Quantum Computation IV Single Qubit Gate
fengxiaot Lv4

The passage introduces the fundamental theory of trapped-ion quantum computing.

Monochromatic field

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

Hatom(S)=12ω0σz=12ω0(eegg)=12ω0(σσσσ)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)

Hmotion(S)=ν(aa+12)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., ax,ay,aza_x, a_y, a_z), since they are three-dimensionally trapped. For example, if the incident beam comes from the xx direction, the ions will consequently oscillates in the xx direction, experiencing the changing of electromagnetic field with respect to xx, thus only axaxa_x^\dagger a_x are effective.

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

Hint(S)=12Ω0(eg+ge)[ei(kx^ωt+ϕ)+ei(kx^ωt+ϕ)]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 x^\hat{x} is the quantized position of ions in the Schrodinger picture, E=E0[ei(kxωt+ϕ)+ei(kxωt+ϕ)]\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 xx direction (which implies the field itself is in the yy or zz direction).

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

Introducing the Lamb-Dicke parameter η=k2mν\eta = k\sqrt{\frac{\hbar}{2m\nu}}, we can rewrite the position operator as

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

and the interaction Hamiltonian

Hint(S)=12Ω0(σ+σ){ei[η(a+a)ωt+ϕ]+ei[η(a+a)ωt+ϕ]}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

Hint(I)=eiH0tHint(S)eiH0t=12Ω0eiHAt(σ+σ)eiHAteiHMt{ei[η(a+a)ωt+ϕ]+ei[η(a+a)ωt+ϕ]}eiHMt\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

Hint(I)=12Ω0(σeiω0t+σeiω0t){ei[η(aeiνt+aeiνt)ωt+ϕ]+ei[η(aeiνt+aeiνt)ωt+ϕ]}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 Ω\Omega is

Ω=Ω01+q/2\Omega = \frac{\Omega_0}{1+q/2}

where qq is a parameter in the Mathieu equation

d2xdξ2+[a2qcos(2ξ)]x=0\frac{\mathrm{d}^2 x}{\mathrm{d}\xi^2}+[a-2q \cos(2\xi)]x=0

Typically, a103a \sim 10^{-3}, q20.015q^2 \sim 0.015, β=a+q2/20.09\beta = \sqrt{a+q^2/2}\sim 0.09, ν=βωRF\nu = \beta \omega_\text{RF}.

Defining the detuning δ:=ωω0\delta := \omega - \omega_0 , the Hamiltonian takes the form under the RWA (Rotating Wave Approximation):

Hint(I)=12Ω[σeiϕeiδteiη(aeiνt+aeiνt)+σeiϕeiδteiη(aeiνt+aeiνt)]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]

Hint(I)=12Ωσeiϕeiδtexp[iη(aeiνt+aeiνt)]+h.c.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 η(a+a)21\eta \sqrt{\langle (a+a^\dagger)^2 \rangle} \ll 1:

exp[iη(aeiνt+aeiνt)]=1+iη(aeiνt+aeiνt)η22(aeiνt+aeiνt)2+ο(η3)\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)

Hint(I)=12Ωσeiϕeiδt[1+iη(aeiνt+aeiνt)]+h.c.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 δ\delta, we can obtain

  • δ=0\delta = 0 , carrier resonance, Hcar(I)=Ω2(σeiϕ+σeiϕ)=Ω2σeiϕ+h.c.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.}
  • δ=ν\delta = -\nu, first red sideband, Hrsb(I)=iηΩ2(σaeiϕσaeiϕ)=iηΩ2σaeiϕ+h.c.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.}
  • δ=ν\delta = \nu, first blue sideband, Hbsb(I)=iηΩ2(σaeiϕσaeiϕ)=iηΩ2σaeiϕ+h.c.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. */

Driven Oscillator

To be continued.

On-resonantly driven harmonic oscillator

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

Hbic=iηΩ2(σaeiϕrσaeiϕr+σaeiϕbσaeiϕb)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

Hint=Hcar+Hrsb+Hbsb=Hcar+HbicH_\text{int} = H_\text{car} + H_\text{rsb} + H_\text{bsb} = H_\text{car}+H_\text{bic}

Hint=Ωcos(νt+ϕ)(σeiϕ++σeiϕ+)+iηΩ2(σaeiϕrσaeiϕr+σaeiϕbσaeiϕb)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 HcarH_\text{car} oscillates rapidly with frequency ν\nu thus obliged to be omitted under RWA, it has larger order of magnitude Ω\Omega compared with sideband terms HbicH_\text{bic} with an order of magnitude ηΩ\eta \Omega. Therefore, the carrier resonance will cause infidelity. We will estimate its effect later. In the following discussion, we still use HintHbicH_\text{int} \approx H_\text{bic}.

Utilizing the relationship between ladder operators and Pauli operators

σ=σx+iσyσ=σxiσyX=12(a+a)P=12i(aa)\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

ϕ+=ϕb+ϕr2ϕ=ϕbϕr2\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:

Hbic=ηΩ[(a+a)cosϕi(aa)sinϕ](σxsinϕ++σycosϕ+)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_+)

Hbic=2ηΩ(Xcosϕ+Psinϕ)(σxsinϕ++σycosϕ+)H_\text{bic} = -2\eta\hbar\Omega (X \cos\phi_- + P\sin \phi_-) (\sigma_x \sin\phi_+ +\sigma_y\cos\phi_+)

Hbic=ηΩ(aeiϕ+aeiϕ)σπ/2ϕ+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 σϕ=σxcosϕ+σysinϕ\sigma_\phi =\sigma_x \cos\phi+\sigma_y \sin\phi is the Pauli matrix on the equatorial plane, with eigenvectors +ϕ\ket{+}_\phi and ϕ\ket{-}_\phi pointing at (cosϕ,sinϕ,0)(\cos\phi,\sin\phi,0) and (cosϕ,sinϕ,0)(-\cos\phi,-\sin\phi,0) respectively on the Bloch sphere.

The evolution operator (in the interaction picture) is

UI=exp(iHbict)=exp[iηΩt(aeiϕ+aeiϕ)σπ/2ϕ+]=D(ασπ/2ϕ+)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 α=iηΩteiϕ\alpha = \mathrm{i}\eta\Omega t \,\mathrm{e}^{\mathrm{i}\phi_-}. /* The displacement operator is D(α)=exp(αaαa)D(\alpha) = \exp(\alpha a^\dagger-\alpha^\star a). */

Off-resonantly driven harmonic oscillator

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

Hbic=iηΩ2(σaeiΔt+iϕrσaeiΔtiϕr+σaeiΔt+iϕbσaeiΔtiϕb)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 ϕbϕbΔt\phi_b \to \phi_b - \Delta t and ϕrϕr+Δt\phi_r \to \phi_r +\Delta t. And ϕ+,ϕ\phi_+, \phi_- are consequently time-dependent

ϕ+=ϕb+ϕr2ϕ+ϕ=ϕbϕr2ϕΔt\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

Hbic=ηΩ(aeiΔtiϕ+aeiΔt+iϕ)σπ/2ϕ+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 σϕ=σxcosϕ+σysinϕ\sigma_\phi =\sigma_x \cos\phi+\sigma_y \sin\phi is the Pauli matrix on the equatorial plane, with eigenvectors +ϕ\ket{+}_\phi and ϕ\ket{-}_\phi pointing at (cosϕ,sinϕ,0)(\cos\phi,\sin\phi,0) and (cosϕ,sinϕ,0)(-\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

UI=exp(i0tHintdt1220tdt10t1dt2[Hint(t1),Hint(t2)]+)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 ϕ+\phi_+ to ±π/2\pm \pi/2, which will lead to σπ/2ϕ+=σx\sigma_{\pi/2-\phi_+} = \sigma_x . Or, we fix ϕ+\phi_+ to 0, which will lead to σπ/2ϕ+=σy\sigma_{\pi/2-\phi_+} = \sigma_y . The reason is that σϕ\sigma_\phi does not commute with itself, yielding a mix of σx,σy,σz\sigma_x,\sigma_y,\sigma_z when calculating commutator [Hint(t1),Hint(t2)][H_\text{int}(t_1),H_\text{int}(t_2)].

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

exp(i0tHintdt)=exp(ηΩΔ[aeiϕ(1eiΔt)aeiϕ(1eiΔt)]σx)\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(1220tdt10t1dt2[Hint(t1),Hint(t2)])=exp[i(ηΩΔ)2(ΔtsinΔt)]\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 eA+B=eAeBe[A,B]/2\mathrm{e}^{A+B} = \mathrm{e}^A \mathrm{e}^B \mathrm{e}^{-[A,B]/2}, we find the commutator term e[A,B]/2\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.

UI=exp(ηΩΔ[aeiϕ(1eiΔt)aeiϕ(1eiΔt)]σx)exp[i(ηΩΔ)2(ΔtsinΔt)]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

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

α(t)=ηΩΔ(1eiΔt)eiϕΦ(t)=(ηΩΔ)2(ΔtsinΔt)\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, ηΩ/Δ\eta \Omega / \Delta controls the norm and eiϕ\mathrm{e}^{\mathrm{i}\phi_-} controls the phase. The ion’s motion encloses a cycle in the phase space, with a period 2π/Δ2\pi / \Delta.
  • For the global phase, at time t=2πN/Δt = 2\pi N/\Delta, the accumulated phase is 2πN(ηΩΔ)22\pi N \left(\frac{\eta\Omega}{\Delta}\right)^2.
  • The Pauli operator acting on the internal state space makes the displacement spin-dependent. For +x\ket{+}_x , the motional state will be displaced by α(t)\alpha(t) . While for x\ket{-}_x , the motional state will be displaced by α(t)-\alpha(t) .


  • On-resonant bichromatic field

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

    UID(α(t)σϕ),αηΩtU^\mathrm{I} \sim D(\alpha(t)\sigma_\phi), \, |\alpha| \sim\eta\Omega t

  • Off-resonant bichromatic field

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

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

    αηΩΔ(1eiΔt)Φ(t)=(ηΩΔ)2(ΔtsinΔt)|\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

Old Scheme

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

exp[iη(aeiνt+aeiνt)]=1+iη(aeiνt+aeiνt)η22(aeiνt+aeiνt)2+o(η3)\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)

Hint(I)=12Ωσeiϕeiδt[1η22(2aa+1)+iη(aeiνt+aeiνt)η22(a2ei2νt+a2ei2νt)]+h.c.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 δ\delta, we can obtain

  • δ=2ν\delta = -2\nu, second red sideband, Hrsb2=14η2Ω(σa2eiϕ+σa2eiϕ)=14η2Ωσa2eiϕ+h.c.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.}
  • δ=2ν\delta = 2\nu, second blue sideband, Hbsb2=14η2Ω(σa2eiϕ+σa2eiϕ)=14η2Ωσa2eiϕ+h.c.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

Hbic2=14η2Ω(σa2eiϕr+σa2eiϕr+σa2eiϕb+σa2eiϕb)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 ϕ+=(ϕb+ϕr)/2,ϕ=(ϕbϕr)/2\phi_+ = (\phi_b + \phi_r)/2,\phi_- = (\phi_b - \phi_r)/2 , then the Hamiltonian takes the form

Hbic2=12η2Ω(a2eiϕ+a2eiϕ)(σxcosϕ+σysinϕ+)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

UI=exp(iHbic2t)=exp[i2η2Ωt(a2eiϕ+a2eiϕ)σϕ+]=S(ξσϕ+)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 SS is the squeezing operator

S(ξ)=exp(12ξa212ξa2),ξ=i2η2ΩteiϕS(\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_-}

New Scheme

New scheme include 4 beams of light, 2 pairs of bichromatic field.

  • A bichromatic field with detuning ±(ν+Δ)\pm (\nu+\Delta) generates

    Hbic±(ν+Δ)=ηΩ(aeiΔtiϕ+aeiΔt+iϕ)σπ/2ϕ+H_\text{bic}^{\pm (\nu+\Delta)} = -\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_+}

    Given ϕ=0,ϕ+=±π/2\phi_- = 0, \phi_+ = \pm \pi/2 , the Hamiltonian writes

    Hbic±(ν+Δ)=ηΩ(aeiΔt+aeiΔt)σxH_\text{bic}^{\pm (\nu+\Delta)} = -\eta\hbar\Omega \left(a \mathrm{e}^{\mathrm{i}\Delta t} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t}\right) \sigma_{x}

  • A bichromatic field with detuning ±(νΔ)\pm (\nu-\Delta)

    Hbic±(νΔ)=ηΩ(aeiΔtiϕ+aeiΔt+iϕ)σπ/2ϕ+H_\text{bic}^{\pm (\nu-\Delta)} = -\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_+}

    Given ϕϕeff\phi_- \to \phi_\text{eff}, ϕ+=0\phi_+ = 0 , the Hamiltonian writes

    Hbic±(νΔ)=ηΩ(aeiΔtiϕeff+aeiΔt+iϕeff)σyH_\text{bic}^{\pm (\nu-\Delta)} = -\eta\hbar\Omega \left(a \mathrm{e}^{-\mathrm{i}\Delta t-\mathrm{i}\phi_\text{eff}} + a^\dagger \mathrm{e}^{\mathrm{i}\Delta t+\mathrm{i}\phi_\text{eff}}\right) \sigma_y

The total interaction Hamiltonian is

Hint=ηΩ[(aeiΔt+aeiΔt)σx+(aeiΔtiϕeff+aeiΔt+iϕeff)σy]H_\text{int} = -\eta\hbar\Omega \left[ (a \mathrm{e}^{\mathrm{i}\Delta t} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t}) \sigma_{x} + (a \mathrm{e}^{-\mathrm{i}\Delta t-\mathrm{i}\phi_\text{eff}} + a^\dagger \mathrm{e}^{\mathrm{i}\Delta t+\mathrm{i}\phi_\text{eff}}) \sigma_y\right]

The first order term still behaves like a displacement operator

exp(i0tHintdt)D[α1(t)σx+α2(t)σy]\exp\left(-\frac{\mathrm{i}}{\hbar} \int_0^t H_\text{int}\,\mathrm{d}t\right) \sim D[\alpha_1(t)\sigma_x+ \alpha_2(t)\sigma_y]

where α1ηΩΔ(1eiΔt),α2ηΩΔ(1eiΔt)|\alpha_1| \sim \frac{\eta\Omega}{\Delta} (1-\mathrm{e}^{-\mathrm{i}\Delta t}),\, |\alpha_2| \sim \frac{\eta\Omega}{\Delta} (1-\mathrm{e}^{\mathrm{i}\Delta t}) . Therefore, to eliminate the first order term, we set the final time to t=2πN/Δt = 2\pi N / \Delta.

The second order term, however, differs a lot. When calculating the commutator [Hint(t1),Hint(t2)][H_\text{int}(t_1),H_\text{int}(t_2)], only the following terms are non-zero

  1. [(aeiΔt1+aeiΔt1)σx,(aeiΔt2+aeiΔt2)σx][(a \mathrm{e}^{\mathrm{i}\Delta t_1} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t_1}) \sigma_{x}, (a \mathrm{e}^{\mathrm{i}\Delta t_2} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t_2}) \sigma_{x}] , generating a geometric phase Φ1(t)=(ηΩΔ)2(ΔtsinΔt)\Phi_1(t) =\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)
  2. [(aeiΔt1+aeiΔt1)σy,(aeiΔt2+aeiΔt2)σy][(a \mathrm{e}^{\mathrm{i}\Delta t_1} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t_1}) \sigma_{y}, (a \mathrm{e}^{\mathrm{i}\Delta t_2} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t_2}) \sigma_{y}] , generating the same geometric phase Φ2(t)=(ηΩΔ)2(ΔtsinΔt)\Phi_2(t) =\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)
  3. Cross-terms of [σx,σy][\sigma_x,\sigma_y], generating the squeezing operator exp[iη2Ω2ΔtsinΔtΔ2(a2eiϕeff+a2eiϕeff)σz]\exp [\mathrm{i} \eta^2 \Omega^2 \frac{\Delta t - \sin \Delta t}{\Delta^2} (a^2 \mathrm{e}^{-\mathrm{i}\phi_\text{eff}} + a^{\dagger 2} \mathrm{e}^{\mathrm{i}\phi_\text{eff}}) \sigma_z]

so the second-order term is

exp(1220tdt10t1dt2[Hint(t1),Hint(t2)])S[ξ(t)σz]eiΦ(t)\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) \sim S[\xi(t) \sigma_z]\mathrm{e}^{-\mathrm{i}\Phi(t)}

where ξ(ηΩΔ)2(ΔtsinΔt)|\xi| \sim \left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t) . At t=2πN/Δt = 2\pi N / \Delta we have ξη2Ω2Δt|\xi| \sim \frac{\eta^2\Omega^2}{\Delta}t .

Using Baker-Campbell-Hausdorff formula eA+B=eAeBe[A,B]/2\mathrm{e}^{A+B} = \mathrm{e}^A \mathrm{e}^B \mathrm{e}^{-[A,B]/2}, we find the commutator term e[A,B]/2\mathrm{e}^{-[A,B]/2} is actually zero at t=2πN/Δt = 2\pi N / \Delta. However, it is worthwhile to point out that the order of the magnitude of [A,B][A,B] is ηΩ\eta \Omega.


  • Old scheme

    Hintη2Ω(a2+a2)H_\text{int} \sim \eta^2 \hbar\Omega(a^2 +a^{\dagger2})

    UIS[ξ(t)σϕ],ξη2ΩtU^{\mathrm{I}} \sim S[\xi(t) \sigma_\phi],\quad |\xi| \sim \eta^2 \Omega t

  • New scheme

    HintηΩ[(aeiΔt+aeiΔt)σx+(aeiΔt+aeiΔt)σy]H_\text{int} \sim \eta \hbar\Omega \left[ (a \mathrm{e}^{\mathrm{i}\Delta t} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t}) \sigma_{x} + (a \mathrm{e}^{-\mathrm{i}\Delta t} + a^\dagger \mathrm{e}^{\mathrm{i}\Delta t}) \sigma_y\right]

    UID[α1(t)σx+α2(t)σy]S[ξ(t)σz]eiΦ(t)U^{\mathrm{I}} \sim D[\alpha_1(t)\sigma_x+ \alpha_2(t)\sigma_y] S[\xi(t) \sigma_z]\mathrm{e}^{-\mathrm{i}\Phi(t)}

    αηΩΔ(1eiΔt)ξ(ηΩΔ)2(ΔtsinΔt)Φ(t)(ηΩΔ)2(ΔtsinΔt)|\alpha| \sim \frac{\eta\Omega}{\Delta} (1-\mathrm{e}^{-\mathrm{i}\Delta t}) \qquad |\xi| \sim \left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t) \qquad \Phi(t) \sim\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)


Question 1: Could we achieve the squeezing operation with two on-resonant first sideband beams and take ϕ=π/4\phi = \pi /4 ?

HηΩ(a+a)(σx+σy)H \sim\eta\hbar\Omega (a+a^\dagger) (\sigma_x + \sigma_y)

Answer: No. On-resonant fields are time-dependent. [H(t1),H(t2)][H(t_1),H(t_2)] will not appear. The evolution operator is simply UI=exp(iHt)U^\mathrm{I} = \exp(-\frac{\mathrm{i}}{\hbar}Ht).

Question 2: Could we achieve the squeezing operation with two off-resonant first sideband beams and take ϕ=π/4\phi = \pi/4?

Answer: No. Taking a bichromatic field with detuning ±(ν+Δ)\pm (\nu+\Delta) as example

HηΩ(aeiΔt+aeiΔt)(σx+σy)H \sim \eta\hbar\Omega \left(a \mathrm{e}^{\mathrm{i}\Delta t} + a^\dagger \mathrm{e}^{-\mathrm{i}\Delta t}\right)(\sigma_x+\sigma_y)

The first-order term is

exp(i0tHintdt)=exp(ηΩΔ[a(1eiΔt)a(1eiΔt)][σx+σy])D[α(t)(σx+σy)]\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 (1-\mathrm{e}^{-\mathrm{i}\Delta t})-a (1-\mathrm{e}^{\mathrm{i}\Delta t})\right] [\sigma_x+\sigma_y]\right) \sim D[\alpha(t) (\sigma_x+\sigma_y)]

To avoid redundant displacement, we must select t=2πN/Δt = 2\pi N / \Delta.

The second-order term is

exp(1220tdt10t1dt2[Hint(t1),Hint(t2)])=exp[2i(ηΩΔ)2(ΔtsinΔt)]\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[ -2\mathrm{i}\left(\frac{\eta\Omega}{\Delta}\right)^2 (\Delta t - \sin \Delta t)\right]

No a2a^2 or a2a^{\dagger2} is generated.