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An approximation in QM

  1. Jun 10, 2017 #1
    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. jcsd
  3. Jun 10, 2017 #2

    mfb

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    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##.
     
  4. Jun 10, 2017 #3
    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.
     
  5. Jun 11, 2017 #4

    mfb

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    That is more complicated, but it works as well.
     
  6. Jun 11, 2017 #5

    PeroK

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    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.
     
  7. Jun 11, 2017 #6
    Indeed, it's a simpler method.
     
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