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 [itex]H_0 = -\frac{\hbar^2}{2mR^2}\frac{d^2}{d\phi^2}[/itex]
The wavefunction at t=0 is [itex]\psi=ASin\phi[/itex]

Find the mean value of the observable [itex]Sin\phi[/itex]

Homework Equations

The eigenfunction are
[itex]\psi_n = \frac{1}{\sqrt{2\pi}e^{in\theta}}[/itex]
with eigenvalues
[itex]E_n = \frac{n^2\hbar^2}{2mr^2}[/itex]

The Attempt at a Solution

First i find the state of the system
[itex]\psi=ASin\phi =A \frac{e^{i\phi}-e^{-i\phi}}{2}=\frac{\left|1 \right\rangle -\left|-1\right\rangle}{\sqrt{2}}[/itex]
Then i have to calculate
[itex]\left\langle \psi \left| Sin\phi \right| \psi \right\rangle[/itex]
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
[tex]\left\langle \psi \left| Sin\phi \right| \psi \right\rangle[/tex]
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 [itex]Sin\phi = \frac{e^{i\phi}-e^{-i\phi}}{2i}[/itex]
i'm not really sure where to go from here.
What is [itex]e^{i\phi}\left|1\right\rangle[/itex] ?

 

[BruceW]

In the first post, you wrote:

Then i have to calculate
[tex]\left\langle \psi \left| Sin\phi \right| \psi \right\rangle[/tex]
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 [itex]\sin \phi[/itex] and [itex]\cos \phi[/itex] are self-adjoint operators on this Hilbert space!

 

[Dansuer]

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

[itex]Sin\phi \left| n \right\rangle [/itex]

in the [itex]\phi [/itex] basis

[itex]\frac{e^{i\phi}-e^{-i\phi}}{2i} e^{in\phi} = \frac{e^{i(n+1)\phi}-e^{i(n-1)\phi}}{2i}[/itex]



[itex]Sin\phi \left| n \right\rangle =\frac{\left|n+1\right\rangle-\left|n-1\right\rangle}{2i} [/itex]

from this i can calculate the mean values easily.