- #1

- 461

- 16

## Main Question or Discussion Point

The following comes from Griffiths Intro. to QM (2nd Ed) page 53.

We want to solve the Schrödinger Equation for the harmonic oscillator case using a power series method. The details aren't important but you want to solve

##h''(y)-2yh'(y)+(K-1)h=0##

whose recursion formula is

##a_{j+2}=\frac{2j+1-K}{(j+1)(j+2)}a_j##

Griffiths wants to analyze those solutions which aren't normalizable so he considers large values of ##j##. The recursion formula becomes (to large ##j##)

##a_{j+2}=\frac{2}{j}a_j##

Which makes sense, but then he says that this has the approximate solution (from now on is the part where I don't understand)

(1) ##a_{j}\approx \frac{C}{(j/2)!}## where C is a constant

considering large ##y## we get that

(2) ##h(y) = C \sum \frac{1}{(j/2)!}y^j = C \sum \frac{1}{j!}y^{2j}##

So, I consider (1) mysterious and the second equality of (2) ( ##C \sum \frac{1}{j!}y^{2j}##) mysterious as well. Anyone care to help me showing the intermediate left-out steps?

Thanks.

We want to solve the Schrödinger Equation for the harmonic oscillator case using a power series method. The details aren't important but you want to solve

##h''(y)-2yh'(y)+(K-1)h=0##

whose recursion formula is

##a_{j+2}=\frac{2j+1-K}{(j+1)(j+2)}a_j##

Griffiths wants to analyze those solutions which aren't normalizable so he considers large values of ##j##. The recursion formula becomes (to large ##j##)

##a_{j+2}=\frac{2}{j}a_j##

Which makes sense, but then he says that this has the approximate solution (from now on is the part where I don't understand)

(1) ##a_{j}\approx \frac{C}{(j/2)!}## where C is a constant

considering large ##y## we get that

(2) ##h(y) = C \sum \frac{1}{(j/2)!}y^j = C \sum \frac{1}{j!}y^{2j}##

So, I consider (1) mysterious and the second equality of (2) ( ##C \sum \frac{1}{j!}y^{2j}##) mysterious as well. Anyone care to help me showing the intermediate left-out steps?

Thanks.