Shankar 12.6.11 (p. 352)[Assigned as HW9, problem 1, in Ph234]
(1) By combining Eqs. (12.6.48) and (12.6.49) derive the two-term recursion relation for the 3D harmonic oscillator. Argue that if U is to have the right properties near y=0. Derive the quantization condition .
(2) Calculate the degeneracy and parity at each n and compare with Exercise 10.2.3 where the problem was solved in Cartesian coordinates.
(3) Construct the normalized eigenfunctions for n=0 and 1. Write them as linear combinations of the n=0 and n=1 eigenfunctions obtained in Cartesian coordinates.
Solution (from Tudor Costin, Ph234)
Most of this should already be familiar from the last question of HW8. We have
(Eq. 12.6.49)
from which we get
and.
Note that the extraalters the starting index of the sum. If it were not present, we would have to start at n>0 for the sum in the derivatives. Furthermore, the factor (n+l) ensures that there is no divergence in the second derivative. If l=0, the C0 term drops out. If l0, the C0 term corresponds to a non-negative power.
Plugging all of those into the DE (Eq. 12.6.48) results in:
Proceed as before:
Note that theterm in the first sum is zero, so we may start the summation at Next we shift the index on the first sum so thatand the sum now starts at -1:
The last term comes from theterm in the left-hand sum, for which there is no compensating term in the right-hand sum and so must be treated separately once the sums are merged. There are no other yl terms anywhere in the summation, so the only way to satisfy the equality is to set .
The recursion relation that we find is
from which we get
Again all the odd terms will drop out since. This implies that (otherwise the whole series is zero). As before, we argue that the summation must terminate, so the condition on the energy is
, or
b) We now redefine n to beso that . For a fixed n, k can range between zero and n/2 if n is even, or between zero and (n-1)/2 if n is odd. Either way, let us denote this maximum value K.
For a fixed value of k, we have. Compared to the 2D problem from last week, we have m as an additional degree of freedom. For any given l, m can range between -l and +l, so there are 2l+1 degenerate states for a fixed (n,l). The overall degeneracy is
n even: .
n odd: .
So in either case we have.
In 3D, parity takes and. The radial wavefunction is not affected, so the parity operator only acts on. Recall that .
The parity transformation is
As in HW8Q2, the change in introduces a factor of . The change in is harder to deal with, since enters the wavefunction via the associated Legendre polynomials . Since , we analyze (see HW8Q4):
Hence since , and our result follows.
c) For , the entire sum reduces to the constant , so we have
and
To avoid confusion in the notation, let us define and write the wavefunctions in terms of r from now on. The integration measure is, so the normalization condition reads
The full wavefunction is
This is exactly what we would find if we were to express in Cartesian coordinates. Since the ground state is unique, this result is not surprising.
For , there are potentially two terms in the sum. Note that is not allowed (the only way to achieve is by having and ). We have:
Hence andfor some A. Normalize:
and
The full wavefunctions are
Since , we expect that to be related to the state:
As for the other two Cartesian wavefunctions, they must be related to . Recall that and , so the way to combine them is as follows:
and
The proof of this is a (trivial) exercise left to the student. One need only expand the sine and cosine into complex exponentials.