Particle on a Ring: Finding Mean Value of Sin(phi)

[Dansuer]

Homework Statement

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

Homework 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}

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

 

[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)

 

[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 ?

 

[BruceW]

In the first post, you wrote:

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.

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 $\left\langle \psi \left| Sin\phi \right| \psi \right\rangle$

 

[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!

 

[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.