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...


1 Answer 1


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.

For quantum states (i.e. the Schrödinger picture) there's a similar reasoning although one needs an extra step there:

Every (trace-class${}^1$) operator $A$ can be written as a linear combination of four positive semi-definite trace-class operators

As before this fact is what allows for a unique extension of a linear map defined on states to the full operator space. Finally, the:

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. $\square$

1: If you only care about finite-dimensional systems then you may replace "trace-class operator" by "matrix" because the trace class is an inherently infinite-dimensional concept which trivializes for finite dimensions.


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