Solutions of Heisenberg Equations of Motion for Angular Momenta

  • Thread starter Dixanadu
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  • #1
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Hey guys,

Imma type this up in word so its nice and clear!

http://imageshack.com/a/img32/2013/3q8s.jpg [Broken]
 
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Answers and Replies

  • #2
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Your equations look right to me, although I don't have a lot of time to check thoroughly right now. So your equations are coupled differential equations, right? They are of the form

$$y_1' = a y_2$$
$$y_2' = -a y_1$$
$$y_3' = 0$$

where the prime mark denotes differentiation. The last equation should be simplest to solve. What does it mean if a quantity's time derivative is zero? Does it change?

For the other two, I'm a bit rusty on some parts of differential equations, but I would start by taking the time derivative of your first two equations. See if doing that makes a solution clear.
 
  • #3
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damn, sorry posted by accident. Let me type that again
 
  • #4
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So our first equation is:

[itex]\frac{d<L_{x}>}{dt}=\omega_L <L_{y}>[/itex]

If we differentiate this again w.r.t time:

[itex]\frac{d^2 <L_{x}>}{dt^2}=\omega_L \frac{d<L_{y}>}{dt}[/itex]

But we know from our second equation that

[itex]\frac{d<L_{y}>}{dt}=-\omega_L <L_{x}>[/itex]

Substituting gives

[itex]\frac{d^2 <L_{x}>}{dt^2}=-\omega^{2}_{L} <L_{x}>[/itex]

Which has the solution [itex]<L_{x}>=A cos(\omega_L t) + B sin(\omega_L t)[/itex]

But the weird thing is: if we repeat this same thing for the second differential equation, we get the same thing - that is

[itex]\frac{d^2 <L_{y}>}{dt^2}=-\omega^{2}_{L} <L_{y}>[/itex]

So does this basically mean that [itex]<L_{x}>=<L_{y}>[/itex] and that [itex]<L_{z}>[/itex] is constant? I dont see how it can make sense that [itex]<L_{x}>=<L_{y}>[/itex]....
 

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