References in diffusion of quantum state I would like to know if there are books, articles or any other type of references where a (heuristic) derivation of the equation:
\begin{eqnarray}
\textrm{d}|\psi(t)\rangle=-\frac{i}{\hbar}H_{\textrm{s}} |\psi(t)\rangle \textrm{d}t-\frac{1}{2}[L-\ell(t)]^2|\psi(t)\rangle \textrm{d}t+[L-\ell(t)]|\psi(t)\rangle\textrm{d}W_t,
\end{eqnarray}
where: $L$ is a observable; $\ell(t):=\langle \psi(t)|L|\psi(t)\rangle$; $H_{\textrm{s}}$ is the hamiltonian of system and $W_t$ the Wiener process.
 A: The book Quantum state diffusion by Ian Percival is a good reference on the subject as a whole. But I found his derivation of the QSD equation a bit hard to follow, here is an alternative approach
We assume that the evolution of a state vector $ |  \psi \rangle$ is composed of a drift component and a stochastic component as given below
\begin{equation} \label{eq:qsd}
  |  d \psi \rangle = | u \rangle dt +  |  \sigma \rangle (d\xi_R +i d\xi_I)
\end{equation}
Where $d\xi_R$ and $d\xi_I$ are independent ($\mathbb{E}[d\xi_Rd\xi_I]=0$) stochastic processes with mean zero ($\mathbb{E}[d\xi_R]=\mathbb{E}[d\xi_I]=0$). We also normalise $d\xi_R$ and $d\xi_I$ such that $d\xi_{R}^2=d\xi_{I}^2=\frac{dt}{2}$.
We also interpret the density matrix to be the average over pure state density matrices
\begin{equation}
 \rho=\mathbb{E}( |  \psi\rangle \langle  \psi |)
\end{equation}
In order to derive the QSD equation we must assume that the evolution of the density matrix is given by the Lindblad equation (Markovian, Trace preserving, Linear etc)
\begin{equation} \label{eq:lindblad}
 \dot{\rho}= -\frac{i}{\hbar}[H,\rho] +L \rho L^{\dagger}-\frac{1}{2}L^{\dagger} L \rho - \frac{1}{2}\rho L^{\dagger} L .
\end{equation}
If we demand that the norm of our state is conserved by the stochastic evolution given in above and we can derive several constraints on $ |  u \rangle$ and $ |  \sigma \rangle$.
\begin{equation}
 \begin{aligned}
  d \langle \psi | \psi \rangle &= \langle d\psi | \psi \rangle+ \langle  \psi | d\psi\rangle+  \langle  d\psi | d\psi\rangle \\
  &=  \langle  u | \psi  \rangle dt + \langle  \sigma | \psi \rangle (d\xi_R-i d \xi_I) \\
  &\phantom{=}+ \langle   \psi\ u \rangle dt + \langle  \psi| \sigma \rangle (d\xi_R+i d \xi_I) \\
  &\phantom{=}+  \langle  \sigma | \sigma \rangle dt
 \end{aligned}
\end{equation}
Equating the coefficients of the various independent differentials gives
\begin{equation}
 \begin{aligned}
   \langle  \sigma| \sigma \rangle +  \langle  u | \psi \rangle + \langle  \psi | u \rangle &= 0 \\
   \langle  \sigma | \psi \rangle +  \langle   \psi | \sigma \rangle &=0 \\
   \langle   \psi | \sigma \rangle -  \langle  \sigma|\psi \rangle &= 0 .
 \end{aligned}
\end{equation}
Therefore $  \langle  \psi| \sigma \rangle=0$  and $ \langle  \sigma | \sigma \rangle=-2 \text{Re}  \langle  \psi | u \rangle $.
We want to find and expression for $\dot{\rho}$ in terms of $ |  u \rangle,  |  \sigma \rangle \text{ and }  |  \psi \rangle$ so that we may link our stochastic variables to the Lindblad equation
\begin{equation}
 \begin{aligned}
  d \rho &= d( \mathbb{E}[ |  \psi \rangle \langle  \psi |]) \\
  &= \mathbb{E}[ |  d\psi \rangle \langle  \psi | + |  \psi \rangle \langle  d\psi | + |  d\psi \rangle \langle  d\psi | ] \\
  &=\mathbb{E}[ |  u \rangle \langle  \psi | dt+  |  \sigma \rangle \langle  \psi | ( d\xi_R+i d \xi_I)] \\
  &\phantom{\mathbb{E}(}+ |  \psi \rangle \langle  u | dt+  |  \psi \rangle \langle  \sigma | ( d\xi_R-i d \xi_I) \\
  &\phantom{\mathbb{E}(}+ |  \sigma \rangle \langle  \sigma | dt]. \\
 \end{aligned}
\end{equation}
As $d\xi_R$ and $d\xi_I$ are independent with mean zero we may differentiate with respect to time to obtain.
\begin{equation}
 \dot{\rho}=\mathbb{E}(  |  u \rangle \langle  \psi |  + | \psi \rangle  \langle  u | + |  \sigma \rangle \langle  \sigma| ).
\end{equation}
By equating the above equation with the Lindblad equation (\ref{eq:lindblad}) we obtain
\begin{equation}
 \begin{aligned}
  0 &= \mathbb{E}[( |  u \rangle +\frac{i}{\hbar}H |  \psi \rangle +\frac{1}{2}L^{\dagger}L |  \psi \rangle ) \langle  \psi |  \\
  &\phantom{\mathbb{E}(}+ |  \psi \rangle ( \langle  u | -\frac{i}{\hbar} \langle  \psi | H+\frac{1}{2} \langle  \psi | L^{\dagger}L) \\
  &\phantom{\mathbb{E}(}-L |  \psi \rangle \langle  \psi | L^{\dagger} + |  \sigma \rangle \langle \sigma | ]. \\
 \end{aligned}
\end{equation}
We also make the further assumption that
\begin{equation}
 \begin{aligned}
   |  u \rangle =U |  \psi \rangle & &\text{and}& & |  \sigma \rangle =S |  \psi \rangle, 
 \end{aligned}
\end{equation}
where $U$ and $S$ are arbitrary and not necessarily linear operators. This gives us
\begin{equation}
 \begin{aligned}
  0 &= \mathbb{E}[(U+\frac{i}{\hbar}H+\frac{1}{2}L^{\dagger}L) | \psi \rangle  \langle \psi |  \\
  &\phantom{\mathbb{E}(}+ |  \psi \rangle \langle \psi | (U^{\dagger}-\frac{i}{\hbar}H+\frac{1}{2}L^{\dagger}L) \\
  &\phantom{\mathbb{E}(}-L |  \psi \rangle \langle   \psi | L^{\dagger}+S |  \psi \rangle \langle  \psi | S^{\dagger}]. \\
 \end{aligned}
\end{equation}
At first it would appear that we could solve this equation by choosing
\begin{equation}
 \begin{aligned}
  U=-\frac{i}{\hbar}H -\frac{1}{2}L^{\dagger}L& &\text{and}& &S=L, 
 \end{aligned}
\end{equation}
However in this case $ \langle  \psi | \sigma \rangle= \langle  \psi | L | \psi \rangle \neq0$ which fails to satisfy the normalisation conditions. We may modify our solution to be
\begin{equation} \label{eq:AB}
 \begin{aligned}
  U=-\frac{i}{\hbar}H -\frac{1}{2}L^{\dagger}L+A& &\text{and}& &S=L+B.
 \end{aligned}
\end{equation}
From this we obtain the equations;
\begin{align}
 0&=\mathbb{E}[A |  \psi \rangle \langle \psi |+ |  \psi \rangle \langle  \psi | A^{\dagger}+B |  \psi \rangle \langle  \psi |L^{\dagger}+L |  \psi \rangle \langle  \psi | B^{\dagger}+ B |  \psi \rangle \langle  \psi | B^{\dagger}] \\
 0&=  \langle  A \rangle_{\psi}+ \langle  A^{\dagger}\rangle_{\psi}+ \langle  B^{\dagger}B\rangle_{\psi}+ \langle  L^{\dagger}B\rangle_{\psi}+ \langle  B^{\dagger}L\rangle_{\psi}  \\
 0&= \langle  \psi | L+B | \psi \rangle.
\end{align}
Which are solved if we try
\begin{equation}
 \begin{aligned}
  A= \langle  L^{\dagger}\rangle_{\psi}L - \frac{1}{2} \langle  L^{\dagger}\rangle_{\psi} \langle  L\rangle_{\psi}& &\text{and}& &B=- \langle  L\rangle_{\psi}.
 \end{aligned}
\end{equation}
Combining this result with our initial stochastic ansatz we obtain the main result
\begin{equation} \label{eq:QSD}
 \begin{aligned}
   |  d \psi \rangle &= (-\frac{i}{\hbar}H -\frac{1}{2}L^{\dagger}L+ \langle  L^{\dagger} \rangle_{\psi}L - \frac{1}{2} \langle  L^{\dagger}\rangle_{\psi} \langle  L \rangle_{\psi}) | \psi \rangle dt \\
   &\phantom{=}+ (L- \langle  L \rangle_{\psi}) |  \psi \rangle (d\xi_R +i d\xi_I).
 \end{aligned}
\end{equation}
Edit: I found a similar derivation of this equation in "Quantum Trajectories and Measurements in Continuous Time" by Alberto Barchielli and Matteo Gregoratti
