There always has to be time-dependence in some way. In fact, the Schrödinger and Heisenberg-Pictures are just different ways of writing down the same mathematics. There is a magical operator, call it U(t), which creates the time dependent wavefunctions out of the wavefunctions at t=0. So...
x is an operator which is used to calculate the expectation value - the expectation value may change in time, but not the way it is calculated! If the expectation value changes, it is due to the wavefunction which changes in time. In fact, this is the perspective of the so-called "Schrödinger...
Right, but you have conjugate it all and write out the Hamiltonian to see something. And don't forget about "i" just because it's not real, it will get mad if you do to often! :wink:
Ah, I think you were working with what I know as "Ehrenfest's Theorem". The Schrödinger Equation is this: i \hbar \frac{\partial}{\partial t} \Psi(\vec x, t) = \left(
-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(\vec x) \right) \Psi(\vec x, t) .
No, you just have to take the derivative wrt time of the function (\Psi x \Psi^*) and remember that \Psi solves the Schrödinger-Equation (and \Psi^* the "complex conjugate Schrödinger-Equation" ...)
For a QM-Interpretation you have to normalize the statevector, so that
\langle z | z \rangle = 1 . If \langle z_+ | z_+ \rangle = 1 and \langle z_- | z_- \rangle = 1-, this not the case here.
I wouldn't say "amplitude" but "probability": |\langle -z | x \rangle|^2 is the probability...
I think that there is an error in your Lagrangian: Since you are interested in the movement of the cylinder and watch it in the "labatory system", so to say, you can't say that the velocity of the particle in y-direction is R \dot \phi \cos(\phi). \phi is fixed in the "cylinder system", so...