Help Request QHO series method from Griffiths

[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

 

[George Jones]

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?

 

[relativist]

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?

Thanks for your reply. But this leads to

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

 

[George Jones]

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}.

 

[relativist]

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

Thanks once again for your time.

Regards,

Relativist