When I learned quantum mechanics by reading Griffith's book called Introduction to quantum mechanics, I was confused by his description.

In Page 53 of the 2ed edition book, after got the recursion formula $a_{j+2}=\frac{(2j+1-K)}{(j+1)(j+2)}a_j$(First we solved the Schr$\ddot o$dinger equation $\frac{d^2\psi}{d\xi^2}=(\xi^2-K)\psi$, where $K=\frac{2E}{\hbar \omega}$, and the solution is that $\psi(\xi)=h(\xi)e^{-\xi^2/2}$. Then we expressed $h(\xi)$ in the form of power series in $\xi$, e.i $h(\xi)=\sum_{j=0}^\infty a_j\xi^j$.) the author took a approximation at very large $j$, the recursion formula becomes $a_{j+2}\approx\frac{2}{j}a_j$. Up to now, I understand everything. But the next things stuck me.

The author's words as follows,

For at very large $j$, the recursion formula becomes (approximately) $$ a_{j+2}\approx\frac{2}{j}a_j \tag{1} $$ with the (approximate) solution $$ a_j\approx\frac{C}{(j/2)!},\tag{2} $$ for some constant $C$, and this yields (at large $\xi$, where the higher powers dominate) $$ h(\xi)\approx C\sum \frac{1}{(j/2)!}\xi^j \approx C\sum\frac{1}{j!}\xi^{2j}\approx Ce^{\xi^2} \tag{3} $$

The question is that I barely understand that how to deduce equations $(2)$ and $(3)$. In particular, I think the equation $(2)$ is bizarre. And I know if I get $(2)$, I will know how to deduce $(3)$. I hope someone could help me and explain that how to deduce $(2)$.


I see that your problem is you can't deduce equation $(2)$.But note that Hermit polynomial is $h(\xi)=\sum_{j=0}^\infty a_j\xi^j$,and the sum start from 0.So the terms are all even.

Then using $$a_{j+2}\approx\frac{2}{j}a_j$$ we have$$\begin{align}a_j&\approx\frac{2}{j-2}a_{j-2}\\ &\approx\frac{2}{(j-2)(j-4)}a_{j-4}\\ &\approx\frac{1}{(j/2-1)(j/2-2)(j/2-3)}a_{j-6}\\ &\dots\\ &\approx\frac{a_0}{(\frac{j}{2})!} \end{align}$$ where $a_0$ is $C$ in your expression. The rest I'm sure is easy for you to complete.Just put $a_j$ back to $h(\xi)$ and make a variable substitution.

  • $\begingroup$ You're welcome.You just miss something there.BTW,normally we prefer to solve harmonic oscillator problem using ladder operators,it's far easier than Hermit polynomial and have more applications in angular momentum operator,identical partical,etc.And I believe Griffiths put that part before the power series method. $\endgroup$ – Turgon Feb 14 '16 at 6:08
  • $\begingroup$ Your answer is wrong, because your process violets the premise "at large $\xi$". And $C$ is just one constant, not $a_0$. $\endgroup$ – Wang Yun Jul 31 '17 at 12:55

You're trying to solve for a series of the sum

$$h(x) = \sum_{j=0}^\infty a_j x^j $$

and you found that only even terms contribute, with

$$a_{j+2} = \text{something } a_j.$$

Since only even terms contribute, let's make our lives easier and rescale the coefficients as follows: we define $b_n \equiv a_{n/2}$ such that

$$h(x) = \sum_{n=0}^\infty b_n x^{2n}$$

Then for very large $n$ you have

$$b_{n+1} \simeq \frac{1}{n+1} b_n + O(1/n).$$

If you forget about the $1/n$ corrections for now, the above recursion is solved exactly for

$$b_n = \frac{C}{n!}$$

where $C$ is some constant. But that's precisely what you need to show.

  • $\begingroup$ If we know $a_0$, we will get even terms, If we know $a_1$, we will get odd terms. $\endgroup$ – Wang Yun Feb 14 '16 at 1:39
  • $\begingroup$ To be honest, I can't understand your answer. $\endgroup$ – Wang Yun Feb 14 '16 at 2:48

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