Dirac Delta from Continous Eigenfunctions

[Master J]

In the equation for determining the coefficients of eigenfunctions of a continuous spectrum operator, I have trouble understanding the origin of the Dirac delta.

a_f = INTEGRAL a_g ( INTEGRAL F_f F_g ) dq dg

a is the coefficient, F = F(q) is an eigenfunction.

From this it is shown that the first integral (ie. of the eigenfunctions with dq) must be a Dirac delta function, that is, that for f = g it is infinite. Why is this? Landau Lifgarbagez states that it is to prevent the integral with dg from vanishing, but I don't see this.

This would mean that a_f = INTEGRAL a_f (infinity) df ...how is this?


Cheer folks! :)

 

[Bill_K]

This is pretty much the same question as before, right? Only last time it was discrete (Eq 3.5) a_n = Σ a_m ∫ ψ_m ψ_n* dq, and this time it's continuous (Eq 5.3) a_f = ∫ a_f' (∫ψ_f'ψ_f* dq) df'.

In both cases, the reason is (like LL say): "This relation must hold for arbitrary a_f". The only way a_n = Σ a_m (...blah...) can hold for arbitrary a_n is if (...blah...) is a Kronecker delta, δ_mn, and the only way a_f = ∫ a_f' (...blah...) df' can hold for arbitrary a_f is if (...blah...) is a delta function, δ(f-f').

 

[Master J]

That's all good, but it's specifically the infinity part...why must the inner product of the two eigenfunctions when f=g inner product of it with itself) be infinity to satisfy this??

 

[Black Integra]

I can't see what those equation are. But let me say something.

Even though
δ(x)={infinity; x=0}={0; x≠0}

But from the definition:
∫δ(x)dx=1 -----> The area under the 'curve' must be 1:

The Dirac delta function as the limit (in the sense of distributions) of the sequence of Gaussians: Wiki

so, for every single function, there must be a value a that satisfy the equations:
f(a)=∫f(x-a)δ(x)dx

 

[Master J]

\int \Psi _{f} \Psi _{g} dqThought I might as well get used to this Latex thing :D

My question is, why is this integral infinity when f=g ?

Psi is an eigenfunction of a continuous spectrum operator