Quantum Computing III Quantum Gates in Trapped Ions

Trapped-ion quantum computing relies on the controlled interaction between an ion’s internal two-level structure and its quantized motion in the trap. By driving the ion with laser fields near the carrier and motional sidebands, we can engineer spin-dependent forces and geometric phase evolutions that form the essential quantum gates.

Monochromatic field

In Schrodinger picture, the Hamiltonian of the atom’s internal state and motional state is

$$H_\text{atom}^\mathrm{(S)} = \frac{1}{2} \hbar \omega_0 \sigma_z = \frac{1}{2} \hbar \omega_0 (\ket{e}\bra{e} - \ket{g}\bra{g}) = \frac{1}{2} \hbar\omega_0 (\sigma^\dagger\sigma - \sigma\sigma^\dagger)$$

$$H_\text{motion}^\mathrm{(S)} = \hbar \nu \left(a^\dagger a+\frac{1}{2} \right)$$

The quantized motional modes can be in any specific direction (i.e., $a_x, a_y, a_z$), since they are three-dimensionally trapped. For example, if the incident beam comes from the $x$ direction, the ions will consequently oscillates in the $x$ direction, experiencing the changing of electromagnetic field with respect to $x$, thus only $a_x^\dagger a_x$ are effective.

The coupling Hamiltonian describing the interaction between the ion and the electromagnetic field is

$$H_\text{int}^\mathrm{(S)}=\frac{1}{2} \hbar \Omega_0 (\ket{e}\bra{g} + \ket{g}\bra{e}) \left[\mathrm{e}^{\mathrm{i}(k \hat{x} - \omega t +\phi)} + \mathrm{e}^{-\mathrm{i}(k \hat{x} - \omega t +\phi)}\right]$$

where $\hat{x}$ is the quantized position of ions in the Schrodinger picture, $\boldsymbol{E} = \boldsymbol{E}_0 \left[\mathrm{e}^{\mathrm{i}(k x - \omega t +\phi)} + \mathrm{e}^{-\mathrm{i}(kx - \omega t +\phi)}\right]$ is a classical electromagnetic field running in the $x$ direction (which implies the field itself is in the $y$ or $z$ direction).

/* Important: The electromagnetic field is not quantized! */

Introducing the Lamb-Dicke parameter $\eta = k\sqrt{\frac{\hbar}{2m\nu}}$, we can rewrite the position operator as

$$k\hat{x} = \eta \,(a+a^\dagger)$$

and the interaction Hamiltonian

$$H_\text{int}^\mathrm{(S)}=\frac{1}{2} \hbar \Omega_0 (\sigma^\dagger +\sigma)\left\{\mathrm{e}^{\mathrm{i}[\eta(a+a^\dagger) - \omega t +\phi]} + \mathrm{e}^{-\mathrm{i}[\eta(a+a^\dagger) - \omega t +\phi]}\right\}$$

Transforming the into the interaction picture, the Hamiltonian is

$$\begin{aligned} H_\text{int}^\mathrm{(I)} &= \mathrm{e}^{\mathrm{i}H_0 t} H_\text{int}^\mathrm{(S)} \mathrm{e}^{-\mathrm{i}H_0 t} \\ &= \frac{1}{2} \hbar \Omega_0 \,\mathrm{e}^{\mathrm{i}H_\text{A} t} (\sigma^\dagger +\sigma)\mathrm{e}^{-\mathrm{i}H_\text{A} t} \otimes \mathrm{e}^{\mathrm{i}H_\text{M} t} \left\{\mathrm{e}^{\mathrm{i}[\eta(a+a^\dagger) - \omega t +\phi]} + \mathrm{e}^{-\mathrm{i}[\eta(a+a^\dagger) - \omega t +\phi]}\right\} \mathrm{e}^{-\mathrm{i}H_\text{M} t} \end{aligned}$$

Taking advantage of Baker-Campbell-Hausdorff formula, the Hamiltonian is

$$H_\text{int}^\mathrm{(I)} = \frac{1}{2} \hbar \Omega_0 (\sigma^\dagger \mathrm{e}^{\mathrm{i} \omega_0 t} + \sigma \mathrm{e}^{-\mathrm{i} \omega_0 t}) \otimes \left\{\mathrm{e}^{\mathrm{i}[\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) - \omega t +\phi]} + \mathrm{e}^{-\mathrm{i}[\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) - \omega t +\phi]}\right\}$$

Due to micromotion, the Rabi frequency must be corrected. The effective Rabi frequency $\Omega$ is

$$\Omega = \frac{\Omega_0}{1+q/2}$$

where $q$ is a parameter in the Mathieu equation

$$\frac{\mathrm{d}^2 x}{\mathrm{d}\xi^2}+[a-2q \cos(2\xi)]x=0$$

Typically, $a \sim 10^{-3}$, $q^2 \sim 0.015$, $\beta = \sqrt{a+q^2/2}\sim 0.09$, $\nu = \beta \omega_\text{RF}$.

Defining the detuning $\delta := \omega - \omega_0$ , the Hamiltonian takes the form under the RWA (Rotating Wave Approximation):

$$H_\text{int}^\mathrm{(I)} = \frac{1}{2} \hbar \Omega \left[\sigma^\dagger \mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{-\mathrm{i} \delta t} \otimes \mathrm{e}^{\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t})} + \sigma \mathrm{e}^{-\mathrm{i}\phi} \mathrm{e}^{\mathrm{i} \delta t} \otimes\mathrm{e}^{-\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t})} \right]$$

$$H_\text{int}^\mathrm{(I)} = \frac{1}{2} \hbar \Omega \sigma^\dagger \mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{-\mathrm{i} \delta t} \exp \left[\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) \right] + \mathrm{h.c.}$$

The expression of the Hamiltonian can be further simplified in the Lamb-Dicke regime $\eta \sqrt{\langle (a+a^\dagger)^2 \rangle} \ll 1$:

$$\exp \left[\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) \right] = 1 + \mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) - \frac{\eta^2}{2} (a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t})^2 +\omicron(\eta^3)$$

$$H_\text{int}^\mathrm{(I)} = \frac{1}{2} \hbar \Omega \sigma^\dagger \mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{-\mathrm{i} \delta t} \left[1+\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) \right] + \mathrm{h.c.}$$

By changing the detuning $\delta$, we can obtain

• $\delta = 0$ , carrier resonance, $H_\text{car}^\mathrm{(I)} = \frac{\hbar \Omega}{2} (\sigma^\dagger \mathrm{e}^{\mathrm{i}\phi} + \sigma \mathrm{e}^{-\mathrm{i}\phi}) = \frac{\hbar \Omega}{2} \sigma^\dagger \mathrm{e}^{\mathrm{i}\phi} + \mathrm{h.c.}$
• $\delta = -\nu$, first red sideband, $H_\text{rsb}^\mathrm{(I)} = \mathrm{i} \eta \frac{\hbar \Omega}{2} (\sigma^\dagger a \mathrm{e}^{\mathrm{i}\phi}- \sigma a^\dagger \mathrm{e}^{-\mathrm{i}\phi}) =\mathrm{i} \eta \frac{\hbar \Omega}{2} \sigma^\dagger a \mathrm{e}^{\mathrm{i}\phi}+ \mathrm{h.c.}$
• $\delta = \nu$, first blue sideband, $H_\text{bsb}^\mathrm{(I)} = \mathrm{i} \eta \frac{\hbar \Omega}{2} (\sigma^\dagger a^\dagger \mathrm{e}^{\mathrm{i}\phi}- \sigma a \mathrm{e}^{-\mathrm{i}\phi}) =\mathrm{i} \eta \frac{\hbar \Omega}{2} \sigma^\dagger a^\dagger \mathrm{e}^{\mathrm{i}\phi}+ \mathrm{h.c.}$

Bichromatic Field

/* We will set the interaction picture as default. The superscript (I) will be omitted from now on. */

On-resonantly driven harmonic oscillator

Two superimposed laser fields at positive and negative detuning $\pm \nu$ from a common center frequency, respectively, produce a bichromatic light field. The Hamiltonian is

$$H_\text{bic} = \mathrm{i} \eta \frac{\hbar \Omega}{2} (\sigma^\dagger a \mathrm{e}^{\mathrm{i}\phi_r}- \sigma a^\dagger \mathrm{e}^{-\mathrm{i}\phi_r} + \sigma^\dagger a^\dagger \mathrm{e}^{\mathrm{i}\phi_b}- \sigma a \mathrm{e}^{-\mathrm{i}\phi_b})$$

Note: We only consider the superposition of the first sideband Hamiltonians in the expression above. In fact, the complete interaction Hamiltonian should be

$$H_\text{int} = H_\text{car} + H_\text{rsb} + H_\text{bsb} = H_\text{car}+H_\text{bic}$$

$$H_\text{int} = \hbar\Omega \cos (\nu t + \phi_-) (\sigma^\dagger \mathrm{e}^{\mathrm{i}\phi_+} + \sigma \mathrm{e}^{-\mathrm{i}\phi_+}) +\mathrm{i} \eta \frac{\hbar \Omega}{2} (\sigma^\dagger a \mathrm{e}^{\mathrm{i}\phi_r}- \sigma a^\dagger \mathrm{e}^{-\mathrm{i}\phi_r} + \sigma^\dagger a^\dagger \mathrm{e}^{\mathrm{i}\phi_b}- \sigma a \mathrm{e}^{-\mathrm{i}\phi_b})$$

Though carrier resonance term $H_\text{car}$ oscillates rapidly with frequency $\nu$ thus obliged to be omitted under RWA, it has larger order of magnitude $\Omega$ compared with sideband terms $H_\text{bic}$ with an order of magnitude $\eta \Omega$. Therefore, the carrier resonance will cause infidelity. We will estimate its effect later. In the following discussion, we still use $H_\text{int} \approx H_\text{bic}$.

Utilizing the relationship between ladder operators and Pauli operators

$$\begin{gathered} \sigma^\dagger = \sigma_x + \mathrm{i} \sigma_y \qquad \sigma = \sigma_x - \mathrm{i} \sigma_y \\ X = \frac{1}{2} (a+a^\dagger) \qquad P = \frac{1}{2\mathrm{i}}(a-a^\dagger) \end{gathered}$$

and defining the following parameters

$$\phi_+ = \frac{\phi_b + \phi_r}{2} \qquad \phi_- = \frac{\phi_b - \phi_r}{2}$$

the Hamiltonian of bichromatic light field could take these alternative forms:

$$H_\text{bic} = - \eta \hbar \Omega \left[(a+a^\dagger) \cos \phi_- -\mathrm{i} (a-a^\dagger)\sin\phi_-\right] (\sigma_x \sin\phi_+ +\sigma_y\cos\phi_+)$$

$$H_\text{bic} = -2\eta\hbar\Omega (X \cos\phi_- + P\sin \phi_-) (\sigma_x \sin\phi_+ +\sigma_y\cos\phi_+)$$

$$H_\text{bic} = -\eta\hbar\Omega \left(a \mathrm{e}^{-\mathrm{i}\phi_-} + a^\dagger \mathrm{e}^{\mathrm{i}\phi_-}\right) \sigma_{\pi/2-\phi_+}$$

where $\sigma_\phi =\sigma_x \cos\phi+\sigma_y \sin\phi$ is the Pauli matrix on the equatorial plane, with eigenvectors $\ket{+}_\phi$ and $\ket{-}_\phi$ pointing at $(\cos\phi,\sin\phi,0)$ and $(-\cos\phi,-\sin\phi,0)$ respectively on the Bloch sphere.

The evolution operator (in the interaction picture) is

$$U^\mathrm{I} = \exp\left(-\frac{\mathrm{i}}{\hbar}H_\text{bic}t\right) = \exp\left[\mathrm{i}\eta\Omega t (a \mathrm{e}^{-\mathrm{i}\phi_-} + a^\dagger \mathrm{e}^{\mathrm{i}\phi_-}) \sigma_{\pi/2-\phi_+}\right] = D(\alpha \sigma_{\pi/2-\phi_+} )$$

where $\alpha = \mathrm{i}\eta\Omega t \,\mathrm{e}^{\mathrm{i}\phi_-}$. /* The displacement operator is $D(\alpha) = \exp(\alpha a^\dagger-\alpha^\star a)$. */

Off-resonantly driven harmonic oscillator

Two superimposed laser fields at positive and negative detuning $\pm (\nu+\Delta)$ from a common center frequency, respectively, produce a bichromatic light field. The Hamiltonian is

$$H_\text{bic} = \mathrm{i} \eta \frac{\hbar \Omega}{2} (\sigma^\dagger a \mathrm{e}^{\mathrm{i} \Delta t +\mathrm{i}\phi_r}- \sigma a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t -\mathrm{i}\phi_r} + \sigma^\dagger a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t+\mathrm{i}\phi_b}- \sigma a \mathrm{e}^{\mathrm{i}\Delta t-\mathrm{i}\phi_b})$$

which is equivalent to a transformation with $\phi_b \to \phi_b - \Delta t$ and $\phi_r \to \phi_r +\Delta t$. And $\phi_+, \phi_-$ are consequently time-dependent

$$\phi_+ = \frac{\phi_b + \phi_r}{2}\to \phi_+ \qquad \phi_- = \frac{\phi_b - \phi_r}{2} \to \phi_- - \Delta t$$

The interaction Hamiltonian takes the form

$$H_\text{bic} = -\eta\hbar\Omega \left(a \mathrm{e}^{\mathrm{i}\Delta t-\mathrm{i}\phi_-} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t+\mathrm{i}\phi_-}\right) \sigma_{\pi/2-\phi_+}$$

where $\sigma_\phi =\sigma_x \cos\phi+\sigma_y \sin\phi$ is the Pauli matrix on the equatorial plane, with eigenvectors $\ket{+}_\phi$ and $\ket{-}_\phi$ pointing at $(\cos\phi,\sin\phi,0)$ and $(-\cos\phi,-\sin\phi,0)$ respectively on the Bloch sphere.

Since the Hamiltonian is time-dependent, we must use the Magnus expansion to calculate the evolution operator

$$U^\mathrm{I} = \exp\left(-\frac{\mathrm{i}}{\hbar} \int_0^t H_\text{int}\,\mathrm{d}t -\frac{1}{2\hbar^2} \int_0^t \mathrm{d}t_1 \int_0^{t_1} \mathrm{d}t_2 \, [H_\text{int}(t_1),H_\text{int}(t_2)] +\cdots\right)$$

Now, we must fix $\phi_+$ to $\pm \pi/2$, which will lead to $\sigma_{\pi/2-\phi_+} = \sigma_x$ . Or, we fix $\phi_+$ to 0, which will lead to $\sigma_{\pi/2-\phi_+} = \sigma_y$ . The reason is that $\sigma_\phi$ does not commute with itself, yielding a mix of $\sigma_x,\sigma_y,\sigma_z$ when calculating commutator $[H_\text{int}(t_1),H_\text{int}(t_2)]$.

We choose $\sigma_x$ here. Then the first-order term and the second-order term are

$$\exp\left(-\frac{\mathrm{i}}{\hbar} \int_0^t H_\text{int}\,\mathrm{d}t\right) = \exp \left( \frac{\eta\Omega}{\Delta} \left[a^\dagger \mathrm{e}^{\mathrm{i}\phi_-} (1-\mathrm{e}^{-\mathrm{i}\Delta t})-a \mathrm{e}^{-\mathrm{i}\phi_-}(1-\mathrm{e}^{\mathrm{i}\Delta t})\right] \sigma_x\right)$$

$$\exp\left(-\frac{1}{2\hbar^2} \int_0^t \mathrm{d}t_1 \int_0^{t_1} \mathrm{d}t_2 \, [H_\text{int}(t_1),H_\text{int}(t_2)]\right) = \exp \left[ -\mathrm{i}\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)\right]$$

Using Baker-Campbell-Hausdorff formula $\mathrm{e}^{A+B} = \mathrm{e}^A \mathrm{e}^B \mathrm{e}^{-[A,B]/2}$, we find the commutator term $\mathrm{e}^{-[A,B]/2}$ is actually zero. The evolution operator is simply the product of the first order term and the second order term.

$$U^\mathrm{I} = \exp \left( \frac{\eta\Omega}{\Delta} \left[a^\dagger \mathrm{e}^{\mathrm{i}\phi_-} (1-\mathrm{e}^{-\mathrm{i}\Delta t})-a \mathrm{e}^{-\mathrm{i}\phi_-}(1-\mathrm{e}^{\mathrm{i}\Delta t})\right] \sigma_x\right) \exp \left[ -\mathrm{i}\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)\right]$$

The evolution operator can be decomposed into a displacement operator and a global phase

$$U^I = D[\alpha(t) \sigma_x] \, \mathrm{e}^{-\mathrm{i}\Phi(t)}$$

$$\alpha(t)= \frac{\eta\Omega}{\Delta} (1-\mathrm{e}^{-\mathrm{i}\Delta t}) \mathrm{e}^{\mathrm{i}\phi_-} \qquad \Phi(t) =\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)$$

• For the displacement parameter, $\eta \Omega / \Delta$ controls the norm and $\mathrm{e}^{\mathrm{i}\phi_-}$ controls the phase. The ion’s motion encloses a cycle in the phase space, with a period $2\pi / \Delta$.
• For the global phase, at time $t = 2\pi N/\Delta$, the accumulated phase is $2\pi N \left(\frac{\eta\Omega}{\Delta}\right)^2$.
• The Pauli operator acting on the internal state space makes the displacement spin-dependent. For $\ket{+}_x$ , the motional state will be displaced by $\alpha(t)$ . While for $\ket{-}_x$ , the motional state will be displaced by $-\alpha(t)$.

Classical Picture: Driven Oscillator

Now we explain that, under the interaction between a bichromatic light field and an ion, the effect on the ion’s motion is equivalent to the forced vibration of a classical harmonic oscillator. Because the direction of the force on the ion depends on its internal state, this effect is also called a spin-dependent force.

First, let us review the model of a driven oscillator. Consider an undamped oscillator driven by a periodic external force $F(t)=F\cos(\omega t+\phi)$. Its equation of motion is

$$m\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + kx=F\cos(\omega t+\phi)$$

Define the natural frequency $\omega_0=\sqrt{k/m}$ and $f=F/m$. Then the equation becomes

$$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + \omega_0^2 x = f\cos(\omega t+\phi)$$

This is an inhomogeneous second-order differential equation. To simplify, assume both initial position and velocity are zero, the solutions are:

• Resonant case $\omega=\omega_0$: the position is $x(t)=\frac{f}{2\omega_0} t \sin\omega_0 t$, and the momentum is $p(t)=\frac{f}{2\omega_0}(\sin\omega_0 t + \omega_0 t \cos\omega_0 t)$. The position and momentum oscillate while growing linearly in time.

• Detuned case $\omega\neq \omega_0$: the position is $x(t)=\frac{f}{\omega_0^2-\omega^2}[\cos\omega t - \cos\omega_0 t]$, and the momentum is $p(t)=\frac{f}{\omega_0^2-\omega^2}[\omega_0\sin\omega_0 t - \omega\sin\omega t]$. The oscillator’s position and momentum only oscillate periodically.

Plotting the trajectories in phase space yields the two distinct behaviors. One spirals outward from the origin, similar to the displacement parameter $\alpha$ behaving like $\mathrm{i}\eta\Omega t$. The other repeatedly loops around the origin, similar to $1-\mathrm{e}^{-\mathrm{i}\Delta t}$.

However, these pictures are not fully clear because they include the oscillator’s dynamical phase, such as the $\sin\omega_0 t$ factor in $t\sin\omega_0 t$. This causes the resonantly driven oscillator to follow a spiral rather than a straight line in phase space. By applying a low-pass filter that removes the intrinsic oscillation at frequency $\omega_0$ - analogous to transforming from the Schrödinger picture to the interaction picture in quantum mechanics - the resulting trajectories exactly match the expressions for $\alpha$.

We may also describe the system using Lagrangian mechanics. The Lagrangian is

$$L = \frac{1}{2} m\dot{x}^2 - \frac{1}{2} k x^2 + xF(t)$$

The Hamiltonian is

$$H = \frac{p^2}{2m} + \frac{1}{2} k x^2 - xF(t)$$

After canonical quantization,

$$H^{(\mathrm{S})} = \frac{\hat{p}^2}{2m} + \frac{1}{2} k \hat{x}^2 - \hat{x}F(t)$$

Substituting $\hat{x}\to\sqrt{\frac{\hbar}{2m \omega_0}} (a+a^\dagger)$ and $\hat{p} \to \sqrt{\frac{\hbar}{2m \omega_0}} \mathrm{i} (a-a^\dagger)$, we obtain

$$H^{(\mathrm{S})} = \hbar \omega_0 \left(a^\dagger a+\frac{1}{2} \right) - \frac{F}{m} \sqrt{\frac{\hbar}{2m \omega_0}} (a+a^\dagger) \cos(\omega t +\phi)$$

Define $\eta=\sqrt{\frac{\hbar}{2m\omega_0}}$ and $\Omega=F/m$. Transforming into the interaction picture and applying the rotating-wave approximation yields

$$H^{(\mathrm{I})} = - \frac{1}{2}\eta\hbar\Omega \left(a \mathrm{e}^{\mathrm{i}\Delta t-\mathrm{i}\phi_-} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t+\mathrm{i}\phi_-}\right)$$

This is exactly the motional part of the effective Hamiltonian used for bichromatic light–ion interaction. We may therefore conclude: A bichromatic light field acting on an ion is equivalent to applying a periodic driving force to its motion, with the driving frequency equals the frequency of the light field. The only difference is that the direction of this force depends on the ion’s spin state. If an ion in state $\ket{\uparrow}$ is displaced in phase space by $\alpha$ under the bichromatic drive, then an ion in state $\ket{\downarrow}$ will be displaced by $-\alpha$. That’s why it’s called spin-dependent force.

Summary

• On-resonant bichromatic field

  $$H_\text{bic} \sim \eta\hbar\Omega (a+a^\dagger) \sigma_\phi$$

  $$U^\mathrm{I} \sim D(\alpha(t)\sigma_\phi), \, |\alpha| \sim\eta\Omega t$$

• Off-resonant bichromatic field

  $$H_\text{bic} \sim \eta\hbar\Omega (a \mathrm{e}^{\mathrm{i}\Delta t} +a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t}) \sigma_\phi$$

  $$U^\mathrm{I} \sim D[\alpha(t)\sigma_\phi] \, \mathrm{e}^{-\mathrm{i}\Phi(t)}$$

  $$|\alpha| \sim \frac{\eta\Omega}{\Delta} (1-\mathrm{e}^{-\mathrm{i}\Delta t}) \qquad \Phi(t) =\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)$$

Second sideband

When considering the second sideband, the Taylor expansion of $\exp \left[\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) \right]$ is

$$\exp \left[\mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) \right]= 1 + \mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) - \frac{\eta^2}{2} (a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t})^2 +o(\eta^3)$$

$$H_\text{int}^\mathrm{(I)} = \frac{1}{2} \hbar \Omega \sigma^\dagger \mathrm{e}^{\mathrm{i}\phi}\mathrm{e}^{-\mathrm{i} \delta t} \left[1-\frac{\eta^2}{2}(2a^\dagger a+1) + \mathrm{i}\eta(a \mathrm{e}^{-\mathrm{i} \nu t}+a^\dagger \mathrm{e}^{\mathrm{i} \nu t}) - \frac{\eta^2}{2} (a^2 \mathrm{e}^{-\mathrm{i} 2\nu t}+a^{\dagger 2} \mathrm{e}^{\mathrm{i} 2\nu t})\right] + \mathrm{h.c.}$$

By changing the detuning $\delta$, we can obtain

• $\delta = -2\nu$, second red sideband, $H_\text{rsb2} = -\frac{1}{4} \eta^2 \hbar \Omega (\sigma^\dagger a^2 \mathrm{e}^{\mathrm{i}\phi} + \sigma a^{\dagger 2} \mathrm{e}^{-\mathrm{i}\phi}) =-\frac{1}{4} \eta^2 \hbar \Omega \sigma^\dagger a^2 \mathrm{e}^{\mathrm{i}\phi} + \mathrm{h.c.}$
• $\delta = 2\nu$, second blue sideband, $H_\text{bsb2} = -\frac{1}{4} \eta^2 \hbar \Omega (\sigma^\dagger a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi} + \sigma a^2 \mathrm{e}^{-\mathrm{i}\phi}) =-\frac{1}{4} \eta^2 \hbar \Omega \sigma^\dagger a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi} + \mathrm{h.c.}$

The bichromatic field constituted by two second sideband light are

$$H_\text{bic2} = -\frac{1}{4} \eta^2 \hbar \Omega (\sigma^\dagger a^2 \mathrm{e}^{\mathrm{i}\phi_r} + \sigma a^{\dagger 2} \mathrm{e}^{-\mathrm{i}\phi_r} + \sigma^\dagger a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi_b} + \sigma a^2 \mathrm{e}^{-\mathrm{i}\phi_b})$$

Still defining $\phi_+ = (\phi_b + \phi_r)/2,\phi_- = (\phi_b - \phi_r)/2$ , then the Hamiltonian takes the form

$$H_\text{bic2} = -\frac{1}{2} \eta^2 \hbar \Omega (a^2 \mathrm{e}^{-\mathrm{i}\phi_-} + a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi_-})(\sigma_x \cos\phi_+ -\sigma_y \sin\phi_+)$$

The evolution operator is thus

$$U^\mathrm{I} = \exp\left(-\frac{\mathrm{i}}{\hbar}H_\text{bic2}t\right) = \exp\left[\frac{\mathrm{i}}{2} \eta^2 \Omega t (a^2 \mathrm{e}^{-\mathrm{i}\phi_-} + a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi_-}) \sigma_{-\phi_+}\right] = S(\xi \sigma_{-\phi_+} )$$

where $S$ is the squeezing operator

$$S(\xi)=\exp \left(\frac{1}{2}{\xi^* a^2 -\frac{1}{2} \xi a^{\dagger 2}}\right) ,\quad \xi = -\frac{\mathrm{i}}{2}\eta^2 \Omega t \mathrm{e}^{\mathrm{i}\phi_-}$$