Pure state and mixed state
See here.
qPCA
Compute
\[ \operatorname{Tr}_{1} \left(e^{-iS\Delta t} \rho \otimes \sigma e^{iS\Delta t} \right) \]where \(\rho\) and \(\sigma\) are two density matrix, \(S\) is the swap operator.
Note that \(e^{-iS\Delta t} = \cos (\Delta t) I_1 \otimes I_2 - i\sin(\Delta t) S\).
First considering that \(\rho\) and \(\sigma\) are density matrix of pure state, i.e, \(\rho = \ket{\psi_1} \bra{\psi_1}, \sigma = \ket{\psi_2} \bra{\psi_2}\),then
\[\begin{aligned} e^{-iS\Delta t} \ket{\psi_1} \ket{\psi_2} \end{aligned} = \cos (\Delta t)\ket{\psi_1} \ket{\psi_2} - i \sin(\Delta t) \ket{\psi_2} \ket{\psi_1}. \]So we have
\[\begin{aligned} & e^{-iS\Delta t} \rho \otimes \sigma e^{iS\Delta t} \\ =& e^{-iS\Delta t} \ket{\psi_1} \ket{\psi_2} \cdot \bra{\psi_1} \bra{\psi_2} e^{iS\Delta t} \\ = & \left[\cos (\Delta t)\ket{\psi_1} \ket{\psi_2} - i \sin(\Delta t) \ket{\psi_2} \ket{\psi_1}\right] \\ \cdot & \left[ \cos (\Delta t)\bra{\psi_1} \bra{\psi_2} + i \sin(\Delta t) \bra{\psi_2} \bra{\psi_1} \right] \\ =& \cos^2 (\Delta t)\ket{\psi_1} \ket{\psi_2}\bra{\psi_1} \bra{\psi_2} + \sin^2(\Delta t)\ket{\psi_2} \ket{\psi_1} \bra{\psi_2} \bra{\psi_1} \\ &- i \sin(\Delta t)\cos (\Delta t)\left[ \ket{\psi_2} \ket{\psi_1}\bra{\psi_1} \bra{\psi_2} - \ket{\psi_1} \ket{\psi_2} \bra{\psi_2} \bra{\psi_1} \right] \end{aligned} \]Note that \(\ket{\psi_1} \ket{\psi_2}\bra{\psi_1} \bra{\psi_2} = \rho \otimes \sigma, \ket{\psi_2} \ket{\psi_1} \bra{\psi_2} \bra{\psi_1} = \sigma \otimes \rho\), so
\[ \operatorname{Tr}_{1}\ket{\psi_1} \ket{\psi_2}\bra{\psi_1} \bra{\psi_2} = \sigma, \operatorname{Tr}_{1} \ket{\psi_2} \ket{\psi_1} \bra{\psi_2} \bra{\psi_1} = \rho. \]Now we consider \(\operatorname{Tr}_{2} \ket{\psi_2} \ket{\psi_1}\bra{\psi_1} \bra{\psi_2}\). We have
\[\begin{aligned} \operatorname{Tr}_{1} \ket{\psi_2} \ket{\psi_1}\bra{\psi_1} \bra{\psi_2} &= \sum_{j} \left( \bra{j}\otimes I \right)\ket{\psi_2} \ket{\psi_1}\bra{\psi_1} \bra{\psi_2}\left( \ket{j}\otimes I \right) \\ &=\sum_{j} \ket{\psi_1}\bra{\psi_2} \otimes (\bra{j} \ket{\psi_2} \bra{\psi_1} \ket{j} ) \\ &= \braket{\psi_1 \mid \psi_2} \ket{\psi_1}\bra{\psi_2} \\ &= \rho\sigma \end{aligned} \]Similarly, $ \operatorname{Tr}_{1} \ket{\psi_1} \ket{\psi_2}\bra{\psi_2} \bra{\psi_1} = \rho \sigma$. So we have
\[\begin{aligned} \operatorname{Tr}_{1} \left(e^{-iS\Delta t} \rho \otimes \sigma e^{iS\Delta t} \right) &= \cos^2 (\Delta t)\sigma + \sin^2(\Delta t)\rho - i \sin(\Delta t)\cos (\Delta t)\left[ \rho, \sigma \right] \\ &=\sigma-i \Delta t[\rho, \sigma]+O\left(\Delta t^2\right) \end{aligned} \] 标签：psi,模拟,rho,Delta,ket,sigma,bra,量子 From： https://www.cnblogs.com/linxiaoshu/p/16917023.html