1
$\begingroup$

In Nielsen and Chuang's Quantum Computation and Quantum information there is an axiomatic definition of the quantum operation (as one of the 3 approaches to quantum operations).

A quantum operation is defined as a map between two sets of density operators satisfying 3 axioms. The axiom 1 states the trace of the output of the operation is less or equal to 1, i.e., $tr(\mathcal{E}(\rho))\leq 1$ which according to the book's explanation includes non-trace-preserving case when $tr(\mathcal{E}(\rho))<1$.

I understand in practical terms when it's useful to consider non-trace preserving maps (non-complete description of a process), but I don't understand how does it fit the definition of the quantum operation. It's defined as a map between spaces of density operators, so the output of the operation must be the density operator and as such it must have a trace equal to 1. Is Axiom 1 an attempt to introduce some more general notion of the quantum operation, that is not a map between spaces of density operators? If so, what is it?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The next paragraph explains this axiom 1. They

"make the convention that $\mathcal{E}$ is to be defined in such a way that $\mathrm{tr}[\mathcal{E}(\rho)]$ is equal to the probability of the measurement ouctcome described by $\mathcal{E}$ occurring."

Put in a different way, one should treat the output as $$\rho^\prime=\frac{\mathcal{E}(\rho)}{P}$$ conditional on the measurement outcome described by $\mathcal{E}$ actually occurring. This actually occurs with probability $P=\mathrm{tr}[\mathcal{E}(\rho)]$. So all of the information is contained in $\mathcal{E}(\rho)$, which is great.

Should another (unspecified) outcome occur, one requires some other quantum operation. Between the lines in this explanation, there is an assumption that all of the possible quantum operations should sum to unity.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The axiom itself is clear, but I'm still confused how it fits the definition of the quantum operation. If $tr[\mathcal{E}(\rho)]<1$, then $\mathcal{E}(\rho)$ is not a density operator, right? But the quantum operation is defined as a map between the sets of density operators, so it should transform original density operator $\rho$ into another density operator, i.e., $\mathcal{E}(\rho)$ must be a density operator and have the trace equal to 1. My guess is that the book somehow refers interchangeably to the quantum operation and its individual elements. $\endgroup$
    – EugeneB
    Jun 12 at 10:15
  • $\begingroup$ @EugeneB correct, then $\mathcal{E}(\rho)$ is not a density operator. The operation is then explicitly not defined as a map between sets of density operators, but as a map with the above properties $\endgroup$
    – Quantum Mechanic
    Jun 13 at 0:04
  • $\begingroup$ @EugeneB the point is that axioms are meant to be useful, not "correct." It's unfortunate when different people use different standards, but you can probably rest assured that someone will explicitly tell you if their maps are not trace preserving $\endgroup$
    – Quantum Mechanic
    Jun 13 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.