The polynomial solutions occur for
|
{ϵ}_{n} = (n + {1\over
2}).
| (7.21) |
The terminating solutions are the ones that contains only even coefficients for even n and odd coefficients for odd n. Let me construct a few, using the relation (7.16). For n even I start with {a}_{0} = 1, {a}_{1} = 0, and for n odd I start with {a}_{0} = 0, {a}_{1} = 1,
Question: Can you reproduce these results? What happens if I start with {a}_{0} = 0, {a}_{1} = 1 for, e.g., {H}_{0}?
In summary: The solutions of the Schrödinger equation occur for energies (n + {1\over 2})ℏω, an the wavefunctions are
|
{ϕ}_{n}(x) ∝ {H}_{n}\left (\sqrt{{mω\over
ℏ}} x\right )\mathop{exp}\nolimits \left (−{mω\over
ℏ} {x}^{2}\right )
| (7.26) |
(In analogy with matrix diagonalisation one often speaks of eigenvalues or eigenenergies for E, and eigenfunctions for ϕ.)
Once again it is relatively straightforward to show how to normalise these solutions. This can be done explicitly for the first few polynomials, and we can also show that
|
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−∞}^{∞}{ϕ}_{{
n}_{1}}(x){ϕ}_{{n}_{2}}(x)\kern 1.66702pt dx = 0\qquad \mathrm{if}\quad {n}_{1}\mathrel{≠}{n}_{2}.
| (7.27) |
This defines the orthogonality of the wave functions. From a more formal theory of the polynomials {H}_{n}(y) it can be shown that the normalised form of {ϕ}_{n}(x) is
|
{ϕ}_{n}(x) = {2}^{−n∕2}{(n!)}^{−1∕2}{\left ({mω\over
ℏπ} \right )}^{1∕4}\mathop{ exp}\nolimits \left (−{mω\over
2ℏ} {x}^{2}\right ){H}_{
n}\left (\sqrt{{mω\over
ℏ}} x\right )
| (7.28) |