- #1
bman!!
- 26
- 0
[SOLVED] Fourier transform of a function such that it gives a delta function.
ok say, if you Fourier transform a delta function G(x- a), the transform will give you something like
∫[-∞ ∞]G(x-a) e^ikx dx
a is a constant
to calculate, which gives you
e^ka (transformed into k space) due to the sifting propety of dirac deltas.
however, you equally turn it around and askfor what function if Fourier transformed gives the dirac delta, you simply compute the inverse transform of G(K-k(0)) (delta function in k space) giving you an integral like:
∫[-∞ ∞] G(k - k(0)) e^-ikx dk which simply enough to compute gives you the answer which is e^-ik(0)x
(note k(0) is simlpy meant to me some constant in k space, i.e. k_0 i just can't get the hang of the ubscripts_
ok, this i understand. it does strike me as a bit hand waviness, so ill get to my point:
say i again use the Fourier procedure to to take the transform of e-ik(0)x which should give an integral looking liike
∫[-∞ ∞](e^-ik(0)x) (e^ikx) dx
now here's the thing. i want to remember this Fourier pair as its a useful result and crops up all the time in QM and optics and all sorts. but i don't see how evaluating this last integral gives a delta function.
cheers
ok say, if you Fourier transform a delta function G(x- a), the transform will give you something like
∫[-∞ ∞]G(x-a) e^ikx dx
a is a constant
to calculate, which gives you
e^ka (transformed into k space) due to the sifting propety of dirac deltas.
however, you equally turn it around and askfor what function if Fourier transformed gives the dirac delta, you simply compute the inverse transform of G(K-k(0)) (delta function in k space) giving you an integral like:
∫[-∞ ∞] G(k - k(0)) e^-ikx dk which simply enough to compute gives you the answer which is e^-ik(0)x
(note k(0) is simlpy meant to me some constant in k space, i.e. k_0 i just can't get the hang of the ubscripts_
ok, this i understand. it does strike me as a bit hand waviness, so ill get to my point:
say i again use the Fourier procedure to to take the transform of e-ik(0)x which should give an integral looking liike
∫[-∞ ∞](e^-ik(0)x) (e^ikx) dx
now here's the thing. i want to remember this Fourier pair as its a useful result and crops up all the time in QM and optics and all sorts. but i don't see how evaluating this last integral gives a delta function.
cheers