# Integrated Information in Relational Quantum Dynamics (RQD)
## Appendix C Proof of Proposition 1 (Canonical Optimal Witness)
### Constrained Stationarity Conditions  
We now derive the necessary conditions that characterize the optimal product state \(\sigma^{*}=\rho_{A}\otimes\rho_{B}\) obtained above. We treat the minimization of \(D_{IS}^{Q}(\rho\|\sigma)\) as a constrained optimization problem on the manifold \(\mathcal{P}_{A|B}\). The constraints are that \(\sigma_{A}\) and \(\sigma_{B}\) are valid density operators: \(\mathrm{Tr}(\sigma_{A})=\mathrm{Tr}(\sigma_{B})=1\) and \(\sigma_{A},\sigma_{B}\geq 0\). The positivity constraints ensure the domain is a compact manifold with boundary, but since the optimum \(\sigma^{*}\) lies in the interior (full-rank case), we can ignore boundary considerations and use the method of Lagrange multipliers for the normalization constraints. Lagrangian setup. Define the Lagrangian function  
\[\mathcal{L}(\sigma_{A},\sigma_{B};\ \alpha,\beta)\ =\ D^{Q}_{JS}(\rho\|\ \sigma_{A} \otimes\sigma_{B})\ +\ \alpha\left(1-\mathrm{Tr}[\sigma_{A}]\right)\ +\ \beta\left(1-\mathrm{Tr}[\sigma_{B}]\right)\,,\]  
where \(\alpha\) and \(\beta\) are real Lagrange multipliers enforcing \(\mathrm{Tr}\sigma_{A}=\mathrm{Tr}\sigma_{B}=1\). (We incorporate the constraints \(\mathrm{Tr}\sigma_{A}=1\), \(\mathrm{Tr}\sigma_{B}=1\) explicitly; the overall trace \(\mathrm{Tr}(\sigma_{A}\otimes\sigma_{B})=1\) then holds automatically.) At the optimum \((\sigma_{A}^{*},\sigma_{B}^{*})\), the gradient of \(\mathcal{L}\) must vanish in all directions tangent to the manifold: the partial derivatives with respect to \(\sigma_{A}\) and \(\sigma_{B}\) are zero (as operators), and partial derivatives with respect to \(\alpha,\beta\) enforce the constraints. Using Equation (A3) and the entropy identities, we can write the differential of \(D^{Q}_{JS}\) for a variation in \(\sigma=\sigma_{A}\otimes\sigma_{B}\). For an arbitrary small change \(\delta\sigma=\delta\sigma_{A}\otimes\sigma_{B}+\sigma_{A}\otimes\delta\sigma _{B}\) (holding \(\rho\) fixed), one finds to first order:  
\[\delta D^{Q}_{JS}(\rho\|\sigma)\ =\ -\frac{1}{2}\mathrm{Tr}\Big{[}\left(\log \tfrac{\rho+\sigma}{2}+I\right)\delta\sigma\Big{]}\ +\ \frac{1}{2}\mathrm{Tr}\big{[}\left(\log\sigma+I\right)\delta\sigma\big{]}\,.\]  
This uses the fact that \(\delta S(X)=-\mathrm{Tr}[\left(\log X+I\right),\delta X]\) and \(S(\sigma_{A}\otimes\sigma_{B})=S(\sigma_{A})+S(\sigma_{B})\). Simplifying, we get the gradient (unconstrained) of the divergence with respect to \(\sigma\):  
\[\
abla_{\sigma}\,D^{Q}_{JS}(\rho\|\sigma)\ =\ -\ \frac{1}{2}\Big{[}\log\Big{(} \frac{\rho+\sigma}{2}\Big{)}\ -\ \log\sigma\Big{]}\,,\]  
which indeed vanishes if and only if \(\log[(\rho+\sigma)/2]=\log\sigma\) (i.e., \(\sigma\) commutes with \(\rho\) and equals \((\rho+\sigma)/2\)). However, this condition cannot be directly applied here because we are restricting \(\sigma\) to be a product state--we must account for the constraints \(\sigma=\sigma_{A}\otimes\sigma_{B}\). Thus, instead of setting the full gradient to zero, we set the projected gradients on each factor to zero, including the Lagrange terms for normalization. Stationarity for each factor.Taking the functional derivative of \(\mathcal{L}\) with respect to \(\sigma_{A}\) (treating \(\sigma_{B}\) as fixed) and setting it to zero, we obtain  
\[\mathrm{Tr}_{B}\Big{[}\Big{(}\log\!\frac{\rho+\sigma^{*}}{2}\Big{)}\ \big{(}I_{A}\otimes\sigma_{B}^{*}\big{)}\Big{]}\ =\ \log\sigma_{A}^{*}\ +\ \lambda_{A}\,I_{A}\,.\] (A4)  
Here \(\mathrm{Tr}_{B}[\cdot]\) denotes partial trace over subsystem \(B\), and \(\lambda_{A}\) is a constant related to the Lagrange multiplier \(\alpha\). Equation (A4) is an operator equation on subsystem \(A\). Intuitively, it says that the reduced operator (on \(A\)) of the \"halfway\" state \((\rho+\sigma^{*})/2\) has the same eigenbasis as \(\sigma_{A}^{*}\) and differs from \(\sigma_{A}^{*}\) only by a scalar shift in its logarithm (ensuring the trace condition).