# QM, the convergence of the harmonic oscillator function.

1. Jan 28, 2015

### Coffee_

1. After finding out that the wave function $\Psi(z) \sim Ae^{\frac{-z^{2}}{2}}$ in the limit of plus or minus infinity Griffiths separates the function into two parts $\Psi(z)=h(z)e^{\frac{-z^{2}}{2}}$

My question will be about a certain aspect of the function $h(z)$

After solving the ODE series method, one finds a certain recursion relationship for the coefficients of h(z) with an even and odd part.

As Griffiths notes for coefficients $a$ of this function, very far up (k large), $a_{k+2} \approx \frac{2}{k} a_{k}$

The next equation suddenly states the following: $a_{k} \approx \frac{C}{(k/2)!}$

I don't understand this step. Let me show you why my reasoning leads me to the wrong conclusion:

Let's start again from what we know: $a_{k+2} \approx \frac{2}{k} a_{k}$. Applying the same thing for $a_{k}$ and plugging in will give: $a_{k+2} \approx \frac{1}{k/2} \frac{1}{k/2 - 1} a_{k-2}$

Keep doing this and I find that $a_{k+2}=\frac{B}{(k/2)!}$ In the book this where I have k+2 stands k and I can't figure out why it's more correct than what I have here.

2. Jan 28, 2015

### BvU

Hello again,

I'm pretty convinced you won't be pleased with this, but hear (read) me out... :)

What matters here isn't the exact value of this ak (or ak+2, for that matter), but the ratio of these coefficients for really big k.

Anything to do with the precise value can be swept in what Griffiths calls C ("for some constant C" -- anything you don't want to be bothered with: throw it in there. As long as it doesn't keep growing with k you're fine).

Mind you, you want to be way above j = K/2 before you come even close to aj+2 / aj $\ \approx\$ 2/j !

(I find I'm unconsciously switching to Griffiths' indices j now, sorry..)

And even after all these years, I find it almost miraculous that from this kind of reasoning one is simply forced to conclude that K can NOT assume arbitrary values, but MUST be exactly equal to 2j + 1 for some j, however big. A hair difference and the power series runs away like $e^{x^2}$. Awesome !

3. Jan 29, 2015

### Coffee_

You were right, I was hoping for having made a stupid reasoning mistake or having overlooked something. The real reason seems to be less elegant than I would have liked but what can you do. Nothing to do with your answer, it was great, thanks.