# Can one always find purifications which preserve equality of statistical mixtures?

When pure states $$|\psi_1⟩$$, $$|\psi_2⟩$$ and $$|\phi_1⟩$$, $$|\phi_2⟩$$ in $$\mathcal{H}_A \otimes \mathcal{H}_B$$ have identical statistical mixtures $$\frac{1}{2}(|\psi_1⟩⟨\psi_1| + |\psi_2⟩⟨\psi_2|) = \frac{1}{2}(|\phi_1⟩⟨\phi_1| + |\phi_2⟩⟨\phi_2|) ,$$ then we know (by linearity of the partial trace) that the reduced states $$\rho_i = \text{tr}_B |\psi_i⟩⟨\psi_i|$$ and $$\sigma_i = \text{tr}_B |\phi_i⟩⟨\phi_i|$$ on the space $$\mathcal{H}_A$$ also have identical mixtures $$\frac{1}{2}(\rho_1 + \rho_2) = \frac{1}{2}(\sigma_1 + \sigma_2)$$. My question concerns the converse of this statement.

Let $$\rho_1$$, $$\rho_2$$ and $$\sigma_1$$, $$\sigma_2$$ be density operators on a finite-dimensional space $$\mathcal{H}$$ which satisfy $$\rho_1 + \rho_2 = \sigma_1 + \sigma_2 .$$ Do there always exist corresponding purifications $$|\psi_1⟩$$, $$|\psi_2⟩$$ and $$|\phi_1⟩$$, $$|\phi_2⟩$$ in $$\mathcal{H} \otimes \mathcal{P}$$ (for some purifying space $$\mathcal{P}$$) of the $$\rho_1$$, $$\rho_2$$ and $$\sigma_1$$, $$\sigma_2$$ which satisfy $$|\psi_1⟩⟨\psi_1| + |\psi_2⟩⟨\psi_2| = |\phi_1⟩⟨\phi_1| + |\phi_2⟩⟨\phi_2| ?$$

More generally, I would like to know about when (if at all) identical mixtures of the form $$\sum_i p_i \rho_i = \sum_i q_i \sigma_i$$ imply the existence of purifications such that $$\sum_i p_i |\psi_i⟩⟨\psi_i| = \sum_i q_i |\phi_i⟩⟨\phi_i|$$.

One data point for the general case (that indicates it's not always possible): $$\rho_1=|0\rangle\langle0|$$, $$\rho_2=|1\rangle\langle1|$$, $$p_1p_2\neq0$$ while $$\sigma_1=p_1\rho_1+p_2\rho_2$$ and $$q_1=1$$. Note that the purifications of the left-hand side are separable, so $$p_1|\psi_1\rangle\langle\psi_1|+p_2|\psi_2\rangle\langle\psi_2|$$ is separable. Meanwhile, the purification of $$\sigma_1$$ must be entangled, so $$q_1|\phi_1\rangle\langle\phi_1|+q_2|\phi_2\rangle\langle\phi_2|$$ is entangled. Thus, the two things are different.

I think I have a similar variant for the special case you're after. Let $$\rho_1=|0\rangle\langle 0|,\quad \rho_2=\frac14(|0\rangle+\sqrt{3}|1\rangle)(\langle 0|+\sqrt{3}\langle 1|), \quad\sigma_1=\frac14(\sqrt{3}|0\rangle+|1\rangle)(\sqrt{3}\langle0|+\langle 1|)$$ This implies that $$\sigma_2=\rho_1+\rho_2-\sigma_1=I/2.$$ As above, the purifications of $$\rho_1$$ and $$\rho_2$$ must be separable, meaning that $$|\psi_1\rangle\langle\psi_1|+|\psi_2\rangle\langle\psi_2|$$ is separable. Thus, our aim is to show that for every possible purification of $$\sigma_1$$ and $$\sigma_2$$, $$\Phi=|\phi_1\rangle\langle\phi_1|+|\phi_2\rangle\langle\phi_2|$$ is entangled. My plan to show this is to take the partial transpose and calculate its determinant, $$\text{det}(\Phi^{T_A})$$. This being negative is a sufficient condition for detecting entanglement.

Let us set $$|\phi_1\rangle=\frac12(\sqrt{3}|0\rangle+|1\rangle)|\gamma\rangle,\quad|\phi_2\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).$$ Note that the only important thing in the range of possible purifications is a relative unitary, which I've incorporated into the arbitrary state $$|\gamma\rangle$$. Note that, strictly, $$|\gamma\rangle$$ could live in a hilbert space of dimension greater than 2 (3 should be sufficient), although in the following I've assumed it's Hilbert space dimension 2.

If you let $$|\gamma\rangle=\cos x|0\rangle+\sin xe^{iy}|1\rangle,$$ then $$\text{det}(\Phi^{T_A})=\frac{1}{32} \left(-\sqrt{3} \sin (2 x) \cos (y)-\cos (2 x)-4\right)$$ (I did this in Mathematica). The largest this can be is $$-1/16$$, i.e. negative. So the state is necessarily entangled. There do not always exist purifications such that $$|\psi_1\rangle\langle\psi_1|+|\psi_2\rangle\langle\psi_2|=|\phi_1\rangle\langle\phi_1|+|\phi_2\rangle\langle\phi_2|$$ (up to the caveat that I haven't considered $$|\gamma\rangle$$ in full generality).

• It's impossible to use this example but with $\sigma_2 = \sigma_1$ to resolve also the special case, right? One could choose two different purifications of $\sigma_1$, both necessarily entangled, whose mixture (with $q_1 = q_2 = 1/2$) is separable.
– Gv26
Jul 22, 2021 at 15:00
• Yes, it doesn't work with the special case, which is why I didn't answer for the special case! I suspect there probably is a variant of this approach that will answer your special case, but haven't had a chance to look carefully yet Jul 22, 2021 at 15:24
• Why do you conclude that a convex combination of entangled states is itself entangled? Counterarguments are quite straightforward; for example equal combinations of the four Bell states Jul 23, 2021 at 15:05
• I don't conclude that a convex combination of entangled states is entangled. I prove that they are entangled by checking the PPT criterion. Jul 23, 2021 at 15:29
• $\sigma_1$ is pure, so the purification is separable. If you take the partial trace of your proposed $|\phi_1\rangle$, you'll get a mixed state outcome unless $|a\rangle=|b\rangle$. Jul 23, 2021 at 15:30

It can be shown by direct calculation that the counterexample provided in the answer of DaftWullie holds in general. As in that answer, define $$\rho_1 = |0⟩⟨0| ,$$ $$\rho_2 = \frac{1}{4}(|0⟩ + \sqrt{3}|1⟩)(⟨0| + \sqrt{3}⟨1|) ,$$ $$\sigma_1 = \frac{1}{4}(\sqrt{3}|0⟩ + |1⟩)(\sqrt{3}⟨0| + ⟨1|) ,$$ and, so that $$\rho_1 + \rho_2 = \sigma_1 + \sigma_2$$ is satisfied, $$\sigma_2 = \rho_1 + \rho_2 - \sigma_1 = \frac{1}{2}(|0⟩⟨0| + |1⟩⟨1|) .$$ Consider purifications of these states $$|\psi_1⟩ = |0⟩ |\alpha⟩ ,$$ $$|\psi_2⟩ = \frac{1}{2}(|0⟩ + \sqrt{3}|1⟩) |\beta⟩ ,$$ $$|\phi_1⟩ = \frac{1}{2}(\sqrt{3}|0⟩ + |1⟩) |\gamma⟩ ,$$ $$|\phi_2⟩ = \frac{1}{\sqrt{2}}(|0⟩ |0⟩ + |1⟩ |1⟩) ,$$ where $$|\alpha⟩$$, $$|\beta⟩$$, and $$|\gamma⟩$$ are arbitrary state vectors in the purifying space (which is also arbitrary). These account for all possible purifications since, by the Schrödinger–HJW theorem, all purifications of a state differ only by a unitary transformation acting on the purifying space.

Suppose for a contradiction that $$|\psi_1⟩⟨\psi_1| + |\psi_2⟩⟨\psi_2| = |\phi_1⟩⟨\phi_1| + |\phi_2⟩⟨\phi_2|$$. Expanding this out gives \begin{alignat}{2} &|0⟩⟨0| \otimes \bigg( |\alpha⟩⟨\alpha| + \frac{1}{4}|\beta⟩⟨\beta| \bigg) & {}+{} &|1⟩⟨0| \otimes \frac{\sqrt{3}}{4}|\beta⟩⟨\beta| \\ {}+{} &|0⟩⟨1| \otimes \frac{\sqrt{3}}{4}|\beta⟩⟨\beta| & {}+{} &|1⟩⟨1| \otimes \frac{3}{4}|\beta⟩⟨\beta| \\ {}={} &|0⟩⟨0| \otimes \bigg( \frac{3}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|0⟩⟨0| \bigg) & {}+{} &|1⟩⟨0| \otimes \bigg( \frac{\sqrt{3}}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|1⟩⟨0| \bigg) \\ {}+{} &|0⟩⟨1| \otimes \bigg( \frac{\sqrt{3}}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|0⟩⟨1| \bigg) & {}+{} &|1⟩⟨1| \otimes \bigg( \frac{1}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|1⟩⟨1| \bigg) . \end{alignat} Since the $$|0⟩$$ and $$|1⟩$$ are orthogonal, we can equate each term on either side of this expression. In particular, from the terms with $$|1⟩⟨0|$$, we must have $$\frac{\sqrt{3}}{4}|\beta⟩⟨\beta| = \frac{\sqrt{3}}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|1⟩⟨0| .$$ From the terms with $$|0⟩⟨1|$$, we must have $$\frac{\sqrt{3}}{4}|\beta⟩⟨\beta| = \frac{\sqrt{3}}{4}|\gamma⟩⟨\gamma| + \frac{1}{2}|0⟩⟨1| .$$ Together, the previous two equalities result in $$|1⟩⟨0| = |0⟩⟨1|$$, which is false. Therefore, for this example, there are no purifications such that $$|\psi_1⟩⟨\psi_1| + |\psi_2⟩⟨\psi_2| = |\phi_1⟩⟨\phi_1| + |\phi_2⟩⟨\phi_2|$$.