# Is Classical-Classical-Quantum state equivalent to Classical-Quantum state?

Suppose we have the following CQ-state between two parties Alice & Bob $$\rho_{A B}^{\otimes n}=\sum_{x^n} p^n\left(x^n\right)\left|x^n\right\rangle\langle\left. x^n\right|^A \otimes \rho_{x^n}^B \tag{1}$$ where $$\left|x^n\right\rangle=\left|x_1\right\rangle \otimes \cdots \otimes\left|x_n\right\rangle, \quad p^n\left(x_n\right)=p\left(x_1\right) \cdots p\left(x_n\right) \; \& \; \rho_{x^n}^B=\rho_{x_1}^B \otimes \cdots \otimes \rho_{x_n}^B$$.

Now suppose there is another classical state chosen by Alice given as $$\left|y^n\right\rangle=\left|y_1\right\rangle \otimes \cdots \otimes\left|y_n\right\rangle \tag{2}$$ So the overall state is $$\hat{\rho}_{A B}^{\otimes n}=\sum_{x^n} p^n\left(x^n\right)\left|y^n\right\rangle\langle\left. y^n\right|^A \otimes \mid x^n\rangle\langle\left. x^n\right|^A \otimes \rho_{x^n}^B \tag{3}$$ Now if we define $$\quad\left|\tilde{x}_i\right\rangle^A=\left|y_i\right\rangle^A\left|x_i\right\rangle^A \quad$$ so $$\left|\tilde{x}^n\right\rangle=\left|y^n\right\rangle\left|x^n\right\rangle$$

$$^{*}$$Further since $$\left|y^n\right\rangle$$ is fined, $$p\left(\tilde{x}_i\right)=p\left(x_i\right)$$ and $$\rho_{x_i}$$ can be considered as a state associated with $$\tilde{x}_i$$.

Therefore changing notation we've $$\tilde{\rho}_{A B}^{\otimes n}=\sum_{\tilde{x}^n} p^n\left(\tilde{x}^n\right)\left|\tilde{x}^n\right\rangle\langle\left.\tilde{x}^n\right|^A \otimes \rho_{\tilde{x}^n}^B \tag{4}$$ which is the expression of a classical-quantum state.

So my question is can we consider the state $$\sum_{x^n} p^n\left(x^n\right)\left|y^n\right\rangle\langle\left. y^n|^A \otimes \mid x^n\rangle\langle\left. x^n\right|^A \otimes \rho_{x^n}^B\right.$$ equivalent to a classical-quantum state for a given $$y^n$$ and apply all the results of information theory to his state as that of any classical-quonlum state.

I'm not sure if my argument in (*) is correct. I would appreciate if you kindly correct me if I am wrong.

Furthermore, if we have the following state: $$\hat{\rho}_{A B}^{\otimes n}=\sum_{x, y} p(x) p(y)\left|y^n\right\rangle\langle\left. y^n\right|^A \otimes \mid x^n\rangle\langle\left. x^n\right|^A \otimes \rho_{x^n}^B \tag{5}$$ is this also equivalent to a classical-quantum state?

• what is going on to get from Eq 3 to Eq 4? Commented Oct 7, 2023 at 18:37

Say you have a pair of sets $$\mathcal{X} = \{x(1), x(2), \dots, x(|\mathcal{X}|)\}$$ and $$\mathcal{Y} = \{y(1), y(2), \dots, y(|\mathcal{Y}|)\}$$. Then we can define random variables $$X, Y$$ taking values in the sets $$\mathcal{X}$$ and $$\mathcal{Y}$$ respectively according to a joint distribution ("mass function") $$p_{XY}(x, y) := \text{Probability}(X=x, Y=y)$$.
In this case, you are always allowed to define another random varaible $$\tilde{X}:= (X, Y)$$ taking values in $$\tilde{\mathcal{X}}:= \mathcal{X} \times \mathcal{Y}$$ with probability $$p_{\tilde{X}}(\tilde{x}):= \text{Probability}(\tilde{X} =\tilde{x})$$. In this case, of course it would be true that \begin{align} \sum_{x, y} p_{XY}(x, y) |x \rangle \langle x| \otimes |y \rangle \langle y| \otimes \rho_{xy}^B = \sum_{\tilde{x}} p_{\tilde{X}}(\tilde{x})|\tilde{x}\rangle \langle \tilde{x}| \otimes \rho_\tilde{x}^B, \tag{1} \end{align} where you could choose $$x := (x_1, \dots, x_n)$$ and $$y = (y_1, \dots, y_n)$$ to recover your setting. Compare to Eqs. $$(3)$$-$$(4)$$, where you changed $$\rho_x^B$$ to $$\rho_{\tilde{x}}^B$$, which is incorrect. Notice how $$\rho_{xy}^B$$ on the LHS is indexed by both $$x$$ and $$y$$ - this associate each possible state prepared in system $$B$$ with a pair of outcomes $$(x, y)$$ sampled in Alice's classical system. This is generally what people associate with a classical-quantum state: Each possible state prepared for Bob is associated with exactly one classical outcome in Alice's register.
If you prepare a state $$\rho_{x}^B$$ that depends only on $$x$$ - basically your Eq. $$(5)$$ - then you can basically factor out the $$Y$$ variable from Alice's system: \begin{align} \sum_{x, y} p_{XY}(x, y) &|x \rangle \langle x|^{A_1} \otimes |y \rangle \langle y|^{A_2} \otimes \rho_{x}^B \\&= \sum_x p_X(x) |x\rangle \langle x|^{A_1} \otimes \left( \sum_y p_{Y|X}(y|x) |y\rangle \langle y|^{A_2} \right) \otimes \rho_x^B. \tag{2} \end{align} In this case, $$A_2$$ contains a classical distribution that Alice samples from, but doesn't use to determine which state $$\rho_x^B$$ is prepared. This isn't really what "classical-quantum state" usually refers to, since there's an extra variable $$Y$$ floating around. But here are a few attempts at operational interpretations of such a state:
• If Bob is doing state discrimination (guess the value of $$x$$ given $$\rho_x^B$$), nothing changes: His ability to succeed is still bounded by the mutual information $$I(A_1:B)$$ via Holevo's theorem.
• If Bob wants to also guess the value of $$y$$, it won't be any harder than guessing $$x$$, since you can show $$$$I(A_1 A_2:B) = I(A_1:B), \tag{3}$$$$ which roughly states the amount of information system $$B$$ contains about the value of $$x$$ (stored in $$A_1$$) doesn't change if you consider system $$A_2$$.
• If Bob wants to guess the value of $$y$$ given $$B$$, the relevant thing to compute would be $$I(A_2:B)$$. This will depend in part on how correlated $$A_1$$ and $$A_2$$ are, for example.