# Non trace-preserving map in axiomatic approach to quantum operations

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?

"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.
• 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. Jun 12 at 10:15
• @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 Jun 13 at 0:04