# Particle on a ring

1. Jul 16, 2014

### Dansuer

1. The problem statement, all variables and given/known data
Consider a particle on a ring with radius R in a plane.
The Hamiltonian is $H_0 = -\frac{\hbar^2}{2mR^2}\frac{d^2}{d\phi^2}$
The wavefunction at t=0 is $\psi=ASin\phi$

Find the mean value of the observable $Sin\phi$
2. Relevant equations
The eigenfunction are
$\psi_n = \frac{1}{\sqrt{2\pi}e^{in\theta}}$
with eigenvalues
$E_n = \frac{n^2\hbar^2}{2mr^2}$
3. The attempt at a solution
First i find the state of the system
$\psi=ASin\phi =A \frac{e^{i\phi}-e^{-i\phi}}{2}=\frac{\left|1 \right\rangle -\left|-1\right\rangle}{\sqrt{2}}$
Then i have to calculate
$\left\langle \psi \left| Sin\phi \right| \psi \right\rangle$
but i don't know how to deal with a function of the operator. I though about expanding it in a taylor series but it does not seem to work.

Any help it's appreciated

2. Jul 17, 2014

### BruceW

try writing out
$$\left\langle \psi \left| Sin\phi \right| \psi \right\rangle$$
in the basis of functions of $\phi$. Then it shouldn't take you too long to convince yourself that the operator $Sin\phi$ becomes something very simple in this basis. hint: what does the operator $\phi$ become in the $\phi$ basis? (p.s. try not to over-think things)

3. Jul 17, 2014

### Dansuer

The ϕ operator in the ϕ becomes the identity operator.
With this in mind, i write $Sin\phi = \frac{e^{i\phi}-e^{-i\phi}}{2i}$
i'm not really sure where to go from here.
What is $e^{i\phi}\left|1\right\rangle$ ?

4. Jul 17, 2014

### BruceW

In the first post, you wrote:
So does this mean that you would know how to calculate $\left\langle \psi \left| \phi \right| \psi \right\rangle$ ? How would you calculate this? Most likely, it will be a similar method to calculate $\left\langle \psi \left| Sin\phi \right| \psi \right\rangle$

5. Jul 18, 2014

### vanhees71

Be careful. The angle operator is not a proper self-adjoint operator on the Hilbert space of periodic square integrable functions, which is the correct Hilbert space for the rotator. It's a good exercise to think about, why this is the case! Another hint: The operators $\sin \phi$ and $\cos \phi$ are self-adjoint operators on this Hilbert space!

6. Jul 18, 2014

### Dansuer

I'll look at the action of the operator on a general eigenstate

$Sin\phi \left| n \right\rangle$

in the $\phi$ basis

$\frac{e^{i\phi}-e^{-i\phi}}{2i} e^{in\phi} = \frac{e^{i(n+1)\phi}-e^{i(n-1)\phi}}{2i}$

$Sin\phi \left| n \right\rangle =\frac{\left|n+1\right\rangle-\left|n-1\right\rangle}{2i}$

from this i can calculate the mean values easily.