# CPTP, Kraus representation and classical registers

What is the best mathematical representation of a quantum system that has some classical registers and some quantum registers?

I'm asking because I'm considering any "physical" process $$\pi()$$ that takes no input, but that outputs both classical and quantum values, like for example a tuple $$(x, \rho)$$, with $$x \in \{0,1\}^n$$ a classical string, and $$\rho$$ a quantum state on $$m$$ qubits, and I'd like to find a kind of Kraus representation of $$\pi$$.

I wanted to use the usual formalism of CPTP, but it's not clear in which Hilbert space lives $$(x, \rho)$$... Indeed, usually, a Hilbert space is stable by superposition, but here the first "register" cannot be in superposition. Of course, I could say that $$x$$ is in fact $$|x\rangle$$ and therefore the output of $$\pi()$$ is strictly contained in a Hilbert space of size $$2^n$$, but then $$\pi()$$ is not simply $$\sigma = \sum_i B_i 1 B_i^\dagger$$, as I'd also need to "measure" $$\sigma$$ to extract a random $$x$$, and then set $$\rho$$ to be the matrix $$\sigma$$ "postselected" on $$x$$.

Is there some better representation for such process that makes the difference between quantum outputs and classical outputs?

• I can also recommend Mark Wilde's great book From Classical to Quantum Shannon Theory, that you find on the arxiv: arxiv.org/abs/1106.1445 It contains large sections that concern themselves with Classical-to-Quantum channels, which should be what you're looking for. Commented Mar 12, 2020 at 20:54

There are some generalizations, though, like this https://en.wikipedia.org/wiki/Quantum_instrument. But in essence the idea is the same as you've described $$-$$ we model classical register with values $$x$$ as a quantum register with orthogonal states $$|x\rangle$$. You can model a measurement in this way, for example. If $$\sum_i B_iB_i^\dagger=I$$ describes a measurement, then $$\Phi(\rho) = \sum_i |i\rangle\langle i| \otimes B_i \rho B_i^\dagger = \sum_i |i\rangle\langle i| \otimes \frac{B_i \rho B_i^\dagger}{ \text{Tr}(B_i \rho B_i^\dagger)} \cdot \text{Tr}(B_i \rho B_i^\dagger)$$ is a corresponding quantum channel where the first quantum register of the output encodes the classical result $$i$$ and the second register contains the quantum state after the measurement.
• And also, I'm thinking that in order to be as general as possible, $B_i$ should apply to a map applied on $\rho$, not $\rho$ directly, and this map can itself be decomposed into a sum using the Kraus decomposition. Commented Apr 1, 2020 at 18:41