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Limit of a complex exponential

  1. May 26, 2014 #1
    Hi,

    BhEN5Ms.jpg

    The two terms should vanish at infinity according to the Quantum textbook of Griffiths, but I don't see how?

    I mean a complex exponential is a periodic function so how can it vanish at infinity?

    If you split up the first term

    exp(ikx) * exp(-ax)

    Take the limit of infinity of this expression

    lim exp(ikx) is not defined
    limit exp(-ax) turns zero

    But 0 * an undefined limit is still undefined, right?

    So help me out, please.
     
    Last edited: May 26, 2014
  2. jcsd
  3. May 26, 2014 #2

    DrClaude

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    Staff: Mentor

    Wrong :tongue2:. ##0 \times \infty## is undefined, but ##0 \times a## where ##|a| < \infty## is always 0.
     
  4. May 26, 2014 #3
    but

    lim (exp(ik-a)*x) to infinity equals lim exp (ikx) times lim exp(-ax) so it is ∞ * 0 ?
     
  5. May 26, 2014 #4

    pasmith

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    Homework Helper

    How do you think [itex]e^{-ax} \cos(kx)[/itex] behaves as [itex]x \to \infty[/itex]?

    If [itex]\lim_{x \to \infty} f(x)[/itex] and [itex]\lim_{x \to \infty} g(x)[/itex] exist, then it follows that [tex]
    \lim_{x \to \infty} f(x)g(x) = \left(\lim_{x \to \infty} f(x) \right)\left(\lim_{x \to \infty} g(x) \right).[/tex] However, even if one or both of [itex]\lim_{x \to \infty} f(x)[/itex] and [itex]\lim_{x \to \infty} g(x)[/itex] do not exist, it might still be the case that [itex]\lim_{x \to \infty} f(x)g(x)[/itex] exists.
     
  6. May 26, 2014 #5

    DrClaude

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    Staff: Mentor

    $$
    0 \leq \left| e^{i k x} \right| \leq 1
    $$
     
  7. May 26, 2014 #6
    So lim exp (ikx) does not exist , but lim exp( |ikx|) does?

    But how make it work in lim (exp(ik-a)x)

    Do you know where this rule can be found or what this rule can be called?


    But then you've already calcultated lim exp (-ax) that turns zero and lim exp (ikx) does not become a number?

    I really don't get it.

    If you calculate the limit as a whole (lim exp(ik-a)), you don't get a * 0 or you do?
     
  8. May 26, 2014 #7
    http://en.wikipedia.org/wiki/Absolute_convergence

    So because of that absolute value (the norm?) that is always 1 it converges? I think I got it.

    e^ikx converges? |e^ikx| = 1 right ? e^ikx * e^-ikx = e ^ (ikx-ikx) = e^0 = 1 right

    but why is the limit not defined?

    Wait, because it's periodic. Limit is not defined, but the max and min stays fixed. And thats why when you multiply with a function that goes to zero at infinity the product of the two functions also does?

    God, I was too tired to make sense out of anything I think.
     
  9. May 26, 2014 #8

    DrClaude

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    Staff: Mentor

    Sorry, I should've been more clear. Indeed
    $$
    \lim_{x \rightarrow \infty} e^{i k x} = \textrm{undefined}
    $$
    but ##0 \leq \left| e^{i k x} \right| \leq 1##, which means it is finite. I use the absolute value since the result is complex, but the important thing is that ##e^{i k x}## stays bounded even as ##x \rightarrow \infty##.

    Therefore, since
    $$
    \lim_{x \rightarrow \infty} e^{-a x} = 0
    $$
    you have
    $$
    \lim_{x \rightarrow \infty} e^{i k x} e^{-a x} = 0
    $$
    The fact that the limit of the first term is undefined doesn't affect the fact that the product of something finite with zero will always be zero.
     
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