Why Are Complex Functions and Hilbert Space Essential in Quantum Mechanics?

[Dathascome]

Hi there,
I took a quantum mechanics class for the first time this past semester, and didn't feel like I fully grasped many aspects of the subject, so decided to try to go through some stuff on my own, and just had a gew questions.
One of them is more a math question: In the book I'm reading (Quantum Mechanics by Zettili) they mention that the space of all complex functions is infinite dimensional, but don't state why, and I'm not quite sure how to show something like this. I know that I need to show that there are an infinite # of linearly independent basis vectors, but I'm not quite sure how to get at this.
My second question is about Hilbert space. This is something else that I don't fully grasp. It seems like something very important to quantum mechanics, but I'm not quite sure why the hilbert space is a good stage for quantum mechanics. The only things I could think of were that maybe it has something to do with how the inner product of some vector in the hilbert space with itself is real (I know that this is the reason why hermitian matrices are important, because the eigenvalues are real, and I thought maybe it was similar), or perhaps something about the whole limit of cauchy sequences coverging.
I guess this is a the same question sort of but what are spaces that are not hilbert spaces and what about them is not useful in the same way that hilbert space is?
So am I onto anything or way off? Any help would be greatly appreciated

 

[HallsofIvy]

The space of all functions is infinite dimensional because it contains the space of all polynomial functions. If the space of all polynomials were finite, that would mean it could be spanned by a finite number of polynomials (every polynomial could be written as a sum of multiples of those polynomials). A finite set of polynomials has a polynomial of highest degree (power) and the space of all sum of multiples of that could not contain polynomials of degree higher than that. But that can't be the set of all polynomials. Since the space of all polynomials must have infinite dimension so must the space of all functions.

The reason why Hilbert spaces are important in quantum theory is that it corresponds to spaces of "operators" and in quantum theory the key things, such as position, momentum, energy, etc. are not numbers, but operators. We need to be able to get numbers out of those: IF we measure the position, we must get a specific number. Of course, by the uncertainty principle, if we did the exact same measurement again, with everything exactly the same, we might get another value. The only single number we can get from the position operator is an "average" postion. But you are right- it is the inner product that allows us to go from operator to number.

 

[marlon]

IF we measure the position, we must get a specific number. Of course, by the uncertainty principle, if we did the exact same measurement again, with everything exactly the same, we might get another value.

That is indeed a very good explanation of the HUP because it eliminates the common misconception that the actual uncertainty arises from the accuracy of the measurement apparatus directly.

The only single number we can get from the position operator is an "average" postion.

What exactly do you mean by average position ? I mean average with respect to what ? Each position measurement will yield an exact number (let us assume that our apparatus is 'perfectly accurate' here) and the fact that we get different numbers resulting from several identical set-ups is the true and only manifestation of the HUP.

regards
marlon

 

[Hurkyl]

Why a Hilbert space? Because it works well.

It's an application of abstract linear algebra: in its early development, we had the Schrödinger picture and the Heisenberg picture, and their space of states were fairly different. (IIRC) But, people eventually figured out the essential features of both state spaces: they were complete inner product spaces (which is the definition of a Hilbert space!). This means:

(1) It's a vector space.
(2) It has an inner product.
Any inner product induces a norm given by |&psi;|² = <&psi;, &psi;>. It has to be real -- try taking the conjugate transpose of it. Anyways, since we have a norm, then we can require
(3) It is complete.
Which means that any Cauchy sequence converges.


Now, there is another, more abstract reason to use Hilbert spaces too. You've been introduced to various operators: it is possible to ignore the state space and just look at how the operators relate to each other. In particular, the operators form something called a C*-algebra. (C-star algebra)

So now that we know we like to study C*-algebras, the next question is how can we represent such a thing. I've read that for whatever reason, the natural thing to do is precisely to represent them as unitary operators on a Hilbert space!

 

[Dathascome]

Thanks for all the help. I see why the space of all functions is infinite dimensional now. As for the rest of the things I asked, i still have some questions.
Just what exactly is it about being both complete and having an inner product that makes hilbert space so relevant to quamtum theory? What is it about other spaces that don't have either or both of these features that makes it not useful in this sense?
How does hilbert space corespond to the space of operator's? And my last and probably silliest question (but I'm going to ask anyway) is why are the key things in quamtum theory operators?
As for the C* stuff I have no idea what that is and can't say much
Thanks again for the help.

 

[Gokul43201]

Inner products are useful because they provide a means of describing an arbitrary vector in terms of components along different basis vectors. So, it is a good thing to have an inner product defined in the space. It is completeness that ensures that such a representation is accurate (ie : that for any function, f, the error in expanding it along a set of base functions u_i goes to 0 as i goes to infinity).