I am currently reading the book "Quantum measurement and control" by Wiseman and Milburn (https://doi.org/10.1017/CBO9780511813948) and something is really bugging me in the chapter on quantum trajectories. Here is the context: they want to derive a stochastic master equation to describe quantum jumps. So first they define the measurement operators $\hat{M}_0$ and $\hat{M}_1$:

\begin{align} \hat{M}_0 &= \hat{I} - \left(\frac{\hat{c}^{\dagger}\hat{c}}{2} + i\hat{H}\right)dt \\ \hat{M}_1 &= \hat{c}\sqrt{dt} \end{align}

$\hat{M}_0$ represents an infinitesimal but not unitarily change and can thus be regarded as a "null result":

\begin{equation} \left| \psi_0(t+dt) \right\rangle = \frac{\hat{M}_0\left|\psi(t) \right\rangle}{\sqrt{\langle \hat{M}_0^\dagger \hat{M}_0 \rangle}} = \left( \hat{1} - dt \left[ i\hat{H} + \frac{1}{2} \hat{c}^\dagger \hat{c} - \frac{1}{2} \langle \hat{c}^\dagger\hat{c} \rangle\right] \right)\left| \psi(t) \right\rangle \end{equation}

$\hat{M}_1$ corresponds to a finite evolution (jump) and is regarded as a "detection":

\begin{equation} \left| \psi_1(t+dt) \right\rangle = \frac{\hat{M}_1\left|\psi(t) \right\rangle}{\sqrt{\langle \hat{M}_1^\dagger \hat{M}_1 \rangle}} = \frac{\hat{c}\left| \psi(t) \right\rangle}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}}) \end{equation}

Then by denoting the number of photodetections up to time $t$ by $N(t)$, the stochastic increment $dN(t)$ obeys

\begin{equation} dN(t)^2 = dN(t) \end{equation}

because $dN$ is either zero or one.

From there it is easy to deduce the stochastic Schrödinger equation (SSE) describing these quantum jumps:

\begin{equation} d | \psi(t)\rangle = \left[ dN(t) \left( \frac{\hat{c}}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}} - 1 \right) + [1 - dN(t)]dt \left( - i\hat{H} - \frac{1}{2} \hat{c}^\dagger \hat{c} + \frac{1}{2} \langle \hat{c}^\dagger\hat{c} \right) \right] \end{equation}

We can simplify this equation by neglecting the term $dN(t)dt$:

\begin{equation} d | \psi(t)\rangle = \left[ dN(t) \left( \frac{\hat{c}}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}} - 1 \right) + dt \left( - i\hat{H} - \frac{1}{2} \hat{c}^\dagger \hat{c} + \frac{1}{2} \langle \hat{c}^\dagger\hat{c} \right) \right] \end{equation}

This is the SSE describing the system. But here is my question: how can we simply pass from this SSE to a stochastic master equation for $\rho$.

Here's what they do in the book. First they define the projector $\hat{\pi}$:

\begin{equation} \hat{\pi}(t) = |\psi(t)\rangle\langle\psi(t)| \end{equation}

then by developing they obtain:

\begin{align} d\hat{\pi} &= |d \psi(t)\rangle\langle\psi(t)| + |\psi(t)\rangle\langle d \psi(t)| + | d \psi(t)\rangle\langle d \psi(t)| \\ &= \left( dN(t) G[\hat{c}] - dt K[i\hat{H} + \frac{1}{2} \hat{c}^\dagger \hat{c}] \right) \hat{\pi}(t) \end{align}

where G and K are super operators defined by

\begin{align} G[r]\rho &= \frac{\hat{r}\rho \hat{r}^\dagger}{Tr[\hat{r}\rho \hat{r}^\dagger]} - \rho \\ H[r]\rho &= \hat{r}\rho + \rho \hat{r}^\dagger - Tr[\hat{r}\rho + \rho \hat{r}^\dagger] \rho \end{align}

What I really don't understand in this development is the first line $d\hat{\pi} = |d \psi(t)\rangle\langle\psi(t)| + |\psi(t)\rangle\langle d \psi(t)| + | d \psi(t)\rangle\langle d \psi(t)|$ . I don't understand where the last term comes from. I would think that the correct form would rather simply be $d\hat{\pi} = |d \psi(t)\rangle\langle\psi(t)| + |\psi(t)\rangle\langle d \psi(t)|$. From that line did the development and it really yields the last line (which is result that I've also encountered in other references but without any explanation). So the last term really seems necessary to obtain the correct answer. This is really a small detail but I am not able to get my head around it. Maybe there is something really fundamental that I don't know about differential too.

So I am looking for either an explanation of that simple line or another general explanation on how to go from a SSE to a stochastic master equation.

  • 1
    $\begingroup$ This might be the product rule for the stochastic Ito calculus: d(ab) = a db + b da + da db. $\endgroup$
    – vivian
    Aug 6, 2019 at 4:27

1 Answer 1


As you stated in your question, the stochastic increment $dN(t)$ obeys

\begin{equation} dN(t)^2 = dN(t) \end{equation}

because $dN$ is either zero or one. Thus, the first term in the SSE, i.e.

\begin{equation} dN(t) \left( \frac{\hat{c}}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}} - 1 \right) |\psi(t)\rangle, \end{equation}

leads to a term

\begin{equation} dN(t) \left( \frac{\hat{c}}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}} - 1 \right) \hat{\rho}(t) \left( \frac{\hat{c}^\dagger}{\sqrt{\langle \hat{c}^\dagger \hat{c} \rangle}} - 1 \right) \end{equation}

in the evolution equation for the density matrix. This will give an order $dt$ result once you perform the ensemble average as $E[dN(t)]= \langle \hat{c}^\dagger \hat{c} \rangle dt$. Note also that the noise increment and the density matrix are statistically independent, i.e. $E[dN(t)\hat{\rho}(t)]=E[dN(t)]E[\hat{\rho}(t)]$.


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