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Experimental realization of two-dimensional synthetic spin–orbit coupling in ultracold Fermi gases

Supplementary materials for “Experimental realization of a two-dimensional syntheticspin-orbit coupling in degenerate Fermi gases”

Lianghui Huang,1, 2 Zengming Meng,1, 2 Pengjun Wang,1, 2 Peng Peng,1, 2

Shao-Liang Zhang,3 Liangchao Chen,1 Donghao Li,1 Qi Zhou‡,3 and Jing Zhang†1, 41State Key Laboratory of Quantum Optics and Quantum Optics Devices,

Institute of Opto-Electronics, Shanxi University, Taiyuan 030006, P.R.China2Collaborative Innovation Center of Extreme Optics, Shanxi University, Taiyuan 030006, P.R.China

3 Department of Physics, The Chinese University of Hong Kong, Shatin, New Territories, Hong Kong4Synergetic Innovation Center of Quantum Information and Quantum Physics,

University of Science and Technology of China, Hefei, Anhui 230026, P. R. China

2D synthetic SOC and the Dirac point

Despite its simple form, equation (2) in the main text leads to significant results of a highly controllable Dirac pointand a 2D synthetic SOC. To see this fact, we first present two lemmas.

Lemma 1, a real 3 × 3 matrix

−Ω12Ω132Ω23

−Ω122 −Ω13

2

−Ω122 −Ω12Ω23

2Ω13−Ω23

2

−Ω132 −Ω23

2 −Ω13Ω232Ω12

, (1)

has two degenerate eigenstates |A〉and |B〉 with zero eigenenergy EA = EB = 0,

|A〉 =1

N1

(− Ω23|1〉 + Ω13|2〉

)

|B〉 =1

N2

(Ω13|1〉 + Ω23|2〉 − Ω12(Ω2

13 + Ω223)

Ω13Ω23|3〉

)(2)

where N1 =√

Ω213 + Ω2

23 and N2 =√

Ω213 + Ω2

23 + Ω212(Ω

213+Ω2

23)2

Ω213Ω

223

are the normalization factors. The third eigenstate|C〉 has the energy

EC = − 12Ω12Ω13Ω23

(Ω212Ω

213 + Ω2

12Ω223 + Ω2

13Ω223). (3)

For positive and negative Ω12Ω13Ω23, |C〉 is the ground or excited state respectively.

Lemma 2, there always exist a finite p0 to satisfy a set of three equations

(p0 − ki)2

2m+ δi = ε − Ωii′Ωii′′

2Ωi′i′′, i = 1, 2, 3 (4)

where ε is a constant, i = i′ = i′′, provided that Ωii′ = 0 for any i, i′, and none of these three momenta ki = k′i is

parallel to other ones so that the moentas k1 − k2, k2 − k3, and k1 − k3, two of which are independent, could span aplane.

These three equations are equivalent to

−p0

m· (k1 − k2) + (δ1 − δ2) = −Ω12Ω13

2Ω23+

Ω12Ω23

2Ω13(5)

−p0

m· (k2 − k3) + (δ2 − δ3) = −Ω12Ω23

2Ω13+

Ω13Ω23

2Ω12(6)

−p0

m· (k3 − k1) + (δ3 − δ1) = −Ω13Ω23

2Ω12+

Ω12Ω13

2Ω23, (7)

only two of which are independent, since the summation of all three become zero on both sides of the equation. Thusit is sufficient to consider only the first two equations.










−Ω12Ω132Ω23

−Ω122 −Ω13

2

−Ω122 −Ω12Ω23

2Ω13−Ω23

2

−Ω132 −Ω23

2 −Ω13Ω232Ω12

, (1)


|A〉 =1

N1

(− Ω23|1〉 + Ω13|2〉

)

|B〉 =1

N2

(Ω13|1〉 + Ω23|2〉 − Ω12(Ω2

13 + Ω223)

Ω13Ω23|3〉

)(2)

where N1 =√

Ω213 + Ω2

23 and N2 =√

Ω213 + Ω2

23 + Ω212(Ω

213+Ω2

23)2

Ω213Ω

223


EC = − 12Ω12Ω13Ω23

(Ω212Ω

213 + Ω2

12Ω223 + Ω2

13Ω223). (3)



(p0 − ki)2


2Ωi′i′′, i = 1, 2, 3 (4)




−p0

m· (k1 − k2) + (δ1 − δ2) = −Ω12Ω13

2Ω23+

Ω12Ω23

2Ω13(5)

−p0

m· (k2 − k3) + (δ2 − δ3) = −Ω12Ω23

2Ω13+

Ω13Ω23

2Ω12(6)

−p0

m· (k3 − k1) + (δ3 − δ1) = −Ω13Ω23

2Ω12+

Ω12Ω13

2Ω23, (7)











−Ω12Ω132Ω23

−Ω122 −Ω13

2

−Ω122 −Ω12Ω23

2Ω13−Ω23

2

−Ω132 −Ω23

2 −Ω13Ω232Ω12

, (1)


|A〉 =1

N1

(− Ω23|1〉 + Ω13|2〉

)

|B〉 =1

N2

(Ω13|1〉 + Ω23|2〉 − Ω12(Ω2

13 + Ω223)

Ω13Ω23|3〉

)(2)

where N1 =√

Ω213 + Ω2

23 and N2 =√

Ω213 + Ω2

23 + Ω212(Ω

213+Ω2

23)2

Ω213Ω

223


EC = − 12Ω12Ω13Ω23

(Ω212Ω

213 + Ω2

12Ω223 + Ω2

13Ω223). (3)



(p0 − ki)2


2Ωi′i′′, i = 1, 2, 3 (4)




−p0

m· (k1 − k2) + (δ1 − δ2) = −Ω12Ω13

2Ω23+

Ω12Ω23

2Ω13(5)

−p0

m· (k2 − k3) + (δ2 − δ3) = −Ω12Ω23

2Ω13+

Ω13Ω23

2Ω12(6)

−p0

m· (k3 − k1) + (δ3 − δ1) = −Ω13Ω23

2Ω12+

Ω12Ω13

2Ω23, (7)


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2

If the two momenta k1 − k2 and k2 − k3 are not parallel to each other, they span a plane. Define this plane asthe x − y plane, the z-component of p0 drops off from these equations, and we could consider a two-dimensionalproblem. The solutions of equations 5 and 6 form two lines on this plane. The intersection then uniquely determinesthe location of p0 on this plane, and the corresponding energy of this degenerate point is ε. Note that regardless ofthe microscopic parameters, there is always a trivial solution of p0 at infinity, which is irrelevant to our discussionshere.

Project the Hamiltonian to the subspace of |A〉 and |B〉, the effective Hamiltonian is,

He =( 〈A|H|A〉〈A|H|B〉

〈B|H|A〉〈B|H|B〉)

=

( 1N2

1(Ω2

23ε1p + Ω213ε2p) 1

N1N2(ε2p − ε1p)Ω13Ω23

1N1N2

(ε2p − ε1p)Ω13Ω231

N22

(Ω2

13ε1p + Ω223ε2p + Ω2

12(Ω213+Ω2

23)2

Ω213Ω

223

ε3p

))

(8)

where εip = (p − ki)2/(2m)+δi. Expand He near the degenerate point p0, and redefine p = p−p0, and ki = ki−p0,then the effective Hamiltonian can be written as:

He =

( 1mN2

1(Ω2

23k1 + Ω213k2) · p 1

mN1N2Ω13Ω23(k1 − k2) · p

1mN1N2

Ω13Ω23(k1 − k2) · p 1mN2

2

(Ω2

13k1 + Ω223k2 + Ω2

12(Ω213+Ω2

23)2

Ω213Ω

223

k3

)· p

)+ ε · I2×2 (9)

where I2×2 is the identity matrix.Dropping the terms that are independent on spin, He can be in general written as

He = (λx1px + λy1py)σx + (λx2px + λy2py)σz (10)

where the coefficients λx1, λx2, λy1, λy2 depend on the microscopic parameters such as ki and Ωii′ .To simplify the notations, we chose a coordinate so that k1y = k2y. Then the effective Hamiltonian (9) can be

written as

He = apxσx + bpxσz + cpyσz + εI (11)

where a, b, c and ε are all constant:

a =1

mN1N2Ω13Ω23 (k1x − k2x)

b =1

2mN21

(Ω2

23k1x + Ω213k2x

) − 12mN2

2

(Ω2

13k1x + Ω223k2x +

Ω212(Ω

213 + Ω2

23)2

Ω213Ω

223

k3x

)

c =1

2mN21

(Ω2

23k1y + Ω213k2y

) − 12mN2

2

(Ω2

13k1y + Ω223k2y +

Ω212(Ω

213 + Ω2

23)2

Ω213Ω

223

k3y

)=

Ω212(Ω

213 + Ω2

23)2

2mN22 Ω2

13Ω223

(k1y − k3y)

When (k1 − k2) × (k1 − k3) = 0, a = 0 and c = 0 are satisfied.We rotate p about the pz axis by ϕ and the spin about the σy axis by θ. If ϕ and θ satisfy

(a2 + b2 − c2) sin 2ϕ + 2bc cos 2ϕ = 0, (a2 − b2 − c2) sin 4θ + 2ab cos 4θ = 0, (12)

the effective Hamiltonian is simplified as

HSO = λxp′xσ′x + λyp′yσ′

z, (13)

where

λx = a cos ϕ cos 2θ − b cos ϕ sin 2θ + c sin ϕ sin 2θ

λy = a sin ϕ sin 2θ + b sinϕ cos 2θ + c cos ϕ cos 2θ (14)

A special case is a perfect symmetric configuration, δ1 = δ2 = δ3, Ω12 = Ω13 = Ω23 so that ki form an equilateraltriangle. The Hamiltonian is HSO = d

2√

3m(pxσx + pyσz) where d = |ki − kj |. It is worth mentioning that choice of

|A〉 and |B〉 in equation (2) is not unique. Because of the degeneracy, there are infinite choices for constructing twoorthogonal eigenstates. A different choice corresponds to a rotation of the planar spin (σx, σz) and leads to a differentexpression of equation (11). However, after proper rotations of both the spin and momentum, equation (13) is alwaysobtained.

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3

Calculation of Raman coupling strength

The Raman coupling strength can be expressed as [1]

Ωij = − I0

2cε0

∑

F ′′,m′′F

〈F j ,mjF |erq|F ′′, m′′

F 〉〈F ′′,m′′F |erq|F i,mi

F 〉∆

, (15)

where, I0 =√

I1 · I2, and the Ii is the intensity of each Raman laser light. c is the speed of light, ε0 is the permittivityof vacuum, e is the elementary charge, and q is an index labeling the component of r in the spherical basis. ∆ isone-photon detuning of Raman lasers. |F i, mi

F 〉 and |F j ,mjF 〉 are two ground hyperfine spin states coupled by a pair

of Raman laser. |F ′′, m′′F 〉 is the middle excited hyperfine spin state in the Raman process.

Here, we consider 40K atoms, the excited states of the laser-atom coupling involve a fine-structure doublet 42P1/2

(D1 line) and 42P3/2 (D2 line) with fine structure splitting of ∼ 3.4 nm as shown in Fig. 1. The Raman couplingstrength can be written with two contributions from D1 line and D2 line [2],

Ωij = − I0

2cε0[

1∆D1

×∑

F ′′1 ,m′′

F

〈F j ,mjF |erq|F ′′

1 ,m′′F 〉〈F ′′

1 ,m′′F |erq|F i,mi

F 〉

+1

∆D2×

∑

F ′′2 ,m′′

F

〈F j ,mjF |erq|F ′′

2 ,m′′F 〉〈F ′′

2 , m′′F |erq|F i, mi

F 〉]. (16)

Here, ∆D1 and ∆D2 are one-photon detuning of Raman lasers relative to the D1 and D2 line respectively. It isrequired in here that ∆D1 and ∆D2 are very larger than the hyperfine energy splitting of the excited state.

The Raman coupling strength for D1 and D2 line can be simplified as

ΩD1,(F ′′,m′′F )

ij = − I0

2cε0

〈F j ,mjF |erq|F ′′,m′′


F 〉∆D1

= − I0

2cε0

C(F ′′,m′′

F )i · C(F ′′,m′′

F )j

∆D1〈J = 1/2|erq|J ′′ = 1/2〉2, (17)

ΩD2,(F ′′,m′′F )

ij = − I0

2cε0

〈F j ,mjF |erq|F ′′,m′′


F 〉∆D2

= − I0

2cε0

C(F ′′,m′′

F )i · C(F ′′,m′′

F )j

∆D2〈J = 1/2|erq|J ′′ = 3/2〉2. (18)

Here, C(F ′′,m′′

F )i is the hyperfine dipole matrix element between the ground |F i,mi

F 〉 and the excited hyperfine state|F ′′,m′′

F 〉 depending on π or σ± transition, and 〈J = 1/2|erq|J ′′ = 1/2〉 and 〈J = 1/2|erq|J ′′ = 3/2〉 are the D1 andD2 transition dipole matrix element respectively, whose values all can be found in [3]. Therefore, the Raman couplingstrength Ω12, Ω13 and Ω23 as a function of one-photon detuning can be obtained from summing contributions fromall the excited hyperfine states of two D-line components using Eqs. 16-18, as shown in Fig. 2a in the main text.

The minus sign in front of Ωi,i′ in equation (2) of the main text is chosen by the convention of decreasing the energyof the system at red detuning, such as that in optical dipole traps induced by red detuning laser, which makes thelight induced energy shift negative. On the other hand, as explained in the previous discussions, Ωi,i′ itself can beeither positive or negative, depending on the detuning we chose.

Experimental data

Since a translation unitary transformation is applied in Eq. 2 in the main text, the three spin states |9/2, 3/2〉 ,|9/2, 1/2〉 and |7/2, 1/2〉 are displaced from the origin by (0,−kr), (0, kr), (−kr, 0) respectively in the (px, py) plane,as shown in Fig. 1 in the main text. In other words, we experimentally investigate the energy-momentum dispersionsin the translated momentum space. As the rf field does not change the momentum of particles, the weight of any spincomponent could be extracted only within the momentum regimes which are occupied by fermions in the initial state





4

ππππ++++σσσσ −−−−σσσσ

FIG. 1: (Color online). Schematic of energy levels of 40K.

before the injection. When the system is initially prepared in the reservoir state |9/2, 5/2〉, the rf field drives atomsto |9/2, 3/2〉. We therefore displace the dispersion of the free state |9/2, 5/2〉 by (0,−kr) in the (px, py) plane, thesame as that of |9/2, 3/2〉, as shown in figure 2. The energy-momentum dispersions of the dress states, as well as theweight of the spin component |9/2, 3/2〉, can be measured in the regime p2

x + (py + kr)2 < k2F , where kF is the Fermi

momentum corresponding to Fermi energy EF of the initial reservoir state |9/2, 5/2〉. kF is equal to ∼ 1.5kr in ourexperiment. Such a regime, which is denoted as regime 1, is highlighted by the four dashed arrows in figure 2 of thissupplementary material.

Alternatively, if the spin state |9/2,−1/2〉 is chosen as the initial reservoir state, the rf field couples atoms tothe |9/2, 1/2〉 state. Thus, the energy-momentum dispersion of the |9/2,−1/2〉 should be displaced by (0, kr) in thetranslated momentum space, the same as |9/2, 1/2〉. The energy-momentum dispersions and the weight of the spincomponent |9/2, 1/2〉 can be measured in regime 2, which is characterized by p2

x + (py − kr)2 < k2F . Such a regime is

highlighted by the four red dashed arrows in figure 2 of this supplementary material. Similarly, if the initial reservoirstate |7/2,−1/2〉, which is coupled by the rf field to the |7/2, 1/2〉, the energy-momentum dispersions and the weightof the spin component |7/2, 1/2〉 can be measured in regime 3, which is characterized by (px + kr)2 + p2

y < k2F .

In the small overlapping area of regime 1, 2 and 3, the weights of all three spin components in the dressed statescan be obtained, similar to the case of one-dimensional spin-orbit coupling explore in reference [4]. However, in ourexperiments, the location of the Dirac point is in general outside this small overlapping area. It is therefore difficult toextract the weight of all the three component near the Dirac point. Nevertheless, a single initial reservoir state, suchas |9/2, 5/2〉 chosen in our experiment, is sufficient to map out the dispersions of the dressed state and the weight ofone of the spin component, near the Dirac point.

In our experiments, we take time-of-flight (TOF) absorption images after the rf field is applied at different frequencies(one of the TOF absorption images is given in figure 3a of this supplementary material). The energy conservationleads to the relation between the energy of the ith dressed state and that of the free reservoir state,

εi(p) = ε0(p) − νrf + Ez, (19)

where εi(p) is the energy of the ith band of the spin-orbit coupled Hamiltonian, ε0(p) = p2/2m is the dispersionof the initial reservoir state |9/2, 5/2〉, νrf is the frequency of the rf field, and Ez is the Zeeman energy differencebetween |9/2, 5/2〉 and |9/2, 3/2〉. In the ideal situation, the solution of this equation is a curve with an infinitesimalwidth in the momentum space. Such a curve may be denoted as C = p∗. Its length is determined by kF of theinitial reservoir state, as explained before. In reality, the finite width of the rf pulse leads to a finite resolution in theenergy space. Thus, the curve representing the solution of equation (19) acquires a finite width, as shown by the rawdata presented in figure 3a. Such a width also depends on d(ε0(p) − εi(p))/dp at p∗. The larger such a derivativeis, the larger the width is. We use a Gaussian shape pulse of the rf field with 450 µs to optimize the resolution in rfspectroscopy. In the TOF image, the full momentum width at half maximum of the atomic density is about 0.4−0.7kr

at the final state |9/2, 3/2〉 of rf spectroscopy. To extract p∗, we use a Gaussian fit to locate the maximum of themeasured atomic density. For a given rf frequency, the curve C = p∗ is shown by the black curve in figure 3b ofthis supplementary material. We have also included a plot, the thick red curve, to represent a finite width (at 0.85





5

−−−−

FIG. 2: (Color online). Spin injection spectroscopy with the different initial reservoir states. The initial reservoir states|9/2, 5/2〉 and |9/2,−1/2〉 with the measured momentum range are shown. The four blue (red) dashed arrows highlight regime1(2) where the weight of |9/2, 3/2〉 (|9/2,−1/2〉) can be mapped out. The initial reservoir state |7/2,−1/2〉 is not plotted inthis figure.

FIG. 3: (Color online). a, Time-of-flight absorption image of spin injection spectroscopy at a given frequency of rf field. b, AGaussian fit to locate the maximum (black line) of the measured atomic density and the, a finite width ∼ 0.25kr (thick redcurve) of the TOF images. Here, νrf = Ez + 3.46Er. The other experimental parameters are the same as the Fig. 3a in maintext.

maximum) of the TOF images ∼ 0.25kr. Using p∗, we obtain the energy of the ith dressed band

εi(p∗) = ε0(p∗) − νrf + Ez. (20)

Since p∗ is a function of νrf , we vary the rf frequency, obtain p∗ and the resultant εi(p∗) for each given rf frequency,as shown in figure 4 of this supplementary material. This allows one to obtain a three-dimensional plot as shownin figure 3a of the main text. It is worth pointing out that, there are several reasons to make the measurements ofsingle particle band structure not exactly agree with the calculations. The main practical issue in experiments is the





6

FIG. 4: (Color online). Spin and momentum resolved spin injection spectra in the (px, py) plane as the function of thefrequencies νrf = Ez + a of rf field. Here, a is labeled in each plot. At each rf frequency, one or two curves in TOF images areobserved, which represent the solutions to equation (19). The thick red curves represent the finite width from the raw data,and the thin black curves represent p∗ obtained from the Gaussian fit of the density. The experimental parameters are thesame as the Fig. 3a in main text. The plots with the frequencies νrf = Ez + a of rf field from 8.88Er to 7.46Er correspondto the single lowest dispersion (Fig. 3a in the main text). The plots with the frequencies of rf field from −0.6Er to −8.41Er

correspond to the two higher energy dispersions touched at a Dirac point (Fig. 3a in the main text).

stability of the magnetic field. To obtain the band structure, we need to repeat the experimental measurements at thedifferent rf frequencies. Thus the stability of the magnetic field (especially the long term stability) will influence themeasured rf spectrum. Due to the finite momentum width, the fitting process will also contribute to the quantitativedisagreement. These issues led to the uncertainty of the energy and momentum in the dispersion curves, which areshown in Fig. 3 a3-a5 and b3-b5 of the main text.

Figure 3 of the main text has depicted cross section plots along the py direction. Here we also include the crosssection plots along the px. The crossings along both directions unambiguously shows the existence of a Dirac pointin the two dimensional momentum space.

[1] T. A. Savard, S. R. Granade, K. M. O’Hara, M. E. Gehm, and J. E. Thomas, Phys. Rev. A 60, 4788 (1999).[2] L. Huang, P. Wang, Z. Fu, and J. Zhang, Acta Optica Sinica. 34(7), 0727002 (2014).[3] The data of 40K can be from T. Tiecke, Ph.D. thesis, University of Amsterdam, 2010. The similar data of 87Rb is from D.

A. Steck, http://steck.us/alkalidata.[4] L. W. Cheuk, et al. Phys. Rev. Lett. 109, 095302 (2012).





7

FIG. 5: (Color online). The cross-section drawings at the Dirac point (Fig. 3 in main text) in the energy-py and px coordinatesrespectively. The arcs represent the results obtained from equation (19), which include the contribution from all momentumstates obtained in TOF (thick red curves in figure 4 of the supplementary material). The black dots represent the resultsobtained from equation (20), which include only εi(p

∗) obtained from the Gaussian fitting (the thin black curves in figure 4 ofthe supplementary material).





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