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 $\sum_i p_i |\psi_i⟩⟨\psi_i| = \sum_i q_i |\phi_i⟩⟨\phi_i|$.


2 Answers 2


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).

  • $\begingroup$ 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. $\endgroup$
    – Gv26
    Jul 22, 2021 at 15:00
  • 1
    $\begingroup$ 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 $\endgroup$
    – DaftWullie
    Jul 22, 2021 at 15:24
  • $\begingroup$ 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 $\endgroup$ Jul 23, 2021 at 15:05
  • $\begingroup$ I don't conclude that a convex combination of entangled states is entangled. I prove that they are entangled by checking the PPT criterion. $\endgroup$
    – DaftWullie
    Jul 23, 2021 at 15:29
  • $\begingroup$ $\sigma_1$ 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 $|a\rangle=|b\rangle$. $\endgroup$
    – DaftWullie
    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 $|\psi_1⟩⟨\psi_1| + |\psi_2⟩⟨\psi_2| = |\phi_1⟩⟨\phi_1| + |\phi_2⟩⟨\phi_2|$.


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