Tunneling ionization of the ${}^{4}F$ and ${}^{6}D$ states of vanadium: Exchange blockade

Using time-dependent density functional theory (TDDFT) calculations, we compare tunneling ionization of the $a{}^{4}F$ ground state and the $a{}^{6}D$ first excited state of vanadium in laser fields of intensities between 1.4 and $4.0\times {10}^{13}\mathrm{W}{\mathrm{cm}}^{-2}$. The calculated ionization yields of the ground state of vanadium were already shown to agree well with experimental results [Chu and Groenenboom, Phys. Rev. A 94, 053417 (2016)]. We find that the tunneling ionization rate of the sextet state is lower than that of the quartet state. This is surprising, since the ionization potential of the sextet is lower than that of the quartet state. This finding, however, is consistent with the experimental observation that niobium, whose ground state is $a{}^6{D}_{1/2}$, has a much smaller ionization yield than vanadium ($a{}^4{F}_{3/2}$), even though their ionization potentials are extremely close [Smits et al., Phys. Rev. Lett. 93, 213003 (2004)]. Our calculations demonstrate the existence of exchange blockade for the higher spin state. It arises from a strong field dynamic effect that mixes the highest and second highest electrons in the same set of unoccupied spin orbitals, which causes an isotropic attractive potential that confines the electrons close to the core.

Figure 1

Orbital energy level diagrams for the vanadium $\left(3d^{3}4s^2\right){{}^{4}F}_{9/2}$ and $\left(3d^{4}4s\right){{}^{6}D}_{9/2}$ states.

Figure 2

Calculated ionization yield of the ${{}^{4}F}_{9/2}$ quartet and ${{}^{6}D}_{9/2}$ sextet states of vanadium in 1500-nm 100-fs laser pulses, when induced potentials are not included [see Eq. (3)].

Figure 3

Comparison of the Kohn-Sham potentials (black), the highest spin orbital (blue) and its energy (red) of the ${{}^{4}F}_{9/2}$ and ${{}^{6}D}_{9/2}$ states. The solid lines refer to ${}^{4}F_{9/2}$ and the dotted lines to ${}^{6}D_{9/2}$. The wave functions are shifted vertically such that their zeros occur at the intersections with the corresponding energy levels.

Figure 4

The ionization yield of the ${{}^{4}F}_{9/2}$ and ${{}^{6}D}_{9/2}$ states of vanadium in 1500-nm lasers as a function of the laser intensity, calculated with the all electron TDDFT method [see Eq. (9)].

Figure 5

The $\alpha$ and $\beta$ spin static KS potentials $v_{\sigma }(r,0)$, the highest $\left(4s_{\downarrow }\right)$ and second highest $\left(4s_{\uparrow }\right)$ occupied orbital spin orbitals of the ${}^{4}F_{9/2}$ state and their energies. The wave functions are shifted vertically such that their zeros occur at the intersections with the corresponding energy levels.

Figure 6

The $\alpha$-electron KS potential for the ${{}^{6}D}_{9/2}$ state (dashed line), the highest $\left(4s_{\uparrow }\right)$ (solid line) and second highest $\left(3d_{\uparrow }\right)$ (dotted line) occupied orbitals and their energy levels. The wave functions are shifted vertically such that their zeros occur at the intersections with the corresponding energy levels.

Figure 7

Induced potential, $\mathrm{\Delta }{v}_{\beta }^{\left(l\right)}(r,t_m)P_l(cos\theta )$ for $l=0$ (solid) and $l=1$ (dotted) of the ${{}^{4}F}_{9/2}$ quartet state (black) and $\mathrm{\Delta }{v}_{\alpha }^{\left(l\right)}(r,t_m)P_l(cos\theta )$ for $l=0$ (solid), and for $l=1$ (dashed) of the ${{}^{6}D}_{9/2}$ sextet state (red), at the peak intensity of a $1.54\times {10}^{13}\mathrm{W}{\mathrm{cm}}^{-2}$, 1500-nm laser. $t_m=T/2+3\pi /2\omega$.

Figure 8

Energy diagram for vanadium in a 1500-nm laser of $1.54\times {10}^{13}\mathrm{W}{\mathrm{cm}}^{-2}$. Solid black line: the electronic potential at the peak intensity according to our TDDFT formalism for the quartet ${}^{4}F_{9/2}$. Dotted black line: the same potential for the sextet ${}^{6}D_{9/2}$. Solid red line: the shifted energy level for the quartet ${}^{4}F_{9/2}$. Dotted red line: the shifted energy level for the quartet ${}^{4}F_{9/2}$.