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I need to show the following identity $$\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(k-k')x}dx = \delta(k-k'),$$ where $\delta$ is the Dirac function. In the text it says that the functions $u_k = \frac{1}{\sqrt{2\pi}}e^{ikx}$ represent an orthonormal system and in order to show this it provides the identity above. In case $k=k'$ the right and left hand side of the equation coincide, i.e. $\delta(k') = \infty.$ The quetsion is then why $u_k$ would be orthonormal. In case $k \neq k'$ I do not know how the left hand side will vanish. Can somebody provide a comment or an explanation of how to check that $u_k$ is orthonormal and why there are the difficulties I raised above ?

Many thanks.

user996159
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    In which space(s) are you working? $\delta(k-k^\prime)$ is not a real map. – mathcounterexamples.net Jul 11 '22 at 20:06
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    No, those functions are not orthonormal on $\Bbb R$. Whatever book this is, don't believe what you read there unless you know the proof – David C. Ullrich Jul 11 '22 at 20:08
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    What is the title of the book? And who is the author? – paul garrett Jul 11 '22 at 20:16
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    One would need to define the symbol $\int_{-\infty}^\infty e^{ikx},dx$ to show that it is equivalent to the Dirac Delta distribution $\delta(k)$. I've posted an answer that provides this definition. – Mark Viola Jul 11 '22 at 20:34
  • What was meant is $\frac{1}{2\pi}\int_{-1/\epsilon}^{1/\epsilon}e^{i(k-k')x}dx=\eta_\epsilon(k-k')$ for all $\epsilon\in\Bbb R^+$, with $\eta_\epsilon$ a nascent delta. – J.G. Jul 11 '22 at 20:46

2 Answers2

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One needs to assign a meaning to the symbol $\frac1{2\pi}\int_{-\infty}^\infty e^{i(k-k')x}\,dx$.

Suppose $\psi\in \mathbb{S}$ (i.e. $\psi$ is a Schwartz function). Then, we define the distribution denoted by the symbol $\frac1{2\pi}\int_{-\infty}^\infty e^{i(k-k')x}\,dx$ in terms of the following:

$$\begin{align} \lim_{L\to\infty} \int_{-\infty}^\infty \psi(k') \frac1{2\pi}\int_{-L}^L e^{i(k-k')x}\,dx\,dk'&=\lim_{L\to\infty} \frac1\pi\int_{-\infty}^\infty \psi(k') \frac{\sin(k-k')L}{k-k'}\,dk'\tag1 \end{align}$$

I showed in THIS ANSWER that the limit on the right hand side of $(1)$ is equal to $\psi(k)$. Therefore, we find that in distribution

$$\lim_{L\to\infty} \frac1{2\pi}\int_{-L}^L e^{i(k-k')x}\,dx=\delta(k-k')$$

as was to be shown.

kimchi lover
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Mark Viola
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  • Right. Of course that does not show that the functions $e^{ikt}$ are orthonormal... – David C. Ullrich Jul 11 '22 at 21:29
  • Why the book (quantum mechanics) says the functions are orthonormal ? It is in the context of Fourier transformation. – user996159 Jul 11 '22 at 22:46
  • @DavidC.Ullrich Indeed it does not show orthonormailty on the reals. And as you pointed out, it should not. ;-)) – Mark Viola Jul 12 '22 at 13:38
  • @user996159 I have not read the book to which you refer. But yes, if we interpret the symbol of interest as a Fourier transform, and let $f_{k'}(x)=\frac1{2\pi}e^{-ik'x}$, then, we have

    $$\langle \phi,\mathscr{F}{f_{k'}}\rangle=\langle \mathscr{F}{\phi},f_{k'}\rangle=\mathscr{F}^{-1}{\mathscr{F}{\phi}}=\phi(k')\implies \mathscr{F}{f_{k'}}=\delta(k-k')$$

     – Mark Viola Jul 12 '22 at 14:01
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What you say the book says is simply nonsense; none of the integrals even exist.

The functions are not "orthornomal on $\Bbb R$"; they don't have a chance to be since they're not square-integrable. Of course they are sort of analogous to orthonormal in some sense. Heh, they are orthonormal on the Bohr compactification...

David C. Ullrich
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