Understanding Inner Product in Infinite Dimensional Bases

[amjad-sh]

While I'm reading a book in quantum mechanics, I reached the part "Generalization to infinite dimension".
We know that at infinite dimension many definitions changes.And that what is confusing me!
Take for example the inner product.when we are dealing in finite dimension the definition of inner product in orthonormal basis is shown below. If we deal with infinite dimensional basis the definition of inner product
becomes as shown below.
My Question is: why we can't use the first definition for infinite dimensional case?What is the magical trick that the first definition makes the inner product equals to infinity while the other makes it equal to a specific number due to the use of integration?Do I need to go deeper in integration theory to grasp this or the idea is simpler than that?
thanks

 

[mfb]

The trick is the division by n. I would expect that those f*g can all be of similar size (in particular, not go to zero quickly). Then your sum diverges, but the sum divided by n stays finite and you end up with an integral for the infinite case.

 

[amjad-sh]

OK.
But why we go to the second definition? why in infinite dimension we use the second definition and not the first? what is the reason of inserting Δ in the equation?
I know that the inner product in the first definition diverges, but what permits us to change the first definition to the second definition?why in infinite dimension things change?

 

[bhobba]

But why we go to the second definition? why in infinite dimension we use the second definition and not the first? what is the reason of inserting Δ in the equation?I know that the inner product in the first definition diverges, but what permits us to change the first definition to the second definition?why in infinite dimension things change?

It's a hand-wavey thing to see how the usual definition of inner products from finite spaces can become the integral in infinite dimensional spaces.

Why? Its just the way infinite spaces are - they allow it. And yes deeper integration theory is required - you need Lebesque integration - not for the hand-wavey stuff - but to understand Hilbert spaces. Don't worry about it to start with though.

If you really want to get to the bottom of it you need to study Rigged Hilbert Spaces - but a rigorous treatment of that is quite advanced and not recommended for the beginner.

Thanks
Bill

 

[amjad-sh]

And yes deeper integration theory is required - you need Lebesque integration - not for the hand-wavey stuff - but to understand Hilbert spaces.

.
OK, i will read about this stuff.

If you really want to get to the bottom of it you need to study Rigged Hilbert Spaces - but a rigorous treatment of that is quite advanced and not recommended for the beginner.

.
You mean to get deeper in the hand-wavey stuff or Hilbert space? If you know a way or book that helps me to understand the "hand-wavey stuff" you will be appreciated.
Thanks.

 

[bhobba]

.You mean to get deeper in the hand-wavey stuff or Hilbert space? If you know a way or book that helps me to understand the "hand-wavey stuff" you will be appreciated.

Yes - Chapter 2 - Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/dp/9814578584/?tag=pfamazon01-20

Thanks
Bill

 

[amjad-sh]

OK.
Thank you.