Excitonic Terahertz Absorption in Semiconductors with Effective-Mass

Excitonic terahertz absorption in semiconductors with eﬀective-mass anisotropies

P. Springer,∗ S. W. Koch, and M. Kira Department of Physics and Material Sciences Center, Philipps-Universit¨atMarburg, Renthof 5, 35032 Marburg, Germany (Dated: September 24, 2018) A microscopic approach is developed to compute the excitonic properties and the corresponding terahertz response for semiconductors characterized by anisotropic eﬀective masses. The approach is illustrated for the example of germanium where it is shown that the anisotropic electron mass in the L-valley leads to two distinct terahertz absorption resonances separated by 0.8 meV.

I. INTRODUCTION in two clearly separated exciton resonances in the THz absorption spectrum. Terahertz (THz) spectroscopy has been broadly applied, e.g., to investigate transient photoconductivity [1], inter-molecular vibrations [2], high-harmonic genera- II. THEORY WITH MASS ANISOTROPY tion [3], and the transition energies between excitonic eigenstates in quantum many-body systems [4–6]. For To identify the main consequences of mass anisotropy direct-gap semiconductors with isotropic effective-mass in the THz absorption spectra, we use Ge as a prototype configurations, the excitonic THz excitation dynamics system and a 2-band model to describe the energy dis- and the resulting spectra have been studied extensively persion. Germanium is an indirect semiconductor whose both theoretically [7–10] and experimentally [11–14]. conduction (c) and valence (v) bands are centered around In comparison to direct-gap systems, the correspond- the L and Γ points, respectively, separated by the wave ing investigations in indirect semiconductors such as sili- vector k0, as indicated in the inset of Fig. 1. For excita- con (Si) and germanium (Ge) are more elaborate because tions close to the band gap Eg, it is sufficient to describe optical excitations are accompanied by strong dephasing the electronic energies via [21],

due to intervalley scattering [15] and the indirect exci- 2 2 tons typically involve states characterized by strongly v h ~ k Ek = −Ek = − , (1) anisotropic masses [16]. Experimentally, excitonic fea- 2mh 2 2 tures have been observed in Ge [17] and Si [18] and THz c e X ~ [(k − k0) · ej] studies have been reported recently [19, 20]. Ek = Ek = Eg + , (2) 2me,j Whereas most of the isotropic exciton properties can j be determined analytically [7, 8], even the linear eigen- for the holes (h) and the electrons (e), respectively. The value problem must be solved numerically for anisotropic indices j = {x, y, z} denote the Cartesian components conditions. These subtleties complicate the microscopic and k0 is aligned with the ez direction, as shown in Fig. 1. analysis of the linear and nonlinear optical experiments, Although k in general defines a group of energy minima, and in particular also of the THz absorption measure- 0 we first evaluate the theory for one single k0 and gener- ments. alize the results for multiple k in Sec. IV B. In Ge, the To deal with this problem, we develop in this paper a 0 group of k0 points to the eight L centers. These lie in microscopic approach and an ensuing numerical scheme the center of the hexagonal planes of the truncated octa- to efficiently evaluate the excitonic properties in sys- hedron which defines the first Brillouin zone. tems with anisotropic effective masses. To illustrate the In Ge, the valence band is isotropic with mass mh = scheme, we analyze Ge and show that the THz absorp- 0.33 m while the conduction band masses are m = tion exhibits distinct resonances related to the L-valley 0 e,x me,y ≡ m⊥ = 0.0815 m0 and me,z ≡ mk = 1.59 m0 at electron-mass anisotropy. the L valley [22]. Figure 1 illustrates the directions of The paper is organized as follows: In Sec. II, we extend the anisotropic effective mass tensor as an ellipsoid with the generalized Wannier equation to systems with mass the xy-plane mass given by m and with m in the z- anisotropy and discuss the system Hamiltonian and basic ⊥ k arXiv:1602.02972v1 [cond-mat.mtrl-sci] 9 Feb 2016 direction. THz absorption equations. In Sec. III, we present an efficient numerical scheme to obtain the radial solutions of the Wannier equation. We analyze the modifications A. System Hamiltonian of the selection rules and the THz absorption spectra for different polarizations in Sec. IV. For the example The many-body Hamiltonian is Hˆ = Hˆ + Vˆ + Hˆ of Ge, we then show that the mass anisotropy results 0 THz where the non-interacting electrons are described by [23]

ˆ X h c † v † i H0 = Ekaˆc,kaˆc,k + Ekaˆv,kaˆv,k , (3) ∗ [email protected] k

! !

K exciton states φλ and their binding energies Eλ have to be computed from the generalized Wannier equation [23, 27]

R ˜ R e h X R 0 Eλφλ (k) = Ekφλ (k) − 1 − fk − fk V|k−k0|φλ (k ) , k0 (7) Energy

e(h) Momentum where fk is the electron (hole) distribution. Further Coulomb correlation eﬀects, such as excitation induced dephasing [28, 29], could be included via complex scattering matrices [30], but are omitted here for simplicity. Non-vanishing carrier distributions renormalize the electron–hole pair energy

2 2 ˜ X ~ [k · ej] X e h Ek = − V|k−k0| fk0 + fk0 , (8) 2µj j k0

−1 −1 after we have introduced a reduced mass µj = me,j + Figure 1. Schematic illustration of the system. In the xy- m−1. Since two of the three µ are identical in Ge, the plane (dark grey), the effective electron mass is denoted by h j energy dispersion (8) simplifies to m⊥ (white arrow), in z-direction it is mk (red arrow). The THz field (red) is either polarized parallel or perpendicular 2k2 ~2k2 (blue arrows) to the z-axis which is aligned with k0. The ˜ ~ ⊥ k X e h Ek = + − V|k−k0| fk0 + fk0 , (9) inset schematically depicts the band structure of Ge. 2µ⊥ 2µ k k0

−1 −1 −1 † with µ⊥(k) = m⊥(k) + mh and the momentum k = with Fermionic creation (annihilation) operatorsa ˆλ,k (k⊥, kk), both being decomposed into directions perpen- (âλ,k) for conduction (λ = c) and valence band (λ = v), dicular (⊥) and parallel (k) to k0, as shown in Fig. 1. respectively. The Coulomb-interaction is given by [24] ˜ In general, Ek is anisotropic for µ⊥ 6= µk. For non- vanishing carrier distributions, the Wannier equation ˆ 1 X X † † V = Vqaˆλ,kaˆν,k0 aˆν,k0+qaˆλ,k−q , (4) defines a non-Hermitian eigenvalue problem with left- 2 0 L R λ,ν k,k ,q and right-handed solutions φλ and φλ , respectively. As shown in Ref. [31], these solutions are connected via containing the usual Coulomb matrix element Vq. For L R e h φλ(k) = φλ (k)/(1−fk −fk ). Due to the mass anisotropy weak THz fields, the light-matter coupling follows λ and the fk dependence, Eq. (7) cannot be solved ana- from [25] lytically. In Sec. III, we therefore present a method to numerically determine the anisotropic exciton wave func- ˆ X λ † R HTHz = − jk · ATHzaˆλ,kaˆλ,k , (5) tions φλ . λ,k Once the exciton wave functions are known, we can directly evaluate the THz absorption via the susceptibility with the current-matrix elements in the effective-mass ν ν ν ν ? approximation, X Sλ(ω)nλ − [Sλ(−ω)nλ] χ(ω) = 2 , (10) 0ω ( ω + iγJ ) λ,ν ~ |e| k X (k − k0) · ej jh = − ~ , je = −|e| e , (6) k m k ~ m j h j e,j derived in Ref. [23]. The susceptibility defines the linear absorption α(ω) = ω/crIm[χ(ω)] yielding the THz Elliott and a THz field ATHz(t) ≡ A(t)eA. Due to the mass formula, where cr is the speed of light within the medium. e anisotropy in jk, the THz interaction is sensitive to the Equation (10) also contains a decay constant γJ for the polarization eA of the applied field. THz current, as well as a THz response function

β ν ν X (Eβ − Eλ)Jλ Jβ Sλ(ω) = , (11) B. Anisotropic Excitons Eβ − Eλ − ω − iγ β ~

ν To compute the THz probe absorption spectrum, we where γ is the dephasing constant, and nλ assign exciton have to specify the initial many-body state of the semi- correlations. The transition-matrix element conductor. Here, we consider a situation where the L- ν X L ? R point electrons and the Γ -point holes have formed bound Jλ = φλ(k) j(k)φν (k) (12) electron–hole pairs, i.e. excitons [26]. The corresponding k 3

R 1.0 First of all, we expand φλ into spherical harmonics 0, 0 { } 2, 0 ∞ l R X X m { } φλ (k) = Rλ,l,m(k)Yl (θ, ϕ) . (13) )(norm.) 0.5 l=0 m=−l k ( Inserting Eq. (13) into Eq. (7) and projecting spherical

,l,m (a) harmonics yields an eigenvalue problem for the radial

GS 0.0 part alone R 2 2 0.0 0.4 0.8 1.2 ~ k (1) EλRλ,l,m(k) = l,mRλ,l,m(k) Wave vector ka0 2µz ∞ 3.4 Z 0 0 2 0 e h − dk (k ) Vk,k0 fk0 + fk0 Rλ,l,m(k) 3.3 0 ∞ (meV) Z 1 3.2 e h 0 0 2 l 0 − 1 − fk − fk dk (k ) Vk,k0 Rλ,l,m(k ) ! (b) 0

GS 3.1 "

E 2 2 ~ k X (2) + l+ξ,mRλ,l+2ξ,m(k) 2 4 6 8 2µz ξ=±1 Number of included l states (3) + l,m+ξRλ,l,m+2ξ(k) Figure 2. Frame (a) shows the different radial parts R # GS,l,m X (4) of the ground state whose dominant character is {l = 0, m = + l+σ,ξσm+(σ−1)Rλ,l+2σ,m+2ξ(k) , 0} (shaded area). The only other non-vanishing component σ=±1 has {2, 0} symmetry (solid line). The Bohr radius is a0 = (14) 30.98 A.˚ Frame (b) shows the computed transition energy EGS→1 (circles) between the two energetically lowest states e(h) e(h) when fk = fk is radially symmetric. The as a function of maximally included l states. The dashed line (j) acts as a guide to the eye. mass anisotropy introduces coupling strengths l,m and l Coulomb-matrix elements Vk,k0 given in App. B and C, respectively. If the carrier distributions are not radially symmetric, they can be expanded similarly to Eq. (13). e h contains the reduced current j(k) = jk + jk · eA. Equation (14) couples only identical l and m elements Since both the Wannier equation and j(k) are generally and those shifted by ±1. At the same time, the expan- anisotropic in systems with anisotropic effective masses, sion (13) typically provides a fast convergence in terms the THz absorption will depend on the direction of de- of included l states. Therefore, this eigenvalue problem tection, which allows us to directly monitor the mass can be solved with only a few l and m combinations. For anisotropy via α(ω). isotropic electron masses, Eq. (14) reduces to the usual form of the radial Wannier equation. For Ge-type systems, Eq. (14) simplifies significantly (3) (4) because l,m = l,m = 0 for two identical masses, as III. ANISOTROPIC EIGENVALUE PROBLEM shown in App. B. Then, the anisotropic coupling links only states with identical m. To identify excitons based on their main orbital character, we label them accord- In the case of isotropic effective masses, the radial and ing to that {l, m} component which produces the largest angular dependency in the Wannier equation can be sep- P m 2 k |Rλ,l,m(k)Yl (θ, ϕ)| . For example, a 1s exciton is arated such that Eq. (7) can be solved as an effective one dominated by a {0, 0} component. dimensional problem. [30, 32, 33]. However, since the ro- tation symmetry is broken by the mass anisotropy, the radial and angular degrees of freedom are coupled leading IV. EXCITON MASS ANISOTROPY IN to a significantly more complicated configuration. GERMANIUM In the following, we develop an efficient scheme to deal with the general solution of the Wannier equation in To demonstrate the overall implications of the mass e h anisotropic media. For this purpose, we allow for all µj anisotropy, we consider weak excitations (fk = fk = 0) in Eq. (8) to be different. In the case of Ge, we then in Ge. In practice, we solve Eq. (14) numerically for a simply take two of the masses as equal. finite number of l and m states. Figure 2(a) presents 4

state from a spherically symmetric s-wave (isotropic) to Continuum a squeezed, slightly peanut-shaped wave function. The 0.0 squeezing is explained by the heavier mass in the kk direction. The corresponding distortions become signiﬁ- cant only in the presence of strong anisotropy. We also −1.1 observe that anisotropy squeezes 2s (not shown) and 2p −1.4 states to very diﬀerent shapes compared to their isotropic counterparts, which explains why they become energeti- −1.8 cally non-degenerate.

E? A. Single-Valley THz Absorption

Ek l = 0 Even though the rapid inter-valley equilibration [34] l = 1 may make it experimentally difficult to confine the exci- Binding energy (meV) l = 2 tons into a single valley, it is instructive to first study the l = 3 THz response for such a situation and then generalize it for multi-valley excitations, as done in Sec. IV B.√ Here, −5.2 3π we assume that the L valley is located at k0 = a ez 0 ±1 ±2 which contains the lattice constant a and is aligned with Magnetic quantum number m the z-axis as mentioned in Sec. II. To study well defined cases, we assume that the THz Figure 3. Binding energy of exciton states in Ge correspond- field is polarized either along the z-axis, eA = ez ≡ ek, 2 ing to their dominant character. Each column represents or perpendicular to it, eA = xex + yey ≡ e⊥ with x + 2 a magnetic quantum number while the l state are denoted y = 1. The THz response (10) to e⊥ polarized light by different line styles. If excitons are exclusively populated is identical for any (x, y) combination. Therefore, we in the ground state, the energetically lowest possible transi- may chose y = 0 and x = 1 without loss of generality. tion for parallel and perpendicular THz field polarizations are The corresponding ek and e⊥ polarized fields excite the marked with arrows. system via

e~kz e~kx jk(k) = − , j⊥(k) = − , (15) the dominant radial parts RGS,l,m(k) of the ground state µk µ⊥ (λ = 0 ≡ GS), showing that the {l = 0, m = 0} fraction peaks roughly ten times higher than a {2, 0} component. which yields the transition-matrix elements Therefore, the ground state exciton has dominantly a ∞ {0, 0} character. To verify the convergency, Fig. 2(b) Z k e~ 3 X ? shows the transition energy EGS→1 between the two en- Jλ,ν = − dk k Rλ,l,m(k) µz ergetically lowest states as a function of the number of 0 l,m included l states. As we can see, converged results are m m × gl Rν,l+1,m(k) + gl−1Rν,l−1,m(k) , (16) obtained already for seven l states. The exciton energy structure is shown in Fig. 3 for five and different m states. For isotropic systems, the first ex- ∞ Z cited {l = 0, m = 0} and the lowest {l = 1, m = 0} and ⊥ e~ 3 X ? {l = 1, m = ±1} states define the energetically degener- Jλ,ν = − dk k Rλ,l,m(k) µx ate 2s and 2p states, respectively. The mass anisotropy 0 l,m h clearly removes this degeneracy, producing a 0.415 meV × hmR (k) + hm−1R (k) (0.362 meV) separation between the 2s and the lowest l ν,l+1,m+1 l−1 ν,l−1,m−1 {l = 1, m = 0} ({l = 1, m = ±1}) states. −m −m−1 i − hl Rν,l+1,m−1(k) − hl−1 Rν,l−1,m+1(k) , To visualize the effects of the effective-mass anisotropy, (17) Fig. 4 compares a collection of exciton wave functions R φλ (k) for an isotropic (left column) and an anisotropic between the exciton states λ and ν where (right column) case. Each wave function represents the energetically lowest state for the dominant symmetry s m (l − m + 1)(l + m + 1) annotated on the right-hand side. The anisotropic re- gl = , (18) sults correspond to Ge masses and the isotropic case (2l + 3)(2l + 1) s is computed with mk = m⊥ = 1.59 m0. As expected, m 1 (l + m + 1)(l + m + 2) the isotropic case produces the well-known hydrogen-like hl = − , (19) wave functions. Anisotropy already modifies the ground 2 (2l + 1)(2l + 3) 5

0.4 Isotropic Anisotropic (a)

0.2

0.5 { ) (scaled) 0 , ! ( 0.0 0 k }

↵ 0.0 −0.5 1s 1.0 (b)

0.5 {

1 0.5 , ) (scaled)

0.0 0 ! } ( −0.5 ? 0.0 2p ↵ 0.5 1.0 1.5 !/

0.5 { ~ k 1 E 0 , a

0.0 1 ? Figure 5. Comparison of the THz absorption spectrum in } k bulk Ge for parallel (a) and perpendicular (b) polarization −0.5 2p at a single L valley. Excitons are assumed to only populate the ground state. The energy axis is normalized to the transition energy Ek of the lowest possible transition for parallel polarization, see Fig. 3. 0.5 { 2 ,

0.0 0 } Wave vector and the vectorial sum has been replaced by an integral −0.5 3d in the usual way [24]. Equations (16) and (17) imply diﬀerent selection rules ν caused by the mass anisotropy. In the ek direction, Jλ

0.5 {

2 is non-zero only for transitions involving equal m com-

, ponents while the l components diﬀer by ±1. For the 0.0 1

} e⊥ polarized exciton, m components couple to m ± 1 and l to l ± 1. Assuming that only the exciton ground −0.5 3d state is occupied, the energetically lowest THz transition with non-vanishing transition-matrix element will involve the dominantly {0, 0} and {1, 0} states for ek excitation.

0.5 {

2 However, an e⊥ excitation involves {0, 0} and {1, ±1}

, components. Due to the energy diﬀerence between the 0.0 2

} {1, 0} and {1, ±1} states, the transition energy to the lowest excited state is E = 3.38 meV for e polarized −0.5 k k 3d excitation and E⊥ = 4.16 meV for e⊥ polarized excitation. Both transitions are illustrated in Fig. 3 by arrows. −0.5 0.0 0.5 −0.5 0.0 0.5 Since Ek and E⊥ differ by 0.78 meV, the mass anisotropy Wave vector k a0 produces direction-dependent THz resonances. However, k a realistic experiment typically contains excitations in all R Figure 4. Comparison of the wave functions φλ (k) for L valleys, which mixes ek and e⊥ responses, as discussed l ≤ 2 using isotropic (left column, mk = m⊥ = 1.59 m0) and in Sec. IV B. anisotropic (right column, mk = 1.59 m0, m⊥ = 0.0815 m0) A configuration where initially the 1s exciton state masses. The dominant character of the individual rows are is populated can be realized at sufficiently low temper- indicated on the right. Additionally, the respective symmetry atures and long after a weak optical excitation, yield- family in more familiar terms 1s, 2p, and 3d is displayed in each row. Each wave function is the energetically lowest for ing abundant exciton formation [35]. We then have its respective dominant symmetry. nλ,ν = δλ,ν δλ,GSnGS where nGS is the ground-state exciton density. Choosing γ = γJ = 0.3 meV, we compute the THz absorption αk(ω) and α⊥(ω) for parallel and perpendicular polarizations, respectively, according 6

12 =1 1.0 3 =2 ⇥8 =3

(meV) 8

=4 E

4 (a) 0.5 ) (scaled) ↵ (!)

! ? 1.0 (

↵ ↵ (!) k

(norm.) 0.0

| 0.5 , 0.5 1.0 1.5

GS (b) !/

J ~ | 0.0 Ek

5 10 15 20 Figure 7. The full absorption spectrum (shaded area) ac- m /m cording to Eq. (20) has been scaled by a factor of 3/8 for k ? visibility. It is compared to the absorption for parallel (solid) and perpendicular (dashed) polarization from Fig. 5. Figure 6. (a) Energetic difference ∆Eλ = Eλ − E0 between the four lowest excited states. (b) THz-transition matrix- element from the ground state to the corresponding end states λ = 2 state whereas λ = 3 remains dark for both polar- indicated in (a). For the end state λ = 3, neither parallel nor izations because it has the same s-like symmetry as the perpendicular polarization allow a transition from the ground ground state. As a measure of the overlap of the partic- state. Hence, the matrix element vanishes. ipating wave functions, |JGS,λ| monotonically decreases for parallel polarization while it increases in the perpendicular case. Consequently, the corresponding THz ab- to Eqs. (10)–(12), (16), and (17). Figures 5(a) and (b) sorption appears stronger in perpendicular polarization show αk(ω) and α⊥(ω), respectively. As we can see, the as observed in Fig. 5. absorption spectra exhibit two distinctly different reso- nance energies, corroborating the symmetry-based argu- ments about the selection rules. At the same time, the B. Multi-Valley THz Absorption peak heights differ by a factor of 2.54, and the parallel direction yields a less asymmetric tail towards high Since Ge experiments typically generate excitons in all energies. of its L valleys, it is important to consider how this as- To identify general trends of THz-transition ener- pect alters the results of Sec. IV A. Any arbitrary THz gies and oscillator strengths, we study the binding en- field polarization addresses each L point with a different ergies and transition-matrix elements as a function of mixture of perpendicular and parallel directions. Hence, anisotropy mk/m⊥ while mk is kept constant. Fig- the total susceptibility that defines the overall absorption ure 6(a) shows the λ-to-GS transition energy ∆Eλ = must be a tensor. It is composed out of the individual Eλ − EGS for the first four excited states. With van- susceptibility tensors for each valley and diagonal with ishing anisotropy (mk/m⊥ = 1), the first three excited the components χ1,1 = χ2,2 = χ⊥ and χ3,3 = χk for the excitons (λ = {1, 2, 3}) become energetically degenerate, single L valley discussed in Sec. IV A. We show in App. D as expected for pure 2p and 2s excitons. With increasing that not only the total tensor but also its elements remain anisotropy, this degeneracy is lifted. As a typical trend, diagonal, and the overall absorption becomes isotropic ∆Eλ changes fast close to mk/m⊥ = 1 while it saturates 4 with increasing mk/m⊥. α(ω) = 2α (ω) + α (ω) . (20) 3 ⊥ k The saturation can be attributed to the fact that de- creasing m⊥ (or increasing m /m⊥) freezes the motion As a simple interpretation of Eq. (20), k in the parallel direction compared to the perpendicular 1/3 2α⊥(ω) + αk(ω) defines the mean over three direction. We also can deduce from the THz-Elliott for- cartesian directions, and the factor four accounts for the mula that JGS,λ determines the strength of absorption for absorption of the four unique L valleys. the λ-to-GS transition. Figure 6(b) shows the transition- Figure 7 compares the total THz absorption (shaded matrix elements from the ground to the first four excited area) with its αk (solid line) and α⊥ (dashed line) con- states. The selection rules determine that transitions to stituents. Even though α(ω) is a mixture of all directions, λ = 1 and λ = 4 states occur only for the ek polarization. it still exhibits a residual double-peak structure due to Conversely, e⊥ excitation yields only transitions to the the mass anisotropy. Therefore, a THz measurement in 7

Ge can directly detect the mass anisotropy in THz ab- tion reads sorption spectra as long as the dephasing does not greatly R 0 ˜ R 0 exceed the anisotropic splitting (0.78 meV). Eλφλ (k, k ) = Ek,k0 φλ (k, k ) e h X R 0 − 1 − fk0 − fk Vqφλ (k − q, k − q) , q V. CONCLUSIONS (A1)

We have presented an efficient numerical method in which the electron–hole pair energy is defined as to solve the Wannier equation for systems with mass anisotropy based on an expansion into spherical harmon- ˜ e h X e h Ek,k0 = Ek0 + Ek − Eg − Vq fk0−q + fk−q . (A2) ics. We use Ge as a prototype system to illustrate the q main consequences of the mass anisotropy for the exciton wave functions and the THz absorption. We find that The solutions of Eq. (A1) define the wave functions of the anisotropy lifts the degeneracy of excitons with the an indirectly bound electron–hole pair and their binding same l quantum number. We find excitonic wave function energies. To further simplify the problem, we introduce modifications that can be traced back to the coupling of the concept of diagonal excitons, e.g. the momentum dif- 0 different l and m components in our expansion in terms ference is assumed to be constant: k = k + k0. Equa- of spherical harmonics. tions (A1) and (A2) then reduce to Eqs. (7) and (8), The wave function distortions result in modified se- respectively. lection rules for THz induced transitions, depending on the polarization of the applied THz field. For polarization parallel to the Γ → L direction, transitions involve Appendix B: Coupling Strengths states whose l differs by ±1 while m remains constant. For perpendicular polarizations, we find that both l and The coupling strengths in Eq. (14) emerge when the m are changed by ±1 between the states. As a conse- kinetic part of the Wannier equation is projected on to quence, two distinct THz resonances can be observed, spherical harmonics, e.g. by computing each assigned to one of the polarization directions. Our calculations indicate that these resonances should be ob- Z m ? R servable in good quality THz-absorption experiments. Ik ≡ dΩ [Yl (Ω)] Ekφλ (k) , (B1) Besides excitonic features, future experiments could also explore how mass anisotropy modifies properties of where the unrenormalized part of Eq. (8) in spherical more complicated quasiparticles such as biexcitons [36, coordinates is 37] and dropletons [38]. In particular, the presented approach can be used as a starting point to also compute 2k2 E = ~ the relevant correlation functionals to determine the en- k 2 ergetically possible configurations [39]. sin2(θ) cos2(ϕ) sin2(θ) sin2(ϕ) cos2(θ) × + + , µx µy µz (B2) ACKNOWLEDGMENTS and the solid angle is given by Ω ≡ (θ, ϕ). The recurrence This work is funded by the German Science Foundation formula for the associated Legendre polynomials [41] are (DFG) through the Research Training Group Function- very useful to solve these integrals. Defining the masses alization of Semiconductors (GRK 1782) and via Grants No. KI 917/3-1 and No. SFB 1083. 1 1 1 1 1 2 ± = ± , ± = + ± , (B3) µxy µx µy µxyz µxy µz

Appendix A: Generalized Wannier Equation for leads to the following coupling strengths Indirect Excitons (1) µz µz l,m = Nl + Ll + − Mm − 3 , µxyz µxyz Analogous to direct semiconductors, the exciton wave √ function in an indirect semiconductor is determined by (2) µz NlJl,m the homogenous solution to the semiconductor Bloch l,m = − − , µxyz 2(2l + 1) equations (SBE) [40]. However, we now have to allow (3) µz NlJl,m for transitions between diﬀerent momentum states of the l,m = − − , µxy 2 participating carriers. Repeating the derivation of the √ SBE in Hartree–Fock approximation for the interband (4) µz Nl D † E l,m = − Bl,m , (B4) 0 0 4(2l + 1) polarization pk,k = aˆv,k aˆc,k , the homogeneous solu- µxy 8

m containing the coeﬃcients due to the orthonormality of Yl . At the same time, 1 Eq. (C4) yields Nl = ,Ll = l(l + 1) , Z m 0 (2l − 1)(2l + 3) l 0 L 3 Yl (Ω ) Vk,k0 = dΩ V|k−k0| , (C7) 3 2π Y m(Ω) Y l M = (m + 1)(m − 1) ,B2 = (l + m + j) , m l,m where the right-hand side must also be independent of j=0 m and Ω. Hence, we can chose them at our convenience. rh ih i 2 2 2 2 The simplest choice m = 0 and θ = ϕ = 0 yields Jl,m = (l + 1) − m l − m . (B5) π 2 Z 0 0 l e 0 sin(θ )Pl(cos(θ )) Vk,k0 = 2 dθ 2 , (C8) 4π 0r k2 + (k0) − 2kk0 cos(θ0) Appendix C: Coulomb-Matrix Element 0

with the Legendre polynomials Pl. Not having to intro- Inserting Eq. (13) into the Coulomb part of Eq. (7) duce an additional numerical grid is the main advantage yields of Eq. (C8) over Eq. (C5). We note that Pl can always Pl j X R 0 be written in the form Pl(x) = j=0 al,jx with the ex- Ik ≡ V|k−k0|φλ (k ) k0 pansion coefficients [41] Z 1 L 3 3 0 m 0 l (l + j − 1) 0 l 2 = 2π d k V|k−k |Yl (Ω ) a = 2 . (C9) l,j j l ∞ Z X 0 0 2 0 Thus, the remaining integral in Eq. (C8) becomes = dk (k ) Rn,l,m(k ) l,m π 0 1 Z cosj(θ0) Z I ≡ d cos(θ0) 0 L 3 m 0 j 0 × dΩ V|k−k0|Yl (Ω ) , (C1) K 1 − Q cos(θ ) 2π 0 ∞ 2 X Q2n−1 where we have first replaced the vectorial sum by an inte- = , (C10) gral which is then implemented in spherical coordinates KQj+1 2n − 1 n=p+1 k = (k, Ω). To evaluate the solid angle integral, we need the explicit form of the Coulomb potential 2 0 2 2kk0 with K ≡ k + (k ) , 0 ≤ Q ≡ K ≤ 1, and p = j/2 Z (p = (j +1)/2) if j is even (odd). For high orders in j and 1 3 −ik·r ik0·r V 0 = d r V (r)e e , (C2) small Q, I can show numerical instabilities due to limited |k−k | L3 j machine precision. In this case, the series in Eq. (C10) 2 where V (r) = e contains the background dielectric has to be terminated approproiatly. Finally, the Coulomb 4π0rr renormalization of the kinetic part of Eq. (7), constant r, and r = 16 in Ge [22]. Using Eq. (C2) and the plane-wave expansion [42] X e h Ik ≡ V|k−k0| fk0 + fk0 , (C11) −ik·r X l ? m ? m k0 e = 4π (−i) jl (kr)[Yl (Ω)] Yl (Θ) , (C3) l,m is projected to spherical harmonics again by substituting the vectorial sum with an integral in spherical coordi- with r = (r, Θ), the Ω0 integral in Eq. (C1) can be ex- nates. If we make use of the freedom to define k = kez pressed via under the integral, we obtain Z 0 L 3 m 0 m l ∞ dΩ V|k−k0|Yl (Ω ) ≡ Yl (Ω)Vk,k0 , (C4) Z ? Z 2π h m0 i 0 0 2 dΩ Yl0 (Ω) Ik = dk (k ) where 0 ∞ e h 0 Z × fk0 + fk0 Vk,k0 Rλ,l,m(k) , l 2 2 ? 0 V 0 = dr r V (r)j (kr)jl(k r) , (C5) (C12) k,k π l 0 if we assume radially symmetric carrier distributions. containing the spherical Bessel functions jl. In this for- mat, the projection of Eq. (C1) to spherical harmonics is particularly easy since Appendix D: Susceptibility Tensor ∞ Z ? Z h m0 i 0 0 2 0 l The macroscopic current can be decomposed as dΩ Yl0 (Ω) Ik = dk (k ) Rn,l,m(k )Vk,k0 , X X X λ † 0 J = J = j aˆ aˆ , (D1) n kn λ,kn λ,kn (C6) n n λ,kn 9 where kn is in the vicinity of the n-th L valley. The for the valley with n = 1. For all valleys, we find center of the four unique L valleys are 1 √ √ X χ(ω) = χ⊥ (en,x ⊗ en,x + en,y ⊗ en,y) L1 = 0 0 1 , L2 = − 2 − 6 1 , 3 n √ √ √ 1 1 4 L3 = −2 2 0 −1 , L4 = − 2 6 1 . (D2) + χ e ⊗ e = (2χ + χ )1 , (D3) 3 3 k n,z n,z 3 ⊥ k

Let en,j (j = {x, y, z}) denote the cartesian unit vectors where en,z is aligned with Ln. Then, the tensorial suscep- where the last step follows as the explicit directions (D2) tibility becomes χ⊥(e1,x ⊗e1,x +e1,y ⊗e1,y)+χke1,z ⊗e1,z are applied.

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