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In Griffiths' book page 53, when we derive the solution of the quantum harmonic oscillator by using the power series way, we have: $$a_{j+2} = \frac{2j+1-K}{(j+1)(j+2)}\, a_{j} .$$ And for large $j$, we have: $$a_{j+2}\approx\frac{2}{j}\,a_j.$$ Up to this point I totally agree (one just takes the limit).

However, the subsequent derivation of solution $a_{j}$ and $h(\xi)$ I attached from the Griffiths' textbook are very confusing.

  • How did it go from $a_{j+2}\approx\frac{2}{j}a_j$ to the solution of $a_{j}$?
  • Also, how do the second and third approximations work in $h(\xi)$?

My questions are mainly mathematical. I very much hope someone can provide a derivation or refer a link where these questions may already be answered.

enter image description here enter image description here

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    $\begingroup$ So... We're basically just going to ignore that this is the pdf snapshot. Hmm. $\endgroup$
    – eqb
    Commented Aug 5, 2014 at 7:07
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    $\begingroup$ @eqb What do you mean? You didn't expect me to write down those equations if I can take a snapshot, right? $\endgroup$
    – Lawerance
    Commented Aug 5, 2014 at 7:09
  • $\begingroup$ @eqb School hasn't started yet so I can't ask my professor. Sorry guys. $\endgroup$
    – Lawerance
    Commented Aug 5, 2014 at 7:09
  • $\begingroup$ You don't need to be sorry when you have a question! It is perfectly fine. $\endgroup$
    – Ellie
    Commented Aug 5, 2014 at 7:50

1 Answer 1

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Looking at even indices, from $a_{2j_0+2} \approx \frac{1}{j} a_{2j_0}$, we get : $$a_{2j_0+2k} \approx \frac{(j_0-1)!}{(j_0+k-1)!}a_{2j_0} = \frac{1}{(j_0+k)!} C \tag{1}$$, with $C=(j_0+k)\,(j_0-1)! \,\,a_{2j_0}$

So, taking $n= 2(j_0+k)$, we have finally $a_{n} \approx \frac{1}{(n/2!)} \, C$

Finally, $h_{even}(\xi) \approx \sum\limits_{n= j_0 + 1}^{+\infty}a_{2n} \xi^{2n}= C\sum\limits_{n= j_0 + 1}^{+\infty}\frac{1}{n!} (\xi^2)^n $

The behaviour of $h_{even}(\xi)$ for high $\xi$, is then the same as the behaviour of $C\sum\limits_{n= 0}^{+\infty}\frac{1}{n!} (\xi^2)^n = C e^{\xi^2}$

You have the same demonstration for $h_{odd}(\xi)$, so finally, the behaviour of $h(\xi)$ for high $\xi$ is $C e^{\xi^2}$, too.

UPDATE (to answer to the OP's comments)

a) You have : $a_{2j_0+2} \approx \frac{1}{j} a_{2j_0}$, then you apply the formula to $(j_0+k)$ : $a_{2j_0+2k} \approx \frac{1}{j_0+k-1} a_{2j_0+2k-2}$, so finally, you have : $a_{2j_0+2k} = \frac{1}{(j_0+k-1)(j_0+k-2)...(j_0)}a_{2j_0}$. You multiply numerator and denominator by $(j_0+k)(j_0-1)!$, and you get the formula $(1)$

b) The identity is $e^y = \sum\limits_{i=0}^{+\infty}\dfrac{y^n}{n!}$

c) You have to choose some particular $j_0$ as a departure point for the recurrence relation described in a). I used this notation to show that $j_0$ is not a variable, it is fixed.

d) The identity described in b) is true for all $y$, because the serie is convergent for all $y$

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  • $\begingroup$ In your first line, you claim $a_{2j_0 +2k} \approx \frac{(j_0-1)!}{(j_0+k-a)!}\,a_{2j_0}$. How do you get this approximation? In your fourth line, you have the left side of the summation equal to the right side $e^{\xi^2}$. I vaguely remember that this is a calculus identity. Can you refer me to a link to this identity? $\endgroup$
    – Lawerance
    Commented Aug 5, 2014 at 16:55
  • $\begingroup$ In your first equation you wrote $a_{2j_0+2} \approx \frac1j a_{2j_0}$. Should the $j$ here be $j_0$? $\endgroup$
    – Lawerance
    Commented Aug 5, 2014 at 17:02
  • $\begingroup$ For the fourth line approximation, if what you use is Taylor approximation, $e^x = \sum \frac{x^n}{n!} $, it is wrong because $x -> 0$ when you can use the approximation. But here as the book claims, $\xi$ is very large. So, you can't use the Taylor approximation. $\endgroup$
    – Lawerance
    Commented Aug 5, 2014 at 17:24
  • $\begingroup$ @Lawerance : I update the answer, to answer to your comments (see section UPDATE) $\endgroup$
    – Trimok
    Commented Aug 6, 2014 at 8:29
  • $\begingroup$ Thanks so much, indeed, with my 200% thankfulness. Great illustration and mathematical proof. I will refer my classmates to this link if they do have questions about this proof once the quarter starts. Thanks again! $\endgroup$
    – Lawerance
    Commented Aug 6, 2014 at 16:36

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