Choosing the eigenvector stochastically

[Mentz114]

The equation $\hat{O}|\psi\rangle \rightarrow \alpha_n|\mathbf{e}_n\rangle$ where $|\mathbf{e}_n\rangle$ is an eigenvector of the operator and $\alpha_n$ is its eigenvalue, is central in the QT formalism. This is as much as we can get from quantum theory but an ideal instrument should enable us to estimate the $p_n\equiv |\alpha_n|^2$ from the observed frequencies of different outcomes.

When there are 2 eigenstates it seems straightforward to define a stochastic evolution operator thus $\hat{D}=diag( e^{\lambda(p_0-s)t}, e^{-\lambda(p_0-s)t})$
where $s$ is a random variate (nearly) uniformly distributed in $(0,1)$. $\hat{D}$ selects which eigenstate to grow and which to shrink with the proportions $p$ and $1-p$. $\hat{D}$ evolves with time to having only one non-null eigenvector.
$\hat{D}$ has to be considered part of the apparatus and the acquisition of $s$ could be from a random phase or from another special interaction perhaps with an internal state.

With more than 2 dimensions $\hat{D}$ has the form

\begin{align}
\hat{D} &= \left[ \begin{array}{cccc}
e^{\lambda(p_0-s)t} & 0 & 0 & 0 \\\
0 & e^{-\lambda(p_0-s)(p_0+p_1-s)t} & 0 & 0 \\\
0 & 0 & e^{-\lambda(p_0+p_1-s)(p_0+p_1+p_2-s)t} & 0 \\\
0 & 0 & 0 & e^{\lambda(s-(p_0+p_2+p_3))t}
\end{array} \right]
\end{align}For any $s$ only one exponent is positive.

 

[azntoon]

All other terms get exponentially small.It is not clear to me how to extend this formalism to more than 3 eigenstates.