我们现在研究由于$SU(2)$自旋旋转对称而不是$S_z$守恒的约束。一个绕轴$\mathbf{n}$旋转$\phi$的自旋转动$\mathcal{U}_{\mathbf{n}}^\phi$作用在二分量旋量上为
\begin{pmatrix}
\psi_\uparrow\\
\psi_\downarrow\\
\end{pmatrix}\rightarrow\mathcal{U}_{\mathbf{n}}^\phi\begin{pmatrix}
\psi_\uparrow\\
\psi_\downarrow
\end{pmatrix},\quad \mathcal{U}_{\mathbf{n}}^{\phi}=e^{-i\frac{\phi}{2}\mathbf{\sigma}\cdot\mathbf{n}}
绕$S_x,S_y$方向的转动$\phi$的变换为
\mathcal{U}_{S_x}^\phi\begin{pmatrix}
\psi_{\uparrow}\\
\psi_{\downarrow}
\end{pmatrix}=\begin{pmatrix}
\cos\frac{\phi}{2}&-i\sin\frac{\phi}{2}\\
-i\sin\frac{\phi}{2}&\cos\frac{\phi}{2}
\end{pmatrix}\begin{pmatrix}
\phi_\uparrow\\
\phi_\downarrow
\end{pmatrix}=\begin{pmatrix}
\cos\frac{\phi}{2}\psi_\uparrow-i\sin\frac{\phi}{2}\psi_\downarrow\\
-i\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow
\end{pmatrix}\mathcal{U}_{S_y}^\phi\begin{pmatrix}
\phi_\uparrow\\
\phi_\downarrow
\end{pmatrix}=\begin{pmatrix}
\cos\frac{\phi}{2}&-\sin\frac{\phi}{2}\\
\sin\frac{\phi}{2}&\cos\frac{\phi}{2}
\end{pmatrix}\begin{pmatrix}
\phi_\uparrow\\
\phi_\downarrow
\end{pmatrix}=\begin{pmatrix}
\cos\frac{\phi}{2}\psi_\uparrow-\sin\frac{\phi}{2}\psi_\downarrow\\
\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow
\end{pmatrix}我们有SU(2)下，各个费米子算符变换形式
\begin{split}
&\mathcal{U}_{S_x}^\phi\psi_\uparrow\mathcal{U}_{S_x}^{-\phi}=\cos\frac{\phi}{2}\psi_\uparrow-i\sin\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_x}^\phi\psi_\downarrow\mathcal{U}_{S_x}^{-\phi}=-i\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_x}^\phi\psi_\uparrow^\dagger\mathcal{U}_{S_x}^{-\phi}=\cos\frac{\phi}{2}\psi_\uparrow+i\sin\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_x}^\phi\psi_\downarrow^\dagger\mathcal{U}_{S_x}^{-\phi}=i\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow
\end{split}\begin{split}
&\mathcal{U}_{S_y}^\phi\psi_\uparrow\mathcal{U}_{S_y}^{-\phi}=\cos\frac{\phi}{2}\psi_\uparrow-\sin\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_y}^\phi\psi_\downarrow\mathcal{U}_{S_y}^{-\phi}=\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_y}^\phi\psi_\uparrow^\dagger\mathcal{U}_{S_y}^{-\phi}=\cos\frac{\phi}{2}\psi_\uparrow-\sin\frac{\phi}{2}\psi_\downarrow\\
&\mathcal{U}_{S_y}^\phi\psi_\downarrow^\dagger\mathcal{U}_{S_y}^{-\phi}=\sin\frac{\phi}{2}\psi_\uparrow+\cos\frac{\phi}{2}\psi_\downarrow
\end{split}所以我们有
\mathcal{U}_{S_x}^\phi\Psi\mathcal{U}_{S_x}^{-\phi}=\mathcal{U}_{S_x}^\phi\begin{pmatrix}
\psi_\uparrow\\
\psi_\downarrow^\dagger
\end{pmatrix}\mathcal{U}_{S_x}^{-\phi}=\begin{pmatrix}
\cos\frac{\phi}{2}\psi_\uparrow-i\sin\frac{\phi}{2}\psi_\downarrow\\
i\sin\frac{\phi}{2}\psi_\uparrow^\dagger+\cos\frac{\phi}{2}\psi_\downarrow^\dagger
\end{pmatrix}=\cos\frac{\phi}{2}\Psi-i\sin\frac{\phi}{2}\Psi^c
\mathcal{U}_{S_y}^\phi\Psi\mathcal{U}_{S_y}^{-\phi}=\mathcal{U}_{S_y}^\phi\begin{pmatrix}
\psi_\uparrow\\
\psi_\downarrow^\dagger
\end{pmatrix}\mathcal{U}_{S_y}^{-\phi}=\begin{pmatrix}
\cos\frac{\phi}{2}\psi_\uparrow-\sin\frac{\phi}{2}\psi_\downarrow\\
\sin\frac{\phi}{2}\psi_\uparrow^\dagger+\cos\frac{\phi}{2}\psi_\downarrow^\dagger
\end{pmatrix}=\cos\frac{\phi}{2}\Psi-\sin\frac{\phi}{2}\Psi^c
这样$\mathcal{U}_{S_x}^\phi$和$\mathcal{U}_{S_y}^\phi$把$\Psi$光滑的转动到$\Psi^c$. 特别地是绕$S_x$或者$S_y$转动$\pi$的作用为一个分离的PH变换，$\Psi\rightarrow-i\Psi^c$或者$\Psi\rightarrow-\Psi^c$. 也就是说，如果我们把方程\eqref{class a}解释为粒子数守恒的系统，那么$\mathcal{U}_{S_i}^\pi S_z\mathcal{U}_{S_i}^{-\pi}=-S_z,(i=x,y)$可以被视为一个电荷共轭$\mathcal{C} Q\mathcal{C}^{-1}=-Q$. $\pi$的旋转$\mathcal{U}_{S_i}^\pi$是PH变换的一个例子，它的平方等于$-1$. 与$Class D$的PH变换相反。对于单粒子哈密顿$H$，$\pi$的旋转对称$\mathcal{U}_{S_i}^\pi$导出条件
\rho_2H^T\rho_2=-H
简单推导一下
不失一般性，假设哈密顿有绕$S_x$轴转动$\pi$的对称性.
\begin{split}
\hat{H}&=\mathcal{U}_{S_x}^\pi\hat{H}\mathcal{U}_{S_x}^{-\pi}=i(\Psi^{c})^\dagger \mathcal{U}_{S_x}^\pi H\mathcal{U}_{S_x}^{-\pi}(-i\Psi^c)\\
&=-i\rho_2(\Psi)^T\mathcal{U}_{S_x}^\pi H\mathcal{U}_{S_x}^{-\pi}i\rho_2(\Psi^\dagger)^T\\
&=(\Psi^T)\rho_2H\rho_2(\Psi^\dagger)^T\\
&=-\Psi^\dagger \rho_2H^T\rho_2\Psi=\Psi^\dagger H\Psi
\end{split}所以得到$\rho_2^TH^T\rho_2=-H$
满足这个条件的哈密顿集合叫做Class C.
我们注意到对于哈密顿二次型，$\pi-$转动对称$\mathcal{U}_{S_i}^\pi$的守恒实际上对应一个全体的$SU(2)$自旋转动对称。这是由于对于任意绕$S_x$或$S_y$的SU(2)转动，哈密顿$\hat{H}$是$\Psi^\dagger H\Psi$和$\Psi^{c\dagger}H\Psi^c$的叠加
\begin{split}
\Psi&\rightarrow\cos\frac{\phi}{2}\Psi-i\sin\frac{\phi}{2}\Psi^c\\
\Psi^\dagger&\rightarrow\cos\frac{\phi}{2}\Psi^\dagger+i\sin\frac{\phi}{2}\Psi^{c\dagger}\\
\hat{H}=\Psi^\dagger H\Psi&\rightarrow\cos^2\frac{\phi}{2}\Psi^\dagger H\Psi+\sin^2\frac{\phi}{2}\Psi^{c\dagger}H\Psi^{c}
\end{split}这里$\Psi^\dagger H\Psi^c=\Psi^{c\dagger}H\Psi=0$(自旋$\frac{1}{2}$系统的Kramers简并)。由于
\Psi^{c\dagger}H\Psi^c=-i\rho_2\Psi^THi\rho_2(\Psi^\dagger)^T=\Psi^T\rho_2 H\rho_2(\Psi^\dagger)^T=-\Psi^\dagger\rho_2 H^T\rho_2\Psi=\Psi^\dagger H\Psi
所以我们得到
\hat{H}=\Psi^\dagger H\Psi\xrightarrow{SU(2)}\cos^2\frac{\phi}{2}\Psi^\dagger H\Psi+\sin^2\frac{\phi}{2}\Psi^{c\dagger}H\Psi^{c}=\Psi^\dagger H\Psi=\hat{H}
BdG哈密顿在全部的$SU(2)$自选转动对称下是不变的
最后在$S_z$守恒的系统里添加$TRS$会导致$\Psi^\dagger H\Psi\xrightarrow{\mathcal{T}}\Psi^T\rho_2H^*\rho_2(\Psi^\dagger)^T=-\Psi^\dagger\rho_2 H^\dagger\rho_2\Psi=H$
得到$\rho_2H^\dagger\rho_2=-H$. 结合PHS可以得到
\rho_2 H^T\rho_2=-H,H^*=H
这定义了Class CI