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Normalized equation for particle in a ring

  1. Oct 16, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Normalized equation for particle in a ring, where V=0 on a ring of radius 'a' and infinite everywhere else.

    2. Relevant equations


    3. The attempt at a solution
    Replcing x by rθ,
    $$-\frac{\hbar^2}{2I}\frac{\partial^2\psi}{\partial\theta^2}=E\psi$$
    By guess, I found out that ##Ae^{ik\theta}## is an eigenfunction of the hamiltonian.
    To find ##A##, I used:
    $$\int_0^{2\pi}|\psi|^2rd\theta=1$$
    But in the solution, they used
    $$\int_0^{2\pi}|\psi|^2d\theta=1$$So the normalization constants were different. Which one is correct?
     
  2. jcsd
  3. Oct 16, 2016 #2

    PeroK

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    You are not integrating a function of ##r, \theta## over two dimensions. It's a simple one-dimensional integral for ##\theta##.
     
  4. Oct 16, 2016 #3

    Titan97

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    the particle moves along a ring. so the elemental length should be rdtheta
     
  5. Oct 16, 2016 #4

    PeroK

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    You've defined your ##\psi## in terms of ##\theta##. ##|\psi|^2## is the probability density function in terms of ##\theta##, not in terms of ##l## where ##l## is the arc length along the ring.

    Normalisation means that the total proability is 1, so the simple integral of ##|\psi(\theta)|^2 d\theta## is required.

    If you transformed ##\psi## into a function of arc length, ##l##, then you would integrate ##|\psi(l)|^2 dl##.

    To illustrate this, a constant pdf for ##\theta## would be ##p(\theta) = \frac{1}{2 \pi}##, whereas a constant pdf for ##l## would be ##p(l) = \frac{1}{2 \pi a}##
     
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