For a bipartite system: $\mathcal{H}=\mathcal{H}_{a}\otimes\mathcal{H}_{b}$ described by a density operator $\hat{\rho}_{ab}$, I can promote it to a vector in the Liouville space, $|\hat{\rho}_{ab}\rangle\rangle$. The Schmidt decomposition for such a vector would be: $$ |\hat{\rho}_{ab}\rangle\rangle=\sum_{n}\gamma_{n}|\hat{\sigma}_{n}\rangle\rangle\otimes|\hat{\tau}_{n}\rangle\rangle $$ where $\{|\hat{\sigma}_{n}\rangle\rangle\}$ and $\{|\hat{\tau}_{n}\rangle\rangle\}$ will be orthonormal (in terms of the Hilbert-Schmidt inner product) basis of $\mathcal{D}(\mathcal{H}_{a})$ and $\mathcal{D}(\mathcal{H}_{b})$ respectively. Is there any argument to relate those bases to the physical reduced density matrix for the subsystems A and B?

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    $\begingroup$ Can you clarify what kind of relation you envision? You mean similar as the Schmidt basis is the eigenbasis of the reduced density matrix? You could define a "super-reduced density matrix" (or reduced super-density-matrix?) and make an analogous statement, of course. $\endgroup$ Commented Jan 31 at 9:27
  • $\begingroup$ I was thinking that since the linear state vector space, the linear superoperator space and the Liouville space are isomorphic there must be some relationship between the "super-reduced density matrix" and the reduced density matrix acting on the state vector space. However I cannot see the physical differences between them. Could it be possible that the reduced density matrix captures the quantum dynamics of the subsystem state by itself, while the "super-reduced density matrix" captures the dynamics of the same subsystem within a larger framework of the joint system? $\endgroup$
    – Oscarcillo
    Commented Feb 6 at 12:50

1 Answer 1


I don't think there's any general relation to be found here, and my argument is the entangled (as easily seen via PPT) state $$ \rho=\begin{pmatrix} \frac{1}{2} & 0 & \frac{1}{8} & \frac{1}{8} \\ 0 & 0 & 0 & 0 \\ \frac{1}{8} & 0 & \frac{1}{4} & \frac{1}{8} \\ \frac{1}{8} & 0 & \frac{1}{8} & \frac{1}{4} \end{pmatrix} $$ Applying the Schmidt decomposition to the vectorized state $|\rho\rangle\!\rangle$ yields $$ |\rho\rangle\!\rangle=\frac58|\sigma_1\rangle\!\rangle\otimes|\tau_1\rangle\!\rangle+\frac14|\sigma_2\rangle\!\rangle\otimes|\tau_2\rangle\!\rangle+\frac18|\sigma_3\rangle\!\rangle\otimes|\tau_3\rangle\!\rangle $$ with \begin{align*} |\sigma_1\rangle\!\rangle=\begin{pmatrix}\sqrt{\frac{2}{3}}\\0\\\frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\end{pmatrix}=|\tau_1\rangle\!\rangle\quad ,\quad |\sigma_2\rangle\!\rangle=\begin{pmatrix}-\frac{1}{\sqrt{3}}\\0\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{pmatrix}=|\tau_2\rangle\!\rangle\quad ,\quad |\sigma_3\rangle\!\rangle=\begin{pmatrix}0\\0\\-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix}= |\tau_3\rangle\!\rangle\,. \end{align*} Now the reduced states of $\rho$ read $$ {\rm tr}_A(\rho)=\begin{pmatrix} \frac{3}{4} & \frac{1}{8} \\ \frac{1}{8} & \frac{1}{4} \end{pmatrix}\quad\text{ and }\quad {\rm tr}_B(\rho)=\begin{pmatrix} \frac{1}{2} & \frac{1}{8} \\ \frac{1}{8} & \frac{1}{2} \end{pmatrix} $$ but $|\sigma_j\rangle\!\rangle,|\tau_j\rangle\!\rangle$ all feature a zero entry and, more gravely, the same holds true for any linear combination of these vectors. Thus neither $|{\rm tr}_A(\rho)\rangle\!\rangle$ nor $|{\rm tr}_B(\rho)\rangle\!\rangle$ can be expressed (linearly) using the Schmidt basis vectors $|\sigma_j\rangle\!\rangle,|\tau_j\rangle\!\rangle$.


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