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Background

Assume we have a density matrix $\rho$ of a sub-ensemble. However, we have an imperfect measuring instrument. While it does perform a measurement, we do not know exactly when it performs the measurement. If we want to perform the measurement at $t= 0$. Then, the measuring instrument performs the measurement in the time interval $(- \epsilon, \epsilon)$. The probability of it happening within this interval at time $t$ is given by $p(t) \delta t$. Obviously, $ \int_{- \epsilon}^\epsilon p(t) d t = 1$. The density operator post measurement at time $t$ is given by:

$$\rho_{M} = \Big ( \frac{P_i U ( t) \tilde \rho U^\dagger ( t) P_i}{\text{Tr} P_i U ( t) \tilde \rho U^\dagger ( t) } \Big )$$

With being the $P_i$ is an arbitrary projection operator $\tilde \rho$ being the density matrix at time $0$ (assuming no measurement) and $U$ being the unitary operator. The resultant density matrix when the non-ideal measurement device is included is given by at time $\epsilon$ is $\rho(\epsilon)$ is given by:

\begin{equation} \rho (\epsilon) = p( - \epsilon + \delta t) U(2 \epsilon - \delta t ) \Big ( \frac{ P_i U (-\epsilon + \delta t) \tilde \rho U^\dagger (-\epsilon + \delta t) P_i }{\text{Tr} (P_i U (-\epsilon + \delta t) \tilde \rho U^\dagger (-\epsilon + \delta t) )} \Big) U^\dagger (2 \epsilon - \delta t ) \delta t \end{equation} $$+ $$

$$p( - \epsilon + 2 \delta t) U(2 \epsilon - 2\delta t ) \Big ( \frac{ P_i U (-\epsilon + 2\delta t) \tilde \rho U^\dagger (-\epsilon + 2\delta t) P_i }{\text{Tr} P_i U (-\epsilon + 2\delta t) \tilde \rho U^\dagger (-\epsilon + 2\delta t) } \Big)U^\dagger (2 \epsilon - 2\delta t ) \delta t $$

$$+ $$

$$ \vdots$$

$$ + $$

\begin{equation} p( \epsilon - \delta t) U(\delta t ) \Big ( \frac{P_i U (\epsilon - \delta t) \rho ' U^\dagger (\epsilon - \delta t) P_i}{\text{Tr} P_i U (\epsilon - \delta t) \rho ' U^\dagger (\epsilon - \delta t)} \Big) U^\dagger ( \delta t ) \delta t \end{equation}

Now, in the limit $\delta t \to 0$ we get an integral:

\begin{equation} \implies \rho(\epsilon) = \int_{- \epsilon}^\epsilon p(z) U( \epsilon - z) \Big ( \frac{P_i U ( z) \tilde \rho U^\dagger ( z) P_i}{\text{Tr} P_i U ( z) \tilde \rho U^\dagger ( z) } \Big ) U^\dagger ( \epsilon - z) dz \end{equation}

Let us see what happens as $\epsilon \to 0$. The first thing we notice is this would imply:

\begin{equation} \epsilon \to 0 \implies p(z) \to \delta (z) \end{equation}

Substituting this result in our integral we get the normal result of an ideal measurement at $t = 0$:

\begin{equation} \lim_{\epsilon \to 0}\rho( \epsilon) = \frac{ P_i \tilde \rho P_i}{Tr P_i \tilde \rho } \end{equation}

Question

Consider the von Neumann entropy $S(\rho)$ of the system. Is there a way to find the probability distributions which ensure the von Neumann entropy always increases?

$$ S(\rho(-\epsilon)) \leq S(\rho(\epsilon)) $$

Try to solve for as general as possible?

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