# Normalization of eigenfunction to Dirac-delta function

In the first chapter of Principles of Quantum Mechanics by R. Shankar, he describes finding the eigenvalues and eigenfunctions of the operator $K=-iD=-i\frac{d}{dx}$. For context, he does this:

What I don't understand is how he arrived at $A=1/\sqrt{2\pi}$. It seems to be because (since this is an infinite-dimensional space) we want to normalize to the Dirac delta function, but I don't understand why $$\frac1{2\pi}\int_{-\infty}^\infty e^{-i(k-k')x}dx=\delta(k-k').\tag{*}$$ He doesn't really explain this. How does he normalize it?

• Delta Dirac is a well- defined tempered distribution. As of such, there is for it a well-defined Fourier transformation of a function. $\mathfrak{F}(f(x)) = \delta (k-k')$ Nov 16 '17 at 11:30
• If your question is just about the normalization of the last formula (*), then it is a pure Mathematics question, explained in Fourier theory. Nov 16 '17 at 11:33
• Ok, I probably don't have the mathematical machinery yet. I will take this for granted for now until later. Thank you.
– hddh
Nov 16 '17 at 11:35

For a function $\:f\left(x\right)\:$ the Fourier transform is defined as
$$\overset{\boldsymbol{\sim}}{f}\left(s\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!f\left(x\right)e^{isx}\mathrm dx \tag{01}$$ This transformation is invertible, that is $$f\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\overset{\boldsymbol{\sim}}{f}\left(s\right)e^{\boldsymbol{-}ixs}\mathrm ds \tag{02}$$
With $\:f\left(x\right)=\delta\left(x\right)\:$ equation (01) yields $$\overset{\boldsymbol{\sim}}{\delta}\left(s\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\delta\left(x\right)e^{isx}\mathrm dx =\dfrac{1}{\sqrt{2\pi}} \tag{03}$$ that is the Fourier transform of the $\:\delta\:$ function is the constant $\:1/\sqrt{2\pi}$ and equation (02) gives $$\delta\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\dfrac{1}{\sqrt{2\pi}}e^{\boldsymbol{-}ixs}\mathrm ds =\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}ixs}\mathrm ds \tag{04}$$
There are also other alternative ways to reach this result, for example using expressions of the $\:\delta\:$ function as limit of "proper" functions.