:26) + + − − + + − − (cid:27) V. OPTICAL PROPERTIES OF
which has the same form as the corresponding formula NONCENTROSYMMETRIC METAL
for the normal state (23).
Expanding the integrandup to the first order in ∂ξ±q A. Dispersion law
∂k
we obtain for the current given by eqn. (19):
To derive the light dispersion law we start from the
e2γ2
jg(ω,q)=ie 0¯hq [I1 +(h¯ω)2I3 ]A (ω,q), (45) Maxwell equations
i ijl c m lm lm j
4π
rotB= j , (51)
d3k n(ξ )−n(ξ ) ∂ξ ∂ξ c
Il1m =Z (2π¯h)3kˆl(cid:20)− (ξ++−ξ−)2− (cid:18)∂km+ + ∂k−m(cid:19) rotE=−1c∂∂Bt (52)
1 ∂n(ξ ) ∂ξ ∂n(ξ ) ∂ξ
+ + − −
+ + , (46)
supplied by the density of current expression
ξ −ξ ∂ξ ∂k ∂ξ ∂k
+ − (cid:18) + m − m(cid:19)(cid:21)
ǫ ∂E
j= +σE+λrotE. (53)
I3 = d3k kˆ −3n(ξ+)−n(ξ−) ∂ξ+ + ∂ξ− 4π ∂t
lm (2π¯h)3 l (ξ −ξ )4 ∂k ∂k
Z (cid:20) + − (cid:18) m m(cid:19) Thefirsttermherecorrespondstothedispersionlesspart
1 ∂n(ξ+) ∂ξ+ ∂n(ξ−) ∂ξ− of the displacement current. The second one is the con-
+ + .(47)
(ξ −ξ )3 ∂ξ ∂k ∂ξ ∂k ductivity current written at infrared frequency region
+ − (cid:18) + m − m(cid:19)(cid:21)
6
ω > v /δ, ωτ > 1, where the current is locally related Then repeating all the calculations we come to the same
F
with an electric field, δ is the skin penetration depth. results (57)-(60) modified by the substitution
The last one is the gyrotropy current
icν
λ → Λ=λ− . (62)
jg =λrotE. (54) ω
Asbeforewediscussthemetalwiththecubicpointsym- We remind that the superconducting state current den-
metry. sity given by eqn. (61) is worth to use at the high
frequencies ω > v /δ where the inequality ¯hω >> ∆
Eliminating the magnetic induction, we obtain F
is certainly valid. Here, ∆ is superconducting energy
ǫ ∂2E 4πσ∂E 4πλ∂rotE gap. In the low frequency limit ¯hω < ∆ one should also
∇2E= + + . (55) take into account the London density of current j =
c2 ∂t2 c2 ∂t c2 ∂t L
−(c/4πδ2)(A −¯hc∇ϕ/2e). The interplay between the
L
Taking solution for the circularly polarized wave Londoncurrentj and the Drude currentj =σ(ω)E is
L D
discussed in the textbook.16
E=(xˆ±iyˆ)E ei(kr−ωt) (56)
0
we come to the dispersion relation B. Magnetic moment
ǫω2 4πiσω 4πiλωk
k2 = + ± . (57) The magnetization in gyrotropic media is
c2 c2 c2
1
Itisworthtobenotedthatforamediawithtimereversal M= λE, (63)
2c
breakingonehastosubstitutehereσ →σ =σ ±iσ ,
± xx xy
where σ is the Hall conductivity.
xy such that rotation of the magnetization is equal to one
In neglect the gyrotropy term the complex index of
half of gyrotropy part of the current density
refraction
1
ck j =c rotM. (64)
g
N = =n+iκ 2
ω
The relationship between the density of gyrotropy cur-
is expressed through the diagonal part of complex con-
rent and the magnetization is a generalproperty of non-
ductivity σ =σ′+iσ′′ by means of the usual relations
centrosymmetricmaterials(seealso13). Bothofthemcan
be obtained from the gyrotropy term in action
4πσ′′ 4πσ′
n2−κ2 =ǫ− , 2nκ= .
ω ω 1
− dtd3r(λE)B.
2c
The gyrotropy term leads to the difference in the refrac-
Z
tionindicesofclockwiseandcounterclockwisepolarized
Byvariationofactioninrespectof−Band−A/c,taking
light. In the first order in respect to λ = λ′ +iλ′′ the
into accountthat λ is anodd function of derivative∂/∂t
refraction index is
and making use the definitions E = −(1/c)∂A/∂t and
N± =n+iκ± 2πiλ. (58) B=rotA, we cometo M givenby eqn.(63) andjg given
by eqn.(54) correspondingly.
c
All these considerations are valid both for the normal
Hence, the differences in the real and imaginary parts of and as well for the superconducting state. However, in
the refraction indices of circularly polarized lights with the latter case the gyrotropy action
the opposite polarization are
1 ¯hc
4πλ′′ Sg =− dtd3r (λE)B+ν A− ∇ϕ B .
∆n=n −n =− , (59) 2c 2e
+ − Z (cid:26) (cid:20) (cid:21) (cid:27)
c (65)
contains one extra term which is absent in the normal
state due to the gauge invariance. The corresponding
4πλ′
∆κ=κ −κ = . (60) expressionsforthe magneticmomentandgyrotropycur-
+ −
c
rent are
In the superconducting state the gyrotropy current
1 1 ¯hc
(54) has more general form given by eqn. (4). Hence, M= λE+ ν A− ∇ϕ , (66)
2c 2c 2e
we should use the more general formula for the current (cid:20) (cid:21)
ǫ ∂E
j= +σE+λrotE+νB. (61) j =λ rotE+νB. (67)
g
4π ∂t
7
C. Boundary conditions D. Reflection coefficient and the Kerr effect
To consider the problem of light reflection normally The equations (70) and (76) allow express the ampli-
incident to the flat surface of noncentrosymmetric metal tudes of reflected wave through the amplitude of the in-
weneedtofindtherelationsbetweenthewaveamplitude cident wave. We have for reflection coefficient
propagating inside (z >0) the material
E± 1−N±∓ 2πiΛ
R± = 2 = c , (77)
Ein =E eiω(Nz/c−t) (68) E± 1+N±± 2πiΛ
0 1 c
and the amplitudes of incident and reflected waves out- where the refraction index is
side it
2πiΛ
N =n+iκ± . (78)
Eout =E eiω(z/c−t)+E e−iω(z/c+t). (69) ± c
1 2
Now one can rewrite eqn. (77) in more habitual form
We have
Eizn=o =Eozu=t0 R± = 1−N˜±, (79)
1+N˜±
that is
where an effective refraction index is
E =E +E . (70)
0 1 2 4πiΛ
N˜ =n+iκ± , (80)
±
AtthesametimefromthedifferenceoftheMaxwellequa- c
tions (52) inside and outside of material we obtain
and the effective differences in the real and imaginary
partsoftherefractionindicesofcircularlypolarizedlights
1 ∂
(rotEin−rotEout)z=0 =− (Bin−Bout)z=0. (71) with the opposite polarization are
c∂t
8πΛ′′
The difference of the magnetic inductions at the bound-
∆n˜ =n˜ −n˜ =− , (81)
+ −
ary is given by the jump of magnetzation c
(Bin−Bout) =4πM . (72)
z=0 z=0 8πΛ′
∆κ˜ =κ˜ −κ˜ = . (82)
+ −
In the stationary magnetic field parallelto the surface c
of the metal Hout =H xˆ this equation yields
x Makingusethesedefinitionswecanapplythestandard
procedure22tocalculatetheKerrrotationforlinearlypo-
2πκ
Bxint(z =0)=Hx, Byint(z =0)= c Aiynt(z =0). larized light normally incident from vacuum to the flat
(73) boundary of a medium. The light is reflected as ellip-
In the normal state where κ = 0 the boundary condi- tically polarized with the major axis rotated relative to
tions add nothing special to the centrosymmetric case. the incident polarization by an amount
In the superconducting state the solution of the London
(1−n2+κ2)∆κ˜+2nκ∆n˜
equations supplied by these boundary conditions results θ = . (83)
in quite unusual helical field distribution found in the (1−n2+κ2)2+(2nκ)2
paper.14
Forthelightincidenttothemetallicsurfaceusing(71),
VI. THE KERR ROTATION
(72), and (66) we obtain
4π ∂ TofindtheKerrrotationinthenormalstateletussub-
(rotEin−rotEout) =− λ −cν Ein (74)
z=0 2c2 ∂t z=0 stitute theeqn. (28)ineqns. (81),(82). Wefind∆κ˜ =0
(cid:18) (cid:19)
and∆n˜ expresses throughratioof the light frequency to
Substituting here eqns. (68), (69) we come to the band splitting 2γ k as
0 F
2π 2α ¯hω
zˆ×(NE −E +E )= ΛE (75) ∆n˜ =− . (84)
0 1 2 0
c 3πγ k
0 F
For the combinationsE± =Ex±iEy of the electric field Here, α=e2/¯hc is the fine structure constant.
component this relation can be rewritten as Welimitourselvesbythefrequenciesnotexceedingthe
band splitting γ k . Although the latter is not known
0 F
2πiΛ
N±± E±−E±+E± =0 (76) for many noncentrosymmetric materials, one can expect
c 0 1 2 it is about hundred Kelvin or in the frequency units
(cid:18) (cid:19)
8
∼ 1013rad/sec.23 As an example we consider the sit- and wave vector in noncentrosymmetric metal. The ob-
uation when the frequency of light is of the order of this tained general formula valid both in the normal and in
value and larger than the quasiparticles scattering rate the superconducting state was applied to the calculation
(clean limit): 1 << ωτ < ω τ, where ω = 4πne2/m∗ ofobservablephysicalpropertiesinthefrequencyinterval
p p
istheplasmafrequency. Inthisfrequencyregionthereal smaller than the band splitting ¯hω < γ k . The latter
0 F
p
and imaginary part of conductivity are σ′ ≈ ω2/4πω2τ in its turn was supposed to be smaller than the Fermi
p
and σ′′ ≈ ωp2/4πω. Then, one can find 2nκ ≈ ωp2/ω3τ energy γ0kF < εF. The calculations was performed in
and κ2 −n2 ≈ ω2/ω2. Thus, for the Kerr angle we ob- the clean case ωτ > 1, that is, in particular, important
p
tain to neglect the vortex corrections. We did not discuss
the anomalousskineffect assuming thatthe wavelength
2α ¯hω2 does not exceed the skin penetration depth δ > v /ω.
θ ≈− . (85) F
3πγ k ω2τ In the normal state the current contains the gyrotropic
0 F p
partwhichisoddfunctionofthewavevectorandthefre-
So, the Kerr angle in noncentrosymmetric metals can quency. It presents a sort of displacement current orig-
have measurable magnitude, in particular if we compare inating of band splitting in noncentrosymmetric metal.
it with the Kerr angle of the order of 6×10−8 rad mea- In the superconducting state there is an additional part
sured in the superconducting Sr RuO by the Stanford of the gyrotropy current proportional to magnetic field.
2 4
group.24 The temperature dependence of gyrotropy conductivity
For ∆n˜ in the superconducting state we obtain in the superconducting state was found. As an example
theKerrrotationforthepolarizedlightreflectedfromthe
2α ¯hω 1 2γ0kF surface of noncentrosymmetricmetal with cubic symme-
∆n˜ =− 1− Y(T) − (1−Y(T)) .
π 2γ k 3 3h¯ω try is calculated.
(cid:20) 0 F (cid:18) (cid:19) (cid:21)
(86) Acknowledgments
Finally,fortheKerrangleinthesamefrequencyinterval
as for the normal state we have
2αω ¯hω 1 2γ k
θ =− 1− Y(T) − 0 F (1−Y(T)) . One of the authors (V. P. M.) is indebted to E. Kats
πω2τ 2γ k 3 3h¯ω
p (cid:20) 0 F (cid:18) (cid:19) (cid:21) andL. Falkovskyfor the numeroushelpful discussionsof
(87) naturalopticalactivityandtechnicalproblemsrelatedto
its calculation.
VII. CONCLUSION
Thefinancialsupportofanotherauthor(Y.Y.)bythe
We have presented here the derivation of the current Global COE program (G10) from Japan Society for the
responsetotheelectromagneticfieldwithfinitefrequency Promotion of Science is gratefully acknowledged.
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