**2. Theory**

#### *2.1. Theoretical Description of Spin Polarization in Two-Color Multiphoton Ionization*

Consider the photoionization of an atom by two spatially and temporally overlapping electromagnetic pulses of XUV and IR radiations. Both pulses are circularly polarized and collinear, propagating along the *z*-axis. To describe the interaction of these pulses with the atom, we use the strong field approximation (SFA) [24,25]. We suppose that the XUV pulse is comparatively weak so that its interaction with electrons can be considered in the first-order perturbation theory, and we use the rotating wave approximation. The

IR pulse is rather strong 1011–1013 W/cm2, but not strong enough to distort the bound atomic states. For a description of the atomic wave function, we use a single-active-electron approximation. The final continuum states of the emitted electron are described by the nonrelativistic Volkov wave functions [26]. Note that we ignore the influence of the magnetic field of the IR pulse on the spin orientation of the emitted electron. As it was shown in paper [27], this influence is generally rather small. In this case, the amplitude of the photoionization can be written using the time-dependent distorted wave approximation as follows (we use atomic units throughout the paper unless otherwise indicated) [28]

$$\mathcal{A}\_{\vec{k},m\_s,m\_\vec{\boldsymbol{\eta}}} = -i \int\_{-\infty}^{\infty} dt \,\,\vec{\mathcal{E}}\_X(t) \langle \psi\_{\vec{k}}(\vec{r},t) \chi\_{m\_s} \,|\,\hat{d} \,|\,\phi\_{\vec{j},m\_\vec{\boldsymbol{\eta}}}(\vec{r}) \rangle e^{i(E\_b - \omega\_X)t} \,. \tag{1}$$

Here, E ˜ *X*(*t*) is the envelope of the XUV pulse electric field, *ωX* is its mean frequency, and *Eb* is the ionization potential. The matrix element *ψk*(*r*, *<sup>t</sup>*)*<sup>χ</sup>ms* | ˆ*d* | *φj*,*mj*(*r*)- describes a transition from the initial state of the atomic electron *φj*,*mj*(*r*) with the total angular momentum and its projection *j*, *mj*, to the final continuum state in the IR field *ψk*(*r*, *<sup>t</sup>*)*<sup>χ</sup>ms* with the momentum *k* and spin state *χms*, with *ms* being the projection of the spin on the *z* axis, and ˆ *d* being the dipole operator. Note that we ignore the spin-orbit interaction in the continuum.

For a circularly polarized XUV beam with the polarization vector ±*X* , the dipole operator is given by

$$\hat{d}^{\pm} = \left(\tilde{\epsilon}\_X^{\pm} \vec{r}\right) = -\sqrt{4\pi/3} \, r Y\_{1,\pm1}(\vec{r}) \,, \tag{2}$$

where plus and minus signs correspond to right- and left-circularly polarized XUV photons, respectively, and *Yl*,*<sup>m</sup>* is a spherical harmonic. The wave function *ψk*(*r*, *t*) in Equation (1) describes the "dressed" photoelectron in the laser field, which is characterized by the final (asymptotic) momentum *k*. Within the SFA, the wave function of the photoelectron is represented by the non-relativistic Volkov wave function [26]:

$$\psi\_{\vec{k}}(\vec{r},t) = \exp\left\{i[\vec{k} - \vec{A}\_L(t)]\vec{r} - i\Phi(\vec{k},t)\right\}.\tag{3}$$

Here,

˜

$$\Phi(\vec{k},t) = -\frac{1}{2} \int\_t^\infty dt' \left[\vec{k} - \vec{A}\_L(t')\right]^2\tag{4}$$

with *A L*(*t*) being the vector potential of the laser field, which we define as *A L*(*t*) = ∞*t dt*E*L*(*t*) (E*L*(*t*) is the IR laser electric field vector). For circularly polarized IR laser light, E *L*(*t*) is

$$\mathcal{E}\_L(t) = \frac{1}{\sqrt{2}} \mathcal{E}\_L(t) \left[ \mathfrak{k} \cos \omega\_L t \pm \mathcal{G} \sin \omega\_L t \right],\tag{5}$$

where E *L*(*t*) is the envelope of the laser pulse, *ωL* is its mean frequency, *x*<sup>ˆ</sup>(*y*<sup>ˆ</sup>) is a unit vector along the *x* (*y*) axis, and the plus (minus) sign corresponds to the right- (left-) circularly polarized IR light.

Consider the matrix element in Equation (1) and uncouple the spin angular momentum in the initial state wave function using Clebsch–Gordan coefficients:

$$<\langle \psi\_{\overline{k}}(\vec{r},t)\chi\_{m\_s} \mid \hat{d} \mid \phi\_{\overline{j},m\_{\overline{j}}}(\vec{r})\rangle = \langle \psi\_{\overline{k}}(\vec{r},t)\chi\_{m\_s} \mid \hat{d} \mid \sum\_{m\_0,m'\_s} \left( l\_0 m\_{0\prime} \not\ge m'\_{\overline{k}} \mid jm\_{\overline{j}} \right) \phi\_{l\_0,m\_0}(\vec{r})\chi\_{m'\_{\overline{i}}} \rangle \,, \tag{6}$$

Here *φ*0,*m*0 (*r*) describes the initial state of the electron with the orbital angular momentum 0 and its projection *m*0. Taking into account that the dipole operator does not act on the spin variables, the matrix element is reduced to

$$
\langle \psi\_{\vec{k}}(\vec{r},t)\chi\_{m\_s} \mid \vec{d} \mid \phi\_{j,m\_j}(\vec{r}) \rangle = \langle \psi\_{\vec{k}}(\vec{r},t) \mid \vec{d} \mid \phi\_{l\_0,m\_0}(\vec{r}) \rangle \left(l\_0 m\_0 \, \frac{1}{2} m\_s \mid j m\_j \right) \,. \tag{7}
$$

Substituting this expression into the amplitude (1), one obtains

$$\mathcal{A}\_{\overleftarrow{k},m\_{\sf s},m\_{\sf j}} = \left(l\_0 m\_{0\sf s}, \frac{1}{2} m\_{\sf s} \mid j m\_{\sf j}\right) \delta\_{m\_0, m\_{\sf j} - m\_{\sf s}} \mathcal{M}\_{\overleftarrow{k},m\_{\sf j}} \,\tag{8}$$

where *<sup>δ</sup><sup>m</sup>*,*m* is the Kronecker symbol and

$$\mathcal{M}\_{\vec{k},m\_0} = -i \int\_{-\infty}^{\infty} dt \,\,\vec{\mathcal{E}}\_X(t) \langle \psi\_{\vec{k}}(\vec{r},t) \mid \hat{d} \mid \phi\_{l\text{-}m\_0}(\vec{r}) \rangle e^{i(E\_b - \omega\_X)t} \,. \tag{9}$$

The probability for the photoelectron to have a certain projection *ms* is proportional to the square of the amplitude (8) averaged over projections *mj*:

$$\mathcal{W}\_{m\_s}^{\vec{j}}(\vec{k}) = \frac{1}{2j+1} \sum\_{m\_{\vec{j}}} |\left(l\_0 m\_{0\prime} \,\_2^1 m\_s \, | \, j m\_{\vec{j}}\right) \delta\_{m\_0 m\_{\vec{j}} - m\_s} \mathcal{M}\_{\vec{k}, m\_0}|^2 \,. \tag{10}$$

In the following, we consider the *z*-component of the polarization vector, *Pjz* which is parallel to the photon beam direction. This is the largest component in the considered energy range [3]. Additionally, this is the only non-zero component which characterizes the angle-integrated spin polarization of the total photoelectron flux. The degree of spin polarization (*z*-componen<sup>t</sup> of the polarization vector) is usually defined as the ratio

$$P\_z^{\vec{j}}(\vec{k}) = \frac{\mathcal{W}\_{1/2}^{\vec{j}}(\vec{k}) - \mathcal{W}\_{-1/2}^{\vec{j}}(\vec{k})}{\mathcal{W}\_{1/2}^{\vec{j}}(\vec{k}) + \mathcal{W}\_{-1/2}^{\vec{j}}(\vec{k})}. \tag{11}$$

In particular, for the photoionization of p atomic shell (*l*0 = 1), *j* = 1/2, *mj* = ±1/2 and *j* = 3/2, *mj* = ±1/2, ±3/2. Then for *j* = 1/2 one has

$$\mathcal{W}\_{1/2}^{1/2}(\vec{k}) = \frac{1}{2} \left\{ |\frac{1}{\sqrt{3}} \mathcal{M}\_{\vec{k},0}|^2 + |\frac{\sqrt{2}}{\sqrt{3}} \mathcal{M}\_{\vec{k},-1}|^2 \right\},\tag{12}$$

$$\mathcal{W}^{1/2}\_{-1/2}(\vec{k}) = \frac{1}{2} \left\{ |\frac{\sqrt{2}}{\sqrt{3}} \mathcal{M}\_{\vec{k},1}|^2 + |\frac{1}{\sqrt{3}} \mathcal{M}\_{\vec{k},0}|^2 \right\}.\tag{13}$$

For *j* = 3/2, one has

$$\mathcal{W}\_{1/2}^{3/2}(\vec{k}) = \frac{1}{4} \left\{ |\mathcal{M}\_{\vec{k},1}|^2 + |\frac{\sqrt{2}}{\sqrt{3}} \mathcal{M}\_{\vec{k},0}|^2 + |\frac{1}{\sqrt{3}} \mathcal{M}\_{\vec{k},-1}|^2 \right\},\tag{14}$$

$$\mathcal{W}\_{-1/2}^{3/2}(\vec{k}) = \frac{1}{4} \left\{ |\frac{1}{\sqrt{3}} \mathcal{M}\_{\vec{k},1}|^2 + |\frac{\sqrt{2}}{\sqrt{3}} \mathcal{M}\_{\vec{k},0}|^2 + |\mathcal{M}\_{\vec{k},-1}|^2 \right\}. \tag{15}$$

Thus the spin polarization *Pz* for the case of *j* = 1/2 is

$$P\_z^{1/2}(\vec{k}) = \frac{|\mathcal{M}\_{\vec{k},-1}|^2 - |\mathcal{M}\_{\vec{k},1}|^2}{|\mathcal{M}\_{\vec{k},0}|^2 + |\mathcal{M}\_{\vec{k},1}|^2 + |\mathcal{M}\_{\vec{k},-1}|^2} \,. \tag{16}$$

while for *j* = 3/2 the polarization is

$$P\_z^{3/2}(\vec{k}) = \frac{1}{2} \frac{|\mathcal{M}\_{\vec{k},1}|^2 - |\mathcal{M}\_{\vec{k},-1}|^2}{|\mathcal{M}\_{\vec{k},0}|^2 + |\mathcal{M}\_{\vec{k},1}|^2 + |\mathcal{M}\_{\vec{k},-1}|^2} \,. \tag{17}$$

Note that in the considered model, *P*3/2 *z* (*k*) = −12*P*1/2 *z* (*k*), as it is in the single photon ionization [4].

For calculating the amplitudes <sup>M</sup>*k*,*m*<sup>0</sup> we expand the continuum wave function *ψk* in partial waves and apply the dipole selection rules. Then the matrix element of the dipole

operator for circularly polarized light, Equation (2), and for a particular projection *m*0 can be written as

$$d\_{\vec{k}\_0, m\_0}^{\pm} = d\_{\ell\_0 + 1, m\_0 \pm 1} \Upsilon\_{\ell\_0 + 1, m\_0 \pm 1}(\theta\_{0\prime} \,\phi\_0) e^{i\delta\_{\ell\_0} + 1}$$

$$+ d\_{\ell\_0 - 1, m\_0 \pm 1} \Upsilon\_{\ell\_0 - 1, m\_0 \pm 1}(\theta\_{0\prime} \,\phi\_0) e^{i\delta\_{\ell\_0 - 1}} \,. \tag{18}$$

Here *<sup>d</sup>*0±1,*<sup>m</sup>*0±<sup>1</sup> are the partial dipole amplitudes for the transitions from the initial state (0, *m*0), and *<sup>δ</sup>*0±<sup>1</sup> are the photoionization phases. The angles (*<sup>θ</sup>*0, *φ*0) give the direction of electron emission from the atom before propagation in the optical laser field. These angles are connected with the detection angles (*θ*, *φ*) after propagation in the IR field by the relations:

*<sup>θ</sup>*0(*t*) = arccos(*kz*/*k*0(*t*)), (19) exp(*<sup>i</sup>φ*0(*t*)) = (*kx* − *ALx*(*t*)) + *i*(*ky* − *ALy*(*t*)) (*k*20(*t*) − *k*2*z* )1/2 ,

where *<sup>k</sup>*20(*t*)=(*<sup>k</sup>* − *A L*(*t*))2.

In the particular case of p-subshell ionization (0 = 1, *m*0 = 0, ±1) in the absence of the IR field, s and d partial waves contribute. Then, collecting Equations (9), (18) and (19), one can obtain for a circularly polarized XUV pulse and a circularly polarized IR pulse the following expression

$$\mathcal{M}\_{\vec{k},m\_0} = -i \int\_{-\infty}^{\infty} dt \tilde{\mathcal{E}}\_X(t) \left[ d\_{2,m\_0 \pm 1} \mathcal{Y}\_{2,my \pm 1}(\theta\_0(t), \phi\_0(t)) e^{i\delta\_d} \right. \tag{20}$$

$$+ d\_{0,0} \delta\_{m\_0 \pm 1,0} \mathcal{Y}\_{0,0} e^{i\delta\_s} \left[ \right. \left. e^{i\Phi(\vec{k},t)} e^{i(E\_b - \omega\_\chi)t} \right. \tag{20}$$

where *δ<sup>m</sup>*,<sup>0</sup> is a Kronecker symbol.

Suppose that the XUV pulse is right-circularly polarized (upper sign in Equation (18)), then

$$\mathcal{M}\_{\vec{k},0} = -i \int\_{-\infty}^{\infty} dt \mathcal{E}\_X(t) d\_{2,1} \mathcal{Y}\_{2,1}(\theta\_0(t), \phi\_0(t)) e^{i\delta\_d} e^{i\Phi(\vec{k},t)} e^{i(\mathbb{E}\_b - \omega \chi)t},\tag{21}$$

$$\mathcal{M}\_{\vec{k},1} = -i \int\_{-\infty}^{\infty} dt \mathcal{E}\_X(t) d\_{2,2} \, \mathcal{Y}\_{2,2}(\theta\_0(t), \phi\_0(t)) \epsilon^{i\delta\_d} e^{i\Phi(\vec{k},t)} \epsilon^{i(E\_b - \omega \chi)t},\tag{22}$$

$$\mathcal{M}\_{\vec{k},-1} = -i \int\_{-\infty}^{\infty} dt \mathcal{E}\_X(t) \left[ d\_{2,0} \mathcal{Y}\_{2,0}(\theta\_0(t), \phi\_0(t)) e^{i\delta\_d} + d\_{0,0} \mathcal{Y}\_{0,0} e^{i\delta\_s} \right] e^{i\Phi(\vec{k},t)} e^{i(E\_b - \omega \chi)t} \,. \tag{23}$$

#### *2.2. Choice of Parameters and Details of Calculations*

In particular calculations for Xe atoms, the matrix elements *d*2,2, *d*2,1 and *d*2,0 and phases *δd* and *δs* were calculated using the Herman–Skillman potential [29] in the nonrelativistic single electron model. To test the results, we calculated the spin polarization of photoelectrons by expression (16) but for a negligibly small IR field. The results agree well with more advanced calculations by Cherepkov [3] within the time-independent RPAE.

The XUV pulse was assumed to be of a Gaussian shape:

$$\bar{\mathcal{E}}\_X(t) = \exp[-(t - t\_0)^2 / (2\tau\_X^2) \, , \tag{24}$$

where *t*0 determines the delay of the XUV pulse relative to the optical pulse (OP), and *τX* determines its duration. The duration FWHM was taken to be 23 fs. The duration of the OP is 60 fs, and we suppose that the XUV pulse is at the middle of the optical pulse.

The fine structure spitting of the 5p state in Xe is 1.3 eV. Therefore, in order to avoid overlapping of the two series of sidebands from 5p3/2 and 5p1/2 ionization, the optical photon energy is set equal to 3.1 eV as it was chosen in a recent experiment [15]. At this photon energy, the two combs of sidebands are clearly separated, which makes it possible to determine the spin polarization for each peak [15]. For the illustrations below, we chose the electron emission angle of 90◦, where the number of sidebends is maximal. We also present the angle-integrated polarization.

The calculated cross section and spin polarization were convoluted with a Gaussian function which imitates the energy resolution of about 0.7 eV in a possible experiment.

In the following, we assume that the XUV pulse is right circularly polarized, while the helicity of the optical pulse is changing. Below, we present the calculated spin polarization component *P*1/2 *z* for the photoionization of the Xe(5*p*1/2) state. We remind that the spin polarization for the second component of the spin-orbit doublet Xe(5*p*3/2) can be easily obtained multiplying the calculated results by −1/2. If in an experiment the spin-orbit doublet is not resolved, the spin polarization of the photoelectrons is negligibly small [4].

#### **3. Results for Ionization of Xe and Discussion**

Figure 1a shows the results of the calculations for the Xe(5*p*1/2) ionization at the XUV photon energy of 38.4 eV (photoelectron energy without optical field is 25 eV) and emission angle of 90◦. At this energy, the calculated photoelectron spin polarization in the absence of the optical field is *P*1/2 *z* = 0.35. The dressing laser intensity is 5 × 10<sup>11</sup> W/cm<sup>2</sup> and photon energy is 3.1 eV. At this intensity, there are two noticeable sidebands. We remind that the XUV pulse is right-circularly polarized. When the OP is also right-circularly polarized, the electron polarization at the sidebands becomes smaller (green solid line) than at the central line. On the contrary, for the left-circular polarization of optical beam, the spin polarization at the sideband is larger than at the central line. The polarization of the central line is increasing (decreasing) when the OP is right- (left-) circularly polarized in comparison with the case of the negligible optical field. We note that the calculated polarization is practically constant along the spectral line and changes abruptly to another value at the other line.

**Figure 1.** The photoelectron spin polarization component *P*1/2 *z* for the XUV photon energy *h*¯ *ωX* = 38.4 eV and the emission angle of 90◦. Red line: left-circularly polarized dressing optical pulse (OP), green line: right-circularly polarized OP. The thin solid line shows electron spectrum in arbitrary units. The OP intensity is panel (**a**) 0.5 × 10<sup>12</sup> W/cm2, panel (**b**) 5 × 10<sup>12</sup> W/cm2. The dashed straight line shows the spin polarization when the OP is absent.

Figure 1b shows similar values but for larger intensity 5 × 10<sup>12</sup> W/cm2. With the increase in intensity, the number of sidebands is increasing. Qualitatively, the polarization results are similar to the previous case. Note that the shift of polarization for the central line increases. From the other side, the polarization of the sidebands exhibits a weak dependence on the laser field intensity.

The angular distributions of the spin polarization for the same parameters as in Figure 1b are shown in Figure 2 (upper panel) for the central line (CL), first high-energy

sideband (SB1) and first low-energy sideband (SB-1). One can see that in all cases, the spin polarization component *P*1/2 *z* has a minimum at the emission angle of 90◦. The minimal value of polarization is different for left- and right-circularly polarized optical pulses, as it is seen also in Figure 1b. In Figure 2 (lower panel), the angular distribution of the electron yield is shown for the same parameters as in the upper panel. Interestingly, the behavior of the polarization near angles 0◦ and 180◦ is drastically different for the central line, where *P*1/2 *z* → 1, and for the first high-energy sideband at the left-polarized IR pulse and first low-energy sideband at the right-polarized IR pulse, where *P*1/2 *z* → 0.

**Figure 2.** Angular distribution of the spin-polarization component *P*1/2 *z* (**a**) and of the electron yield (**b**) for the XUV photon energy of 38.4 eV (the photoelectron energy without optical field is 25 eV) and the OP intensity 5.0 × 10<sup>12</sup> W/cm2. Solid lines correspond to the left-circularly polarized OP, dashed lines, to the right-circularly polarized. CL denotes central line. SB1—high-energy sideband. SB-1—low-energy sideband.

This behavior may be explained using conservation of the angular momentum projection. Indeed, in our non-relativistic model, the projection of the orbital angular momentum is conserved. For the central line and right-circularly polarized XUV pulse, this gives *Mi* + *me* = +1, where *Mi* and *me* are projections of the ion and the electron orbital angular momenta. If the electron is emitted at zero angle, *me* = 0 and therefore, *Mi* = 1. Then in the final state <sup>2</sup>*P*1/2, the projection of spin *Si* must be *Ms* = −1/2. Since in photoionization the spin projection is also conserved, *Ms* + *se* = 0 and therefore *se* = 1/2, i.e., *P*1/2 *z* = +1. For the first high-energy sideband and left-circularly polarized optical photon, the sum of the projections of XUV and laser photons is zero, and therefore *Mi* + *me* = 0. In the

forward emission, *me* = 0, thus *Mi* = 0 and the spin projection may be *Ms* = ±1/2 with equal probability, which means that *P*1/2 *z* = 0. Similarly one can show that for the first low-energy sideband and the right- circularly polarized OP photon, the polarization is also zero in the forward direction. We note that the behavior of the spin polarization in forward and backward directions, although interesting, is practically not very important since the cross section of the sideband formation at emission angles 0◦ and 180◦ is negligibly small for circularly polarized light (see lower panel of Figure 2).

Figure 3 shows the polarization for the electron energy of 35 eV and OP intensity 1 × 10<sup>12</sup> W/cm2. All other parameters are the same as in Figure 1. For this energy, the electron polarization without laser field is larger [3], *P*1/2 *z* ≈ 0.7. When the optical radiation is switched on, the polarization for sidebands is changing similarly to the previous case.

**Figure 3.** The same as in Figure 1 but for the XUV photon energy 48.4 eV (the photoelectron energy without optical field is 35 eV) and the OP intensity is 1.0 × 10<sup>12</sup> W/cm2. Red line—left-circularly polarized OP, green line—right circularly polarized OP.

The total angle-integrated polarization for sideband electrons behaves qualitatively similarly to that for the emission angle of 90◦. As an example, the angle integrated polarization is shown in Figure 4 for the electron energy of 25 eV and laser intensity of 5.0 × 10<sup>12</sup> W/cm2.

**Figure 4.** Total spin polarization *P*1/2 *z* (solid lines) and spin polarization at 90◦ (dashed lines) at the XUV photon energy of 38.4 eV. Red lines: left-circularly polarized OP, green lines: right-circularly polarized OP. The IR intensity is 5.0 × 10<sup>12</sup> W/cm2. The thin solid line shows the electron spectrum in arbitrary units.

The general behavior of the total (angle integrated) spin polarization is similar to that of the polarization at 90◦ emission angle. However, the total polarization is larger than at 90◦, since the angular distribution of the polarization *P*1/2 *z* has a minimum in the direction perpendicular to the beam.

Qualitatively, the above discussed behavior of the spin polarization of the photoelectrons in the dressing laser field can be explained as follows. If one assumes that both the XUV pulse and the laser pulse are sufficiently long (i.e., contain many oscillations of the electric field), then following the procedure described in paper [30], one can obtain analytical expressions for the matrix elements <sup>M</sup>*k*,*m*. We also make additional approximations. First, we take into account that we consider electron emission at 90◦. Thus we can neglect the contribution of the matrix element <sup>M</sup>*k*,<sup>0</sup> (Equation (21)), containing the spherical function *Y*2,1 which is small around 90◦. In the matrix element <sup>M</sup>*k*,<sup>−</sup><sup>1</sup> (Equation (23)), we neglect the term containing *d*2,0*Y*2,0, since in the considered energy range *d*2,0 << *d*0,0 (i.e., the contribution of the d partial wave is much smaller than the contribution of the s partial wave). Then, using the Jacobi–Anger expansion in terms of Bessel functions, the matrix elements <sup>M</sup>*k*,<sup>−</sup><sup>1</sup> and <sup>M</sup>*k*,+<sup>1</sup> can be presented as a sum of contributions corresponding to different sidebands (for the derivation of these expressions, see the Appendix A)

$$\mathcal{M}\_{\tilde{k},-1} = -i \frac{1}{\sqrt{4\pi t}} \, d\_{00} \sum\_{n=-\infty}^{+\infty} \, \tilde{\mathcal{E}}\_X^{(n)} i^n \, J\_n(q) \, . \tag{25}$$

$$\mathcal{M}\_{\vec{k},1} = -i \frac{\sqrt{15}}{\sqrt{32\pi}} d\_{22} \sum\_{n=-\infty}^{+\infty} \mathcal{E}\_{X}^{(n)} \, \vec{i}^{n} \left[ J\_{n}(q) \pm \frac{2A\_{L}}{k} J\_{n}'(q) \right]. \tag{26}$$

Here *Jn*(*q*) and *Jn*(*q*) are Bessel functions and their derivatives, *q* = *kAL ωL* , where *AL* is the amplitude of the laser vector potential, and

$$\tilde{\mathcal{E}}\_{X}^{(n)} = \int dt \bar{\mathcal{E}}\_{X}(t) \exp\left[i(E\_{b} - \omega\_{X} + k^{2}/2 + n\omega\_{L})t\right].\tag{27}$$

For a long pulse, the right-hand side of the last equation is close to a delta function, which expresses the energy conservation. Each term in the sums (25) and (26) presents the contribution of the *n*-th sideband. If one neglects the interference between different terms and the contribution from the neighboring terms, then the squares of the matrix elements (25) and (26) for the sideband number *n* are

$$|\mathcal{M}\_{\vec{k},-1}(n)|^2 = \frac{|d\_{0,0}|^2}{4\pi} \left[|l\_n(q)|^2\right],\tag{28}$$

$$\left|\mathcal{M}\_{\vec{k},1}(n)\right|^2 = \frac{15|d\_{2,2}|^2}{32\pi} \left[f\_n(q) \pm \frac{2A\_L}{k} l\_n'(q)\right]^2. \tag{29}$$

Ignoring small terms of the order ( *ALk* )2, one can write the spin polarization *<sup>P</sup>jz*(*n*) of the *n*-th sideband:

$$P\_z^{1/2}(n) \approx \frac{1 - \alpha(1 \pm 4\frac{A\_L}{k}\frac{l\_n'(q)}{f\_n(q)})}{1 + \alpha(1 \pm 4\frac{A\_L}{k}\frac{l\_n'(q)}{f\_n(q)})},\tag{30}$$

where *α* = 158 |*d*2,2|2/|*d*0,0|2. Taking into account that *α* << 1, one can finally obtain the following simple approximate equation for the electron spin polarization at the *n*-th sideband:

$$P\_z^{1/2}(n) \approx 1 - 2\alpha \left[ 1 \pm 4 \frac{A\_L}{k} \frac{l\_n'(q)}{f\_n(q)} \right]. \tag{31}$$

Upper (lower) sign in Equation (31) corresponds to the right- (left-) circularly polarized laser pulse. (We remind that the XUV pulse is assumed to be right-circularly polarized.) For sidebands with *n* = 0, and for small *q*, Equation (31) can be rewritten as

$$P\_z^{1/2}(n) \approx 1 - 2\alpha \left[1 \pm 4\frac{|n|\omega\_L}{k^2}\right].\tag{32}$$

It follows that for the right-circularly polarized IR pulse, the electron polarization diminishes with the increase in |*n*|, while for the left polarized pulse, it increases, as we see in Figures 1 and 3. Additionally, the polarization of the sidebands does not depend on the IR laser intensity.

For the central line, the situation is different. Here, *J*0(*q*)/*J*0(*q*) = −*J*1(*q*)/*J*0(*q*) and at small *q*

$$P\_z^{1/2}(n=0) \approx 1 - 2\alpha \left[1 \mp 2\frac{A\_L^2}{\omega\_L}\right].\tag{33}$$

Thus, the electron polarization in the central line is larger for pulses with the same helicity and smaller when the pulses have different helicity, and the difference of polarizations increases with the increase in laser intensity. This is clearly seen in Figures 1 and 3.
