# Increasing the von Neumann entropy despite the measurement?

## 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?