Ab initio spin-resolved photoemission and electron pair emission from a Dirac-type surface state in W(110)

The electronic structure calculations were performed within the framework of the local spin-density approximation to density-functional theory. Our computations rely on the layer Korringa–Kohn–Rostoker (KKR) method for spin-dependent systems [31, 32]. In this relativistic approach, spin–orbit coupling is accounted for by solving the Dirac equation. The site-dependent potentials have been obtained by full-potential augmented plane waves (FLAPW) calculations⁶ and subsequently transferred to our KKR code. Since all the results presented in this paper rely on the computational setup used earlier, it is sufficient to sketch its main ingredients here; for details see [15].

The W(110) surface is treated as a semi-infinite system, in which the potential in the outermost three layers differs from that of the bulk-like layers (with bulk potential). The top layer is relaxed inward by Δd = −3% with respect to the bulk interlayer distance, in agreement with low-energy electron diffraction (LEED) and surface x-ray diffraction analyses [33, 34]. The image-potential barrier is taken as a smooth interpolating function [35].

From the layer-resolved Green function G_ll of the entire system, we calculate the spectral density N_l(E,k) for layer l at energy E and surface-parallel wavevector k

$$\begin{equation} N_{l}(E, \boldsymbol{k}) = -\frac{1}{\pi} \lim_{\eta \to 0^{+}} \Im {\rm Tr}\,G_{ll}(E + {\mathrm{i}} \eta, \boldsymbol{k}). \end{equation} \tag{ 1 }$$

By taking appropriate partial traces, the spectral density can further be decomposed with respect to angular momentum and spin projection. When discussing the spin texture of the Dirac surface state, we rely on spin differences

$$\begin{equation} S_{l \mu}(E, \boldsymbol{k}) \equiv S_{l}(E, \boldsymbol{k}; \uparrow_{\mu}) - S_{l}(E, \boldsymbol{k}; \downarrow_{\mu}) \end{equation} \tag{ 2 }$$

of spin-dependent spectral densities that are resolved with respect to the chosen spin quantization axis μ.

SARPES spectra were calculated within the relativistic one-step model of photoemission [31, 36, 37], using our layer KKR computer code. Spin–orbit coupling, boundary conditions and the transition matrix elements between the initial state and the time-reversed spin-polarized LEED state (that describes the outgoing electron) are fully taken into account. The result of the computation is the spin density matrix ${\boldsymbol{\underline{\underline{\rho}}}}$ of the photoelectron that depends on the kinetic energy E and the detection direction of the photoelectron (expressed in terms of the wavevector k). It gives the intensity $I = {\rm tr} {\boldsymbol{\underline{\underline{\rho}}}}$ and the spin polarization $s_{\mu } = \boldsymbol {\mu } \cdot {\rm tr} ({\boldsymbol{\underline{\underline{\sigma}}}} {\boldsymbol{\underline{\underline{\rho}}}}) / I$ of the chosen spin quantization axis $\boldsymbol{\mu}$ ( ${\boldsymbol{\underline{\underline{\sigma}}}}$ vector of Pauli matrices).

Excitation of the DSS by linearly polarized light has been investigated earlier [16]. In this paper, we focus on circularly polarized light (helicity σ_±) with a photon energy of 21.22 eV (He_I); this particular energy was chosen because it is available both in synchrotron radiation facilities and from rare-gas discharge lamps in a laboratory. We investigate three setups—${\mathcal{A}}$, ${\mathcal{B}}$ and ${\mathcal{C}}$—with decreasing degree of symmetry. While setup ${\mathcal{A}}$ is for normal incidence, ${\mathcal{B}}$ is for off-normal incidence within the $\overline {\mathrm {H}}$– $\overline {\Gamma }$– $\overline {\mathrm {H}}$ mirror plane of the surface ([001] direction). In both cases we focus on electron detection in the $\overline {\mathrm {H}}$– $\overline {\Gamma }$– $\overline {\mathrm {H}}$ mirror plane in which the surface state shows almost linear and strong dispersion. Setup ${\mathcal{C}}$ is for off-normal incidence and electron detection in [$\overline {1}11$] direction ($\overline {\mathrm {S}}$– $\overline {\Gamma }$– $\overline {\mathrm {S}}$) which is not a mirror plane. The surface state shows strong dispersion in this plane as well [10]. It becomes 'flattened' along $\overline {\mathrm {N}}$– $\overline {\Gamma }$– $\overline {\mathrm {N}}$ [11]. For ${\mathcal{B}}$ and ${\mathcal{C}}$ the polar angle of incidence ϑ_ph is chosen as 45°.

In setups ${\mathcal{A}}$ and ${\mathcal{B}}$, the spin polarization of the Dirac surface state is fully aligned perpendicular to the mirror plane. Considering as an example an xz mirror plane (s in table 1), s_y does not change sign upon reflection and, hence, is allowed non-zero, in contrast to s_x and s_z. The photoemission process lowers this symmetry restriction because the electric field A of the light has to be considered in addition (A in table 1). For circularly polarized light incident within the xz plane, the electric field vector can be expressed as

$$\begin{equation} \boldsymbol{A} = (A_{x}, A_{y}, A_{z}) \propto (\cos\vartheta_{\mathrm{ph}}, \pm {\mathrm{i}}, \sin\vartheta_{\mathrm{ph}}). \end{equation} \tag{ 3 }$$

In particular, A_y changes sign upon reflection m_xz: left-handed circular polarized light is turned into right-handed circular polarized light, and vice versa (σ_± → σ_± in table 1). As a consequence, the total photocurrent I is unchanged with respect to reversal of the light helicity: there is no circular dichroism [38]. Furthermore, all spin components of the photoelectron are allowed non-zero but s_x and s_z change sign upon helicity reversal while s_y does not. In a schematic notation, (I;s_x,s_y,s_z;σ_±) = (I; − s_x,s_y,−s_z;σ_±) holds.

In setup ${\mathcal{C}}$, both light incidence and electron detection are not within a mirror plane of the surface, leading in general to circular dichroism; furthermore, all Cartesian components of the photoelectron's spin polarization s may be non-zero.

We employed a relativistic KKR-type formalism, which has been presented in detail in earlier work [22, 39]. The basic ingredients are four one-electron states |i〉 (with i = 1,2,3,4), which are obtained as solutions of the Dirac equation with a complex potential and are characterized by energies E_i, surface-parallel momenta k_i and Rashba component spin labels σ_i = ± . In the (e,2e) process, the Coulomb interaction induces a transition from an initial two-electron state |1,2〉, which is an antisymmetrized product of the primary electron state |1〉 and the valence electron state |2〉, to a final two-electron state representing the two emitted electrons. |3,4〉 is an approximate solution of a two-particle Dirac equation with the electron–electron interaction approximated by a screened Coulomb interaction. Being of course also antisymmetric with respect to interchanging the two electrons, it involves essentially products of the two one-electron states |3〉 and |4〉 and a Coulomb correlation factor. For details we refer to [39]. An important property of the (e,2e) process is the conservation of energy and parallel momentum:

$$\begin{equation} E_{1} + E_{2} = E_{3} + E_{4}, \quad \boldsymbol{k}_{1} + \boldsymbol{k}_{2} = \boldsymbol{k}_{3} + \boldsymbol{k}_{4} + \boldsymbol{g}, \end{equation} \tag{ 4 }$$

where g is a surface reciprocal lattice vector.

The spin-resolved (e,2e) reaction cross sections 'intensities' can be expressed in a golden rule form with Coulomb matrix elements between the above initial and final two-electron states. Denoted by Iσ1σ3σ4, they involve a sum over the valence electron spin labels σ2 and depend explicitly on the spin labels of the incoming and the two outgoing electrons. For parallel spins (σ3 = σ4), the Iσ1σ3σ4 contain both exchange and Coulomb interaction between the two emitted electrons, whereas for antiparallel spins, they contain only the Coulomb interaction. Further, they are in general functions of Ei and ki, with i = 1,3,4. For fixed Ei and ki, 4 implies definite valence electron energy and parallel momentum, i.e. one can map the dispersion relation of the valence electron states.

As a prerequisite for discussing the electron spectroscopies, we recall briefly the salient properties of the DSS. We focus on the directions in which this state disperses strongly [11]. Along $\overline {\mathrm {H}}$– $\overline {\Gamma }$– $\overline {\mathrm {H}}$ (left in figure 1), the dispersion is almost linear, while it becomes slightly parabolic along $\overline {\mathrm {S}}$– $\overline {\Gamma }$– $\overline {\mathrm {S}}$ (right in figure 1). Its Dirac point shows up at E_F–1.25 eV and $\overline {\Gamma }$, in agreement with experiment [10, 12]. The orbital decomposition of the top layer's spectral density yields that the surface state belongs to the Σ₁ (single-group) representation⁷, with the by far largest contribution from d_z² orbitals.

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The spin texture is of Rashba-type, as dictated by point group and time-reversal symmetry, namely in-plane and perpendicular to k (this spin-polarization component is usually termed 'the Rashba component'). Time-reversal symmetry dictates a spin reversal upon turning k into −k (cf the red–blue asymmetry in panels (c) and (f) of figure 1). The degree of spin polarization is about 90% (in absolute value) close to the Dirac point. It exceeds that of the Dirac surface state in the topological insulator Bi₂Te₃ (about 60% [27, 41]). Within this respect, the Dirac surface state of W(110) is closer to modeled Dirac states [42] than those in bichalcogenide topological insulators, thus lending itself for investigations of spin–orbit effects in electron spectroscopies from strongly spin-textured surface states [25, 43].

In the following, we discuss the evolution of photoelectron's spin texture and of circular dichroism when lowering the symmetry of the SARPES setup (${\mathcal{A}} \to {\mathcal{B}} \to {\mathcal{C}}$). We also show how to disentangle the spin polarization from the initial state from that due to optical orientation.

Due to the symmetry relation (I;s_x,s_y,s_z;σ_±) = (I; − s_x,s_y,−s_z;σ_±) that holds for setups ${\mathcal{A}}$ and ${\mathcal{B}}$, it is sufficient to present only results for one helicity (here: σ₊); the numerical results are fully in line with these transformations. For normal incidence (setup ${\mathcal{A}}$, left in figure 2), the intensity distribution is symmetric about $\overline {\Gamma }$, i.e. I(k) = I(− k). The optical orientation tilts the electron spin toward the photon spin (i.e. along −z) [24]; it is so strong that s_z exceeds the Rashba component s_y and the very weak s_x by far. To be more specific, s_z is as large as 80% close to the Dirac point and s_x reads 40% for s_y at (E,k) = (− 0.91 eV, −0.156 a⁻¹₀); (note the decreasing color saturation in (b) via (c)–(d)). This means that optical orientation can (almost) fully align the electron's spin to the photon spin even if the initial state's spin polarization is perpendicular to the photon spin.

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When changing the polar angle of incidence from 0° (setup ${\mathcal{A}}$) to 45° (setup ${\mathcal{B}}$) in the xz plane, the intensity from the DSS increases (compare the DSS in figure 2(a) with that in (e)). Due to the off-normal incidence, one would expect an intensity asymmetry in the polar angle of detection ($\vartheta _{\mathrm {e}} \leftrightarrow -\vartheta _{\mathrm {e}}$ or $\boldsymbol {k} \leftrightarrow -\boldsymbol {k}$). This effect is apparently minute for this surface state. The s_z component of the spin polarization becomes reduced but s_x becomes sizably large ((f), (h)). But more strikingly, the Rashba component is 'restored' (compare figure 2(g) with figure 1(c)); it shows the largest degree of spin polarization of all Cartesian components. More precisely: s = (− 43%,+87%, −19%) at (E,k) = (− 1.15 eV, 0.045 a⁻¹₀) in (f)–(h). This implies that the degree of spin polarization is larger for the photoelectron (|s| = 99%) than for the initial state (s_y = |s| = + 85% at the same (E,k) in figure 1(c)).

Setup ${\mathcal{A}}$ and in particular setup ${\mathcal{B}}$ are suited to investigate the Rashba component of the surface state's spin texture. Although optical orientation affects the photoelectron's spin, the initial state's spin polarization can be distinguished from that due to optical orientation because there is no dichroism. In case of non-zero dichroism, the analysis can be extended as follows.

Spin-resolved experiments and calculations for dichroism provide a set of four spectra for a chosen spin quantization axis μ:

$$\begin{equation} \fl I(\sigma_{+}, \uparrow_{\mu}) = \frac{1}{2} \left[ 1 + P_{\mu}^{\mathrm{is}} + P_{\mu}^{\mathrm{oo}} \right] I(\sigma_{+}), \quad I(\sigma_{+}, \downarrow_{\mu}) = \frac{1}{2} \left[ 1 - P_{\mu}^{\mathrm{is}} - P_{\mu}^{\mathrm{oo}} \right] I(\sigma_{+}), \end{equation} \tag{ 5a }$$

$$\begin{equation} \fl I(\sigma_{-}, \uparrow_{\mu}) = \frac{1}{2} \left[ 1 + P_{\mu}^{\mathrm{is}} - P_{\mu}^{\mathrm{oo}} \right] I(\sigma_{-}), \quad I(\sigma_{-}, \downarrow_{\mu}) = \frac{1}{2} \left[ 1 - P_{\mu}^{\mathrm{is}} + P_{\mu}^{\mathrm{oo}} \right] I(\sigma_{-}) \end{equation} \tag{ 5b }$$

with the spin-integrated photocurrents I(σ_±) ≡ I(σ_±,↑_μ) + I(σ_±,↓_μ). Here, we have decomposed the photoelectron's polarization into a part associated with the spin polarization of the initial state (is) and a part due to the optical orientation (oo), P_μ(σ_±) ≡ P^is_μ ± P^oo_μ. P^is_μ does not change sign upon helicity reversal but P^oo_μ does. Thus, the differences

$$\begin{equation} D^{\mathrm{cd}} \equiv I(\sigma_{+}, \uparrow_{\mu}) + I(\sigma_{+}, \downarrow_{\mu}) - I(\sigma_{-}, \uparrow_{\mu}) - I(\sigma_{-}, \downarrow_{\mu}), \end{equation} \tag{ 6a }$$

$$\begin{equation} D_{\mu}^{\mathrm{oo}} \equiv I(\sigma_{+}, \uparrow_{\mu}) - I(\sigma_{+}, \downarrow_{\mu}) - I(\sigma_{-}, \uparrow_{\mu}) + I(\sigma_{-}, \downarrow_{\mu}), \end{equation} \tag{ 6b }$$

$$\begin{equation} D_{\mu}^{\mathrm{is}} \equiv I(\sigma_{+}, \uparrow_{\mu}) - I(\sigma_{+}, \downarrow_{\mu}) + I(\sigma_{-}, \uparrow_{\mu}) - I(\sigma_{-}, \downarrow_{\mu}) \end{equation} \tag{ 6c }$$

can be expressed as

$$\begin{equation} D^{\mathrm{cd}} = I(\sigma_{+}) - I(\sigma_{-}), \end{equation} \tag{ 7a }$$

$$\begin{equation} D_{\mu}^{\mathrm{oo}} = P_{\mu}^{\mathrm{is}} \left[I(\sigma_{+}) - I(\sigma_{-})\right] + P_{\mu}^{\mathrm{oo}} \left[ I(\sigma_{+}) + I(\sigma_{-})\right], \end{equation} \tag{ 7b }$$

$$\begin{equation} D_{\mu}^{\mathrm{is}} = P_{\mu}^{\mathrm{is}} \left[I(\sigma_{+}) + I(\sigma_{-})\right] + P_{\mu}^{\mathrm{oo}} \left[I(\sigma_{+}) - I(\sigma_{-})\right]. \end{equation} \tag{ 7c }$$

If the measure D^cd of the circular dichroism vanishes, then D^oo_μ and D^is_μ allow to disentangle the origin of the spin polarization component μ. To make this point more clear, consider a 'non-dichroic' setup (${\mathcal{A}}$ or ${\mathcal{B}}$). From symmetry considerations and calculated photocurrents, we find I(σ₊,↑_z) = I(σ₋,↓_z) and I(σ₊,↓_z) = I(σ₋,↑_z) but I(σ₊,↑_y) = I(σ₋,↑_y) and I(σ₊,↓_y) = I(σ₋,↓_y). Hence, D^cd = D^oo_y = D^is_z = 0 and D^oo_z ≠ 0 and D^is_y ≠ 0, indicating that s_z is due to optical orientation and s_y is due to the initial state.

We apply this approach to disentangle the spin polarization in the 'dichroic' setup ${\mathcal{C}}$ ($\overline {\mathrm {S}}$– $\overline {\Gamma }$– $\overline {\mathrm {S}}$ azimuth). The circular dichroism D^cd is non-zero (intensity asymmetry of 51% at (E,k) = (− 1.15 eV, 0.05 a⁻¹₀) in figure 3(a)) and asymmetric with respect to k. It is restricted to the DSS (DSS in figure 3(a)) because this is the electronic state with largest degree of spin polarization (figure 1(f)). The 'optical orientation' D^oo_μ provides for both spin components—z (b) and the Rashba component R (c)—a minor background (no change of sign for the DSS), in contrast to the 'initial state' differences D^is_μ (change of sign with k in (d) and (e)). Most strikingly, the absolute value is largest for D^is_R, as seen by the color saturation, suggesting that s_R is mainly due to the initial-state spin polarization.

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Eventually, we briefly address calculated spectra for unpolarized light (not shown here). Because unpolarized light is an incoherent superposition of left- and right-handed circularly polarized light, the photocurrent shows non-zero spin polarization only in the Rashba components; the components brought about by the optical orientation vanish. For the setups ${\mathcal{A}}$ and ${\mathcal{B}}$, the spin-integrated and the s_y-resolved intensities are very similar to those shown in figure 2; s_x and s_z are zero. The similarity holds also for setup ${\mathcal{C}}$: spin-resolved spectra compare well with the data shown in figures 3(d) and (e).

The spin–orbit interaction couples the orbital and the spin degrees of freedom. For the mirror-symmetric setups ${\mathcal{A}}$ and ${\mathcal{B}}$, strict relations between the parity of the spatial part and the spin orientation of a wavefunction hold (confer (8) below). Thus, a measurement of the spin texture by spin-resolved photoemission allows to conclude on the parity of the initial state. It turns out that the spin texture of the Dirac surface state is connected with even orbitals, in agreement with the spectral density in which d_z² orbitals show the largest contribution [15, 16]. While such an entanglement procedure is comparably simple for W(110), it becomes complicated for chalcogenide topological insulators. Nevertheless, Zhu et al [44] have successfully analyzed spin texture and orbital composition of the Dirac surface state in Bi₂Se₃, using an analysis similar to that presented here.

Complementary to photoemission, electron-induced two-electron emission is also a powerful tool for studying the spin–orbit-induced spin texture at surfaces. We demonstrate this for the case of the surface electronic structure of W(110) in the $\overline {\mathrm {H}}$– $\overline {\Gamma }$– $\overline {\mathrm {H}}$ azimuth, for which the scattering plane—xz—is a mirror plane of the semi-infinite crystal. Due to spin–orbit coupling, all four one-electron states have the schematic forms [40]

$$\begin{equation} \mid\!\! \mathrm{even}, \uparrow_{y} \rangle + \mid\!\! \mathrm{odd}, \downarrow_{y} \rangle, \quad \mid\!\! \mathrm{even}, \downarrow_{y} \rangle + \mid\!\! \mathrm{odd}, \uparrow_{y} \rangle, \end{equation} \tag{ 8 }$$

where 'even' and 'odd' indicate spatial parts of even and odd parity with respect to the xz plane of the semi-infinite crystal; the spinor parts are quantized with respect to the Rashba component μ = y.⁸ Since the primary and the two outgoing electron states are predominantly even, a selection rule [45] holds in good approximation: only the even parts of the valence states enter the (e,2e) matrix elements.

The left-hand part of figure 4 shows spectral densities for a bulk layer ((a), (b)) and the topmost surface layer (c)–(f) resolved with respect to spatial symmetry and spin. From panels (c)–(f) it is apparent that the branches of the DSS (labeled A and A' for positive and negative k_2x, respectively) are almost exclusively even and spin-polarized. The surface state branches B, B', C and C' are also strongly spin-polarized but of predominantly odd parity.

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For normally incident primary electrons and emission at equal polar angles (figure 4(g)), the conservation conditions 4 turn the (e,2e) intensities Iσ1σ3,σ4 into functions of the valence electron energy E2 and its parallel momentum component k2x. These are shown for primary spin σ1 = +  in figures 4(h) and (i). The parallel-spin intensity I+++ in panel (h) maps the spin-up even-parity surface state spectral density shown in panel (d), whereas the antiparallel-spin intensity I++− + I+−+ in panel (i) reflects the spin-down even-parity spectral density of panel (f). Since the mirror operation at the yz plane reverses all the spins, the intensities I−−− and I−−+ + I−+− are the mirror images of I+++ and I++− + I+−+, respectively.

The strong spin polarization of the DSS facilitates an investigation of the spin dependence of the exchange-correlation hole. For this purpose, an (e,2e) setup is suitable, in which the two emitted electrons have fixed equal energies E₃ = E₄ and variable surface-parallel momenta, k₃ = −k₄. By virtue of energy and momentum conservation (4) a surface state electron with E₂ and k₂ can then be selected by choosing for the primary electron E₁ = E₃ + E₄ − E₂ and k₁ = −k₂. As a typical example, we select the point (E₂ = E_F − 0.95 eV, k₂ = (+ 0.142,0) a⁻¹₀) on the spin-up branch of the DSS, which is indicated by the green dot in the left-hand part of figure 4. The primary electron with an energy E₁ = 25 eV is incident with parallel momentum k₁ = (− 0.142,0) a⁻¹₀ (i.e. within the $\overline {\mathrm {H}}$– $\overline {\Gamma }$– $\overline {\mathrm {H}}$ azimuth at ϑ₁ = 6.01° and φ₁ = 180°). The emitted electrons have energies of E₃ = E₄ = 9.49 eV and surface-parallel momenta k₃ = −k₄.

The resulting (e,2e) momentum distributions Iσ1σ3σ4(kx,ky) in the plane $\boldsymbol {k} / k \equiv \boldsymbol {k}_{3} / \sqrt {2 E_{3}} = -\boldsymbol {k}_{4} / \sqrt {2 E_{4}}$ (with E3 = E4 in Hartree atomic units) of the outgoing electrons are shown in the left-hand part of figure 5. For primary spin up (σ1 = +), the parallel-spin intensity I+++ exhibits a large central depletion zone which is due to exchange and Coulomb correlation. The antiparallel-spin intensities are by far smaller. For primary spin down (σ1 = −), the antiparallel-spin intensities dominate and show a rather small central 'hole' which is due to Coulomb correlation only. The apparently spin-non-conserving intensities I+−− and I−++ rely on spin–orbit coupling in the primary and outgoing electron states; in fact they vanish in calculations without spin–orbit coupling in these states. The sums of the intensities Iσ1σ3σ4 over the outgoing-electron spin orientations σ3 and σ4 still exhibit, for primary spin σ1 = + , the large exchange-correlation hole and, for σ1 = −, the smaller correlation hole. Exchange and correlation effects can thus be disentangled in experiments without resolving the spins of the emitted electrons.

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In the right-hand part of figure 5, we show analogous momentum distributions obtained for the valence electron surface state with E₂ = E_F − 0.10 eV and k₂ = (0.65,0) a⁻¹₀ (green square in the left-hand part of figure 4), which is selected using primary electrons with E₁ = 25 eV incident at ϑ₁ = 28.61° and φ₁ = 180°. Again, the parallel-spin distributions exhibit an exchange-correlation hole which is larger than the correlation hole in the antiparallel-spin distributions.

A striking difference between the (e,2e) intensities from the DSS and those from the other spin-polarized surface state is that the former are strongest around the k_x-axis, whereas the latter are very small along this axis. This is explained by the different spatial symmetry of the two states (see figure 4). For the outgoing electrons in the xz plane, the above-mentioned selection rule favors states with predominant even parity (like the DSS). For emission in planes other than xz, odd-parity parts of valence states also contribute to the intensities.