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A separable state in $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$ is given by

$$\rho_{s}=\sum_{\alpha,\beta}p(\alpha,\beta)|\alpha\rangle\!\langle\alpha|\otimes|\beta\rangle\!\langle\beta|.$$

Now, my question is, can what can we say about the existence of $\{|\alpha \rangle\}$ and $\{|\beta \rangle\}$ such that all of them are elements from a complete basis (non-unique) in individual subsystem? Whether they exist? or for what kind state they exists (for example, in case zero discord state-they exist, as pointed out by one of the commentators1)

A reason I think such a set of basis exist for all separable state is because separable state space is the convex hull of tesnor products of symmetric rank-$1$ projectors which are affinely independent (Caratheodory's theorem), thus by definifion, there exists a set of linearly independent vectors for each subsystem (mostly, nonunique), let's say those are $\{\alpha\rangle\}$ and $\{|\beta\rangle\}$. Surely, we can add few more vectors to both suitably to span the whole space!! Is it true? Any help is appreciated.

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    $\begingroup$ See Schmidt decomposition ,link: en.wikipedia.org/wiki/Schmidt_decomposition $\endgroup$
    – KP99
    Jul 10, 2021 at 9:38
  • $\begingroup$ Isn't Schmidt decomposition work only for pure states? $\endgroup$
    – WInterfell
    Jul 10, 2021 at 11:14
  • $\begingroup$ You can use Schmidt decomposition technique to determine whether any given state is pure or not. Schmidt rank should be 1 for pure states $\endgroup$
    – KP99
    Jul 10, 2021 at 12:31
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    $\begingroup$ The question is missing details. Are you asking whether this holds for SOME state rho? For ALL states RHO? For ANY such decomposition? For SOME such decomposition? Also, the title does not seem to match the question. $\endgroup$
    – Norbert Schuch
    Jul 13, 2021 at 13:14
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    $\begingroup$ @glS Hm, the question certainly seems to ask: I am doing a separable decomposition of rho using pure state, what properties will those pure states have. Seems rather different from the title. (And your answer over at qc.se also does not seem to be a counterexample to the title question.) -- BTW, one more reason against cross-posts: The clarifying comments only appear in one place. $\endgroup$
    – Norbert Schuch
    Jul 13, 2021 at 13:20

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No, such a decomposition does not always exist.

To see this, we need two facts:

(1) The set of all separable state has the same dimension as the set of all density operators (it is a finite-volume subset) -- that is, it has $\approx D^4$ real parameters (if both systems have dimension $D$).

(2) The family you descibe is fully specified by the $p$ ($\approx D^2$ parameters) and the basis choice for the two bases, which are given by a $D\times D$ unitary matrix (i.e. $\approx D^2$ real parameters) each, i.e., $\approx 3D^2$ parameters in total.

Thus, the ansatz you describe has by far not enough parameters to describe all separable states.

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  • $\begingroup$ huh. Is there a way to get a sense of the type of states that are left stranded in this set difference? $\endgroup$
    – Emilio Pisanty
    Jul 13, 2021 at 20:46
  • $\begingroup$ @Emilio Any random separable state? -- More to the point, the issue is (intuitively) that for a basis $|\alpha\rangle$, $|\alpha\rangle\langle\alpha|$ is not a basis of the space of operators. (This made me think immediately that this should not work.) You would have to use e.g. a SIC-POVM $|\alpha\rangle$. -- Based on this intuition, I would guess sth. like $|0\rangle\langle0|\otimes|0\rangle\langle0|+|+\rangle\langle+|\otimes|+\rangle\langle+|+|1\rangle\langle1\otimes |1\rangle\langle1|$ could be such an example. $\endgroup$
    – Norbert Schuch
    Jul 13, 2021 at 20:52

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