Solutions of Heisenberg Equations of Motion for Angular Momenta

[Dixanadu]

Hey guys,

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

http://imageshack.com/a/img32/2013/3q8s.jpg

 

[MuIotaTau]

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.

 

[Dixanadu]

damn, sorry posted by accident. Let me type that again

 

[Dixanadu]

So our first equation is:

\frac{d<L_{x}>}{dt}=\omega_L <L_{y}>

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

\frac{d^2 <L_{x}>}{dt^2}=\omega_L \frac{d<L_{y}>}{dt}

But we know from our second equation that

\frac{d<L_{y}>}{dt}=-\omega_L <L_{x}>

Substituting gives

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

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

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

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

So does this basically mean that <L_{x}>=<L_{y}> and that <L_{z}> is constant? I don't see how it can make sense that <L_{x}>=<L_{y}>...