# quantum generalisation of random variables

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

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 $$\frac{1}{2^n}I_{2^n}$$.

• 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
– yosh
Nov 16, 2023 at 13:21
• 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 ?
– yosh
Nov 16, 2023 at 13:26