1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Particle on a ring

  1. Jul 16, 2014 #1
    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 [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]
    2. Relevant 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]
    3. 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
     
  2. jcsd
  3. Jul 17, 2014 #2

    BruceW

    User Avatar
    Homework Helper

    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)
     
  4. Jul 17, 2014 #3
    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] ?
     
  5. Jul 17, 2014 #4

    BruceW

    User Avatar
    Homework Helper

    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##
     
  6. Jul 18, 2014 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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!
     
  7. Jul 18, 2014 #6
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Particle on a ring
  1. Particle on a ring (Replies: 1)

  2. Particle on a ring (Replies: 1)

  3. Particle in a ring (Replies: 1)

Loading...