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.


1 Answer 1


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


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