抑制态泄露的drag波

drag波

在Transmon的实验中，比特在驱动\(\left|0\right>\rightarrow \left|1\right>\)跃迁时打入\(w_{01}\)的波，但由于仪器和Transmon非谐性不够大等原因，\(w_{12}\)频率的波幅不为0，从而导致跃迁到2能级引起态泄露，参考文献提出drag波可以抵消以上提及的效应进而抑制态泄露。

推导

可以把驱动下的transmon的哈密顿量写成如下格式：

\[\begin{equation} \begin{aligned} H = w_1 \Pi_1 + w_2 \Pi_2 + E(t)(\sigma_{0,1}^+ + \sigma_{0,1}^-)+ \lambda E(t)(\sigma_{1,2}^+ + \sigma_{1,2}^-) \end{aligned} \end{equation}\]

其中\(\sigma_{i,i+1}^\pm\)是限制在\(\{i,i+1\}\)空间中的升降算符，\(E(t) = E^x (t)\cos(w_dt) + E^y (t)\sin(w_dt)\)，这里假设\(w_{12}\)的驱动和\(w_{01}\)的驱动是线性关系，故假设\(\{12\}\)能级的驱动项乘上一常系数\(\lambda\)，对哈密顿量旋波近似后（\(U= exp(iw_d\Pi_1 + 2iw_d \Pi_2)\)）得到：

\[\begin{equation} \begin{aligned} H^R=\sum_{j=1,2}\delta_{j}\Pi_{j}+\frac{E^x(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{x}+\frac{ E^{y}(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{y} \end{aligned} \end{equation}\]

其中\(\delta_1 = w_1 - w_d\)，\(\delta_2 = \Delta + 2\delta_1\)，\(\Delta\) 是比特的失谐，这里新定义变换幺正算符\(V\)：

\[\begin{equation} \begin{aligned} V = e^S = e^{-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta} \end{aligned} \end{equation}\]

假如比特在做 \(\pi\) 门时\(E^x(t_{start}) = 0, E^x(t_{end}) = 0\)，\(V(t_{start}) = I, V(t_{end}) = I\) 则V表象变换后的态\({\left|\psi\right>}_V (t) = V(t)\left|\psi\right>_{R}\)，既在\(\pi\)门结束时，V表象下的态和旋转表象下的态完全一致，可以到V表象下讨论问题。

将哈密顿量变换导\(V\)下\(H^V = i \dot V V^\dagger + V H^R V^\dagger\)，其中：

\[\begin{equation} \begin{aligned} V H^R V^\dagger \approx H_R +[S,H_R]+\frac{1}{2}[S,[S,R]] \end{aligned} \end{equation}\]

分别求：

\[\begin{equation} [S,H_R] = \sum_{j=1,2}[S,\delta_{j}\Pi_{j}] + \sum_{j=1,2}[S,\frac{E^x(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{x}] + \sum_{j=1,2}[S,\frac{ E^{y}(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{y}] \end{equation}\]

第一项：

\[\begin{equation} \begin{aligned} \sum_{j=1,2}[S,\delta_{j}\Pi_{j}] =& [S,\delta_{1}\Pi_{1}] + [S,\delta_{2}\Pi_{2}] \\ =& [{-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta},\delta_{1}\Pi_{1}] + [{-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta},\delta_{2}\Pi_{2}] \\ =& \frac{-iE^x}{2\Delta}(\delta_{1}[\sigma_{0,1}^y,\Pi_{1}] + \lambda\delta_{1}[\sigma_{1,2}^y,\Pi_{1}] + (\delta_{2}[\sigma_{0,1}^y,\Pi_{2}] + \lambda\delta_{2}[\sigma_{1,2}^y,\Pi_{2}] \\ =& \frac{-iE^x}{2\Delta}(\delta_{1}(-i)\sigma_{0,1}^x + \lambda\delta_{1}i\sigma_{1,2}^x + \delta_{2}0+ \lambda\delta_{2}(-i)\sigma_{1,2}^x) \\ =& \frac{E^x}{2\Delta}(-\delta_{1}\sigma_{0,1}^x + \lambda\delta_{1}\sigma_{1,2}^x - \lambda\delta_{2}\sigma_{1,2}^x) \\ =& \frac{E^x}{2\Delta}(-\delta_{1}\sigma_{0,1}^x + \lambda\delta_{1}\sigma_{1,2}^x - \lambda(2\delta_{1} +\Delta)\sigma_{1,2}^x) \\ =& \frac{-E^x}{2\Delta}(\delta_{1}\sigma_{0,1}^x +\lambda(\delta_{1} +\Delta)\sigma_{1,2}^x) \end{aligned} \end{equation}\]

第二项：

\[\begin{equation} \begin{aligned} &\sum_{j=1,2}[S,\frac{E^x(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{x}] \\=& [S,\frac{E^x(t)}{2}{\sigma}_{0,1}^{x}] +[S,\frac{E^x(t)}{2}\lambda{\sigma}_{1,2}^{x}] \\ =&\frac{E^x(t)}{2}([S,{\sigma}_{0,1}^{x}] +[S,\lambda{\sigma}_{1,2}^{x}]) \\ =&\frac{E^x(t)}{2}([{-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta},{\sigma}_{0,1}^{x}]+[{-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta},\lambda{\sigma}_{1,2}^{x}]) \\ =&\frac{-iE^x(t)^2}{4\Delta}([{\sigma_{0,1}^y+\lambda\sigma_{1,2}^y},{\sigma}_{0,1}^{x}] +[{\sigma_{0,1}^y+\lambda\sigma_{1,2}^y},\lambda{\sigma}_{1,2}^{x}]) \\ =&\frac{-iE^x(t)^2}{4\Delta}([{\sigma_{0,1}^y,{\sigma}_{0,1}^{x}]+\lambda[\sigma_{1,2}^y},{\sigma}_{0,1}^{x}] +\lambda[{\sigma_{0,1}^y,{\sigma}_{1,2}^{x}]+\lambda^2[\sigma_{1,2}^y},{\sigma}_{1,2}^{x}]) \\ =&\frac{-iE^x(t)^2}{4\Delta}((-2i+4i\Pi_1+2i\Pi_2)+\lambda i \sigma_{0,2}^x +\lambda(-i \sigma_{0,2}^x)+\lambda^2(-2i\Pi_1+2i\Pi_2)) \end{aligned} \end{equation}\]

略去常数后：

\[\begin{equation} \begin{aligned} =&\frac{E^x(t)^2}{4\Delta}((4\Pi_1+2\Pi_2)+\lambda \sigma_{0,2}^x+\lambda(-\sigma_{0,2}^x)+\lambda^2(-2\Pi_1+2\Pi_2)) \\=& \frac{E^x(t)^2(4- 2\lambda^2)}{4\Delta}\Pi_1+\frac{E^x(t)^2(2+2\lambda^2)}{4\Delta}\Pi_2 \end{aligned} \end{equation}\]

显然 \(\left[S,H_R\right]\) 第三项为0，故得到 \(\left[S,H_{R}\right]\) 等于：

\[\begin{equation} \begin{aligned} =& \frac{-E^x}{2\Delta}(\delta_{1}\sigma_{0,1}^x +\lambda(\delta_{1} +\Delta)\sigma_{1,2}^x) + \frac{E^x(t)^2(4- 2\lambda^2)}{4\Delta}\Pi_1+\frac{E^x(t)^2(2+2\lambda^2)}{4\Delta}\Pi_2 \\=& \frac{-E^x\delta_1 }{2\Delta}\sigma_{0,1}^x - \frac{E^x \lambda\delta_{1} }{2\Delta}\sigma_{1,2}^x-\frac{E^x\lambda }{2}\sigma_{1,2}^x+\frac{E^x(t)^2(4- 2\lambda^2)}{4\Delta}\Pi_1+\frac{E^x(t)^2(2+2\lambda^2)}{4\Delta}\Pi_2 \end{aligned} \end{equation}\]

由实验中驱动的振幅远小于比特的失谐，既\(E^x \ll \Delta\)，保留哈密顿量中含有\(frac{E^x}{\Delta}\)的0，1次项。\([S,[S,H_R]]\)中\(\frac{E^x}{\Delta}\)一次项有：

\[\begin{equation} \begin{aligned} -&\left[S,\frac{E^x\lambda }{2}\sigma_{1,2}^x\right] \\&=- [-iE^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta,\frac{E^x\lambda }{2}\sigma_{1,2}^x ] \\ &=- \frac{-i{E^x}^2\lambda}{4\Delta}[(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y),\sigma_{1,2}^x ] \\ &=- \frac{-i{E^x}^2\lambda}{4\Delta}([\sigma_{0,1}^y,\sigma_{1,2}^x ]+\lambda[\sigma_{1,2}^y,\sigma_{1,2}^x ]) \\ &=- (\frac{-{E^x}^2\lambda}{4\Delta}\sigma_{0,2}^x+ \frac{-{E^x}^2\lambda^2}{4\Delta}(2\Pi_1 - 2\Pi_2)) \\ &= \frac{ {E^x}^2\lambda}{4\Delta}\sigma_{0,2}^x+ \frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_1 -\frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_2 \end{aligned} \end{equation}\]

综上所述：

\[\begin{equation} \begin{aligned} H^V &= i \dot V V^\dagger + V H^R V^\dagger \\& = \dot E^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta + H_R + [S,H_R] + \frac{1}{2} [S[S,H_R]]... \\& \approx \dot E^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta + \sum_{j=1,2}\delta_{j}\Pi_{j}+\frac{E^x(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{x}+\frac{ E^{y}(t)}{2}\lambda_{j}{\sigma}_{j-1,j}^{y} + \\&\frac{-E^x\delta_1 }{2\Delta}\sigma_{0,1}^x - \frac{E^x \lambda\delta_{1} }{2\Delta}\sigma_{1,2}^x - \frac{E^x\lambda }{2}\sigma_{1,2}^x+\frac{E^x(t)^2(4- 2\lambda^2)}{4\Delta}\Pi_1+\frac{E^x(t)^2(2+2\lambda^2)}{4\Delta}\Pi_2 + \\& \frac{1}{2} (\frac{ {E^x}^2\lambda}{4\Delta}\sigma_{0,2}^x+ \frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_1 -\frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_2 ) +o(\frac{E^x}{\Delta}) \\& = \dot E^x(\sigma_{0,1}^y+\lambda\sigma_{1,2}^y)/2\Delta + \delta_{1}\Pi_{1}+ \delta_{2}\Pi_{2} +\frac{E^x(t)}{2}{\sigma}_{0,1}^{x}+\frac{ E^{y}(t)}{2}{\sigma}_{0,1}^{y} +\frac{E^x(t)}{2}\lambda{\sigma}_{1,2}^{x}+\frac{ E^{y}(t)}{2}\lambda{\sigma}_{1,2}^{y} \\&\frac{-E^x\delta_1 }{2\Delta}\sigma_{0,1}^x - \frac{E^x \lambda\delta_{1} }{2\Delta}\sigma_{1,2}^x - \frac{E^x\lambda }{2}\sigma_{1,2}^x+\frac{E^x(t)^2(4- 2\lambda^2)}{4\Delta}\Pi_1+\frac{E^x(t)^2(2+2\lambda^2)}{4\Delta}\Pi_2 + \\& \frac{1}{2} (\frac{ {E^x}^2\lambda}{4\Delta}\sigma_{0,2}^x+ \frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_1 -\frac{ {E^x}^2\lambda^2}{2\Delta}\Pi_2 ) +o(\frac{E^x}{\Delta}) \end{aligned} \end{equation}\]

整理得：

\[\begin{equation} \begin{aligned} H_V=& (\delta_1+\frac{E^x(t)^2(4- \lambda^2)}{4\Delta})\Pi_1 + (\delta_2 +\frac{E^x(t)^2(2+\lambda^2)}{4\Delta})\Pi_2 \\&+ (\frac{\dot E^x}{2\Delta}+E^y/2)(\sigma_{0,1}^y+\lambda \sigma_{1,2}^y) \\& + \frac{ E^x}{2} (1-\frac{\delta_1}{\Delta}) \sigma_{0,1}^x - \frac{ E^x \lambda \delta_1}{2\Delta}\sigma_{1,2}^x + \frac{ {E^x}^2\lambda}{8\Delta} \sigma_{0,2}^x \end{aligned} \end{equation}\]

当满足\(E^{y}=-\frac{\dot E^{x}}{\Delta}\)以及\(\delta_{1}=\frac{(\lambda^{2}-4){E^x}^2}{4\Delta}\)时，第一行第一项和第二行为0，且最后一行中态泄露项皆为小量可忽略，减少了态泄露。更高阶的结果（算不动了）：

\[\begin{equation} \begin{aligned} E^{x}(t)&=E_{\pi}+\frac{(\lambda^{2}-4)E_{\pi}^{3}}{8\Delta^{2}}-\frac{(13\lambda^{4}-76\lambda^{2}+112)E_{\pi}^{5}}{128\Delta^{4}} \\& E^{y}(t)=-\frac{\dot{E}_{\pi}}{\Delta}+\frac{33(\lambda^{2}-2)E_{\pi}^{2}\dot{E}_{\pi}}{24\Delta^{3}} \\\delta_{1}(t)&=-\frac{(\lambda^{2}-4)E_{\pi}^{2}}{4\Delta}-\frac{(\lambda^{4}-7\lambda^{2}+12)E_{\pi}^{4}}{16\Delta^{3}} \end{aligned} \end{equation}\]

其中\(E_{pi}\) 是驱动比特打入的\(\pi\)脉冲的振幅，而向y通道混入的\(E^y\)波称为drag波。