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What is the quantum information equivalent of a classical probability random variable ? Is it a density matrix or an observable ? If so can someone show me how to write a random variable that follows a uniform law , and how to compute it expectation and probabilities using quatum information quantities. Perhaps my question is not well formulated that because i am still confused about this topic. Thanks

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If I understand your question, the way a classical random variable $X$ with support $\left[2^n\right]=\left\{0,\cdots,2^n-1\right\}$ is represented in quantum information is via a diagonal density matrix $\rho_X$ such that its diagonal entries are equal to the associated probability for $X$. That is, we have: $$\langle x|\rho_X|y\rangle=\begin{cases}0&\text{if }x\neq y\\\mathbb{P}[X=x]&\text{otherwise}\end{cases}$$ where $\langle x|\rho_X|y\rangle$ represents the coefficient at the $x$-th line and $y$-th column of $\rho_X$.

$\rho_X$ represents a quantum state that is prepared in the state $|x\rangle$ with probability $\mathbb{P}[X=x]$. I don't know whether there are some quantities that qualify this quantum state that would lead to $\mathbb{E}[X]$.

However, some quantities defined for quantum states are a generalization of classical quantities, and should match the classical definition for diagonal density matrices, which justifies this choice for representing a classical random variable.

For instance, the von Neumann entropy of $\rho_X$ is: $$S\left(\rho_X\right)=-\mathrm{tr}\left(\rho_X\ln\left(\rho_X\right)\right)$$ Since $\rho_X$ is diagonal, $\ln\left(\rho_x\right)$ is obtained by applying $\ln$ on every diagonal element of $\rho$. We then multiply all these coefficients by those of $\rho_X$ and sum them to get the trace and voilà, we're back to the Shannon entropy of $X$.

In particular, the density matrix representing the uniform distribution on $\left[2^n\right]$ is the maximally mixed state $\frac{1}{2^n}I_{2^n}$.

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  • $\begingroup$ Thank you for your answer, but i am still not entirely satisfied. For example, in the book by Mark M Wilde (chapter 4.1.4) he says that Observable are the generalisation of random variables while density matrices are the generalisation of probability densities. But his notation with the sum of x x x is a little bit disturbing for me. Another thing that doesn't satisfy me with the definition using only density matrices is that bernouilli(p), bernoulli(1-p), rademacher(p) would all have the same density matrix while they are evidently differents $\endgroup$
    – yosh
    Nov 16, 2023 at 13:21
  • $\begingroup$ Another possible issue that i see in this definition is that we can only generalise finite (perhaps discrete too, by considering "infinite" matrices) RV . What would be the density matrix of an exponential law ? $\endgroup$
    – yosh
    Nov 16, 2023 at 13:26

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