I read about the classical-quantum states in the textbook by Mark Wilde and there is an exercise that asks to show the set of classical-quantum states is not a convex set. But I have an argument to show it is a convex set. I wonder whether I made a mistake in my proof.

Here is the definition of classical-quantum states (definition 4.3.5):

The density operator corresponding to a classical-quantum ensemble $\{p_X(x)$, $|x\rangle\langle x|_X \otimes \rho_A^x\}_{x\in \mathcal{X}}$ is called a classical-quantum state and takes the following form: $$\rho_{XA} = \sum_{x \in \mathcal{X}} p_X(x) |x\rangle\langle x|_X \otimes \rho_A^x.$$

My argument about the set of classical-quantum states is convex is as follows. Let $\rho_{XA}$ and $\sigma_{XA}$ to be two arbitrary classical-quantum states. Specifically, we can write $$\rho_{XA} = \sum_{x \in \mathcal{I}_1} p_X(x) |x\rangle\langle x|_X \otimes \rho_A^x,$$ $$\sigma_{XA} = \sum_{x \in \mathcal{I}_2} q_X(x) |x\rangle\langle x|_X \otimes \sigma_A^x,$$ where $\mathcal{I}_1 = \{x: p_X(x) \neq 0\}$ and $\mathcal{I}_2 = \{x: q_X(x) \neq 0\}$.

Then we take the union $\mathcal{I}=\mathcal{I}_1 \cup \mathcal{I}_2$. We define $\rho_A^x$ to be an arbitrary density operator for $x \notin \mathcal{I}_1$ and similarly $\sigma_A^x$ to be an arbitrary density operator for $x \notin \mathcal{I}_2$.

We can then rewrite $\rho_{XA}$ and $\sigma_{XA}$ as $$\rho_{XA} = \sum_{x \in \mathcal{I}} p_X(x) |x\rangle\langle x|_X \otimes \rho_A^x,$$ $$\sigma_{XA} = \sum_{x \in \mathcal{I}} q_X(x) |x\rangle\langle x|_X \otimes \sigma_A^x.$$

Since we are adding zero operators, $\rho_{XA}$ and $\sigma_{XA}$ are not changed.

Then for any $\lambda \in (0,1)$, we want to show $\lambda \rho_{XA} + (1-\lambda) \sigma_{XA}$ is a classical-quantum state. (Note that the trivial case where $\lambda =1$ or $\lambda = 0$ just gives back $\rho_{XA}$ and $\sigma_{XA}$ back, respectively.)

We now define $\xi_{XA} :=\lambda \rho_{XA} + (1-\lambda) \sigma_{XA}.$ $$ \xi_{XA} =\lambda \sum_{x \in \mathcal{I}} p_X(x) |x\rangle\langle x|_X \otimes \rho_A^x + (1-\lambda) \sum_{x \in \mathcal{I}} q_X(x) |x\rangle\langle x|_X \otimes \sigma_A^x\\ =\sum_{x \in \mathcal{I}} |x\rangle\langle x|_X \otimes (\lambda p_X(x) \rho_A^x + (1-\lambda) q_X(x)\sigma_A^x) \\ =\sum_{x \in \mathcal{I}} w_X(x)|x\rangle\langle x|_X \otimes \xi_A^x, $$ where $w_X(x) = \lambda p_X(x) + (1-\lambda) q_X(x)$ and $\xi_A^x = \frac{\lambda p_X(x) \rho_A^x + (1-\lambda) q_X(x)\sigma_A^x}{\lambda p_X(x) + (1-\lambda) q_X(x)}$.

Since $X \in \mathcal{I}$, not both $p_X (x)=0$ and $q_X(x)=0$. For $\lambda \in (0,1)$, we have $w_X(x) \neq 0$ for $x \in \mathcal{I}.$

Also, $\sum_{x \in \mathcal{I}} w_X(x) = \sum_{X \in \mathcal{I}} \lambda p_X(x) + (1-\lambda) q_X(x) = 1$.

Therefore, the state $\xi_{XA}$ is a classical-quantum state. So, I conclude the set of classical-quantum states is convex.

Can anyone point out where I made a mistake?

Or is there a typo in the textbook?


1 Answer 1


Your mistake is that you assume that $\rho$ and $\sigma$ are classical-quantum in the same classical basis on $X$. However, there is no need to do so -- all which is necessary is that there exists such a basis, which can however depend on the state. As soon as you choose a different classical basis for the two states, your argument breaks down.

  • $\begingroup$ Thanks. I see. So, the set of classical-quantum states defined with the same classical basis is convex. But in general, the set of all classical-quantum states are not convex because there is no requirement for the same classical basis. Good to know. $\endgroup$
    – qquery
    Commented Dec 27, 2018 at 23:00
  • 1
    $\begingroup$ @qquery Well, you still need to prove the latter. $\endgroup$ Commented Dec 28, 2018 at 1:04
  • $\begingroup$ Yes. Then we can prove by a counterexample. $\rho_{XA} = \frac{1}{2}|0\rangle\langle 0| \otimes \rho_A^0 + \frac{1}{2}|1\rangle\langle 1| \otimes \rho_A^1$, and $\sigma_{XA} = \frac{1}{2}|+\rangle\langle +| \otimes \sigma_A^+ + \frac{1}{2}|-\rangle\langle -| \otimes \sigma_A^-$. For simplicity, we take $\sigma_A^+ = \rho_A^0$, $\sigma_A^- = \rho_A^1$ and $\rho_A^0 \neq \rho_A^1$. We can verify that $\frac{1}{2} \rho_{XA} + \frac{1}{2} \sigma_{XA}$ is not a classical-quantum state. What do you think? $\endgroup$
    – qquery
    Commented Dec 29, 2018 at 3:25

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