Larmor precession pi/2 out of phase

Therefore, in summary, the given matrix [0,-w;-w,0] is diagonalized to find the equation of motion for \s_+, s_-. This shows that \s_+ and \s_- are out of phase by pi/2 and can be used to show that \s_x, s_y are also out of phase by pi/2.
  • #1
hangainlover
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Homework Statement


I was going over the solution where it says [itex]\s_x, s_y[/itex] are out of phase by pi/2.
They first wrote equation of motion for [itex]\s_x, s_y[/itex] and then did something to the matrix and now the have equation of motion for [itex]\s_+, s_-[/itex]
They rewrite [itex]\s_x, s_y[/itex] in terms of [itex]\s_+, s_-[/itex] to see that [itex]\s_x, s_y[/itex] are out of phase bi pi/2,
I don't get what they did to the matrix [0,-w;-w,0]

Homework Equations



The Attempt at a Solution


I was just trying to understand the solution. when i initially made an attempt, i only noticed [itex]\s_x, s_y[/itex] are precessing .
I have attached the solution
 

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  • #2
.Solution:The equation of motion for \s_x, s_y is \dot \s_x = -w \s_y\dot \s_y = w \s_xWe can rewrite this in matrix form as[\dot \s_x; \dot \s_y] = [0,-w;-w,0][\s_x; \s_y]To find the equation of motion for \s_+, s_-, we diagonalize the matrix [0,-w;-w,0] by writing[\s_x; \s_y] = [1,1;1,-1][\s_+; \s_-]This gives us [\dot \s_x; \dot \s_y] = [0,-w;-w,0][1,1;1,-1][\s_+; \s_-]Simplifying, we get [\dot \s_+; \dot \s_-] = [-w,0;0,w][\s_+; \s_-]Comparing this to the general form of the equation of motion, we see that \s_+ and \s_- are oscillating with frequency w, and they are out of phase by pi/2. We can also see that \s_x, s_y are out of phase by pi/2 by rewriting \s_x = \frac{1}{2}(\s_+ + \s_-)\s_y = \frac{1}{2}(\s_+ - \s_-)
 

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