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)={+∞,x=00,x≠0

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

δij={1,i=j0,i≠j

we have that

δy(x):=δ(x−y)={+∞,x=y0,x≠y

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

Ty:D(R)↦R

such that Ty(φ)=φ(q). It can be checked that informally:

Ty=δy(x)

and

Ty(φ)=φ(y)=<δy,φ>=∫Rφ(x)δy(x)dx.

In this way, δy plays the role of ej when {ei} is an orthogonal basis in a vector space V, since for a vector v∈V its jth coordinate is

coordj(v)=vj=<ej,v>=∑ivi⋅δij.

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

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

in concordance with the bra-ket notation.

The Fourier transform of the Dirac delta is

δa(t)=∫−∞∞e2πiw(t−a)dw==∫−∞∞e−2πiawe2πiwtdw

That is, its Fourier transform is the phase factor e−2πiaw: constant modulus 1, linear phase. This is the extreme case of the Heisenberg principle: perfect localization in time (Δt=0) forces complete delocalization in frequency.
See Wikipedia for a proof.