Delta function and dirac notation

[KostasV]

Hello there !
I found this discussion http://physics.stackexchange.com/qu...a-delta-function-position-space-wave-function about dirac notation and delta function .
The one that answers to the problem says that $<a|a>=1$ and $<a|-a>=0$ .
As far as i know:
1) $<a|-a>=δ(a-(-a))$ which in this case is zero because delta function is zero everywhere except $x=-a$ where it goes to infinity. So i understand why this is zero.
2) $<a|a>=δ(a-a) → ∞$ and not 1 . Am I wrong and this $<a|a>$ is obviously 1. Can you explain me please?

 

[bhobba]

The one that answers to the problem says that $<a|a>=1$ and $<a|-a>=0$ .

You are correct - it is infinity. Strictly speaking its undefined, but is usually taken as infinity. The issue with infinity and why its better to have it undefined is c∂(t) is also zero when t is not zero and infinity when t = 0 for any positive c. So, naively, c'∂(t) = c∂(t) for any positive c or c'. But it acts differently when integrated.

Thanks
Bill

 

[KostasV]

So it is wrong to normalize the function with the two delta functions that he gives using $<a|a>=1$ because it actually goes to infinity.
So how could we normalize $ψ(x)=δ(x-a)+δ(x-a)$ in a correct way ?

 

[bhobba]

So it is wrong to normalize the function with the two delta functions that he gives using $<a|a>=1$ because it actually goes to infinity. So how could we normalize $ψ(x)=δ(x-a)+δ(x-a)$ in a correct way ?

They are distributions - normalisation makes no sense. The square of a delta function doesn't exist.

I suggest getting hold of the following:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

It will help in many areas of applied math not just QM.

Thanks
Bill

 

[KostasV]

Thank you for your response :)