Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Chapter 12, Exercise 12.3.7 (8) – (10).

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In this post, we’ll continue with the solution of the 2-d isotropic harmonic oscillator. In the previous post, we found that the radial equation can be written as

where the dimensionless variables are given by

and has a solution as a power series

The coefficients satisfy the recursion relation

Only for even are non-zero.

In order for to remain finite for large , the series must terminate, which gives the allowable values for the energy as

with

and .

We can now compare the solution obtained in polar coordinates with our earlier solution in terms of rectangular coordinates. First, what are the possible values for for a given energy ? From the relation 7, we can look at even and odd separately. For even , can take values . The first of these values for (that is, for ) each allow two values of such that , namely . If , then we must have . Thus for even the total number of combinations is .

For odd , can take on values , giving a total of possible values. (If this isn’t obvious, write it out for the first few values of odd to see the pattern.) For each of these values of , can take on the two values , thus there are again different combinations. Thus a state with energy has a degeneracy .

We can construct the actual eigenfunctions for a couple of values of by plugging in the appropriate formulas. For there is only one function, which we find by setting . From 4, we have

and from 1 we have

or, in terms of the original variables

The complete solution is given by

so for we have

The constant can be found by normalizing:

This agrees with the earlier result in rectangular coordinates (eqn 26 in this post). This must be the case, since the state is non-degenerate.

For , we have and so we have two solutions:

Again, we normalize

These solutions are linear combinations of the corresponding solutions in rectangular coordinates (converted to polar for comparison):

The combinations are

The parity of the states is found from their behaviour under the transformation (in rectangular coordinates) and . In polar coordinates this is equivalent to the transformation and from 18 and 24 we see that

Thus the parity of is even, and that of is odd. In general, since the dependence enters only through the term , we see that adding to leaves the term unchanged and multiplies the term by , so the parity of state is .