# Can a single-qubit state be nontrivially extended to a non-pure state?

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 $$\rho$$.

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.

• 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? – glS Nov 6 '18 at 9:16
• @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?). – Norbert Schuch Nov 6 '18 at 10:28
• 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 – glS Nov 6 '18 at 10:31
• @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 $\sigma_k$. – Norbert Schuch Nov 6 '18 at 10:34
• Have edited the answer. – Norbert Schuch Nov 6 '18 at 10:40