Hilbert Space,Dirac Notation,and some other stuff

[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.

 

[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 $$g(x)=\theta[x(x-a)]e^{\imath k x}$$ ?

 

[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

 

[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

 

[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.

 

[Ed Quanta]

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

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?

 

[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.

 

[vanesch]

Actually I think you can have uncountably many dimensions too

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.

 

[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.

 

[vanesch]

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.

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.

 

[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.

 

[humanino]

stare at the nice, completely useless spectral mess.

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

 

[selfAdjoint]

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.

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.

 

[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 x_k, if there exists a z in the Hilbert space such that (z, x_k) = 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?

 

[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.

 

[Lonewolf]

Yep, all Hilbert spaces are infinite dimensional...

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?

 

[humanino]

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

 

[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 beginning 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.

 

[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"

 

[Lonewolf]

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

 

[jcsd]

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"

Mathematics of Classical and Quantum Physics Vol. 1 by Byron and Fuller describes a Hilbert space as a 'complete inner product space', this is the defintion of a Hilbert space that I have always understood and it is clear from this defintion that a Hilbert space may be finite-dimensional.

 

[meteor]

Mathematics of Classical and Quantum Physics Vol. 1 by Byron and Fuller describes a Hilbert space as a 'complete inner product space', this is the defintion of a Hilbert space that I have always understood and it is clear from this defintion that a Hilbert space may be finite-dimensional

I'm not an expert in mathematics, but i have a great quantity of mathematical books.
For example, "Introductory real analysis",Kolmogorov,english edition, (also literally):
DEFINITION 5. By a Hilbert space is meant a Euclidean space which
is complete, separable and infinite-dimensional.

But is true taht in QM Hilbert spaces are treated often as finite dimensional, at least in some webpages I have visited

 

[Lonewolf]

All the books I read from mention no condition of finite dimensionality or otherwise. In fact, a well known result is that all finite dimensional inner-product spaces (pre-Hilbert spaces) are in fact necessarily Hilbert spaces.

 

[shchr]

meteor:
I have that book. According to my understanding, that Hilbert space is infinite dimensional means that a function f(x) can be considered as infinite dimensional vector.

 

[humanino]

Who even cares if the Hilbert space must be restricted to finite dimensionality ?
This is semantics with not much purpose. Bourbaki makes no reference to dimensionality, and I think there is really no need to have such a restriction and use special different names for both.
Let us stick to more relevant issues.

 

[humanino]

If you just take time to google it, you will stare many references with names quoting their university professor position, saying "in the case of finite dimensional Hilbert space". Not only wikipedia makes no reference to dimensionality in the definition, so does mathworld.

 

[Haelfix]

Ok..

In quantum mechanics we are usually reffering to an infinite dimensional hilbert space, unless you want to work on a lattice or something like that.. But in terms of a continum our Hilbert spaces are necessarily infinite. The catch is sometimes physicists ignore the hard parts of the spectral theorem and treat it as semi finite dimensional (in a sense). Thats the logical problem I was reffering too.

Mathematically, you are free to invent whatever object you want, define it however you want and investigate its properties. Hence the confusion in convention.

 

[humanino]

Please stop saying $$\left(\mathbb{R}^n,\sqrt{\sum_{i=1}^n x_i^2}\right)$$ is not a Hilbert space. This is probably the first example taught in most lectures. Or address the question in the math section of the forum, and please post a link afterwards. I keep thinking Bourbaki's definitions tend to have some relevance.

 

[vanesch]

Please stop saying $$\left(\mathbb{R}^n,\sqrt{\sum_{i=1}^n x_i^2}\right)$$ is not a Hilbert space.

It is NOT a Hilbert space :tongue2: :tongue2:

Seriously, it should be over the complex numbers, not the reals, no ?

cheers,
patrick.

 

[humanino]

Not necessarilly. It is over some kind of field, among which $$\mathbb{R}$$ or $$\mathbb{C}$$ are at least to me the most relevant example. Yet, I am not certain one could not use another kind of non-commutative field, such as the quaternions $$\mathbb{H}$$. This is not very relevant here.

 

[Lonewolf]

Patrick, I notice you use Berberian's text. According to his exposition, it is a Hilbert space. See 2.8, example 6. It doesn't matter what space you use, as long as it is Cauchy-complete, which the real numbers are under the metric induced by the scalar product. The quaternions are also complete under a similar metric, and so also form a Hilbert space.

I'd like to address selfAdjoint's question in more detail. The existence of a countable dense subset in a Hilbert space is implied by the existence of a countable orthonormal basis, and hence is separable in the topological sense. I could write a proof, but I can't find a margin that will contain it. Instead, I refer you to http://www.mth.kcl.ac.uk/~jerdos/OpTh/w3.pdf , page 4 for a sketch of a proof.

 

[vanesch]

Patrick, I notice you use Berberian's text. According to his exposition, it is a Hilbert space. See 2.8, example 6. It doesn't matter what space you use, as long as it is Cauchy-complete, which the real numbers are under the metric induced by the scalar product. The quaternions are also complete under a similar metric, and so also form a Hilbert space.

Well, in II, paragraph 1, definition 1, it is stated that a pre-hilbert space is a COMPLEX VECTOR SPACE, etc...

In II, paragraph 5, definition 1: A complete pre-Hilbert space is called a Hilbert space.

So this includes finite-dimensional vector spaces, but always over C.
It might be that this is just terminology, which changes from author to author, of course.

cheers,
Patrick.

 

[Lonewolf]

But checking the axioms in definition 1, and taking the real numbers as a subset of the complex numbers, we see that the real numbers do indeed form a pre-Hilbert space. I agree though, different authors do seem to use different criteria for the definition of a Hilbert space. I prefer to take the most general one, myself.

 

[Lonewolf]

Does anybody have anything to say on humanino's question on wavelets? I'd be interested to know too, and I believe wavelets had some of their origin rooted in quantum field theory.

 

[humanino]

I addressed the same question in a french forum, and received the following references :
Wavelet based regularization for Euclidean field theory and stochastic quantization
http://www.worldscibooks.com/mathematics/3066.html
So, apparently the idea is not that bad