Question

21. For a particle of mass, m, moving along a circular path in the xy plane at a fixed distance, r, from the center and with no forces acting on it (V(x)-0), answer the following. Note the similarity to the hydrogen atom. We have an electron moving in the plane of a circle around a nucleus. Note the similarity between the Laplacian below and the azimuthal term in the hydrogen system Write the Schrödinger equation for this system. The Laplacian operator for this system is a. b. Verify that ψ(φ) = Ac" is a solution for this system. c. In this problem, we have a wave going in a circle. For the wave to Forbidden survive, the end of the wave must equal the beginning of the wave. That is urg)-v(φ + 2π) . Apply the appropriate boundary conditions and determine the possible values of m. Normalize the wavefunction. Determine the energy of the system. d. e.

Answer 1

Note1: in the following text wherever h appears it should be treated and read as h(cross) = h/2 $\pi$

Note 2: Normally I am allowed to answer 4 sub-parts only per question. However,for your convienience I have answered (e) part also

Laplacian is given to be $(1/r^2)(d^2/d\phi ^2)$

a. Schrodinger Equation is

$H\Psi =E\Psi$

$(-h^2/2m)(1/r^2)d^{2}(\Psi /d\phi ^2) +V(r,\phi )=E\Psi$

In this potential is zero everywhere and r is a constant

$(-h^2/2mr^2)(1/r^2)(d^{2}\Psi /d\phi ^2) =E\Psi$

$(d^{2}\Psi /d\phi ^2) + \omega ^2\Psi=0$

where $\omega =\sqrt{2mr^2E/h^2}$

b. $\Psi (\phi ) = Ae^{im^{'}\phi }$ here m' is used to distinguish from m

$d\Psi (\phi )/d\varphi = im'Ae^{im^{'}\phi }$

$d^{2}\Psi (\phi )^2/d\varphi = -m'^2Ae^{im^{'}\phi }$

If we substitute in our Schrodinger equation

$-m'^2Ae^{im^{'}\phi }$ + $\omega ^2Ae^{im^{'}\phi } = 0$

Thus above solution satisfies for specific values of m' such that $\omega ^2=m'$

c. Since particle moves in a circle,the wave function must satisfy the condition

$\Psi (\phi )=\Psi (2\pi m'+\phi )$

where m' = (+/-) 1,( +/-) 2 .........................

d. For normalizing

$\int_{0}^{2\pi }\Psi \Psi^{*}d\phi =1$

$A^{2}\int_{0}^{2\pi }e^{im'\phi }e^{-im'\phi }d\phi =1$

$A^{2}(2\pi )=1$

$A=(1/\sqrt{2\pi })$

e. $\omega^{2} ={2mr^2E/h^2} =m'^2$

Thus energy is quantized and hence can take states where m' = (+/-) 1 , (+/-) 2 ..................