# Schmidt decomposition of density operators

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 $$|\Gamma_{k}\rangle\rangle\in\mathcal{H}_{B}\otimes\mathcal{H}_{B}^{\dagger}$$?

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

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 $$|\Gamma_k\rangle\rangle \in \mathcal{H}_B\otimes\mathcal{H}_B^\dagger$$

• Is this also true for the case of entangled states? Commented Feb 28 at 13:23
• Nothing I have written here assumes the state is not entangled Commented Feb 28 at 13:38
• 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? Commented Feb 28 at 14:00
• 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 Commented Feb 28 at 14:29
• I see. Do you have any reference to read more about those constraints? Commented Feb 28 at 14:59