How to Cross an Energy Barrier at Zero Kelvin without Tunneling Effect

This Letter deals with the broad class of magnetic systems having a single or collective spin $S$ with an energy barrier, such as rare-earth elements and their compounds, single molecule magnets with uniaxial anisotropy, and more generally any other anisotropic quantum system made of single or multiple objects with discrete energy levels. Till now, the reversal of the magnetization of such systems at zero kelvin required making use of quantum tunneling with a significant transverse field or transverse anisotropy term, at resonance. Here, we show that another very simple method exists. It simply consists in the application of a particular sequence of electromagnetic radiations in the ranges of optical or microwave frequencies, depending on the characteristics of the system (spin and anisotropy values for magnetic systems). This produces oscillations of the Rabi type that pass above the barrier, thus extending these oscillations between the two energy wells with mixtures of all the $2S+1$ states. In addition to its basic character, this approach opens up new directions of research in quantum information with possible breakthroughs in the current use of multiple quantum bits.

Figure 1

Energies of states $E_m$ [Eq. (2)] for $S=10$, $D=0.1$, and $H_z=0$. We plot $E_m-E_{\min },E_{\min }=E(\pm S)=-D{S}^2$, showing the energy barrier between the states $m>0$ and $m<0$.

Figure 2

Energy levels of the spin $S=10$ as a function of positive $H_z$ for $D=0.1$. Some of the energy differences between the states are depicted by arrows.

Figure 3

Time dependence of the reduced spin components $S_x(t)/S$ (below, thin line) and $S_z(t)/S$ (above, bold line) under the ac field of ${\omega }_{S\to S-1}$ only. $h_{\mathrm{ac}}=0.005$. (In the figure ${\omega }_S$ denotes ${\omega }_{S\to S-1}$.) The initial state is $(S_x/S,S_y/S,S_z/S)=(0,0,1)$. As we have $S=10$, this scheme shows that $S_x$ oscillates between $+2$ and $-2$ and $S_z$ between $S=10$ and $S-1=9$.

Figure 4

Time dependence of the reduced spins components $S_x(t)/S$ (below, thin line) and $S_z(t)/S$ (above, bold line) under the ac field with $h=0.005$ and $H_z=0$. (a) ${\omega }_{S\to S-1}$ and ${\omega }_{S-1\to S-2}$ and (b) ${\omega }_{S\to S-1},\dots ,{\omega }_{S-9\to S-10}$. (c) ${\omega }_{S\to S-1},\dots ,{\omega }_{-S+1\to S}$. In all the cases, the initial state is $(S_x,S_y,S_z)=(0,0,1)$.

Figure 5

Time dependence of the reduced spin components $S_x(t)/S$ (below, thin line) and $S_z(t)/S$ (above, bold line) under the ac field of ${\omega }_{S\to S-1},\dots ,{\omega }_{-S+1\to -S}$ for a noninteger spin $S=19/2$. $H_z=0$ and $h_{\mathrm{ac}}=0.005$. (In the figure ${\omega }_S$ represents ${\omega }_{S\to S-1}$, and so on.) The initial state is $(S_x,S_y,S_z)=(0,0,1)$.

Figure 6

Motions of spin $S=10$ for $H_z=0.1$, $h_{\mathrm{ac}}=0.005$, and the driving ac field $f(t)$ given by Eq. (8). The amplitude $h_{\mathrm{ac}}=0.005$. $⟨S_x(t)⟩$ (thin curve) and $⟨S_x(t)⟩$ (bold curve) are normalized by $S$, The normalized spin length (spin fidelity $s_f$) is given by the bold line (above). The total spin $S(S+1)$ normalized by $S^2$ is given by the blue line.