Physics Friday 46

Quantum Mechanics and Momentum
Part 3: The Fourier Transform, Dirac Delta, and Plancherel’s Theorem

Consider the Fourier transform in one dimension. There are several common conventions for defining it; here, we will use the unitary angular form frequently used in physics: for function f(t) the Fourier transform is
$g(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-\imath\omega{t}}\,dt$ ,
and the inverse transform is
$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\omega)e^{\imath\omega{t}}\,d\omega$

These convert between the time domain and the angular frequency domain. For transforms on a function of a position variable f(x), we have
$g(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-\imath{kx}}\,dx$ ,
and
$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{\imath{kx}}\,dk$
which convert between the position domain and the (angular) wavenumber domain.
For an n dimensional position space with position vector $\mathbf{\vec{x}}$ we find the transform of a function to be in the n dimensional wavenumber space with wavenumber vector $\mathbf{\vec{k}}$
via
$g(\mathbf{\vec{k}})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}f(\mathbf{\vec{x}})e^{-\imath\mathbf{\vec{k}}\cdot\mathbf{\vec{x}}}\,d\mathbf{x}$
and
$f(\mathbf{\vec{x}})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}g(\mathbf{\vec{k}})e^{\imath\mathbf{\vec{k}}\cdot\mathbf{\vec{x}}}\,d\mathbf{k}$

Now, consider the Fourier transform of the Dirac delta “function”:
$\begin{eqnarray}g(k)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(x-x_0)e^{-\imath{kx}}\,dx\\&=&\frac{1}{\sqrt{2\pi}}e^{-\imath{kx_0}}\end{eqnarray}$
and by examination of this, and the inverse transform, we obtain
$\int_{-\infty}^{\infty}e^{-\imath{k}u}\,du=2\pi\delta(k)$ .
(Note that this is not a mathematically rigorous equality, but is valid under integration with a function: $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(k)e^{-\imath{k}u}\,du\,dk=2\pi\int_{-\infty}^{\infty}f(k)\delta(k)\,dk=2\pi{f(0)}$ .)
This relation is useful for proving a number of relations, one of which follows.

For a valid wavefunction of one spatial dimension ψ(x), let us define its Fourier transform φ(k).
Then
$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)e^{-\imath{kx}}\,dx$
The complex conjugate of this is
$\phi^*(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi^*(x)e^{\imath{kx}}\,dx$
where ^* indicates the complex conjugate.
Thus, let us now find $\int_{-\infty}^{\infty}\left|\phi(k)\right|^2\,dk$ :
$\begin{eqnarray}\int_{-\infty}^{\infty}\left|\phi(k)\right|^2\,dk&=&\int_{-\infty}^{\infty}\phi^*(k)\phi^*(k)\,dk\\&=&\normalsize{\int_{-\infty}^{\infty}\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi^*(x')e^{\imath{kx'}}\,dx'\right]\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)e^{-\imath{kx}}\,dx\right]\,dk}\\&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}\psi^*(x')\psi(x)e^{-\imath{k}(x-x')}\,dk\,dx'\,dx\\&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^*(x')\psi(x)\delta(x-x')\,dx'\,dx\\&=&\int_{-\infty}^{\infty}\psi^*(x)\psi(x)\,dx\\\int_{-\infty}^{\infty}\left|\phi(k)\right|^2\,dk&=&\int_{-\infty}^{\infty}\left|\psi(x)\right|^2\,dx\end{eqnarray}$
This is known as Plancherel’s Theorem (or in some physics and engineering sources as Parseval’s Theorem), and we see that if the (position space) wavefunction ψ(x) is normalized, $\int_{-\infty}^{\infty}\left|\psi(x)\right|^2\,dx=1$ , then the corresponding “wavenumber space” wavefunction φ(k) is also normalized.

The reason we want to use such a Fourier transformed function goes back to our de Broglie relations, particularly $p=\hbar{k}$ . Using this, we see that our “wavenumber domain,” via this equation, represents the momentum space basis for our quantum state.

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Tags: Fourier Transform, Friday Physics, Momentum Space, Parseval's Theorem, physics, Plancherel's Theorem, Quantum Mechanics

This entry was posted on November 14, 2008 at 12:03 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

7 Responses to “Physics Friday 46”

Physics Friday 47 « Twisted One 151’s Weblog Says:
November 21, 2008 at 12:11 am | Reply
[...] we can define a Fourier transform of this to the (angular) wavenumber domain, φ(k): Now, as we noted previously, the relation shows that the wavenumber domain represents momentum space. Now, recall that for an [...]
Physics Friday 49 « Twisted One 151’s Weblog Says:
December 5, 2008 at 6:09 pm | Reply
[...] transform applied to a function of time returns a function of (angular) frequency: (introduced here). We also have the deBroglie relation between energy and angular frequency E=ℏω. Thus, [...]
Monday Math 100 « Twisted One 151's Weblog Says:
December 7, 2009 at 12:30 am | Reply
[...] One of the oldest and best known integral transforms is the Laplace transform; it and the related Fourier transform are likely the most commonly used integral transforms. Like the Fourier transform, it transforms a [...]
Monday Math 118 « Twisted One 151's Weblog Says:
May 10, 2010 at 12:05 am | Reply
[...] By twistedone151 The Laplace Transform Part 19: Inverse Laplace Transform Recall from here the Fourier transform and its inverse . Combining these, we see . Now, letting for t<0, the [...]
Monday Math 128 « Twisted One 151's Weblog Says:
July 26, 2010 at 12:14 am | Reply
[...] Math 128 By twistedone151 We have previously discussed the Fourier transform (here and here, especially). In this post, we noted that (using the symmetric angular convention) the [...]
Juho Says:
October 26, 2010 at 11:29 am | Reply
Cool! But how can one combine ω and k in the Fourier Transform and in what conditions? I would like to “derive” Schödinger equation but I need to justify why to use ψ(x,t) that contains both ω and k. So via expection value calculations Schödinger equation can be “derived”.
- twistedone151 Says:
  October 27, 2010 at 12:54 pm | Reply
  One can have both ω and k if your function depends on both time t and space x, as each of the former is the variable for the Fourier transform of the corresponding latter variable. I touch on the time transform, and its relation to the time-dependent Schrödinger equation in part 6 of this series.

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