Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For system of differential equation as follows:\begin{align} \frac{\partial}{\partial t} \begin{pmatrix}\rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{00}\end{pmatrix} &= -\tau i \begin{pmatrix} \rho_{10} - \rho_{01} & \rho_{11} - \rho_{00} \\ \rho_{00} - \rho_{11} & \rho_{01} -\rho_{10} \end{pmatrix} + \tau^2 \begin{pmatrix} \rho_{11} & -\frac{1}{2}\rho_{01} \\ -\frac{1}{2}\rho_{10} & -\rho_{11} \end{pmatrix} \end{align}

Alternative form:

$\rho_{00}'(t)=-\tau i (\rho_{10}(t)-\rho_{01}(t))+\tau^2 \rho_{11}(t)$

$\rho_{01}'(t)=-\tau i (\rho_{11}(t)-\rho_{00}(t))-\tau^2 \frac{1}{2}\rho_{01}(t)$

$\rho_{10}'(t)=-\tau i (\rho_{00}(t)-\rho_{11}(t))-\tau^2 \frac{1}{2}\rho_{10}(t)$

$\rho_{11}'(t)=-\tau i (\rho_{01}(t)-\rho_{10}(t))-\tau^2 \rho_{11}(t)$

Question: How to obtain the analytic solutions ? I tried but failed, it seems the solution is large ? In Mathematica, it gives the solution extremly large with many terms, is it reasonable ?

share|cite|improve this question
Is $\tau$ a parameter or do you mean $\frac{d}{d\tau}$? More to the point, $t$ and $\tau$ are different in your problem? If $\tau$ is a constant your problem is just a constant coefficient system of ODEs which is solved by any number of methods, for example the matrix exponential... if $\tau=t$ then it's harder... – James S. Cook Jun 2 '14 at 15:44
@JamesS.Cook $\tau$ is a constant, I tried matrix method on software Mathematica, it gives the analytic solution extremely large, say more than 30 terms... Is it possible to have simple form solution for the system above ? – Xingdong Jun 2 '14 at 15:48
I fail to see why you express it as a matrix for independent variable, not just the (more natural) vector – vonbrand Jun 3 '14 at 1:29
Well, thirty terms, not too bad, I mean, I'd expect $4$ solutions and the general solution being a linear combination of those as this a linear fourth order ODE. It could be ugly. Minimally, I'd expect like 16 terms because you have 4 independent solutions each begin a 4-vector... unless a bunch of stuff is zero. Also, @vonbrand has a good suggestion, use $x,y,z,w$ instead of the double index variables while your getting your mind wrapped around the solution. Once you sort it out, go back to the matrix which I guess comes from your context. – James S. Cook Jun 4 '14 at 1:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.