Suppose we have two distinct states $\rho,\sigma\in \mathcal{H}_A$. Define the following state

$$\omega = \frac{1}{2}(\rho^{\otimes n} + \sigma^{\otimes n}) \in \mathcal{H}_A^n$$

Let $\mathcal{H}_R\cong \mathcal{H}_A$ and let $n$ copies of $\mathcal{H}_R$ be used to purify $\omega$. After playing around a bit, I think it holds that the purification $\vert\Psi\rangle_{A^nR^n}$ cannot be expressed as some $\vert\psi\rangle_{AR}^{\otimes n}$. I believe even if $|R|>|A|$ is allowed, this still cannot be done i.e. the purification space can be arbitrarily large but the product state structure cannot be acheived for the purification.


To put it more precisely, prove that the following statement is false.

For all quantum states of the form $\omega = \frac{1}{2}(\rho^{\otimes n} + \sigma^{\otimes n}) \in \mathcal{H}_A^n$, there exists a pure product state $\vert\psi\rangle^{\otimes n}_{AR}\in \mathcal{H}^n_{AR}$ such that $\text{Tr}_{R^n}(\vert\psi\rangle\langle\psi\vert^{\otimes n}_{AR}) = \omega$.

On a related note, when is the product state purification possible? For instance, when $\rho = \sigma$, $\omega = \rho^{\otimes n}$ and now, there is a product state that purifies $\omega$. Is there a weaker condition on $\rho$ and $\sigma$ that still allows the purification to be a product state?

If $\rho$ and $\sigma$ satisfy ... , there exists a state a pure product state $\vert\psi\rangle^{\otimes n}_{AR}\in \mathcal{H}^n_{AR}$ such that $\text{Tr}_{R^n}(\vert\psi\rangle\langle\psi\vert^{\otimes n}_{AR}) = \omega$.

  • $\begingroup$ Can you please phrase this in a more precise way, i.e. make clear the order of "there exists psi", "for all N", "given some N", and so on? $\endgroup$ Commented May 14, 2020 at 20:36
  • $\begingroup$ @NorbertSchuch I have edited. I hope the question is clearer now? $\endgroup$ Commented May 14, 2020 at 21:16
  • $\begingroup$ Does my answer answer it? $\endgroup$ Commented May 14, 2020 at 21:45

1 Answer 1


This cannot be true - assuming that I understand the question correctly - since for instance, for $\rho=|0\rangle\langle0|$ and $\sigma=|1\rangle\langle1|$, the measurement outcomes on all $n$ qubits in $\omega$ are perfectly correlated.

On the other hand, on any purification of the form $|\psi\rangle_{AR}^{\otimes n}$, the measurement outcomes on the $A$ systems are entirely uncorrelated.

I'm sure this can be turned into a quantitative argument as well.

  • $\begingroup$ Thank you, that's a nice argument. If I understand correctly, you have also reduced my question to the following: When is a convex combination of product states itself a product state? $\endgroup$ Commented May 14, 2020 at 22:47
  • 1
    $\begingroup$ @user1936752 I guess so. In any case, the convex combination of product state reminds me of the quantum de Finetti theorem, in case this is relevant for you. $\endgroup$ Commented May 14, 2020 at 23:09

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