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7.4 A few solutions

The polynomial solutions occur for

$\displaystyle \epsilon_{n }^{}$ = (n + $ {\frac{1}{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 a0 = 1, a1 = 0, and for n odd I start with a0 = 0, a1 = 1,
H0(y) = 1, (7.22)
H1(y) = y, (7.23)
H2(y) = 1 - 2y2, (7.24)
H3(y) = y - $\displaystyle {\textstyle\frac{2}{3}}$y3. (7.25)

Question: Can you reproduce these results? What happens if I start with a0 = 0, a1 = 1 for, e.g., H0?

In summary: The solutions of the Schrödinger equation occur for energies (n + $ {\frac{1}{2}}$)$ \hbar$$ \omega$, an the wavefunctions are

$\displaystyle \phi_{n}^{}$(x) $\displaystyle \propto$ Hn$\displaystyle \left(\vphantom{\sqrt{\frac{m\omega}{\hbar}}x}\right.$$\displaystyle \sqrt{\frac{m\omega}{\hbar}}$x$\displaystyle \left.\vphantom{\sqrt{\frac{m\omega}{\hbar}}x}\right)$exp$\displaystyle \left(\vphantom{-\frac{m\omega}{\hbar}x^{2}}\right.$ - $\displaystyle {\frac{m\omega}{\hbar}}$x2$\displaystyle \left.\vphantom{-\frac{m\omega}{\hbar}x^{2}}\right)$ (7.26)

(In analogy with matrix diagonalisation one often speaks of eigenvalues or eigenenergies for E, and eigenfunctions for $ \phi$.)

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

$\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \phi_{n_1}^{}$(x)$\displaystyle \phi_{n_2}^{}$(x) dx = 0        if    n1 $\displaystyle \neq$ n2. (7.27)

This defines the orthogonality of the wave functions. From a more formal theory of the polynomials Hn(y) it can be shown that the normalised form of $ \phi_{n}^{}$(x) is

$\displaystyle \phi_{n}^{}$(x) = 2-n/2(n!)-1/2$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/4}$exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$Hn$\displaystyle \left(\vphantom{\sqrt{\frac{m\omega}{\hbar}}x}\right.$$\displaystyle \sqrt{\frac{m\omega}{\hbar}}$x$\displaystyle \left.\vphantom{\sqrt{\frac{m\omega}{\hbar}}x}\right)$ (7.28)


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Next: 7.5 Quantum-Classical Correspondence Up: 7. The Harmonic oscillator Previous: 7.3 Taylor series solution

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk