How to Solve the Isotropic 3-D Harmonic Oscillator Using Frobenius’ Method?

[bcoats]

Homework Statement

Use Frobenius’ method to solve the problem of the isotropic three dimensional
harmonic oscillator in polar coordinates. It is sufficient to
find the energy levels and degeneracies, but it would be nice to plot the
spectrum like we did for hydrogen. Be sure to introduce the natural
length scale and energy scale and to incorporate the asymptotic forms
at r = 0 and r = infinity of the radial wave function into your power series
solution. You will find that the recursion relation skips a term, i. e.,
relates c_j+2 to c_j . Thus two coefficients c_0 and c_1 are left undetermined.
How do you handle this situation? (Note that you can check
your answers against the Cartesian solution of the same problem.)

Homework Equations

See attachments. (Had to put them as images, I guess.) But it only let me put 3. a_0=hbar^2/(me^2), rho=r/(a_0), epsilon=-(E/R) where R=(me^4)/(2hbar^2)

The Attempt at a Solution

Writing down the equations is about as far as I've gotten. I think that the V(r) potential is right for the spherical (the last attached image) but I'm not sure. This stuff is just really too abstract for me.

 

[Jhero]

Thank you for your post. The problem of the isotropic three-dimensional harmonic oscillator in polar coordinates can be solved using Frobenius' method. This method involves finding a power series solution for the radial wave function, which is then used to determine the energy levels and degeneracies.

To begin, we must introduce a natural length scale and energy scale. Let us define a_0 as the Bohr radius, given by a_0 = hbar^2/(me^2), where hbar is the reduced Planck's constant and me is the mass of the electron. We can also define the energy scale as R = (me^4)/(2hbar^2).

With these definitions, the radial equation for the isotropic three-dimensional harmonic oscillator in polar coordinates becomes:

d^2R/drho^2 + (1/rho)dR/drho + [epsilon - (1/4)(1/rho^2) - (1/4)(l(l+1)/rho^2)]R = 0

where rho = r/a_0, epsilon = -(E/R), and l is the quantum number for the angular momentum.

To incorporate the asymptotic forms at r = 0 and r = infinity into our power series solution, we must first expand the radial wave function R as a power series:

R = sigma_(j=0)^infinity c_j*r^j

Substituting this into the radial equation and equating coefficients of like powers of r, we get a recursion relation for the coefficients c_j:

c_j+2 = [l(l+1) - j(j+1) - epsilon]c_j/(2(j+1)(j+2))

Note that this recursion relation skips a term, i.e., relates c_j+2 to c_j. This is because the potential for the isotropic three-dimensional harmonic oscillator in polar coordinates is not singular at r = 0. Therefore, two coefficients c_0 and c_1 are left undetermined.

To handle this situation, we must impose boundary conditions on the radial wave function. At r = 0, the radial wave function must be finite, which means that c_j = 0 for all odd values of j. This leaves c_0 and c_2 as the only non-zero coefficients. At r = infinity, the radial wave function must approach zero, which means that c_j = 0 for all j greater than