= Partial trace in a nonorthogonal product expansion
{title2=$\Psi=\sum_j\xi_j\otimes\eta_j,\quad S_1=\sum_{j,k}\langle\eta_k,\eta_j\rangle|\xi_j\rangle\langle\xi_k|$}
The vectors need not form <orthonormal bases>. With the <inner product> linear in its second entry, expand the rank-one operator $|\Psi\rangle\langle\Psi|$ and take the trace on the second factor. This gives the displayed reduced positive operator. For every <observable> $A$ on the first factor,
$$
\langle\Psi,(A\otimes I)\Psi\rangle=\operatorname{Tr}(AS_1),\qquad\operatorname{Tr}S_1=\|\Psi\|^2.
$$
Positivity also follows directly from $\langle v,S_1v\rangle=\|\sum_j\langle v,\xi_j\rangle\eta_j\|^2$. Thus the physical <density operator> is $S_1/\|\Psi\|^2$ when the nonzero source vector is not normalized. The <Gram matrix> of the second-factor vectors supplies the cross coefficients; replacing those coefficients by Kronecker deltas without checking <orthonormality> changes the subsystem state. This formula is useful when an efficient product expansion uses nonorthogonal vectors, even though the <partial trace> itself is basis-independent.
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