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The Delta Functional (`Delta Function')

The parameter $ \sigma$ in the Gauss function determines its width. A broad wave packet in coordinate ($ x$) space corresponds to a narrow distribution of wave vectors $ k$. What happens in the limit $ \sigma\to 0$ ? In coordinate space, this would correspond to an extremely sharp wave packet around $ x=0$ that could serve as a model for a particle localized at $ x=0$. We define
$\displaystyle \fbox{$ \displaystyle \delta(x)
\equiv\lim_{\sigma\to 0} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}.$}$     (48)

As an ordinary function, this is a somewhat strange mathematical object because it is zero for all $ x\ne 0$, but infinite for $ x=0$. However, it has the useful property that for any (reasonably well-behaving) function $ f(x)$
$\displaystyle \int_{-\infty}^{\infty}dx' \delta(x-x')f(x')=\lim_{\sigma\to 0}
\...
...}dx' f(x') \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-x')^2}{2\sigma^2}} = f(x).$     (49)

For example, multiplication of $ f(x')$ with $ \delta(x')$ and integration over the whole $ x'$-axis gives the value of $ f$ at $ x=0$. Such an operation is called a functional, that is a mapping
$\displaystyle \delta: f \to f(0)$     (50)

that puts a whole function $ f$ to a (complex or real) number. Nevertheless, for historical reasons physicists call this object a delta-function. Remember that $ \delta(x)$ is only defined as in (1.49), that is by integration over a function (`test-function') $ f(x)$.

Another very useful property is the Fourier transform of the Delta-function: We recall our definition

$\displaystyle f(x)$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\int_{-\infty}^{\infty}dk \tilde{f}(k) e^{ikx},\quad
\tilde{f}(k):=\int_{-\infty}^{\infty}dx'{f}(x') e^{-ikx'}$  
$\displaystyle \leadsto f(x)$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi}\int_{-\infty}^{\infty}dk \int_{-\infty}^{\infty}dx...
...infty}dx' \underline{\int_{-\infty}^{\infty}\frac{dk}{2\pi} e^{ik(x-x')}}f(x').$ (51)

Now, comparing with the definition of the Delta function, Eq.(1.49), we recognise
$\displaystyle \fbox{$ \displaystyle \delta(x-x') = \int_{-\infty}^{\infty}\frac{dk}{2\pi} e^{ik(x-x')}.$}$     (52)

The delta function is thus a superposition of all plane waves; the corresponding distribution of $ k$-values in $ k$-space is `extremely broad', that is uniform from $ k=-\infty$ to $ k=+\infty$. Note that we can also obtain the result $ \tilde{\delta}(k)=1$ from the Fourier transform $ \tilde{g}(k)$ of the Gauss function $ g(x)$, Eq. (1.46), in the limit $ \sigma\to 0$.


next up previous contents
Next: * Partial Differential Equations Up: Fourier Transforms and the Previous: Math: Gauss function   Contents
Tobias Brandes 2004-02-04