# Hilbert Space,Dirac Notation,and some other stuff

1. Sep 22, 2004

### Ed Quanta

Ok, so I am a little unsure of how to apply these new concepts I am learning.
Here is a question.

The function g(x)=x(x-a)e^ikx is in a certain Hilbert space
where the finite norm squared equals the integral of the product of Psi's complex conjugate and Psi (dx) is less than infinity.

I must calculate the coefficients of expansion (an) of this function in the series

Psi(x)=summation of n=1 to infinity of (an*(Psi n(x))) where Psi n is the set of basis functions Psi n=(sqrt(2/a))sin(n*pi*x/a)

I am not sure how to do this exactly. If I am unclear, please do tell me and I shall write some more. Peace, and thanks a lot dudes.

Oh by the way, here Psi(x)=g(x) if that wasn't made obvious.

2. Sep 22, 2004

### humanino

The great thing about Hilbert spaces, is that it the same as ordinary vector spaces ! Except that of course, the dimension is allowed to be (countable) infinite, and the scalar product is slightly more involved.

So everything i will write is very easy :
$$\Psi(x)=\sum_n a_n \Psi_n(x)$$
you take the scalar product and use orthonormality :
$$\langle\Psi_n(x)|\Psi(x)\rangle=\sum_m a_m \langle\Psi_n(x)|\Psi_m(x) \rangle=\sum_m a_m \delta^m_n=a_n$$

$$a_n = \int dx\Psi_n^*(x)\Psi(x)$$

At this point I wonder : is it not $$g(x)=\theta[x(x-a)]e^{\imath k x}$$ ?

3. Sep 23, 2004

### Kane O'Donnell

Actually I think you can have uncountably many dimensions too - l2, the set of square summable sequences, is the model for a countable-dimension Hilbert space, whereas L2, the set of measurable Lebesque-square-integrable functions is the model for uncountable-dimension Hilbert spaces. Since in QM we work in subspaces of L2, we're usually talking uncountably many dimensions.

In particular, since the expansion in terms of sin(kx) has k in the reals, there is an uncountably infinite number of basis elements.

Of course, justifying L2 as a Hilbert space is a graduate course in measure theory all in itself, so most of the time we just shrug our shoulders and agree that 'stuff works'. Still, it's an uncountable amount of stuff

Kane

4. Sep 23, 2004

### Kane O'Donnell

Of course (since I should try to actually be relevant for once) in bound-state problems with discrete eigenvalues, as above, where the basis elements are of the form:

sin(n*Pi*x/a)

you get a countable (either infinite or not depending on the potential) number of basis elements making up the bound state space.

Kane

5. Sep 23, 2004

### humanino

Now you were not irrelevant at all. Thank you for the precision. I remembered only that in the case of uncountable-dimension things are much more difficult in theory, but then only mathematicians care. The problem is to make the Hilbert space complete, you have to take the equivalent classes. Anyway, thank you for refreshing my memory.

6. Sep 23, 2004

### Ed Quanta

I see what you did, now do I just integrate the product of the complex conjugate of basis functions (or just the basis functions because they equal their complex conjugate) Psi m= square root(2/a)sin(m*pi*x/a) and g(x) over the region from 0 to a?

7. Sep 23, 2004

### humanino

Yes, you have to do the integration, depending on which is the right definition of $$g(x)$$. I did not do it, I was too lazzy to compute and write The basis functions are real, you can ignore the hermitean conjugate in this case, I was illustrating the general power of the formalism.

Ed, if you want to use latex, you can click on the formula to see the code, or/and train and ask questions [thread=8997]here[/thread] if you please to use it of course.

8. Sep 23, 2004

### vanesch

Staff Emeritus
I'm not sure it is still called a Hilbert space in that case. After all, I thought a Hilbert space had to be Hausdorff-separable, and that implies, if my old memory serves well, the existence of countable bases. Anyways, L2 is countable, because it is the set of equivalence classes of functions which differ by a function which is non-zero only on a carrier of measure zero ; otherwise the inproduct wouldn't be positive definite.

cheers,
Patrick.

9. Sep 23, 2004

### Lonewolf

I think that physicists tend to use Hilbert space to mean one that is separable, while a general Hilbert space admits an uncountable number of vectors in a basis.

10. Sep 23, 2004

### vanesch

Staff Emeritus
In my book "Introduction to Hilbert Space" by Sterling K. Berberian,

A pre-Hilbert space is a vector space over C equipped with a positive definite inproduct (and hence a metric) ;

and it is called a Hilbert space if on top of that it is Cauchy-complete (every cauchy sequence converges).

Finally a "classical Hilbert space" is separable (and countably infinite dimensional).
All classical Hilbert spaces are isomorphic.

You are right. Moreover, I was wrong saying that the separability followed in L2 from the definition of the in-product.

cheers,
patrick.

11. Sep 24, 2004

### Haelfix

Yep, all Hilbert spaces are infinite dimensional (even if physicists sometimes lazily make them seem like they are finite). If they don't have a countable basis, there is not much to do, other than stare at the nice, completely useless spectral mess.

12. Sep 24, 2004

### humanino

:rofl:
I thought I remembered something like that.
Thank you Haelfix !

13. Sep 24, 2004

Staff Emeritus
Patrick, what is the definition of separable for Hilbert spaces? Is it the same as the topological definition of separable? I had always thought so, but some of the comments in the LQG controversy suggested otherwise.

14. Sep 24, 2004

### Lonewolf

Separable Hilbert spaces are indeed separable in the sense of topological and metric spaces, with topology induced from the inner product. The Hilbert space needs only an orthonormal basis to be separable, or equivalently for all (finite or infinite) sequences xk, if there exists a z in the Hilbert space such that (z, xk) = 0, z is necessarily the zero vector.

Are the said comments online anywhere that you could provide a link to, since I'm pretty confident that it's topologically separable?

15. Sep 24, 2004

### humanino

According to wikipedia and my memory, a Hilbert space is separable if and only if it has a countable orthonormal basis. That is also what Haelfix meant I guessed.

16. Sep 24, 2004

### Lonewolf

Is that really true? How about Rn with the inner product as the ordinary scalar product? We know it's complete, since every Cauchy sequence converges with the induced metric, but Rn is finite dimensional?

17. Sep 24, 2004

### humanino

You are rght Lonewolf, it is obviously not true that all Hilbert space are infinite dimensional. That was a misattention I guess.

18. Sep 24, 2004

### humanino

wavelets ?

I have been wondering for a while about this : Hilbert spaces notably "serve to clarify and generalize the concept of Fourier expansion [and] certain linear transformations such as the Fourier transform" (wikipedia again) and Healfix was not totally wrong when he said we physicists are sometimes lazzy with the mathematical available tools, we tend to stick to the ones we are used to. So, what would be wrong if instead of decomposing out operators in an harmonic fashion, we used wavelets ? I could not investigate this enough, because I do not handle those well yet, but my thoughts have been triggered by the fact that, wavelets are especially well-suited for two purposes : signals exhibiting a begining and an end, which is obviously not the case for plane waves ; and signals having a rich content in scales. That is exactly the case of our fields : they are physical fields and "Quantum field theory arose out of our need to describe the ephemeral nature of life" (Zee again). Besides, renormalization is supposed to take care of this scale problem of the fields, whose content blow up at small distances. As discussed in the middle of [thread=44213]this[/thread] recent thread, which unfortunately diverged since then, the fluctuations at small distances might be physical. That could be smoothly taken care by suitable wavelets.

So I guess there is an obvious flaw with not using plane waves, and the following answers will probably try to demonstrate that there is no way out of harmonic analysis. I would rather receive arguments against wavelets, that for plane waves. Anyway, if it inspires you any thought or comment, I would be glad. Thanks.

Last edited: Sep 24, 2004
19. Sep 24, 2004

### meteor

If this an help, the book "Theory of linear operators in Hilbert space", Volume I, N.Akhiezer, english edition, says (literally):
"A Hilbert space H is an infinite dimensional inner product space which
is a complete metric space with respect to the metric generated by the inner product"

Last edited: Sep 24, 2004
20. Sep 24, 2004

### Lonewolf

I guess the infinite-dimensional business must be a physicist thing...