Question about the Hilbert space spanned by a continuum of basis vectors

Question about the Hilbert space spanned by a continuum of basis vectors

Science

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.

a

Joined
08 Oct 06
Moves
24000
05 Mar 10

In my quantum book, it is stated that one property of a Hilbert space is that its vectors are square-integrable. However, when discussing a hilbert space spanned by a continuum of basis vectors, it is stated that the inner product of two basis vectors <Xk, Xk'>=dirac-delta(k'-k), where k and k' are subscripts. This means that the inner product of a basis vector with itself would diverge. According to the definition given earlier about the Hilbert space, the basis vectors could not be constituents of a hilbert space. How then can a hilbert space be spanned by this basis?

silicon valley

Joined
27 Oct 04
Moves
101289
05 Mar 10

what is your quantum book?

j

Joined
22 Jul 08
Moves
25957
05 Mar 10
1 edit

Please delete

a

Joined
08 Oct 06
Moves
24000
05 Mar 10

By Zetttili: http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=quantum+mechanics

silicon valley

Joined
27 Oct 04
Moves
101289
06 Mar 10

Originally posted by amolv06
By Zetttili: http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=quantum+mechanics
Quantum Mechanics: Concepts and Applications - Hardcover (Mar. 23, 2009) by Nouredine Zettili
Buy new: $200.00 $160.00
17 new from $155.74
6 used from $156.62

silicon valley

Joined
27 Oct 04
Moves
101289
06 Mar 10

!!!

silicon valley

Joined
27 Oct 04
Moves
101289
06 Mar 10

i guess i'll look it up on Link. maybe.

a

Joined
08 Oct 06
Moves
24000
06 Mar 10

Too much money, right? Though I got mine for 80. Its not hardback, though.

aw
Baby Gauss

Ceres

Joined
14 Oct 06
Moves
18375
06 Mar 10

Originally posted by amolv06
In my quantum book, it is stated that one property of a Hilbert space is that its vectors are square-integrable. However, when discussing a hilbert space spanned by a continuum of basis vectors, it is stated that the inner product of two basis vectors <Xk, Xk'>=dirac-delta(k'-k), where k and k' are subscripts. This means that the inner product of a basis v ...[text shortened]... ot be constituents of a hilbert space. How then can a hilbert space be spanned by this basis?
You're right! Hilbert spaces aren't the best mathematical way to do with physics since when the spectrum is continuous a lot of integrals diverge.

Another problem with Hilbert spaces is that the only pre-requisite is that the functions be square-integrable in all space. But it doesn't say anything about the derivative, the second derivative or any other "change" in the original function.

The actual mathematical space that you need to work on QM is a rigged Hilbert space. This book is an excellent introduction to a really rigorous mathematical treatment of QM: http://www.amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/9810241054

K

Germany

Joined
27 Oct 08
Moves
3118
06 Mar 10

Originally posted by amolv06
In my quantum book, it is stated that one property of a Hilbert space is that its vectors are square-integrable. However, when discussing a hilbert space spanned by a continuum of basis vectors, it is stated that the inner product of two basis vectors <Xk, Xk'>=dirac-delta(k'-k), where k and k' are subscripts. This means that the inner product of a basis v ...[text shortened]... ot be constituents of a hilbert space. How then can a hilbert space be spanned by this basis?
It's possible as long as the contribution of each basis vector to a vector in the hilbert space is vanishingly small, I think.

aw
Baby Gauss

Ceres

Joined
14 Oct 06
Moves
18375
06 Mar 10

Originally posted by adam warlock
You're right! Hilbert spaces aren't the best mathematical way to do with physics since when the spectrum is continuous a lot of integrals diverge.

Another problem with Hilbert spaces is that the only pre-requisite is that the functions be square-integrable in all space. But it doesn't say anything about the derivative, the second derivative or any ot ...[text shortened]... of QM: http://www.amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/9810241054
http://arxiv.org/abs/quant-ph/0502053 for a free reference on the subject of rigged Hilbert spaces.

a

Joined
08 Oct 06
Moves
24000
06 Mar 10

Many thanks! I'm going to print that article, and try and see if I can find the book in the library. I'll let you know how it goes.

a

Joined
08 Oct 06
Moves
24000
08 Mar 10

a

Joined
08 Oct 06
Moves
24000
08 Mar 10

Originally posted by adam warlock
http://arxiv.org/abs/quant-ph/0502053 for a free reference on the subject of rigged Hilbert spaces.
You jerk! You rigged my Hilbert space. Do you seriously expect me to live in a world that's not in a hilbert space, but a <i>rigged</i> hilbert space? Blasphemy!

In all seriousness, I sincerely appreciate you posting that paper. I read it in its entirety, and I feel like I have an at least workable grasp of the rigged hilbert space concept, as well as a better understanding of dirac's bra-ket notation. I have not had the chance to look through the book yet, but certainly intend to next week over spring break.

aw
Baby Gauss

Ceres

Joined
14 Oct 06
Moves
18375
08 Mar 10

Originally posted by amolv06
You jerk! You rigged my Hilbert space. Do you seriously expect me to live in a world that's not in a hilbert space, but a <i>rigged</i> hilbert space? Blasphemy!

In all seriousness, I sincerely appreciate you posting that paper. I read it in its entirety, and I feel like I have an at least workable grasp of the rigged hilbert space concept, as well as a b ...[text shortened]... the chance to look through the book yet, but certainly intend to next week over spring break.
Well you call them Gelfand triplets or equipped Hilbert spaces yoo.

Ballentine's book is great so be sure to read it.