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Some quantities, such as the entanglement of formation, are defined using a quantity that is minimized over all possible decompositions of a mixed state. A closed form can be found for this in some cases. But what about if no closed form is known? If a have a mixed state for some general $d$-dimensional quantum system, how do I determine all of its possible decompositions?

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  • $\begingroup$ Once you have one decomposition, you have them all by applying unitaries. You can read about it under "unitary freedom of ensemble decomposition" (or similar) in Nielsen&Chuang or en.wikipedia.org/wiki/Density_matrix under "Formulation". $\endgroup$ – Martin Jan 8 '15 at 17:16
  • $\begingroup$ Thanks, but surely this applies only to decompositions into orthogonal states. $\endgroup$ – Matthew Matic Jan 8 '15 at 17:19
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You can find an explanation of how to obtain all ensemble decompositions here: Proving the unitary relation of ensemble decompositions. As Martin said in his comment, they are related by unitaries (or, more precisely, (partial) isometries), regardless of whether the decomposition consists of orthogonal states. Most conveniently, you would start from the eigenvalue decomposition $\rho=\sum p_i\lvert\phi_i\rangle\langle\phi_i\rvert$ and obtain all others by applying isometries, $$\rho=\sum q_i \lvert\psi_i\rangle\langle\psi_i\rvert\ ,$$ where $\sqrt{q_i} \lvert\psi_i\rangle = \sum v_{ij} \sqrt{p_j} \lvert\phi_j\rangle$, and $\sum v_{ij}v_{ij'}=\delta_{jj'}$.

Of course, knowing this parametrization still does not mean that you can compute the entanglement of formation since this typically is a hard optimization problem, and the range of $j$ in the optimal decomposition can be larger than $d$ (but it can be bounded by $d^2$).

Note that if you know one decomposition which is optimal for Entanglement of Formation for a given state $\rho$, you can obtain the optimal decomposition for other states by simply shifting the weights $q_i$ in the optimal decomposition.

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  • $\begingroup$ Could you kindly list some reference about the last paragraph in your answer? Many thanks! $\endgroup$ – Eden Harder Jan 9 '15 at 12:30
  • $\begingroup$ @EdenHarder I don't know where this is written up, but I'm sure it can be found somewhere. The proof is not very difficult, though; if you are interested in the proof, you can post a question and I can post it as an answer. $\endgroup$ – Norbert Schuch Jan 9 '15 at 15:38
  • $\begingroup$ Hi, I post a question here. $\endgroup$ – Eden Harder Jan 10 '15 at 4:25

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