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How do I solve the following coupled linear differential equations $$ \dot{x}(t)=iAy(t)-iBx(t)\\ \dot{y}(t)=iAx(t)+iAz(t)\\ \dot{z}(t)=iAy(t)-iBz(t) $$ for $|x(t)|^2, |y(t)|^2, |z(t)|^2$ ?

given the initial conditions $x(0)=1,y(0)=z(0)=0$ and $|x(t)|^2+|y(t)|^2+|z(t)|^2=1$

Note: A, B are constants and $i=\sqrt{-1}$

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    $\begingroup$ The method of analytical solution is the usual one based on eigendecomposition of the matrix $\begin{bmatrix} -iB & iA & 0 \\ iA & 0 & iA \\ 0 & iA & -iB \end{bmatrix}$. $\endgroup$ – Ian Apr 4 '16 at 14:29
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    $\begingroup$ @ss1729 Ian is telling you to find the eigenvalues and eigenvectors of the above matrix he has produced for you. $\endgroup$ – Mattos Apr 4 '16 at 14:35
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    $\begingroup$ Note that in general it can be hard to find the eigenvalues of a $3 \times 3$ matrix (there is a cubic formula, but it is complicated). In this particular case there is no real difficulty because you can see by inspection that $-iB$ is an eigenvalue. $\endgroup$ – Ian Apr 4 '16 at 14:47
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    $\begingroup$ @ss1729 In the 4x4 case there is an explicit formula for any such matrix, though it is very complicated and quite unstable to evaluate on a computer. Above that, there is in general nothing you can do, because the characteristic polynomial of a matrix can be any polynomial. Indeed there is a so-called "companion matrix" for any given polynomial, which has that polynomial as its characteristic polynomial. $\endgroup$ – Ian Apr 5 '16 at 15:07
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    $\begingroup$ @ss1729 So in view of the Abel-Ruffini theorem (some polynomials of degree 5 and higher have no solution in terms of radicals), you are forced to work approximately, unless there is some very nice simplification (e.g. a large enough number of eigenvalues that you can simply read off at a glance). $\endgroup$ – Ian Apr 5 '16 at 15:08
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you can use Laplace Transform as follow to find the answer.(a fast and easy way to know Laplace Transform is through wikipedia https://en.wikipedia.org/wiki/Laplace_transform) by taking Laplace Transform of above equations we get: $$\dot{x}(t)=iAy(t)-iBx(t)\Rightarrow sX(s)-1=iAY(s)-iBX(s)\\ \dot{y}(t)=iAx(t)+iAz(t)\Rightarrow sY(s)= iAX(s)+iA(s)\\ \dot{z}(t)=iAy(t)-iBz(t)\Rightarrow sZ(s)=iAY(s)-iBZ(s)$$ now we can treat equations in s-domain like a linear system of equations with complex variables and also variable $"s"$ as a constant. so we must solve for $X(s),Y(s),Z(s)$ from equation's. it is easy to prove that: $$X(s)=\frac{iAY(s)+1}{s+iB}\\ iA-A^2Y(s)-s(s+iB)Y(s)-A^2Y(s)=0\\ Z(s)=\frac{iAY(s)}{s+iB}$$ so after finding value of $Y(s)$ from last equation we plug it into the other two and find values of other functions ($X(s),Z(s)$): $$X(s)=\frac{s^2+iBs+A^2}{(s^2+iBs+2A^2)(s+iB)}\\ Y(s)=\frac{iA}{s^2+iBs+2A^2}\\ Z(s)=\frac{-A^2}{(s^2+iBs+2A^2)(s+iB)}$$ now we must take Inverse Laplace Transform from above equations using Partial Fraction Expansion but this depends on the values of constant $A$ and $B$. for a general values of $A$ and $B$ we can get a general complex solution like below assuming that $s^2+iBs+2A^2=(s-\omega_1)(s-\omega_2)$ in which $\omega_1,\omega_2=\frac{-i(B\pm \sqrt{B^2+8A^2})}{2}$ $$X(s)=\frac{\frac{-A^2}{(\omega_1-\omega_2)(\omega_1+iB)}}{s-\omega_1}+\frac{\frac{-A^2}{(\omega_2-\omega_1)(\omega_2+iB)}}{s-\omega_2}+\frac{\frac{1}{2}}{s+iB}\\ Y(s)=\frac{\frac{iA}{\omega_1-\omega_2}}{s-\omega_1}+\frac{\frac{iA}{\omega_2-\omega_1}}{s-\omega_2}\\ Z(s)=\frac{\frac{-A^2}{(\omega_1-\omega_2)(\omega_1+iB)}}{s-\omega_1}+\frac{\frac{-A^2}{(\omega_2-\omega_1)(\omega_2+iB)}}{s-\omega_2}+\frac{\frac{-1}{2}}{s+iB}$$ now we must take Inverse Laplace Transform from above equations using these facts: $$\mathscr L\{e^{at}u(t)\}=\frac{1}{s-a}\iff \mathscr L^{-1}\{\frac{1}{s-a}\}=e^{at}u(t)$$ where $u(t)$ is Unit Step Function A.K.A. Heaviside Step Function. so writing time functions is left to you for example $$y(t) = (\frac{iA}{\omega_1-\omega_2}e^{\omega_1t}+\frac{iA}{\omega_2-\omega_1}e^{\omega_2t})u(t)$$ also you can investigate the last condition $|x(t)|^2+ |y(t)|^2+ |z(t)|^2=1$ holds in Laplace domain using this fact: $$\mathscr L\{f(t)\}=F(s) \Rightarrow \mathscr L\{|f(t)|^2\}=\mathscr L\{f(t)f(t)^*\}=F(s)F(-s)$$ in which $^*$ stands for complex conjugate.

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