# Solutions of Heisenberg Equations of Motion for Angular Momenta

1. Nov 19, 2013

Hey guys,

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

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

Last edited by a moderator: May 6, 2017
2. Nov 19, 2013

### 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.

3. Nov 19, 2013

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

4. Nov 19, 2013

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 dont see how it can make sense that $<L_{x}>=<L_{y}>$....