Quantum Pendulum: Solving in Heisenberg Interp.

[jackdamiels]

Hy

I am trying to solve problem of quantum pendulum in region of unstable equilibrium.
I am doing it in Heiseberg interpretation of QM. The equation of motion that I am getting is
$$\dot{\dot{\theta}} = \omega^2\theta$$,
and the solution is in form of :

$$x (t) = A\cosh(\omega t) + B\sinh (\omega t)$$.

With some starting connditions I can get A i B, that is simple. But problem arose when I am computing standard deviations od for example
$$(\delta x )^2 = <(x - <x>)^2 >$$
I am getting imaginary numbers, and time dependence. Time dependence is OK, because it is Heisenberg picture, but whay imagenery part in this standard deviations. State is :
$$1/{\sqrt{\sigma{sqrt{2\pi}e^{ip_0 x}e^{-\frac{(x-x_0)^2}{4\sigma^2}$$.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[member 553083]

The equation of motion for a quantum pendulum in the region of unstable equilibrium is given by \dot{\dot{\theta}} = \omega^2\theta ,where $\omega$ is the angular frequency.The solution to this equation can be written as x(t) = A\cosh(\omega t) + B\sinh (\omega t) .Using some initial conditions, one can determine the constants A and B.To compute the standard deviation of the position, one needs to compute the expectation value of $(x - <x>)^2$. This will give an expression that contains both real and imaginary parts. The imaginary part is due to the fact that the solution for $x(t)$ contains complex numbers, which means that the expectation value of $(x-<x>)^2$ will also contain complex numbers. The time dependence is expected, since we are working in the Heisenberg picture, where all operators evolve with time.The state of the system can be written as \frac{1}{\sqrt{\sigma\sqrt{2\pi}}e^{ip_0x}e^{-\frac{(x-x_0)^2}{4\sigma^2}}.