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In atomic-orbital-based coupled-perturbed self-consistent field (AO-based CPSCF) theory, one has to solve the equation $FP^1S-SP^1F=S^1PF-FPS^1+SPF^1-F^1PS$ for $P^1$. $F^1$ on the right-hand-side depends on $P^1$, so this equation must be solved iteratively: step one forming the right-hand-side from a guessed $P^1$, step two solving the equation assuming the right-hand-side is a constant and then going back to step one until convergence.

My confusion is about step two. In step two, one solves an equation $AXB-BXA=C$ for $X$. I wonder how to do it and have failed finding anything about it. The closest topic to it I can find is the Sylvester equation $AX+XB=C$, but I wonder if it is relevant.

Knowledge about CPSCF equation: All matrices are square real symmetric; $S$ is positive-definite; $\mathrm{Tr}P^1=0$.

The question is actually about computational chemistry. I cannot put the tag "computational chemistry" here because I don't have enough reputations to create a new tag.

张亦弛
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  • Check this post for an idea about how to proceed: https://mathoverflow.net/questions/6166/is-there-a-name-for-the-matrix-equation-a-x-b-b-x-a-c-x-c-d – Anton Vrdoljak Oct 11 '23 at 04:13
  • If one of the matrices $A$ or $B$ is invertible, say $B$, it seems that it can be transformed into a Sylvester equation by multiplication with $B^{-1}$ form the right and the left: $AXB-BXA=C$, $AX-BXAB^{-1}=CB^{-1}$, $B^{-1}AX-XAB^{-1}=B^{-1}CB^{-1}$, $(B^{-1}A)X+ X(-AB^{-1})=B^{-1}CB^{-1}$. – Gerd Oct 11 '23 at 11:21
  • @Gerd The conversion is valid since $S$ is positive-definite. But now that $B^{-1}A$ and $AB^{-1}$ share the same set of eigenvalues, the equation has more than one solutions and cannot be solved with standard method. – 张亦弛 Oct 11 '23 at 14:55

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