- #1
davidbenari
- 466
- 18
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 constantconsidering 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 constantconsidering 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.