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3. Quantum instruments

3.1. A few words about the quantum formalism

Denote by  ${\textstyle {\mathcal {H}}}$ a complex Hilbert space. For simplicity, we assume that it is finite dimensional. Pure states of a system $S$ are given by normalized vectors of  ${\textstyle {\mathcal {H}}}$ and mixed states by density operators (positive semi-definite operators with unit trace). The space of density operators is denoted by $S$ (${\textstyle {\mathcal {H}}}$). The space of all linear operators in ${\textstyle {\mathcal {H}}}$ is denoted by the symbol ${\textstyle {\mathcal {L}}({\mathcal {H}})}$ . In turn, this is a linear space. Moreover, ${\textstyle {\mathcal {L}}({\mathcal {H}})}$ is the complex Hilbert space with the scalar product, ${\textstyle <A|B>=TrA^{*}B}$. We consider linear operators acting in ${\textstyle {\mathcal {L}}({\mathcal {H}})}$. They are called superoperators.

The dynamics of the pure state of an isolated quantum system is described by the Schrödinger equation:

$i{\tfrac {d}{dt}}\psi (t)={\widehat {H}}\psi (t)(t),\psi (0)=\psi _{0}$

$(3)$

where ${\textstyle {\hat {\mathcal {H}}}}$  is system’s Hamiltonian. This equation implies that the pure state $\psi (t)$ evolves unitarily $\psi (t)={\hat {U}}(t)\psi _{0}$, where  ${\hat {U}}(t)=e^{-it{\hat {\mathcal {H}}}}$ is one parametric group of unitary operators,${\hat {U}}(t):{\mathcal {H}}\rightarrow {\mathcal {H}}$ . In quantum physics, Hamiltonian  ${\textstyle {\hat {\mathcal {H}}}}$ is associated with the energy-observable. The same interpretation is used in quantum biophysics (Arndt et al., 2009). However, in our quantum-like modeling describing information processing in biosystems, the operator  ${\textstyle {\hat {\mathcal {H}}}}$ has no direct coupling with physical energy. This is the evolution-generator describing information interactions.

Schrödinger’s dynamics for a pure state implies that the dynamics of a mixed state (represented by a density operator) is described by the von Neumann equation:

${\frac {d{\hat {\rho }}}{dt}}(t)=-i[{\hat {\mathcal {H}}},{\hat {\rho }}(t)],{\hat {\rho }}(0)={\hat {\rho }}_{0}$

$(4)$