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?

Originally posted by amolv06 By Zetttili: http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=quantum+mechanics

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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

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.

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.

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.

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.