# What is the domain of the dual map of a quantum channel?

Possibly a naive question...if the dual map of a quantum channel gives the evolution of the system in the Heisenberg picture by acting on observables, and observables are self-adjoint operators on the state space, does that mean that the domain of the dual map of a quantum channel is the subspace of self-adjoint bounded linear maps from the state space to itself and not the entire space of bounded linear maps on the state space?

Also, if a quantum channel acts on density operators, and density operators are required to have trace 1, then how can we define a quantum channel as a map $$B(H) \to B(H)$$ for a Hilbert space $$H$$? This is how I have always seen it defined...

The key here is linearity; a (real-)linear map $$\Phi$$ defined on all self-adjoint bounded operators on $$H$$ has a unique extension to all of $$B(H)$$. This is due to the fact that every $$B\in B(H)$$ can be written as the linear combination of two self-adjoint operators via $$B=\Big(\frac{B+B^*}2\Big)+i\Big(\frac{B-B^*}{2i}\Big)\tag1$$ so \begin{align*} \Phi':B(H)&\to B(H)\\ B&\mapsto \Phi\Big(\frac{B+B^*}2\Big)+i\,\Phi\Big(\frac{B-B^*}{2i}\Big) \end{align*} is well-defined, linear (straightforward computation), and an extension of the original map as $$\Phi'\equiv\Phi$$ on all self-adjoint operators.
Every (trace-class$${}^1$$) operator $$A$$ can be written as a linear combination of four positive semi-definite trace-class operators
Proof. First decompose $$A$$ into self-adjoint (trace-class) operators as in (1), and then use that every self-adjoint (trace-class) operator $$A'$$ can be written as the difference of two positive semi-definite operators through its spectral decomposition: $$A'=\sum_j a_j|g_j\rangle\langle g_j|=\underbrace{\sum_{j:a_j\geq 0}a_j|g_j\rangle\langle g_j|}_{A'_+\geq0}-\underbrace{\sum_{j:a_j<0}|a_j|\,|g_j\rangle\langle g_j|}_{A'_-\geq0}$$ The final step is to turn these into states by normalizing them: \begin{align*} A&=\frac12\big(A+A^*\big)_+-\frac12\big(A+A^*\big)_--\frac i2\big(i(A-A^*)\Big)_++\frac i2\Big(i(A-A^*)\Big)_-\\ &=\tfrac{{\rm tr}((A+A^*)_+)}2\tfrac{(A+A^*)_+}{{\rm tr}((A+A^*)_+)}-\tfrac{{\rm tr}((A+A^*)_-)}2\tfrac{(A+A^*)_-}{{\rm tr}((A+A^*)_-)}\\ &\qquad\qquad\qquad-\tfrac{i{\rm tr}((i(A-A^*))_+)}2\tfrac{(i(A-A^*))_+}{{\rm tr}((i(A-A^*))_+)}+\tfrac{i{\rm tr}((i(A-A^*))_-)}2\tfrac{(i(A-A^*))_-}{{\rm tr}((i(A-A^*))_-)}\,. \end{align*} NB: of course if any of the $${\rm tr}((\ldots)_\pm)$$ is zero then we could disregard them from the start; just to make sure we're not dividing by zero. $$\square$$