Dirac delta

Is not a function, but a distribution (functional analysis). In the context of distribution (functional analysis) it is defined as:

< δ, φ >= φ (0)

Informally, Dirac delta function is usually written as

δ (x) = {\begin{cases} + \infty, & x = 0 \\ 0, & x \neq 0 \end{cases}

More on the Dirac delta: it picks out the value of $φ$ at 0 ignoring the others, so in some sense it is a continuous version of the Kronecker delta! That is, in the same sense that

δ_{i j} = {\begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}

we have that

δ_{y} (x) := δ (x - y) = {\begin{cases} + \infty, & x = y \\ 0, & x \neq y \end{cases}

Even more on this, consider for simplicity $D (R)$ (see distribution (functional analysis)). For every $y \in R$ we can consider

T_{y} : D (R) \mapsto R

such that $T_{y} (φ) = φ (q)$ . It can be checked that informally:

T_{y} = δ_{y} (x)

and

T_{y} (φ) = φ (y) =< δ_{y}, φ >= \int_{R} φ (x) δ_{y} (x) d x .

In this way, $δ_{y}$ plays the role of $e_{j}$ when ${e_{i}}$ is an orthogonal basis in a vector space $V$ , since for a vector $v \in V$ its $j$ th coordinate is

c o o r d_{j} (v) = v_{j} =< e_{j}, v >= \sum_{i} v_{i} \cdot δ_{i j} .

In the same sense that $δ_{i j}$ , with $i$ running from 1 to $n$ , are the coordinates of $e_{j}$ in the basis ${e_{i}}$ , the expression $δ_{y} (x)$ represents the "coordinates" of $δ_{y}$ in the basis ${δ_{x}}$ . The notation is not appropriate, and indeed this last sentence should be:

the expression $δ_{y} (x)$ represents the "coordinates" of $| y ⟩$ in the basis ${δ_{x}}$

in concordance with the bra-ket notation.

The Fourier transform of the Dirac delta is

δ_{a} (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{i w (t - a)} d w =

= \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- i a w} e^{i w t} d w

See Wikipedia for a proof.