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Consider a generic single-qubit state $$\rho=\lambda_1\lvert \lambda_1\rangle\!\langle \lambda_1\rvert+\lambda_2\lvert \lambda_2\rangle\!\langle \lambda_2\rvert\in\mathcal H_S.$$ I am interested in understanding what are the possible extensions of $\rho$, that is, the states $\tilde\rho\in\mathcal H_{SE}$ such that $\operatorname{tr}_E(\tilde\rho)=\rho.$ If is relatively easy to find the general structure of extensions that are pure, but less so in the more general case of non-pure extensions.

In particular, is it possible to have a non-trivial extension of $\rho$ which is not a purification?

By non-trivial here I mean that it must also decrease the amount of uncertainty associated with $\rho$. This means no trivial extensions of the form $\tilde\rho=\rho\otimes\sigma$, and no extensions built by simply attaching a set of orthonormal states to the eigenvectors of $\rho$, that is, no extensions of the form $\tilde\rho=\sum_k \lambda_k \lvert\lambda_k\rangle\!\langle\lambda_k\rvert\otimes\sigma_k$ with $\lambda_k$ the eigenvalues of $\rho$.

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Sure. Just take any random purification with a large purifying space $\mathbb C^2\otimes C^d$, and trace the $\mathbb C^d$ component.

To give a randomly made up example, $$ \rho = \left(\begin{matrix} .25 & .20 & .10 & .05 \\ .20 & .25 & .00 & .05 \\ .10 & .00 & .25 & -.15 \\ .05 & .05 & -.15 & .25 \end{matrix}\right) $$ is a purification of the state $$ \rho_A = \left(\begin{matrix} .50 & .05 \\ .05 & .50 \end{matrix}\right)\ . $$

That the example is not compatible with the special forms $\tilde\rho$ you give above can be straightforwardly checked from the eigenvalues of $\rho$, which are incompatible with the forms $\tilde\rho$ you give above -- for both those $\tilde\rho$, it holds that the eigenvalues of $\rho$ can be written as a sum of two eigenvalues of $\tilde\rho$ each, which can be easily tested not to be the case.

To explain the last argument in more detail:
Let $\tilde\rho=\sum \lambda_k |\lambda_k\rangle\langle\lambda_k|\otimes\sigma_k$ (which includes the first purification if all $\sigma_k$ are equal). Denote by $\mu_i(\sigma_k)$ the eigenvalues of $\sigma_k$. Then, the eigenvalues of $\tilde\rho$ are $$ \tau_{i,k} = \lambda_k\,\mu_i(\sigma_k)\ . $$ Thus, we have that $$ \sum_i\tau_{i,k} = \lambda_k $$ (as $\mathrm{tr}\,\sigma_k=1)$, i.e., each two (where "two" is the dimension of the purification) eigenvalues of $\tilde\rho$ add up to an eigenvalue $\lambda_k$ of $\rho$.
It can be easily checked that this property does not hold for the example.


Note that this richness of extensions is exactly a problem in computing the squashed entanglement, where one optimizes over (non-pure) extensions of arbitrary dimensions.

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  • $\begingroup$ ah, I think I see it now, thanks. Just a few notes. I guess you meant to write "can be written as product of two eigenvalues (which still does not seem quite accurate, but I get what you mean)? Also, do you know of some reference discussing the general structure of state extensions? $\endgroup$ – glS Nov 6 '18 at 9:16
  • $\begingroup$ @glS Indeed, product. And I agree it is a bit ambigous, but with a bit of thought it should be clear? (I can write a formula if you think it helps. I admit that I only came up with that argument while I was typing up the answer -- originally I just said: "It's random, so it will not have those properties."). Not sure where to find sth. about this. You can try papers on squashed entanglement (or Christandl's PhD thesis?). $\endgroup$ – Norbert Schuch Nov 6 '18 at 10:28
  • $\begingroup$ yes yes it's not a problem. I would say that the spectrum of $\rho\otimes\sigma$ is given by the products of the eigenvalues of $\rho$ and $\sigma$, and for the other case we have a straightforward generalisation of this. I'll check that out, thanks $\endgroup$ – glS Nov 6 '18 at 10:31
  • $\begingroup$ @glS Wait, no, I should have thought first (not enough coffee?). I do indeed mean sum: If you trace the extension, you see that the eigenvalues are $\lambda_k\,\mathrm{tr}\sigma_k$, which are indeed the sum of two eigenvalues of the extension (namely $\lambda_k\,\mu_i(\sigma_k)$, with $\mu_i(\sigma_k)$ the eigenvalues of $\sigma_k$. $\endgroup$ – Norbert Schuch Nov 6 '18 at 10:34
  • $\begingroup$ Have edited the answer. $\endgroup$ – Norbert Schuch Nov 6 '18 at 10:40

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