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Suppose I have a state in Bell basis. For example \begin{equation} \rho = \begin{pmatrix} \rho_{11} &0 & 0 & \rho_{14} \\ 0 &\rho_{22} & \rho_{23} & 0 \\ 0 &\rho_{32} & \rho_{33} & 0 \\ \rho_{41} &0 & 0 & \rho_{44} \nonumber \end{pmatrix} \end{equation} and another state as \begin{equation} \rho' = \begin{pmatrix} \rho'_{11} &0 & 0 & \rho'_{14} \\ 0 &\rho'_{22} & \rho'_{23} & 0 \\ 0 &\rho'_{32} & \rho'_{33} & 0 \\ \rho'_{41} &0 & 0 & \rho'_{44} \nonumber \end{pmatrix} \end{equation} Provided that both the state has same purity, how can I rotate $\rho$ to get $\rho'$.

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  • $\begingroup$ Why aren't either of your example states pure? It looks like they've been hit by parity measurements. e.g. like this circuit $\endgroup$ – Craig Gidney Jun 19 '16 at 0:51
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I don't think identical purity by itself suffices to make $\rho$ and $\rho'$ equivalent, which implies identical spectrum. Assuming this is what you are looking for and both states are known, one general approach is to diagonalize both $\rho$ and $\rho'$, then find the basis transformation from the eigenstates of one to the other. For instance, if
$$ \rho = U D U^\dagger, \;\;\;\rho' = V D V^\dagger $$ where $D$ is diagonal and $U$, $V$ are unitary, $U^\dagger U = U U^\dagger = V^\dagger V = V V^\dagger = I$, then $W = V U^\dagger$ transforms $\rho$ to $\rho'$, $$ \rho' \equiv V D V^\dagger = (VU^\dagger) U D U^\dagger(VU^\dagger)^\dagger = W\rho W^\dagger $$ Recall that $U$ and $V$ have as columns the eigenvectors of $\rho$, respectively $\rho'$, and note that in your particular case the eigenvalue problem decouples nicely, so $W$ is easy to find.

Equivalently, you could solve directly for a unitary $W$ such that $$ \rho' = W\rho W^\dagger $$

If you are further interested in the expression of $W$ in terms of Pauli matrices, treat it as an element of the $(2\times 2)\otimes (2\times 2)$ Hilbert space of operators/matrices with trace scalar product, and take as orthonormal basis the direct products $\frac{1}{2}{\hat\sigma}_\mu \otimes {\hat\sigma}_\nu$, $\mu,\;\nu = 0,1,2,3$, ${\hat\sigma}_0 = {\hat I}$. Then $W$ expands as $$ W = \frac{1}{2}\sum_{\mu,\nu}{w_{\mu\nu}{\hat\sigma}_\mu \otimes {\hat\sigma}_\nu} $$ with coefficients $w_{\mu\nu}$ given by corresponding scalar products: $$ w_{\mu\nu} = \frac{1}{2}Tr\left({\hat\sigma}_\mu \otimes {\hat\sigma}_\nu W \right) $$

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