Hello all,(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to understand the analytic method for the solution of the (quantum) harmonic oscillator equation.

[tex]-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^{2}x^{2}\psi = E\psi[/tex]

Let

[tex]\zeta = \sqrt{\frac{m\omega}{\hbar}}x[/tex]

[tex]K = \frac{2E}{\hbar\omega}[/tex]

We consider

[tex]\psi(\zeta) = h(\zeta)e^{-\zeta^{2}/2}[/tex]

The equation transforms to,

[tex]\frac{d^2 h}{d\zeta^2} - 2\zeta\frac{dh}{d\zeta} + (K-1)h = 0[/tex]

Assuming the power series solution,

[tex]h(\zeta) = \sum_{j=0}^{\infty}a_{j}\zeta^{j}[/tex]

we get the recursion formula,

[tex]a_{j+2} = \frac{2j+1-K}{(j+1)(j+2)}a_{j}[/tex]

For large j,

[tex]a_{j+2} \cong \frac{2}{j}a_{j}[/tex]

So,

[tex]a_{j+2} \cong \frac{2}{j}\frac{2}{(j-1)}\frac{2}{(j-2)}\ldots\frac{2}{1}a_{1}[/tex]

which simplifes to

[tex]a_{j+2} \cong \frac{2^j}{j!}a_{1}[/tex]

But the solution given is

[tex]a_{j} = \frac{C}{(j/2)!} = C \frac{1}{(j/2)}}\frac{1}{((j/2)-1)}\ldots\ldots[/tex]

I think I am missing something here. What am I doing wrong? Also, I know that the [itex](j/2)![/itex] is just notation but what if the "large j" include some possibly odd values? In that case it wouldn't even be defined. Thanks for your help.

EDIT: 800th post...yay

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Minor mathematical query from Griffiths [QM]

**Physics Forums | Science Articles, Homework Help, Discussion**