Recent content by Hargoth

Since the mass is attached to a spring, the length l of the pendulum is not constant.

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...

No, you have to conjugate \Psi as well. Therefore \Psi \Rightarrow \Psi^* .

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" ...)

Write down the derivative and think of the Schrödinger-Equation, then. ;) There's still an i (imaginary unit) missing, I think ...

Are you sure your \frac{d}{dx} in the first equation isn't \frac{d}{dt} and some i is missing?

Yeah, I just wanted to say that your probability of finding z in state z from the equation above would be 2^2=4 , so you have to normalize.

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...

Yeah, but if | z_+ \rangle, | z_- \rangle are basekets of the Hilbert space you consider, your equation would be a definition of | z \rangle

What is | z \rangle ?

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...

Yeah, thx. Well, if you get the solution (by yourself or from the classes ;) ), I'd like to hear it. :smile: