# Help Request QHO series method from Griffiths

1. Sep 14, 2010

### relativist

I Uwould be grateful if anybody can kindly help me with this particular doubt that cropped when I was working through Introduction to Quantum Mchanics by Griffiths ( Griffiths page 66 )

My question is how is the following obtained :

aj = C / (J/2)! ( Please read j as subscript of a )

I would be grateful to anybody who who can clarify my doubt.

Thanks

Relativist

2. Sep 14, 2010

### George Jones

Staff Emeritus
In my copy of Griffiths, this is on pages 53-54.

What do you get when use j = 2 in

$$a_{j+2} = \frac{2}{j} a_j?$$

j = 4? j = 6? j = 8?

3. Sep 14, 2010

### relativist

a10 = a2 / 4! when it should be a10 = a2 / 5! according to aj = C / (j / 2)!. Can you please tell me how this can be approximately coorect when we are considering large values of j. I worked for j = 100 and found a2 / 49 ! when it should be a2 / 50.

Regards,

Relativist

4. Sep 14, 2010

### George Jones

Staff Emeritus
Okay, how about reworking the argument in Griffiths as follows.

Consider the series

$$h \left( x \right) = \sum^\infty_{j=0} a_j x^j$$

with

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

for large $j$. Now consider

$$e^{x^2} = \sum^\infty_{n=0} \frac{x^{2n}}{n!} = \sum^\infty_{n=0} c_{2n} x^{2n},$$

so, for large $j$,

$$c_{j+2} = \frac{2}{j} c_j.$$

Therefore, asymptotically, $h \left( x \right)$ looks like $$e^{x^2}$$.

5. Sep 14, 2010

### relativist

Thanks very much for the clear explanation. I can see it now.

Thanks once again for your time.

Regards,

Relativist