# Optimized Unconventional Geometric Gates in Superconducting Circuits
## 3 Quantum Gates and Robustness  
To realize an arbitrary unconventional geometric quantum gate, we consider a two-level system consisting of qubit states \(\{\ket{0},\ket{1}\}\). One choice of the auxiliary bases can be  
\[\ket{\tilde{\psi}_{1}(t)} =\cos\frac{\alpha(t)}{2}\ket{0}+\sin\frac{\alpha(t)}{2}e^{i\beta( t)}\ket{1},\] \[\ket{\tilde{\psi}_{2}(t)} =-\sin\frac{\alpha(t)}{2}e^{-i\beta(t)}\ket{0}+\cos\frac{\alpha(t )}{2}\ket{1}, \tag{6}\]  
and these states are required to fulfill the cyclic condition \(\ket{\tilde{\psi}_{i}(0)}=\ket{\tilde{\psi}_{i}(T)}\). Accordingly, the geometric phase is obtained as  
\[\gamma_{g2}(t^{\prime})=-\gamma_{g1}(t^{\prime})=i\int_{0}^{t}\bra{\tilde{ \psi}_{2}(t^{\prime})}\ket{d_{t}}\tilde{\psi}_{2}(t^{\prime})\rangle\,dt^{ \prime}=\frac{1}{2}\int_{0}^{t}[1-\cos\alpha(t)]\dot{\beta}(t)\,dt; \tag{7}\]  
when the cyclic evolution condition is satisfied, the above integral can be converted into an area integral using [PERSON]'s formula. At this point the geometric phase is dependent on the solid angle enclosed by the closed path. For brevity, we redefine \(\gamma_{g2}(t)=\gamma_{g}(t)\), and \(f_{2}(t)=-f_{1}(t)=f(t)=(1+\eta)\gamma_{g}(t).\) The evolution operator for the auxiliary states above can be written as  
\[U(T)=e^{-if(T)}\ket{\tilde{\psi}_{1}(0)}\bra{\tilde{\psi}_{1}(0)}+e^{if(T)} \ket{\tilde{\psi}_{2}(0)}\bra{\tilde{\psi}_{2}(0)}=e^{-if(T)\mathbf{n}\cdot\mathbf{ \sigma}}, \tag{8}\]where \(n=(\sin\alpha_{0}\cos\beta_{0},\sin\alpha_{0}\sin\beta_{0},\cos\alpha_{0})\) with \(\alpha_{0}=\alpha(0)\), \(\beta_{0}=\beta(0)\), and \(\sigma\) is the Pauli vector. The evolution operator \(U(T)\) represents an arbitrary one-qubit gate with an arbitrary rotating axis \(n\) and an arbitrary rotating angle \(2f(T)\). Now, we start to construct the one-qubit T-gate and H-gate and discuss the effect of the ratio on the robustness of the quantum gates. The parameters for constructing quantum gates can be set as  
\[T-gate:(\alpha_{0},\beta_{0})=(0,0),\,f(T)=\pi/8,\] \[H-gate:(\alpha_{0},\beta_{0})=(\pi/4,0),\,f(T)=\pi/2. \tag{9}\]  
The exponential maps of the T-gate and H-gate are \(U(T)=\exp{(-i\pi\varphi_{z}/8)}\) and \(U(T)=\exp{(-i\pi(\varphi_{z}+\sigma_{x})/2\sqrt{2})}\), respectively. In order to make the phase \(f(T)\) positive, we choose a counterclockwise evolution direction of the quantum state, i.e., the derivative of the azimuth angle is \(\hat{\beta}(t)>0\). The evolution path of the quantum state is shown in Figure 1a. First, we can take the path parameters \((\alpha(t),\beta(t))\) starting from the north pole \((0,0)\) to the point A \((\pi/4,0)\) along the geodesic \(\beta(t)=0\) for \([0,T_{1})\). For this path, the Hamiltonian can be obtained, \(H(t)=\frac{1}{2}\Omega(t)[e^{i(\beta(t)+\pi/2)}\ket{1}\bra{0}+\text{H.c}]\), with the Rabi frequency being \(\Omega(t)=\dot{\alpha}(t)>0\). The Rabi frequency area is calculated as \(\int_{0}^{T_{1}}\Omega(t)\,dt=\pi/4\). Then, the quantum state evolves from point A \((\pi/4,0)\) to \((\pi/4,\pi/[2(1+\eta)(2-\sqrt{2})])\) along the latitude line \(\alpha(t)=\pi/4\), for \((T_{1},T_{2})\), with \(\eta<\frac{\sqrt{2}}{2-\sqrt{2}}(\approx 2.41)\). For this path, the Hamiltonian reads \(H(t)=\frac{1}{2}\Delta(t)(\ket{1}\bra{1}-\ket{0}\bra{0})+\frac{1}{2}\Omega(t)[ e^{i(\beta(t)+\pi)}\ket{1}\bra{0}+\text{H.c}]\), where the Rabi frequency and the detuning are written as  
\[\Omega(t) =\frac{\sqrt{2}}{2}[\frac{\sqrt{2}}{2}(1+\eta)-\eta]\hat{\beta}(t),\] \[\Delta(t) =-2[1+\frac{\sqrt{2}}{2}(1+\eta)]\sin^{2}(\frac{\pi}{8})\hat{ \beta}(t). \tag{10}\]  
Note that we took \(\eta<\frac{\sqrt{2}}{2-\sqrt{2}}\) above just to make the Rabi frequency \(\Omega(t)\) positive. Accordingly, the Rabi frequency area is \(\int_{T_{1}}^{T_{2}}\Omega(t)\,dt=\frac{\sqrt{2}}{2}[-\eta+\frac{\sqrt{2}}{2}( 1+\eta)](\beta(T_{2})-\beta(T_{1}))\). Meanwhile, the ratio must be \(-1<\eta\) to ensure that \(\hat{\beta}(t)>0\).