# Failure to see the validity of an approximation to DiffEq.

1. Dec 8, 2015

### davidbenari

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.

2. Dec 9, 2015

### Buzz Bloom

Hi david:

I can help with the first question, but there is something I don't understand regarding your second question.

Regarding
Just confirm that (1) is the solution to the recursion equation
aj+2=2/j aj
by substituting (1) and a modified (1) for j+2 into the above equation.

Regarding the the second mystery, I don't understand the relationship between the constants aj and the function h(y) of the differential equation.

Regards,
Buzz

3. Dec 9, 2015

### davidbenari

Buzz:

I guess I forgot to mention that we're supposing the solution $h(y)$ is of the form $h(y)=\sum a_j y^j$. I.e. a power series.

I'm relatively confused about what you said here. It would be helpful if you could be more explicit. Specifically I'm not sure how this solves the recursion relation.

Thanks!

4. Dec 9, 2015

### Buzz Bloom

Hi david:

aj+2 = (2/j) aj
aj ≈ C/(j/2)!
aj+2 ≈ C/((j+2)/2)!​

Now one needs only to show that
C / ((j+2)/2)! ≈ C (2/j) / (j/2)!​
This can be more easily seen by cancelling the Cs and examining the reciprocals.
((j+2)/2)! ≈ (j/2)! / (2/j) = (j/2)! × (j/2)
((j/2)+1)! = (j/2)! × (1+J/2) ≈ (j/2)! × (j/2)​
Cancelling the (j/2)!s gives
(1+J/2) ≈ (j/2)​
which is a reasonable approximate equality for sufficiently large j.

Regards,
Buzz