Unveiling pseudospin and angular momentum in photonic graphene .... momentum-ma ...... of sublattice A gives rise to the vortex generation in sublattice B.
Supplementary Information: Unveiling pseudospin and angular momentum in photonic graphene
Daohong Song1, Vassilis Paltoglou2, Yi Zhu3, Sheng Liu4, Daniel Gallardo4, Liqin Tang1, Jingjun Xu1, Mark Ablowitz5, Nikolaos K. Efremidis2, and Zhigang Chen1,4
1
The MOE Key Laboratory of Weak-Light Nonlinear Photonics, and TEDA
Applied Physics Institute and School of Physics, Nankai University, Tianjin 300457, China 2 3
Department of Applied Mathematics, University of Crete, 71409 Heraklion, Crete, Greece
Zhou Pei-Yuan Center for Applied Mathematics, Tsinghua University, Beijing, 100084, China
4
Department of Physics and Astronomy, San Francisco State University, San Francisco, CA 94132 5
Department of Applied Mathematics, University of Colorado, 526 UCB, Boulder, CO 80309
1.
Experimental setup for optical induction and probing the photonic graphene
The experimental setup (see Fig. S1) relies on the optical induction method1-4 which leads to a honeycomb pattern of refractive index change in a nonlinear photorefractive crystal. We probe the pseudospin states by launching three or two interfering beams as the probe to the honeycomb lattice (HCL), and monitor the output transverse intensity pattern and phase of the probe exiting the lattice. To do so, we use a beam from an argon-ion laser operating at 488nm wavelength and split it into two beams, one being ordinarily polarized for “writing” the HCL pattern into the crystal, and the other being extraordinarily polarized for probing the pseudospin states. The writing beam passes through a rotating diffuser, turning into partially spatial incoherent, before it is sent through a specially designed amplitude mask3. The mask generates either three interfering beams that together generate a triangular lattice interference pattern, or six interfering beams that generate a honeycomb lattice pattern. Such intensity patterns remain invariant during propagation through the crystal. For the experiment of Fig. 3, the HCL was generated by employing the self-defocusing nonlinearity on the triangular intensity pattern4, as used in our previous work with graphene edge states5. In this case, the refractive index pattern is the “negative” of the intensity pattern of the lattice-inducing beam. For the experiment of Fig. 4, the HCL was generated by employing the self-focusing nonlinearity on the honeycomb intensity pattern, so in this latter case, the refractive index pattern matches the intensity pattern directly2,3. In both cases, a honeycomb pattern of waveguide arrays (photonic graphene) is established, with the same structure as shown in the left insert in Fig. S1. The probe beams, created either by interfering three beams to a triangular pattern as for Fig. 3 or by interfering two beams to a linear fringe-like pattern as for Fig. 4, are affected by the induced HCL and propagate under the influence thereof. In our experiment, the probe beams are appropriately focused and fine-tuned so they are aimed into the desired Dirac points in the first Brillouin zone of the HCL as shown in the right insert in Fig. S1.
2.
Numerical simulation and asymptotic calculations of the Dirac equation
In this Appendix, we elaborate on the theoretical analysis we used to obtain the results presented in Fig. 5, especially about the derivation, numerical and asymptotic approach of the Dirac equation.
Figure S1 1: Experimentaal setup for optical o inductiion of honeyccomb photonicc lattice (top-left insert), annd for observing the pseudospiin-mediated vo ortex generatio on with an apppropriately coonfigured and momentum-maatched m. PBS: Polarizzing beam spliitter; SBN: stro ontium barium niobate. To seelectively excitte the two subllattices probe beam of the indu uced photonic graphene, g the probe p beam is launched l into eeither three K ppoints or only the top two K points of the first Brillouin zonee (as illustrated d in top-right in nsert).
When n the coupled d mode theory y can be appllied, typicallyy when the reefractive-indeex contrast is large for the ph hotonic lattices, the opticaal field Ψ (r, Z ) associatedd with the paaraxial equatiion Eq. (1) ccan be A, B expressed d as a sum of the t localized modes w m , n (r ) at each siite accordinglly:
(
)
Ψ (r, Z ) = e −iβ Z (ψ A ) m,n (Z) w mA ,n (r ) + (ψ B ) m ,n (Z) w mB ,n (r ) . m,n
(S1)
In the abo ove equation (ψ A, B ) m ,n (Z) are site and z -dependent coefficients. Each mode w mA,,Bn obeys
1 EwmA,,nB + ∇ 2 wmA,,nB + Vloc wmA,,nB = 0 , 2
(S2)
where E is i the mode propagation p constant, Vloc corresponds c tto an isolatedd waveguide aat site (m, n) in an elsewise uniform u mediium with indeex n0 . In the above equatiions we introdduced the norrmalized quanntities
X = x / L0 , Y = y / L0 and Z = z / z0 , with z0 = k0 L20 , such that r = ( X , Y ) . The potenttial is normalizeed to V0 = ( k0 L0 ) 2 Δn0 / n0 . Note that the total optiical field couuld also be exxpressed usinng the Bloch mo odes associateed with the periodic p poten ntial. For exam ample if the ffield coincidees with a partticular
(
)
Bloch mo ode, we hav ve Ψ (r, Z ) = e − iβ (k ) Z eik ⋅( r + Z z )u (k ; r, Z ) , where u has the pperiodicity oof the
potential. A similar equation holds for the Bloch modes with the total potential profile replacing Vloc in (S2). We assume that the overlap integral between the elements of the set = {wmA,n , wmB,n } is small due to the fact that the optical potential wells are deep and well separated, and we also assume that the local potentials Vloc at each site are identical. Inserting the Ansatz (S1) into the paraxial Schrödinger equation [Eq. (1) in the main text] and using A B equation (S2), multiplying either with wm′ ,n′ or wm′ ,n′ and integrating over the transverse plane, we
derive the coupled mode equations:
i i
d (ψ A ) m, n dZ d (ψ B ) m, n dZ
= (Δβ )(ψ A ) m, n − t ((ψ B )m −1,n + (ψ B )m ,n −1 + (ψ B ) m ,n )
，
(S3)
= (Δβ )(ψ B ) m, n − t ((ψ A )m +1, n + (ψ A ) m, n +1 + (ψ A ) m ,n )
where Δβ = β − βc . The propagation constant (energy in the context of the Schrödinger equation) β c is the diagonal matrix element of the potential difference V (r ) − Vloc with respect to the set , i.e.
β c = wmA,n | V (r ) − Vloc | wmA,n , and t is the off diagonal matrix element of the potential V (r ) between neighboring basis functions, i.e. t = wmA,n | V (r ) | wmB,n ) . An excitation of the lattice is realized by setting
( = ( (ψ
) )e
(ψ A ) m ,n = (ψ A ) m ,n ei ( β − βc ) Z e (ψ B ) m ,n
i ( β − βc ) Z B ) m ,n e
− iQ⋅R mA ,n
(S4)
− iQ⋅R mB ,n
in the coupled mode equations, where Q is some wavevector of the inverse lattice. These equations then take the form
i i
d (ψ A ) m ,n dZ d (ψ B ) m ,n dZ
+ tf A (ψ B ) = 0 ，
(S5)
+ tf B (ψ A ) = 0
where
f A (ψ B ) = (ψ B ) m −1,n e + iQ⋅d3 + (ψ B ) m ,n −1 e + iQ⋅d2 + (ψ B ) m,n e + iQ⋅d1 f B (ψ A ) = (ψ A ) m +1,n e − iQ⋅d3 + (ψ A ) m ,n +1 e − iQ⋅d2 + (ψ A ) m ,n e − iQ⋅d1
.
(S6)
In the continuous limit and for excitations near the Dirac cones, the coupled mode equations become the linear Dirac equations describing massless particles, i.e.,
i∂ Zψ A + D(∂ − μψ B ) = 0 i∂ Zψ B − D(∂ μψ A ) = 0
where ∂ μ = ∂ X + i μ∂ Y = eiμφ ∂ r +
,
(S7)
iμ ∂φ , μ = (−1)m = ±1 , m = 0, … ,5 is the index of the six r
Dirac points, and D = − 3t / 2 . In the main text and the simulation for Figs. 5a, 5b, we set D = 1 , and for convenience we also dropped the tilde. We follow the same convention in the following sections.
2.1 Solution for a Gaussian input By combining Eqs. (S7), we obtain the following wave equation
∂ 2ψ ∇ψ − 2 =0, ∂Z 2
(S8)
where ψ stands for ψ A and ψ B . Thus both coefficients describe a wave propagating on the corresponding sublattice. A connection between the spinor elements is established only at the input facet since the initial conditions must satisfy the Dirac equations. The solution can be written in integral form using the Fourier transform of ψ as
ψ (r, Z ) =
d 2k (φ+ e−i β ( k ) Z + φ−eiβ ( k ) Z )eik⋅r ， 2 (2π )
(S9)
where β = k and the coefficients φ± are related to the Fourier transform of the initial conditions
1 i φ = {ψ ( Z = 0)} as φ± = φ ± φZ . We split the two fields as follows β 2
ψ A (r, Z ) = (ψ A )+ (r, Z ) + (ψ A ) − (r, Z ) ψ B (r, Z ) = (ψ B )+ (r, Z ) + (ψ B ) − (r, Z )
(S10)
and derive integral expressions assuming, for example, a Gaussian input on sublattice A and no input on sublattice B. The initial conditions for the fields and their spatial Fourier transform at the input facets are
ψ A = e− r and
2
/(4 r02 )
φ A = 4π r0 2e− k r0
2 2
(S11)
ψ B = 0 φB = 0 ，
(S12)
where φ A = {ψ A } , φB = {ψ B } , 2r0 is the width of the Gaussian beam, ψ A and ψ B are functions of the transverse position vector ( r , Z = 0 ), and φ A and φB are functions of the transverse wave vector ( k , Z = 0 ). The initial conditions for the derivatives of the fields (from the Dirac equations (S7)) are
(ψ A ) Z = 0 (φ A ) Z = 0
(S13)
and
(ψ B ) Z = −i∂ ±ψ A (φB ) Z = 4π kr02e− k r0 e± iθ 2 2
，
(S14)
where (φ A ) Z = {(ψ A ) Z } , (φB ) Z = {(ψ B ) Z } , the polar coordinates of the position vector and the wavevector are r = (r , φ ) and k = ( k , θ ) respectively. This dependence of the derivative on the angle variable will introduce vorticity into the launched beam, which initially carried no angular momentum. The amplitudes of the Fourier transform are
(φ A )± = 2π r02e− k r0 = Ce− k r0 , 2 2
2 2
(S15)
and
(φB )(±μ ) = ( μ )2π ir02e − k r0 eiμθ = ( μ )iCe− k r0 eiμθ , 2 2
2 2
(S16)
where μ = ±1 are for the K / K ′ -points, respectively, and C = 2π r02 . Thus the solution is
(ψ A ) ± (r , Z ) = C
dk − k 2 r02 iβ Z ke e J 0 ( kr ) , 2π
ψ B( μ ) ± (r , Z ) = ( μ )Ceiμφ
dk − k 2 r02 iβ Z ke e J1 ( kr ) , 2π
(S17) (S18)
where J 0 (kr ) is the zeroth-order Bessel function of the first kind, and J1 (kr ) the first-order Bessel function of the first kind. The above integrals cannot be calculated analytically, therefore, we will derive asymptotics in the large kr-regime (kr>1).
2.2 Large kr asymptotes for finding ψ A in optical graphene We will approximate the integral
∞
I κ = dk ke − r0 k
2 2
+ i ( −1)κ Dzk
0
2π
0
dθ ikr sin(θ ) e 2π
(S19)
with κ = 0,1 , first by utilizing the stationary phase method for the integral over the angle θ , i.e. for the integral J 0 ( kr ) =
dθ ikr sin(θ ) , e 2π
2π
0
(S20)
and then by applying the steepest descent method on the resulting integrals after the first step.
2.2.1 Stationary Phase Method
The well-known result for the stationary phase approximation of the zeroth-order Bessel function of the first kind is J 0 (kr )
1 2π kr
i kr − π4 − i kr − π4 +e = e
1 2π kr
π i ( −1)σ kr − 4 e . σ =0,1
(S21)
2.2.2 Steepest Descent Method: (a) Determination of saddle points/ Direction of steepest descent curve Next we apply the steepest descent method to the integrals of (S19) which are written as
Iκ = Iκσ σ
/ 2π r with ∞ 2 Iκσ = dk fσ (k )er0 φκσ ( k ) . 0
(S22)
The various functions appearing in the integrals are
φκσ (k ) = − k 2 + i (−1)κ σ
fσ (k ) = k e − i ( −1)
π /4
z r k + i (−1)σ 2 k 2 r0 r0 ，
(S23)
.
( n −1) (k ) = 0 for the derivatives of φ , we calculate the saddle points, and from From the conditions φ
φ ( n ) (k ) ≠ 0 we find the order of these points. We derive that for each integral ( Iκσ ), we have one first-
kκσ , with kκσ = i order (n = 2) saddle point p w
(−1)κ z + (−1)σ r . The diirections of ssteepest descent is 2r0 2
determineed from the co onditions
Im{φκσ (k ) − φκσ (kκσ )} = 0 Re{φκσ (k ) − φκσ (kκσ )} < 0.
((S24)
From thee Taylor expansion of φκσσ , the abovee conditions lead to ϕ = 00, π . Becausse there exists the branch point k = 0 , care c must be taken not to o include thee saddle poinnts along thee branch cutt. The dle points lie on the imaginary axiss. The point s with (κ , σ ) = (0, 0) annd (1,1) lie oon the four sadd positive/n negative imaaginary axiss, respectiveely, while thhe remainingg saddle poinnts (κ , σ ) = (0,1) or (1, 0) moves m from the positivee to the negaative imagin ary axis or vvice versa aas one crossees the surface of o the cone z = r . Since for each integral i therre is only a single secoond-order brranch point, wee introduce a branch cut from k = 0 to infinity aalong the neggative real axxis. Applyinng the condition ns (S23), we get Im{φκσ (k ) − φκσ (kκσ )} = −2k R k I + 2kκσ k R = 0 or
k R = 0 or − 2k I + 2κσ = 0 k I = kκσ
(S25)
k -pplane. k0 standds either for k + or for k− . The wiggled line Figure S2 2: Integration n contour in complex c along the negative real axis represen nts the branch h cut.
2.2.3 Steeepest Descentt Method: (b b) Integral along the conttour. The integrral (S22) alon ng path (1) (seee Fig.S2)) caan be written as R 2 2 dt d π Iκσ(1) = − ′ fσ (k )er0 (φκσ ( kκσ )−t ) − fσ (kκσ )e r0 φκσ ( kκσ ) , 0 2r0 κ φκσ
(S26)
where we assumed that fσ (k ) fσ (kκσ ) and extended the integration from R to ∞ . For each (κ , σ ) we must calculate the quantities fσ (kκσ ) and φκσ (kκσ ) separately. After doing the algebra we split the sum
κ Iκ
into two parts I + and I − , where the sign in the subscripts refers to two opposite group velocities.
The results are
I
(1) +
1 =− 2 2r0
Z +r − e r
( Z + r )2 4 r0 2
(S27)
,
and I
(1) −
1 =− 2 2r0
− |Z −r| θ ( r − Z )e r
( Z − r )2 4 r0 2
(S28)
,
where θ ( r − Z ) is the step function. For the path (2) we integrate k from 0 to k μσ . Since we have k = ±it , depending on the position of the saddle point, the integration is now from 0 to the corresponding limits for the variable t . We again examine the different (κ , σ ) cases separately. For example, if (κ , σ ) = (0, 0) , we have k00 = it+ , where
t+ = ( Z + r ) / (2r0 2 ) and φ00 (t ) = t . Thus the integral becomes t+ 2 2 2 2 2 I00(2) = i dt t e r0 ( t − 2t+ t ) = i t+3/ 2 2 F2 (t+ )e − r0 t+ , 0 3
(S29)
1 5 7 where 2 F2 (t+ ) stands for the hypergeometric function 2 F2 ,1; , ; t+2 r0 2 . Similar expressions 2 4 4
are obtained for the other cases. If we consider again the sum
3/2
I
(2) −
κ Iκ , the result turns to be
− 4 F (( Z − r ) / (2r0 )) Z − r ( ) θ Z r e =− 2 2 − 2 3 2π r 2r0
( Z − r )2 4 r02
.
(S30)
We note that the contributions from paths (3) and (4) to the integral are negligible for all combinations of
(κ , σ ) .
2.3 Large kr asymptotes for finding ψ B in optical graphene
We will approximate the integral ∞
I κ′ = dk ke − r
k + i ( −1)κ DZk
2 2
0
2π
0
dθ i ( kr sin(θ ) −θ ) e 2π
(S31)
again with κ = 0,1 , following the same procedure as in section 2.2, first utilizing the stationary phase method for the first-order Bessel function of the first kind, and then applying the steepest descent method on the resulting integrals. After the first step the integral Iκ′ can be written ′ Iκ′ = Iκσ σ
/ 2π r with ∞ 2 ′ Iκσ = dk gσ (k )e r0 φκσ ( k ) ,
(S32)
0
and the function gσ appearing in the integral can be written as σ
gσ (k ) = (−i) k (−1)σ e− i ( −1)
π /4
.
(S33)
Comparing with the asymptotical approach presented above for finding ψ A , we notice the only difference lies in these factors used for calculating integrals. Thus, we omit the details here but rather point out that we must evaluate similar integrals on the same contour. Again only the contributions from path (1) and (2) are significant. The results are similar to those obtained in the previous case and are summarized in the following section.
2.4 Summary
In summary, when the sublattice A is initially excited, the above analysis leads to
ψ A (r, Z) = F+ (r )e− (Z + r )
2
/(4 r02 )
+ F− (r )e − (Z −r )
2
/ ( 4 r02 )
(S34)
where F+ = −
1 Z +r , 2 r
(S35)
and F− = F> + F< with
F> = −
1 r−Z θ (r − Z ) 2 r
(S36)
3/2
F (( Z − r ) / (2r0 )) Z − r θ (Z − r ) . F< = − 2 2 2 3r0 π r 2r0
From (S34) and (S35), we can clarify some points: (a) The amplitude depends on r −1/ 2 , which means that the intensity depends on r −1 . Thus we have diffractionless propagation accompanied by energy redistribution over the area of the annulus (characteristic of conical diffraction). (b) The waves propagate outward/inward with their respective group velocities. Note also that the r −1/ 2 -dependence of the fields can be explained by the following simple argument. From the
continuity equation, it follows that the quantity (IS) is preserved. Thus the intensity (I) is inversely proportional to the area (S) of the annulus. For large values of r, this area equals 2π rdr and therefore the intensity has a r −1 -dependence, hence the field has a r −1/ 2 -dependence. Similarly, the result for the field ψ B ( μ ) (r , Z) is
ψ B ( μ ) (r, Z) = μ eiμφ G+ (r )e − (Z + r )
2
/(4 r02 )
+ G− (r )e − (Z − r )
2
/(4 r02 )
(S37)
with
G+ = (−i) F+ , G> = iF> , G< = (−i ) F
