4
$\begingroup$

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...

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

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

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