## Appendix A: the general solution to the evolution operatorWe intend in this appendix to give a self-contained proof of Magnus' exponential solution to the equation for the evolution operator. The original proof is to be found in Magnus' paper(1954). Here we will give a short but complete proof introducing in the meanwhile the necessary tools used while working with Magnus' expressions and to get acquainted with algebraic manipulations inherent to problems involving non-commutative operators. Magnus makes use of an original technique to manage the derivative of the exponential of a linear operator (a square matrix in our case). He transforms this derivative into an algebraical expression, and uses it to solve the differential equation. In what follows where with . With this notation, it is easy to show by a straightforward calculation that where we remind that the exponential of an operator Following Magnus, we extend the previous notation to a polynom in an evident form: where And we are ready to demonstrate the first two formulas which we will use hereafter. ## Formula 1This formula arises from the straightforward calculation of its left-hand-side. We begin calculating the effect of on the left of the exponential: Next we multiply on the left by : Now, so Formula 1 is demonstrated. ## Formula 2The left-hand-side of Formula 2 is obtained by multiplying the left-hand-side of Formula 1 on the left by and on the right by . If we do the same multiplications in the right-hand-side of Formula 1 we obtain or expanding the right-hand-side where which can be seen to correspond to the expansion of Next, Magnus demonstrates what can be called
Now we can obtain the following equivalent expressions where we have used notation (A2). Next we can separate indexes and write If we suppose now that , immediately we obtain that The inverse implication is completely equivalent. With the Inversion Lemma and Formula 2, we have all the instruments to solve the equation for the evolution operator
Let us suppose , then By using Formula 2 with and , one obtains which compared with Eq. (A3) gives We can now apply the Inversion Lemma with the same substitutions in
to obtain and This expression can be finally expanded using the following power series where the has the given values. To integrate the resulting equation for we start at t=0, where to satisfy boundary conditions. Introducing this solution into the equation we obtain a new solution: We can iterate the procedure to obtain , , ... as The solution is obtained as the limit of this series: © European Southern Observatory (ESO) 1999 Online publication: December 22, 1998 |