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Section 12.3 Definition of the Fourier Transform

Consider the (square integrable) function \(f(x)\text{;}\) its Fourier transform is defined by:

\begin{equation} {\cal F}(f) =\tilde{f} (k)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx}\, f(x)\, dx\tag{12.3.1} \end{equation}

The inverse of the Fourier transform is given by:

\begin{equation} {\cal{F}}^{-1}(\tilde{f}) = f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk\tag{12.3.2} \end{equation}

To show that the inverse Fourier transform is indeed the inverse operation, start with the right-hand-side of the inverse Fourier transform and insert the definition of the Fourier transform

\begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk \amp= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \left[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x')\, e^{-ikx'}\, dx'\right]\, e^{ikx}\, dk\notag\\ \amp= \int_{-\infty}^{\infty} f(x') \left[\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-x')}\,dk\right]\, dx'\notag\\ \amp = \int_{-\infty}^{\infty} f(x') \delta(x-x')\, dx'\notag\\ \amp = f(x)\tag{12.3.3} \end{align}

where in the second to the last line, we have used the integral representation of the delta function, see Section 11.6, to evaluate the expression in the square brackets.


  • We strongly suggest the convention of putting the exponential to the left of the function in the integrand of the Fourier transform (and to the right for the inverse Fourier transform) to highlight the relationship between the Fourier Transform and the quantum mechanics notion of finding the projection of a quantum wavefunction \(f(x)\) onto a plane wave (Notice the complex conjugate of the plane wave that turns the ket into a bra).

  • We always use the convention of putting a factor of \(\frac{1}{\sqrt{2\pi}}\) into the definitions of both the Fourier Transform and its inverse (see below) to make the operations symmetric in this way. This convention is NOT universal; use caution when using other resources.