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For the function $$h(\xi) = C \sum \frac 1 {(j/2)!} \xi^j$$ Griffith's makes the following approximation at large $\xi$:

$$h(\xi) = C \sum \frac 1 {(j/2)!} \xi^j \approx C \sum\frac 1 {j!} \xi^{2j} \approx C e^{\xi^2}$$

Any explanations how the author did this? As in, what's his logic?

Reference: Introduction to Quantum Mechanics by Griffith's. Page 40, Ch.2 for Edition 1 and Page 54, Ch.2 for Edition 2.

Sidd
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  • What do you mean "why"? Because it is convenient for arriving at a simpler result later on almost certainly! I'm not sure what the actual questiion here is. – ACuriousMind Sep 24 '15 at 12:31

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Griffiths puts in the approximation signs $\approx$ because the argument is generally handwavy, but the inner relation is actually exact:

$$h(\xi) \approx C \sum \frac 1 {(j/2)!} \xi^j = C \sum\frac 1 {j!} \xi^{2j} = C e^{\xi^2}.$$

The reason for this is that the sum over $j$ is in steps of two, because the symmetry of the equation allows you to have separate even and odd series. The inner step is just a relabelling of the series, and the final step is a simple calculation.

The real mettle of the approximation is saying that the recursion of the series, $$ a_{j+2}=\frac{2j+1-K}{(j+1)(j+2)}a_j $$ can be reduced to the approximate form $a_{j+2}\approx \frac2j a_j$ at large $j$. If you're more precise then you use that intuition to show that, if you define $a_j$ as above with (say) $a_0=1$ and $a_1=0$, and you define as a standard $b_{j+2}=\frac2j b_j$ with $b_0=1$ and $b_1=0$, then there exist an index $J$ and a constant $c$ such that for all $j\geq J$ you have $a_j>c b_j$, i.e. the series you're interested in is bounded below by your standard, which produces a divergent function. Once you show that, then it follows that $$ h(\xi)>\sum_{j=0}^{J-1}(a_j-b_j)\xi^j+c\sum_{j=0}^{J-1}b_j\xi^j =\sum_{j=0}^{J-1}(a_j-b_j)\xi^j+ce^{\xi^2} $$ which diverges at large $\xi$.

E.P.
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I don't think he's trying to say that the two sums are approximately equal. I believe what he's TRYING to say is that both sums diverge similarly, i.e. both are exponential in $\xi$. Really, the sum on the left is more like half the sum on the right. But this is all just part of Griffith's heuristic argument to show that only certain SHO energies lead to normalizable wavefunctions, so a handwaving "they both diverge at a similar rate" is all you need.

Jahan Claes
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  • Still, why the specific choice of terms? If you look at the transition from $\frac 1 {(j/2)!}$ to $\frac 1 {j!}$, you see that the power of the $\xi$ also doubles. That shows that he is indeed using some specific appropriate terms to make his point. I am interested in the reasons behind that selection. In other words, why didn't he keep the power of $\xi$ the same, etc. – Sidd Sep 24 '15 at 04:31
  • @Sidd He just wanted to get something that could be written as an exponential. – Jahan Claes Sep 24 '15 at 17:08
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    @Sidd Here's another way to say it. He's just trying to show that the series grows exponentially for large $\xi$, thus the wavefunction isn't normalizable. If he drops half the terms in the summation and the series STILL grows exponentially, then the original series had to grow at LEAST exponentially. – Jahan Claes Sep 24 '15 at 18:38