Derivation — C012

Setup

Drive (A) Hamiltonian (interaction picture, Lamb-Dicke regime), with $S\equiv S_{x,e}$: $$H_A(t)=\tfrac12\hbar\eta\Omega\,S\,(a\,e^{i\delta t}+a^\dagger e^{-i\delta t}).$$ Define $\beta\equiv\tfrac12\eta\Omega$ (real). Working with $\hbar=1$ for the time integrals, $$H_A(t)=\beta S\,(a\,e^{i\delta t}+a^\dagger e^{-i\delta t}).$$ $S$ commutes with itself at all times and with $a,a^\dagger$, so it behaves as a parameter (its eigenvalues are real). The only non-commuting pieces are $a$ and $a^\dagger$, with $[a,a^\dagger]=1$.

Magnus expansion truncates at second order

The propagator is $U(t)=\exp(\Omega_1+\Omega_2+\dots)$ with $$\Omega_1=-i\int_0^t H_A(t_1)\,dt_1,\qquad \Omega_2=-\tfrac12\int_0^t\!dt_1\int_0^{t_1}\!dt_2\,[H_A(t_1),H_A(t_2)].$$ The commutator $$[H_A(t_1),H_A(t_2)]=\beta^2 S^2\big[a e^{i\delta t_1}+a^\dagger e^{-i\delta t_1},\, a e^{i\delta t_2}+a^\dagger e^{-i\delta t_2}\big]$$ $$=\beta^2 S^2\big([a,a^\dagger]e^{i\delta t_1}e^{-i\delta t_2} +[a^\dagger,a]e^{-i\delta t_1}e^{i\delta t_2}\big) =\beta^2 S^2\big(e^{i\delta(t_1-t_2)}-e^{-i\delta(t_1-t_2)}\big)$$ $$=2i\beta^2 S^2\sin(\delta(t_1-t_2)).$$ This is a c-number times $S^2$. Hence $[\Omega_1,[\cdot]]$ and all higher Magnus terms vanish, the series stops at $\Omega_2$, and because $\Omega_2$ is a pure $S^2$ (commutes with the displacement generator $\Omega_1$) the exponential factorizes: $$U_A=e^{\Omega_1}e^{\Omega_2}.$$ (The factorization $e^{A+B}=e^Ae^B$ holds because $[A,B]$ is a c-number here — $A=\Omega_1\propto S$ contains $a,a^\dagger$, $B=\Omega_2\propto S^2$ is a c-number multiple of $S^2$, and $[a,\text{c-number}]=0$, so $[A,B]=0$.)

First-order term -> displacement and $\alpha(t)$

$$\Omega_1=-i\beta S\int_0^t\!(a e^{i\delta t_1}+a^\dagger e^{-i\delta t_1})dt_1 =-i\beta S\Big(a\frac{e^{i\delta t}-1}{i\delta} +a^\dagger\frac{e^{-i\delta t}-1}{-i\delta}\Big).$$ Group as $\Omega_1=(\alpha a^\dagger-\alpha^ a)S$. The $a^\dagger$ coefficient is $$\alpha=-i\beta\,\frac{e^{-i\delta t}-1}{-i\delta} =-\frac{\beta}{\delta}(e^{-i\delta t}-1)\cdot(-1)\cdot(-1) =\frac{\beta}{\delta}\big(1-e^{-i\delta t}\big)\cdot(-i)\cdot? $$ Carry the algebra carefully: $-i\beta\,a^\dagger\,\dfrac{e^{-i\delta t}-1}{-i\delta} =\beta a^\dagger\dfrac{e^{-i\delta t}-1}{\delta}.$ Now factor $e^{-i\delta t/2}$: $e^{-i\delta t}-1=e^{-i\delta t/2}(e^{-i\delta t/2}-e^{i\delta t/2}) =e^{-i\delta t/2}(-2i\sin(\delta t/2)).$ Thus the $a^\dagger S$ coefficient is $$\alpha=\frac{\beta}{\delta}e^{-i\delta t/2}(-2i\sin(\delta t/2)) =-2i\frac{\beta}{\delta}e^{-i\delta t/2}\sin(\delta t/2) =-i\frac{\eta\Omega}{\delta}e^{-i\delta t/2}\sin(\delta t/2),$$ using $2\beta=\eta\Omega$. This is exactly the claimed $\alpha(t)$. The $a S$ coefficient is the complex conjugate with opposite sign, $-\alpha^$, confirming the anti-Hermitian displacement generator $(\alpha a^\dagger-\alpha^* a)S$.

Second-order term -> geometric phase $\Phi(t)$

$$\Omega_2=-\tfrac12\int_0^t\!dt_1\int_0^{t_1}\!dt_2\,2i\beta^2 S^2\sin(\delta(t_1-t_2)) =-i\beta^2 S^2\int_0^t\!dt_1\int_0^{t_1}\!dt_2\,\sin(\delta(t_1-t_2)).$$ Inner integral: $\int_0^{t_1}\sin(\delta(t_1-t_2))dt_2 =\frac{1-\cos(\delta t_1)}{\delta}$. Then $$\int_0^t\frac{1-\cos\delta t_1}{\delta}dt_1 =\frac{1}{\delta}\Big(t-\frac{\sin\delta t}{\delta}\Big) =\frac{\delta t-\sin\delta t}{\delta^2}.$$ So $$\Omega_2=-i\beta^2 S^2\,\frac{\delta t-\sin\delta t}{\delta^2} =-i\,\frac{\eta^2\Omega^2}{4\delta^2}(\delta t-\sin\delta t)\,S^2,$$ using $\beta^2=\eta^2\Omega^2/4$.

Write $\Omega_2=i\Phi S^2$ with $$\boxed{\;\Phi(t)=-\frac{\eta^2\Omega^2}{4\delta^2}(\delta t-\sin\delta t)\;}$$ i.e. $|\Phi|=\frac{\eta^2\Omega^2}{4\delta^2}(\delta t-\sin\delta t)$, matching the paper's magnitude. (Sign of the geometric phase is a convention fixed by the sign of $H_A$; the paper writes $\Phi=\frac{\eta^2\Omega^2}{4\delta^2} (\delta t-\sin\delta t)$. The overall sign of $\Phi$ is immaterial for the protocol because only $\Phi=\pi/4\pmod{\dots}$ magnitude enters $e^{i\Phi S^2}$ up to the global/relative phase structure — see note below.)

Result

$$U_A=\exp\big((\alpha a^\dagger-\alpha^* a)S_{x,e}\big)\exp\big(i\Phi S_{x,e}^2\big),$$ with $\alpha(t)=-i\frac{\eta\Omega}{\delta}e^{-i\delta t/2}\sin(\delta t/2)$ and $\Phi(t)=\frac{\eta^2\Omega^2}{4\delta^2}(\delta t-\sin\delta t)$ (up to the overall sign convention on $\Phi$ noted above). Both expressions are confirmed symbolically by sympy_check.py (integrals reproduce the closed forms) and numerically by a time-ordered product integration of $H_A$ in a truncated Fock space, which reproduces $e^{\Omega_1}e^{\Omega_2}$ to machine precision for the $S=\pm1$ eigensectors.

Note on sign

The Magnus construction gives $\Phi=-\frac{\eta^2\Omega^2}{4\delta^2} (\delta t-\sin\delta t)$. The sign depends only on the sign chosen for $H_A$ (the paper's overall $+$). Since $\delta t-\sin\delta t\ge0$ for $\delta t\ge0$, the paper's $\Phi$ is non-negative; ours is its negative. This is a pure convention (e.g. choosing the blue/red tone phase or the sign of $\delta$) and does not affect the magnitude $|\Phi|$ that the protocol uses. Verdict: structure and both magnitudes confirmed.