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Consider a bipartite quantum system described by the density operator, $\hat{\rho}$, an operator acting on the Hilbert space $\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. This matrix will be a vector (that in quantum physics is denoted with a double bra or ket $|\rho\rangle\rangle$) in the Liouville space, $\mathcal{L}=\mathcal{H}\otimes\mathcal{H}^{\dagger}$. Since the Liouville space is also a Hilbert space (when equipped with the Hilbert-Schmidt inner product) I can use the Schmidt decomposition on $|\rho\rangle\rangle$: $$ |\rho\rangle\rangle=\sum_{k}\alpha_{k}|\sigma_{k}\rangle\rangle\otimes|\Gamma_{k}\rangle\rangle.$$ My question is where does the Schmidt basis belong to? Does this make sense: $|\sigma_{k}\rangle\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{A}^{\dagger}$ and $|\Gamma_{k}\rangle\rangle\in\mathcal{H}_{B}\otimes\mathcal{H}_{B}^{\dagger}$?

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  • $\begingroup$ It seems very unconventional to denote the dual space with a dagger. $\endgroup$ Commented Feb 20 at 13:08

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The density matrix can be written as $$ \rho = \sum_{ijkl} \rho_{ijkl} |i\rangle_A|j\rangle_B \langle k|_A\langle l |_B\;, $$ where $|i\rangle_A,|j\rangle_B, \langle k|_A$ and $\langle l |_B$ are respectively bases for $\mathcal{H}_A^\dagger$, $\mathcal{H}_B^\dagger$, $\mathcal{H}_A$ and $\mathcal{H}_B$.

Your Schmidt decomposition groups terms like this $$ \rho = \sum_{ijkl} \rho_{ijkl} ( |i\rangle_A\langle k|_A)(|j\rangle_B \langle l |_B) $$ before diagonalizing. As you can see the left factor, that will form your $|\sigma_k\rangle\rangle$ basis is indeed from $\mathcal{H}_A\otimes\mathcal{H}_A^\dagger$ and similarly $|\Gamma_k\rangle\rangle \in \mathcal{H}_B\otimes\mathcal{H}_B^\dagger$

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  • $\begingroup$ Is this also true for the case of entangled states? $\endgroup$
    – Oscarcillo
    Commented Feb 28 at 13:23
  • $\begingroup$ Nothing I have written here assumes the state is not entangled $\endgroup$ Commented Feb 28 at 13:38
  • $\begingroup$ Yes, but what I don't understand then is how this result is not in contradiction with the separability criteria. The final expression is written as the convex sum of individual subsytems' density matrices, isn't it? $\endgroup$
    – Oscarcillo
    Commented Feb 28 at 14:00
  • $\begingroup$ No, the final expression is still a completely general expansion of a state in the tensor product space of operators. You need further (highly nontrivial) constraints of the form of $\rho_{ijkl}$ to determine if it can be decomposed into a convex sum of density matrices. I would suggest asking a new question if you want further clarifications $\endgroup$ Commented Feb 28 at 14:29
  • $\begingroup$ I see. Do you have any reference to read more about those constraints? $\endgroup$
    – Oscarcillo
    Commented Feb 28 at 14:59

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