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I have a matrix $\hat{\rho}$ such that $$\frac{d}{dt}\hat{\rho} = \frac{i}{\hbar}[\hat{\rho},\hat{H}]$$ And the question asks me to find the time dependence $\hat{\rho}$(t).

The general solution to this differential equation is $$\hat{\rho} = \exp\left(\frac{i}{\hbar}Ht\right)\hat{\rho(0)}\exp\left(-\frac{i}{\hbar}Ht\right)$$ Where $\hat{\rho(0)} = \left( \begin{array}{ccc} \frac{1}{2} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{2} \\ \end{array} \right)$ and $H=AI+BY$, where $Y= \left( \begin{array}{ccc} 0 & -i \\ i & 0 \\ \end{array} \right)$.

So I think what I have to do is substitute $H$ into the general solution for $\hat{\rho}$, but I'm finding that a bit difficult to actually evaluate. If I Taylor expand the matrix exponential it's an infinite series, so I'm not sure how to deal with that - is there a way to make $\hat{\rho(0)}$ into an exponential as well and simplify it that way?

I also considered writing $\rho = \frac{1}{2}(I+a(t)X+b(t)Y+c(t)Z)$, and if I substitute that back into the original differential equation then I could find the coefficients by evaluating the commutator. But I'd still first need to write it out with
$$\hat{\rho} = \exp\left(\frac{i}{\hbar}Ht\right)\hat{\rho(0)}\exp\left(-\frac{i}{\hbar}Ht\right)$$ So my original problem with the exponential remains.

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$\newcommand{mtx}[4]{\left(\begin{array}{cc}#1 & #2 \\ #3 & #4\end{array}\right)}$

Split the problem into small pieces

Diagonalizing $H$

$$ H =\mtx{A}{0}{0}{A} + \mtx{0}{-iB}{iB}{0} = \mtx{A}{-iB}{iB}{A} \tag{1} $$

The diagonal form of $H$ is

$$ H = U\Lambda U^{\dagger} \tag{2} $$

with

$$ \Lambda = \mtx{A-B}{0}{0}{A+B} ~~~\mbox{and}~~~ U = \frac{1}{\sqrt{2}}\mtx{i}{-i}{1}{1} \tag{3} $$

I want to emphasize that $U^\dagger = U^{-1}$

Exponentiating $H$

\begin{eqnarray} \exp\left(\frac{it}{\hbar}H\right) &\stackrel{(2)}{=}& \exp\left(\frac{it}{\hbar}U\Lambda U^{\dagger}\right) \\ &=& U \exp\left(\frac{it}{\hbar}\Lambda\right)U^{\dagger} \\ &=& \frac{1}{2}\mtx{i}{-i}{1}{1}\mtx{e^{it(A-B)/\hbar}}{0}{0}{e^{it(A+B)/\hbar}} \mtx{-i}{1}{i}{1} \\ &=& (\cdots) \tag{4} \end{eqnarray}

I will leave the rest to you, it is just multiplying three matrices. Simple stuff

Putting everything together

With the result in (4), your initial conf. $\rho(0)$ and a similar version of (4) for the case $\exp(-it H/\hbar)$ you can compute your final result

$$ \rho(t) = \exp\left(\frac{it}{\hbar}H\right) \rho(0) \exp\left(-\frac{it}{\hbar}H\right) \tag{5} $$

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  • $\begingroup$ Thanks! I would never have thought of diagonalising! In the matrix U though, where does the $1/\sqrt{2}$ come from? I don't get that factor. $\endgroup$ – user13948 Jan 20 '18 at 17:29
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    $\begingroup$ @Karacoreable It is just a normalization factor, to ensure that the eigenvectors all have norm 1 $\endgroup$ – caverac Jan 20 '18 at 17:35

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