# An approximation in QM

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1. Jun 10, 2017

### Tspirit

1. The problem statement, all variables and given/known data
In the Griffiths book <Introduction to QM>, Section 2.3.2: Analytic method (for The harmonic oscillator), there is an equation ($\xi$ is very large)
$$h(\xi)\approx C\sum\frac{1}{(j/2)!}\xi^{j}\approx C\sum\frac{1}{(j)!}\xi^{2j}\approx Ce^{\xi^{2}}.$$
How to understand the meaning of the third approximately equal sign?

2. Relevant equations
I know the Taylor series
$$e^{x}=\frac{x^{n}}{n!}$$

3. The attempt at a solution
If take place of $\xi$ with $2\xi$, the equation may hold. However, can it be like this? What is the series of $e^{\xi^{2}}$ in real? I don't want the approximation. Thank you.

2. Jun 10, 2017

### Staff: Mentor

If the sum runs from j=0 to infinity, it is not an approximation, it is exact. It is the Taylor series of $e^x$ with $x=\xi^2$.

Note that $\xi^{2j} = (\xi^2)^j$.

3. Jun 10, 2017

### Tspirit

Thank you! I have found the problem I met. In fact, I want to prove the following equation $$e^{x^{2}}=\sum\frac{x^{2n}}{n!},$$ but didn't succeed at first. Now I can. This is the process.
Adopting the Taylor expansion of the left, with the derivertives
$$(e^{x^{2}})^{(0)}=e^{x^{2}},(e^{x^{2}})^{(1)}=e^{x^{2}}2x,(e^{x^{2}})^{(2)}=e^{x^{2}}(4x^{2}+2),(e^{x^{2}})^{(3)}=e^{x^{2}}(8x^{3}+12x),(e^{x^{2}})^{(4)}=e^{x^{2}}(16x^{4}+48x^{2}+12) ...,$$
$$e^{x^{2}}=1+0+2x^{2}/{2!}+0+12x^{4}/{4!}+...,$$ which is exactly equal to the right expression $$1+x^2/(1!)+x^4/(2!).$$
Actually, the zero terms in the Taylor expansion stopped me from calculating continuously in the beginning, making me think they are not equal.

4. Jun 11, 2017

### Staff: Mentor

That is more complicated, but it works as well.

5. Jun 11, 2017

### PeroK

You could simply let $y = x^2$, then:

$e^y = \sum\frac{y^{n}}{n!}$

And

$e^{x^2} = \sum\frac{(x^2)^{n}}{n!} = \sum\frac{x^{2n}}{n!}$

In fact, you don't really need to introduce the intermediate variable $y$. You can simply write down this last equation directly.

6. Jun 11, 2017

### Tspirit

Indeed, it's a simpler method.