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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?

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  • $\begingroup$ what is going on to get from Eq 3 to Eq 4? $\endgroup$
    – forky40
    Oct 7, 2023 at 18:37

1 Answer 1

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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 \begin{equation} I(A_1 A_2:B) = I(A_1:B), \tag{3} \end{equation} 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.

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