continuous evolution

Question

I have a discrete evolution equation such that $$\rho(t+\tau) = M_0 \rho(t)M_0^\dagger+M_1 \rho(t)M_1^\dagger$$ $M_0\;\&\;M_1$ are two operators such that $tr(M_0^\dagger M_1)=0$ i.e. they are orthogonal and $M_0^\dagger M_0+M_1^\dagger M_1 =1$. I want to convert this discrete evolution equation into a continuous evolution equation. Please give some hint.

Answer 1

For $\tau \rightarrow 0$ you can use Taylor series.

$\rho(t+\tau) = \rho(t)+ \frac{d \rho}{dt}\tau$

Therefore

$\frac{d \rho}{dt} = \frac{1}{\tau}\left[ M_0 \rho(t)M_0^\dagger+M_1 \rho(t)M_1^\dagger - \rho(t) \right]$

Since there is only one variable we don't need partial derivative. You can calculate the RHS for $\tau \rightarrow 0$ (numerically or analytically) and that will be your answer.

In case $M_0,M_1$ also depends on both $t$ and $\tau$, it might be possible to go analytically. Let consider $\tau=dt$.

$\rho(t+dt) = \rho(t)+ \frac{d \rho}{dt}|_{t}dt$

$M_0(t+dt) \rho(t)M_0^\dagger(t+dt) = M_0(t) \rho(t)M_0^\dagger(t) + [dM_0/dt \rho(t) M_0(t)^\dagger + M_0 \rho(t) dM_0^\dagger/dt] dt$

$M_1(t+dt) \rho(t)M_1^\dagger(t+dt) = M_1(t) \rho(t)M_1^\dagger(t) + [dM_1/dt \rho(t) M_1(t)^\dagger + M_1 \rho(t) dM_1^\dagger/dt] dt$

Now let say, $\rho(t)$ commutes with both $M_0$ and $M_1$. In that case the above two terms will give you

$[M_0M_0^\dagger+M_1M_1^\dagger]\rho(t)+[dM_0/dt M_0(t)^\dagger + M_0 dM_0^\dagger/dt + dM_1/dt M_1(t)^\dagger + M_1 dM_1^\dagger/dt] \rho(t)dt$

$\Rightarrow \rho(t) + \frac{d}{dt}[M_0M_0^\dagger+M_1M_1^\dagger]\rho(t)dt \Rightarrow \rho(t)$

So your $\rho$ will be constant in time.

Otherwise you have to keep all the terms.