Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Chapter 8. Section 8.6, Exercise 8.6.3.

Given the propagator for the harmonic oscillator, it is possible to work backwards and deduce the eigenvalues and eigenfunctions of the Hamiltonian, although this isn’t the easiest way to find them. We’ve seen that the propagator for the oscillator is

where is some function of time which is found by doing a path integral. Shankar cheats a bit by just telling us what is:

To deduce (some of) the energy levels, we can compare the propagator with its more traditional form

where is the th energy level. In position space this is

We can try finding the energy levels as follows. We take , which is equivalent to taking the end time to be a multiple of a complete period of the oscillator, so that the particle has returned to its starting point. In that case, 1 becomes

If we can expand this quantity in powers of , we can compare it with the series 4 and read off the energies from the exponents in the series. To do this, we write

To save writing, we’ll define the symbol

so that

We can now expand the last factor using the binomial expansion to get

In terms of the original variables, we get

Comparing with 4, we find energy levels of

These correspond to . The odd energy levels are missing because the corresponding wave functions are odd functions of and are therefore zero at , so the corresponding terms in 4 vanish. The numerical coefficients in 11 give us for .

To get the other energies, as well as the eigenfunctions, from a comparison of 1 and 4 is possible, but quite messy, even for the lower energies. To do it, we take as before, but now we take . That is, we start the oscillator off at some location and then look at it exactly one period later, when it has returned to the same position. The propagator 1 now becomes

We now need to expand this in a power series in , which gets very messy so is best handled with software like Maple. Shankar asks only for the first two terms in the series (the terms corresponding to and ) but even doing this by hand can get very tedious. The result from Maple is, for the first two terms:

Comparing this with 4, we can read off:

To check this, we recall the eigenfunctions we worked out earlier, using Hermite polynomials

The first two Hermite polynomials are

Plugging these into 23 and comparing with 20 and 22 shows we got the right answer.

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JosephThese posts are very helpful, am a Masters student & benefiting a lot……thanks.