# [Physics] Normalization of eigenfunction to Dirac-delta function

dirac-delta-distributionsfourier transformhilbert-spacenormalizationquantum mechanics

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?

For a function $$f(x)$$ the Fourier transform is defined as:
$$$$\overset{\boldsymbol{\sim}}{f}\left(k\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!f\left(x\right)e^{ikx}\mathrm dx \tag{1}$$$$ This transformation is invertible, that is: $$$$f\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\overset{\boldsymbol{\sim}}{f}\left(k\right)e^{\boldsymbol{-}ikx}\mathrm dk \tag{2}$$$$
With $$f\left(x\right)=\delta\left(x\right)$$, equation (1) yields: $$$$\overset{\boldsymbol{\sim}}{\delta}\left(k\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\delta\left(x\right)e^{ikx}\mathrm dx =\dfrac{1}{\sqrt{2\pi}} \tag{3}$$$$ That is, the Fourier transform of the $$\delta$$-function is the constant $$1/\sqrt{2\pi}$$. Equation (2) gives: $$$$\delta\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\dfrac{1}{\sqrt{2\pi}}e^{\boldsymbol{-}ikx}\mathrm dk =\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}ikx}\mathrm dk \tag{4}$$$$
This is sometimes called the integral definition of the $$\delta$$-function.
Exchanging the roles of $$k$$ and $$x$$ in equation (4), the definition becomes: $$$$\delta\left(k\right)=\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}ikx}\mathrm dx \tag{5}$$$$
Replacing $$k$$ in equation (5) with $$k-k'$$ we arrive at: $$$$\delta\left(k-k'\right)=\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}i(k-k')x}\mathrm dx \tag{6}$$$$ which is the equation that Shankar used.