A complex integral from the text

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The discussion revolves around the integral of the exponential function involving complex numbers, specifically the equation ∫ e^(i(p-p')x/ħ)dx = 2πħδ(p-p'). Participants explore the implications of this integral, noting that evaluating it directly leads to undefined limits due to oscillatory behavior at infinity. The conversation also touches on related integrals, such as ∫ e^(2ikx)dx, which equals zero for k ≠ 0, highlighting inconsistencies with earlier statements. The importance of understanding Fourier transforms in this context is emphasized, as they relate to the properties of these integrals. Overall, the discussion seeks clarity on the mathematical foundations of these complex integrals.
quasar987
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In my book QM book, Gasiorowicz says that

\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx = 2\pi \hbar \delta(p-p')

Where does that come from? I mean, set i(p-p')/h = K. Then the solution is

\frac{\hbar e^{i(p-p')x/\hbar}}{i(p-p')}

and evaluate at infinity, it doesn't exists as the limit of cos and sin at infinity do not exist.

There must be something in the fact that the integral is complex but I haven't studies complex analysis yet so go easy on me please.
 
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What, no one?
 
This takes a bit of work. Are you familiar with Fourier transforms?

Regards,
George
 
Just the basics. Forget it then. How about this one?

\int_{-\infty}^{\infty}e^{2ikx}dx = 0
 
quasar987 said:
\int_{-\infty}^{\infty}e^{2ikx}dx = 0

This isn't consistent with your first post in this thread. To see this, set p'=0, \hbar =1, and x = 2u in the integral in your first post.

Regards,
George
 
Well they are respectively equation 3-40 and 3-54 in Gasiorowizc. :frown:
 
It's okay if k \neq 0.

<br /> \int_{-\infty}^{\infty}e^{2ikx}dx = \frac{1}{2} \int_{-\infty}^{\infty}e^{iku}du = \pi \delta (k)<br />

This equals zero if k \neq 0.

Regards,
George
 
K! k is proportional to energy so k = 0 is kind of a trivial case. All good.


:frown: --> :smile:
 
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