Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Chapter 10, Exercises 10.2.2 – 10.2.3.

We’ve seen that the 3-d isotropic harmonic oscillator can be solved in rectangular coordinates using separation of variables. The Hamiltonian is

The solution to the Schrödinger equation is just the product of three one-dimensional oscillator eigenfunctions, one for each coordinate. That is

Each one-dimensional eigenfunction can be expressed in terms of Hermite polynomials as

with the functions for and obtained by replacing by or and by or . We also saw earlier that in the 3-d oscillator, the total energy for state is given in terms of the quantum numbers of the three 1-d oscillators as

and that the degeneracy of level is .

Since the Hermite polynomial has parity (that is, odd (even) polynomials are odd (even) functions), the 3-d wave function has parity .

We can write the one state and three states in spherical coordinates using the standard transformation

Using the notation , we have, using and :

We can check that these are the correct spherical versions of the eigenfunctions by using the Schrödinger equation in spherical coordinates, which is

The spherical laplacian operator is

You can grind through the derivatives by hand if you like, but I just used Maple to do it, giving the results

In two dimensions, the analysis is pretty much the same. In the more general case where the masses are equal, but , the Hamiltonian is

A solution by separation of variables still works, with the result

The total energy is

For a given energy level , there are ways of forming out of a sum of 2 non-negative integers, so the degeneracy of level is .

The one state and two states are

To translate to polar coordinates, we use the transformations

so we have

Again, we can check this by plugging these polar formulas into the polar Schrödinger equation, where the 2-d Laplacian is

The results are

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Yasin ŞaleHi Sir,

Good job!

But passage from Eqn. (18) to Eqn. (20) is not clear enough, for people who are unfamiliar with these objects. I think you’d better to remark that $ \omega_x = \omega_y $, before or after Eqn. (20). Unless, you must write total energy of 2d oscillator as $ E=\hbar\omega_x(n_x+\frac{1}{2})+ \hbar\omega_y(n_y+\frac{1}{2}) $.

Thanks for your clear, brief and instructive articles.

gwrowePost authorI’ve covered the solution by separation of variables in the post in the link “3-d isotropic harmonic oscillator” in the first line. The solution uses the same method whether or not , giving equation 20 for the energies.