**Appendix A**

Let derive expressions (25) and (26) for the case of a long IR pulse. Suppose that the IR pulse is circularly polarized, then the pulse electric field may be presented as

$$\mathcal{E}\_L(t) = \mathcal{g}(at)\frac{\mathcal{E}\_L}{\sqrt{2}} \left[ \hat{\mathbf{x}}\cos(\omega\_L t) \pm \hat{\mathbf{y}}\sin(\omega\_L t) \right],\tag{A1}$$

where E ¯ *L* is the field amplitude and upper (lower) sign corresponds to right- (left-) circularly polarized IR field. The corresponding vector potential is

$$A\_L(t) = \mathcal{g}(at)A\_L[\mathfrak{X}\sin(\omega\_L t) \mp \mathcal{g}\cos(\omega\_L t)]\tag{A2}$$

with *AL* = −E ¯ *<sup>L</sup>*/√<sup>2</sup>*ωL*. Here, and in Equation (A1), we introduced an auxiliary function *g*(*x*) which is smooth, equal to unity at small *x* and tends to zero limit at large |*x*|. It allows us to calculate the integral (4) when *t* → ±<sup>∞</sup>, and *α* → 0. In the following, we assume

that *k AL* and ignore the quadratic term *A*2*L* in Equation (4). In this approximation, taking into account that *kx* = *k* sin *θ* cos *φ* and *ky* = *k* sin *θ* sin *φ*, the Volkov phase can be presented as

$$\Phi(\vec{k},t) = \frac{k^2}{2}t + \frac{kA\_L}{\omega\_L}\sin\theta\cos(\phi \mp \omega\_L t). \tag{A3}$$

Substituting this expression into Equations (22) and (23) and using the Jacobi–Anger expansion,

$$\exp\left(i\kappa\cos\alpha\right) = \sum\_{n=-\infty}^{+\infty} i^n \exp\left(in\alpha\right) I\_n(\kappa) \,. \tag{A4}$$

where *Jn*(*κ*) is the Bessel function, one obtains for M( *k*, −<sup>1</sup>)

$$\mathcal{M}\_{\tilde{k},-1} = -i \sum\_{n=-\infty}^{+\infty} \int\_{-\infty}^{\infty} dt \,\mathcal{E}\_{\mathcal{X}}(t) d\_{00} \mathbf{y}\_{0,0} i^n \exp(in\mathfrak{p}) \tag{A5}$$

$$\times \exp(\mp in\omega\_L t) f\_n(q) \exp\left[i(E\_b + \frac{k^2}{2} - \omega\_{\mathcal{X}})t\right]$$

$$= -i d\_{00} \frac{1}{\sqrt{4\pi t}} \sum\_{n=-\infty}^{+\infty} i^n \exp(\mp in\mathfrak{p}) \, f\_n(q) \, \mathcal{E}\_{\mathcal{X}}^{(n)},$$

where *q* = *kAL* sin *θ*/*<sup>ω</sup>L* and the following notation is introduced:

$$\mathcal{E}\_X^{(n)} = \int\_{-\infty}^{\infty} dt \mathcal{E}\_X(t) \exp\left[i\left(E\_b - \omega\_X + \frac{k^2}{2} + n\omega\_L\right)t\right].\tag{A6}$$

Similarly for M( *k*, 1) one obtains

$$\mathcal{M}\_{\tilde{k},1} = -i \sum\_{n=-\infty}^{+\infty} \int\_{-\infty}^{\infty} dt \,\mathcal{E}\_X(t) d\_{2,2} \, Y\_{2,2}(\theta\_0, \phi\_0) i^n \exp(in\phi) \tag{A7}$$

$$\times \exp(\mp in\omega\_L t) f\_n(q) \exp\left[i(E\_b + \frac{k^2}{2} - \omega\_X)t\right].$$

The spherical harmonic *<sup>Y</sup>*2,2(*<sup>θ</sup>*0, *φ*0) can be expressed in terms of angles *θ*, *φ* using Equation (19) as follows:

$$Y\_{2,2}(\theta\_0, \phi\_0) \equiv \sqrt{\frac{15}{32\pi}} \sin^2 \theta\_0 \exp(i2\phi\_0)$$

$$\approx \sqrt{\frac{15}{32\pi}} \Big[ \sin^2 \theta \exp(2i\phi) \pm 2i \frac{A\_L}{k} \sin \theta \exp(i\phi) \exp(\pm i\omega\_L t) \tag{A8}$$

$$+ 2 \frac{A\_L}{k} \sin^3 \theta \exp(2i\phi) \sin(\omega\_L t \mp \varphi) \Big].$$

Here, the upper (lower) sign corresponds to the right (left) circular polarization of the IR field, and we kept only linear terms in *AL*/*k*, which is considered to be small *AL*/*k* 1. Substituting this expression into Equation (22), one obtains the matrix element <sup>M</sup>*k*,<sup>1</sup>

$$\mathcal{M}\_{\tilde{k},1} = -i d\_{2,2} \sqrt{\frac{15}{32\pi}} \sum\_{n=-\infty}^{+\infty} i^n \exp(\mp in\phi) I\_n(q) \tag{A9}$$

$$\times \left[ \sin^2 \theta \exp(2i\phi) \dot{\mathcal{E}}\_X^{(n)} \pm i \frac{A\_L}{k} \sin \theta (2 - \sin^2 \theta) \exp(i\phi) \,\, \dot{\mathcal{E}}\_X^{(n \pm 1)} \right.$$

$$\pm i \frac{A\_L}{k} \sin^3 \theta \,\, \exp(3i\phi) \mathcal{E}\_X^{(n \mp 1)} \Big].$$

This expression can be rewritten by rearranging the terms in the sums as

$$\mathcal{M}\_{\vec{k},1} = -i d\_{2,2} \sqrt{\frac{15}{32\pi}} \sum\_{n=-\infty}^{+\infty} \mathcal{E}\_X^{(n)} i^n \exp[i(2\mp n)\phi] \tag{A10}$$

$$\times \left[ \sin^2 \theta f\_n(q) + \frac{A\_L}{k} (2 - \sin^2 \theta) \, J\_{n \mp 1}(q) \right.$$

$$-\frac{A\_L}{k} \sin^3 \theta \, J\_{n \pm 1}(q) \right].$$

From Equations (A5) and (A10), by setting *θ* = 90◦ and *φ* = 0 and using identity *Jn*−<sup>1</sup>(*q*) − *Jn*+<sup>1</sup>(*q*) = <sup>2</sup>*Jn*(*q*), one obtains equations

$$\mathcal{M}\_{\tilde{k},-1} = -i \frac{1}{\sqrt{4\pi}} \, d\_{00} \sum\_{n=-\infty}^{+\infty} \, \tilde{\mathcal{E}}\_X^{(n)} i^n \, f\_n(q) \, . \tag{A11}$$

$$\mathcal{M}\_{\vec{k},1} = -i \frac{\sqrt{15}}{\sqrt{32\pi}} d\_{22} \sum\_{n=-\infty}^{+\infty} \mathcal{E}\_X^{(n)} \, i^n \left[ J\_n(q) \pm \frac{2A\_L}{k} J\_n'(q) \right],\tag{A12}$$

which coincide with Equations (25) and (26).
