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lecture_35 (dragged) 2

# lecture_35 (dragged) 2 - d dt Ψ = AΨ Ψ(0 = Ψ has the...

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MA 36600 LECTURE NOTES: MONDAY, APRIL 20 3 To see why this su ces, first recall the Product Rule: d dt x = d dt Ψ x 0 = d dt Ψ x 0 = [ A Ψ ] x 0 = A x . As for the initial condition: x (0) = Ψ (0) x 0 = I x 0 = x 0 . We now show that Ψ ( t ) satisfies the initial value problem above. We have the expansion Ψ ( t ) = k =0 A k t k k ! = I + A t + A 2 t 2 2 + A 3 t 3 6 + · · · = Ψ (0) = I . This function has the derivative d dt Ψ = 0 + A + A 2 t + A 3 t 2 2 + · · · = k =1 A k t k 1 ( k 1)! = A · k =1 A k 1 t k 1 ( k 1)! = A Ψ . Using these ideas, it is easy to see that in general the initial value problem
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Unformatted text preview: d dt Ψ = AΨ , Ψ (0) = Ψ has the unique solution Ψ ( t ) = exp ( A t ) Ψ . Keep in mind that the order oF the matrices here does matter! The matrix Ψ exp ( A t ) is not a solution to the initial value problem....
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