# Is the set of classical-quantum states convex?

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?

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.
• 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? – qquery Dec 29 '18 at 3:25