Refering to the motivation given by @markmitchison, The issue with the derivation motivated above is that, if we try to compute the expectation values of some powers of time evolved operator $\hat{O}(t)$, it wont agree in general with the correct expectation value. For example if we first solve for $\hat{O}(t)$ and then we try to evaluate $\langle\hat{O}(t)\hat{O}(t)\rangle$ or something it wont be correct. I dont know exactly what is the reason for this, but there might be something to do with the fact that the open system dynamics of Hisenberg operators should also include noise terms.
Let me give you one specific example where I have encountered this. Lets take Hamiltonian
\begin{align}
\hat{H} = \Delta \sum_{n=-\infty}^{\infty} \left( |n\rangle \langle n+1| + h.c \right),
\label{eq:TightBinding}
\end{align}
Let us define operators as,
\begin{align*}
\hat{X}=d\hat{N}=d\sum_{n=-\infty}^{\infty}n|n\rangle \langle n|, \hspace{0.2cm}
\hat{K}=\sum\limits_{n=-\infty}^{\infty}|n\rangle \langle n+1|,
\hspace{0.2cm}
\hat{K}^{\dagger}=\sum_{n=-\infty}^{\infty}|n+1\rangle \langle n|,
\end{align*}
These operators satisfy,
\begin{align*}
[\hat{K},\hat{N}]=\hat{K},
\hspace{0.5cm}
[\hat{K}^{\dagger},\hat{N}]=-\hat{K}^{\dagger},
\hspace{0.5cm}
[\hat{K},\hat{K}^{\dagger}]=0.
\end{align*}
In the form of these operators, the Hamiltonian becomes,
\begin{align}
\hat{H}= \Delta (\hat{K}+\hat{K}^{\dagger}) \label{eq:TightBinding2},
\end{align}
Coming to the continuous measurement part, the non-selective master equation for the dynamics of the system is,
\begin{align}
\dot{\hat{\rho}} = -\frac{i}{\hbar}[\hat{H},\hat{\rho}] - \frac{\Gamma}{2}[\hat{X},[\hat{X},\hat{\rho}]]
\label{NSmasterequation},
\end{align}
Where $\Gamma$ is the parameter that controls the strength of the measurement. We will be interested in the dynamics of the mean position $\langle\hat{X}\rangle$ and the variance of the position $\sigma_x=\sqrt{\langle\hat{X}^2\rangle-\langle\hat{X}\rangle^2}$. We can write down the equivalent adjoint master equation for any explicitly time-independent system operator $\hat{O}$ as,
\begin{align}
\dot{\hat{O}} = -\frac{i}{\hbar}[\hat{O},\hat{H}] - \frac{\Gamma}{2}[\hat{X},[\hat{X},\hat{O}]]
\label{NSadjointmasterequation}
\end{align}
Let us write down the equation of motion for $\hat{X}$. We have,
\begin{align}
\dot{\hat{X}} &= -\frac{i}{\hbar}[\hat{X},\hat{H}] - \frac{\Gamma}{2}[\hat{X},[\hat{X},\hat{X}]] \nonumber\\
&= -\frac{id\Delta}{\hbar}[\hat{N},(\hat{K}+\hat{K}^{\dagger})]+0\nonumber\\
&= -\frac{id\Delta}{\hbar} (\hat{K}^{\dagger}-\hat{K}).
\label{xopequation}
\end{align}
Now the time evolution of the operator $\hat{K}$ is,
\begin{align}
\dot{\hat{K}} &= -\frac{i}{\hbar}[\hat{K},\hat{H}] - \frac{\Gamma}{2}[\hat{X},[\hat{X},\hat{K}]] \nonumber\\
&=0-\frac{\Gamma d^2}{2}[\hat{N},[\hat{N},\hat{K}]] \nonumber\\
&=-\frac{\Gamma d^2}{2}\hat{K} \nonumber
\end{align}
From the above equation we can see that $\hat{K}^{\dagger}(t) - \hat{K}(t) = e^{-\frac{\Gamma d^2}{2}t}(\hat{K}^{\dagger}(0)-\hat{K}(0))$. Using this solution we get,
\begin{align}
\dot{\hat{X}} &= -\frac{id\Delta}{\hbar} e^{-\frac{\Gamma d^2}{2}t}(\hat{K}^{\dagger}(0)-\hat{K}(0))\nonumber\\
\implies \hat{X}_t &= \hat{X}_0 + \frac{2i\Delta}{\hbar\Gamma d}\left(\hat{K}^{\dagger}_0-\hat{K}_0\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)
\label{xopequation2}
\end{align}
From the above equation we get the time evolution of $\langle\hat{X}\rangle_t$ as,
\begin{align}
\langle\hat{X}\rangle_t &= \langle\hat{X}\rangle_0 + \frac{2i\Delta}{\hbar\Gamma d}\left(\langle\hat{K}\rangle^{\dagger}_0-\langle\hat{K}\rangle_0\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)
\label{xopmeanequation}
\end{align}
Now let us look at the time evolution of the $\langle X^2\rangle_t$ of the initially localized wavefunction.
\begin{align}
\hat{X}_t^2 = \hat{X}_0^2-\left(\frac{4\Delta^2}{\hbar^2d^2\Gamma^2}\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)^2(\hat{K}_0^{\dagger}-\hat{K}_0)^2 + \left(\frac{2i\Delta}{\hbar\Gamma d}\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)\{\hat{X}_0,(\hat{K}_0^{\dagger}-\hat{K}_0)\}
\label{xop2equations}
\end{align}
For the expectation values we get,
\begin{align}
\langle\hat{X}^2\rangle_t = \langle\hat{X}^2\rangle_0-\left(\frac{4\Delta^2}{\hbar^2d^2\Gamma^2}\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)^2\langle(\hat{K}^{\dagger}-\hat{K})^2\rangle_0 + \left(\frac{2i\Delta}{\hbar\Gamma d}\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)\langle\{\hat{X},(\hat{K}^{\dagger}-\hat{K})\}\rangle_0
\label{xop2meanequations}
\end{align}
For the initial condition $|0\rangle$ we have, $\langle\hat{X}\rangle, \langle(\hat{K}^{\dagger}-\hat{K})\}\rangle_0=0$, $\langle\hat{X}^2\rangle_0=0$ and $\langle(\hat{K}^{\dagger}-\hat{K})^2\rangle_0=-2$ and we get,
\begin{align}
\langle\hat{X}^2\rangle_t = \left(\frac{8\Delta^2}{\hbar^2d^2\Gamma^2}\right)\left(e^{-\frac{\Gamma d^2}{2}t}-1\right)^2
\label{xop2meanequations2}
\end{align}
It turn out that this expression is incorrct. The correct experssion can be derived by writing down the closed form equation of motion for the moments we have,
\begin{align}
\frac{d}{dt}\begin{pmatrix}
\langle \hat{N}^2\rangle\\
\langle \{\hat{N},\hat{K}\}^{\dagger} \rangle -\langle \{\hat{N},\hat{K}\}\rangle\\
\langle \hat{K}^2 - \hat{K}\hat{K}^{\dagger}\rangle\\
\langle \hat{K}^2 \rangle
\end{pmatrix} &= \begin{pmatrix}
0 & -i\Delta & 0 & 0\\
0 -\frac{\Gamma}{2} & -4i\Delta & 0\\
0 & 0 & 0 & -\Gamma\\
0 & 0 & 0 & -2\Gamma
\end{pmatrix} \begin{pmatrix}
\langle \hat{N}^2\rangle\\
\langle \{\hat{N},\hat{K}\}^{\dagger} \rangle -\langle \{\hat{N},\hat{K}\}\rangle\\
\langle \hat{K}^2 - \hat{K}\hat{K}^{\dagger}\rangle\\
\langle \hat{K}^2 \rangle
\end{pmatrix}
\end{align}
With the initial condition, $ \langle \hat{N}^2\rangle_0 = \langle \{\hat{N},\hat{K}\}^{\dagger} \rangle -\langle \{\hat{N},\hat{K}\}\rangle = \langle \hat{K}^2 \rangle = 0$ and $ \langle \hat{K}^2 - \hat{K}\hat{K}^{\dagger}\rangle = 1$ we get,
\begin{align}
\langle \hat{N}^2\rangle_t = \frac{8\Delta^2}{\Gamma^2}\left[\Gamma t - 2\left(1 - e^{-\frac{1}{2}\Gamma t}\right) \right]
\label{xop2meanequations3}
\end{align}
It was a long time ago that I did this calculations and I still do not know the exact reason as to why the first approach gives incorrect expressions.
