Quantum-classical crossover in a nanospin system embedded in a Josephson $\varphi _0$ junction

Abstract

The quantum–classical crossover of the escape rate is studied in a nanomagnetic Josephson $\varphi _0$ junction within the framework of the spin-coherent-state path integral method. The nonlinear perturbation approach is employed to obtain the crossover boundary separating the first- and the second-order crossovers. The region for the first-order crossover is greatly suppressed by the bias current applied to the junction as well as the external magnetic field. These features can be tested using existing experimental techniques.

Appendices

Appendix A: Expressions for the coefficients in the formulation of the problem

In this appendix, we summarize the coefficients for Eqs. (7) and (8) as:

$$\begin{aligned} \Phi _1= & {} \tilde{E}_{\theta \theta } \csc \theta _0,\nonumber \\ \Phi _2= & {} \frac{1 }{2}\left( \tilde{E}_{\theta \theta \theta }- 2 \Phi _1 \cos \theta _0\right) \csc \theta _0, \nonumber \\ \Phi _3= & {} \frac{1}{2} \tilde{E}_{\theta \phi \phi } \csc \theta _0,\nonumber \\ \Phi _4= & {} \frac{1}{6} \left( \tilde{E}_{\theta \theta \theta \theta }- 3 \Phi _2 \cos \theta _0\right) \csc \theta _0+ \frac{\Phi _1}{2} q(\theta _0) ,\nonumber \\ \Phi _5= & {} \frac{1}{2} \left( \tilde{E}_{\theta \theta \phi \phi } - 2 \Phi _3 \cos \theta _0\right) \csc \theta _0\end{aligned}$$

(A1)

$$\begin{aligned} \Theta _1= & {} - \tilde{E}_{\phi \phi } \csc \theta _0, \nonumber \\ \Theta _2= & {} -2 \Phi _3 -\Theta _1 \cot {\theta _0} ,\nonumber \\ \Theta _3= & {} -\frac{1}{6} \tilde{E}_{\phi \phi \phi \phi } \csc \theta _0,\nonumber \\ \Theta _4= & {} -\Phi _5 +(\Theta _2 -\Phi _3) \cot \theta _0+ \frac{\Theta _1}{2} q(\theta _0), \end{aligned}$$

(A2)

where $q(\theta ) = 1- \cot \theta + 2 \cot ^2\theta$, $E_\mathrm{eff} = \tilde{E}$, $\tilde{E}_{\theta \theta }=[\partial ^2 \tilde{E}/\partial \theta ^2]_{\theta =\theta _0,\phi =0}$, $\tilde{E}_{\theta \phi \phi }=[\partial ^3 \tilde{E}/\partial \theta \partial \phi ^2 ]_{\theta =\theta _0,\phi =0}$, and so on.

Let us first discuss the system in which $\delta \theta$ is real and $\delta \phi$ is imaginary to the first order in perturbation theory. Then, we can substitute $\delta \theta \simeq a \theta _1 \cos (\omega \tau )$ and $\delta \phi \simeq i a \phi _1 \sin (\omega \tau )$ into Eqs. (7) and (8) and obtain the relation:

$$\begin{aligned} \frac{\theta _1}{\phi _1} = \frac{\rho \omega }{\Phi _1} = \frac{\Theta _1}{\rho \omega }, \end{aligned}$$

(A3)

which gives the oscillation frequency $\omega =\omega _0=\sqrt{\Phi _1 \Theta _1}/\rho$. That is, the oscillation frequency in this order dose not shift.

To find the change in the oscillation frequency, we need to investigate Eqs. (7) and (8) in terms of order higher than a. A simple analysis shows that $\omega$ starts to change from $\omega _0$ at the order of $a^3$. Hence, plugging $\delta \Omega$, which contains terms up to $O(a^3)$, into Eqs. (7) and (8), we find Eq. (9) for the shift of the oscillation frequency, where:

$$\begin{aligned} t_1= & {} \frac{\Phi _3}{2 \Theta _1} - \frac{\Phi _2}{2 \Phi _1},\nonumber \\ t_2= & {} \frac{\Phi _3}{6\Theta _1}+\frac{\Phi _2}{6\Phi _1}+\frac{\Theta _2}{3\Theta _1},\nonumber \\ f_1= & {} 0\nonumber \\ f_2= & {} \frac{\Phi _1 \Phi _3 }{3 \Theta _1} + \frac{\Phi _1 \Theta _2}{6 \Theta _1} + \frac{\Phi _2}{3}, \end{aligned}$$

(A4)

which are related to the terms of order of $a^2$ as:

$$\begin{aligned} \delta \theta\simeq & {} a \delta \theta _1 + a ^2 \delta \theta _2 + a^3 \delta \theta _3,\nonumber \\ \delta \phi\simeq & {} a \delta \phi _1 + a ^2 \delta \phi _2 + a^3 \delta \phi _3, \end{aligned}$$

(A5)

with

$$\begin{aligned} \delta \theta _1= & {} \theta _1 \cos (\omega \tau ), \nonumber \\ \delta \theta _2= & {} \theta ^{2}_{1} [t_1 + t_2 \cos (2 \omega \tau ) ], \nonumber \\ \delta \phi _1= & {} i \phi _1 \sin (\omega \tau ), \nonumber \\ \delta \phi _2= & {} i \theta ^{2}_{1} [f_1 + f_2 \sin (2 \omega \tau ) ]. \end{aligned}$$

(A6)

Appendix B: Parameters in the absence of the longitudinal magnetic field

The parameters for the crossover in Sect. 3.1 can be obtained using Eqs. (15) and (A1)–(A2):

$$\begin{aligned} \Phi _1= & {} 2 h_y + 2 i_x \cot \theta _0+ 2 \cos (2 \theta _0) \csc \theta _0, \nonumber \\ \Phi _2= & {} \frac{1}{2} \csc ^2 \theta _0\left[ 3 i_x + 4 \cos \theta _0+ i_x \cos (2 \theta _0) + h_y \sin (2 \theta _0) \right] , \nonumber \\ \Phi _3= & {} \cot \theta _0(h_y + 2 k \sin \theta _0), \nonumber \\ \Phi _4= & {} \frac{1}{3} \big [ 2 h_y + 5 (i_x + \cos \theta _0) \cot \theta _0+ 3 h_y \cot ^2 \theta _0\nonumber \\&\quad + 6 (i_x + \cos \theta _0) \cot ^3 \theta _0+ \sin \theta _0\big ], \nonumber \\ \Phi _5= & {} - h_y \csc ^2 \theta _0- 2 k \sin \theta _0, \end{aligned}$$

(B1)

$$\begin{aligned} \Theta _1= & {} -2 \Phi _3 \tan \theta _0, \nonumber \\ \Theta _2= & {} -2 k \cos \theta _0, \nonumber \\ \Theta _3= & {} -\frac{1}{6} \Theta _1, \nonumber \\ \Theta _4= & {} k \sin \theta _0. \end{aligned}$$

(B2)

Inserting them into Eq. (A4), we have the coefficients for $\delta \Omega$ on the order of $a^2$ given by:

$$\begin{aligned} t_1= & {} \frac{ 3 i_x + 4 \cos \theta _0+ i_x \cos ( 2 \theta _0) + h_y \sin (2 \theta _0)}{8 \sin \theta _0\left[ i_x \cos \theta _0+ \cos (2 \theta _0) + h_y \sin \theta _0\right] } \nonumber \\&\quad -\frac{ \cot \theta _0(h_y + 2 k \sin \theta _0)}{4(h_y + k \sin \theta _0) }, \nonumber \\ t_2= & {} - \frac{ 3 i_x + 4 \cos \theta _0+ i_x \cos (2 \theta _0) + h_y \sin (2 \theta _0)}{ 24 \sin \theta _0\left[ i_x \cos \theta _0+ \cos (2 \theta _0) + h_y \sin \theta _0\right] } \nonumber \\&\quad - \frac{ \cot \theta _0(h_y + 2 k \sin \theta _0)}{12(h_y + k \sin \theta _0) } + \frac{ k \cos \theta _0}{3(h_y + k \sin \theta _0) },\nonumber \\ f_2= & {} -\frac{1}{6} \csc ^2 \theta _0\big [ 4 i_x + 5 \cos \theta _0+ 2 i_x \cos (2 \theta _0) \nonumber \\&\quad + \cos (3 \theta _0) + 2 h_y \sin (2 \theta _0) \big ]. \end{aligned}$$

(B3)

Using Eqs. (B1), (B2), and (B3), we have the expression for $g(i_x, h_y, k, \theta _0)$ in Eq. (20) and the oscillation frequency around the top of the energy barrier, Eq. (21), to third order.