Rigged Hilbert space, separable space, domain of CSCO, mapping

1. Jan 30, 2013

Petro z sela

Suppose that we have rigged Gilbert space Ω$\subset$H$\subset$Ω$\times$ (H is infinite-dimensional and separable).
1. Is the Ω a separable space?
2. Is the Ω$\times$ a separable space?
3. Consider the complete set of commuting observables (CSCO) which contain both bounded and unbounded operators. Eigenvectors of CSCO span the space of states. Is this space of states Ω or H?
4. I know that unbounded closed operators have domain Ω and map Ω into Ω. And what about bounded (compact) operators: Ω into Ω, Ω into H or somewhat else?
5. In what space do CSCO form algebra?
Thanks

2. Jan 30, 2013

dextercioby

In short.

3. Ωx.
4. H into H, because a compact operator is bounded everywhere hence continuous, so it can be uniquely extended to all vectors of H.
5. Both Ω and Ωx, but for different mappings of course. The (anti)dual mappings in case of Ωx.

For 1. and 2. I'll have to check.

3. Jan 30, 2013

Petro z sela

3. You say Ω$\times$. It means that I can decompose any vector from Ω$\times$ over the eigenvectors of CSCO. Is this true for vectors from H\Ω? I don't think so, because nuclear spectral theorem allow us to decompose only vectors from Ω over the eigenvectors of unbounded operator (suppose CSCO contain unbounded operator).
4-5. Ok. Bounded operators have domain H, and unbounded have domain Ω. I am not sure but if they form CSCO algebra they must have one common invariant domain.

4. Jan 30, 2013

dextercioby

3. Yes I agree with your last statement, vectors from H\Ω bear no significance to physics and are not interesting mathematically.
4-5. What are you not sure about ?

5. Jan 30, 2013

Petro z sela

4-5. Operators forming CSCO algebra must have one common invariant domain. You said both Ω and Ω$\times$. Which of them? I think must be one domain.

6. Jan 30, 2013

dextercioby

3. The eigenvectors must necessarily be in the topological (anti)dual space, if there's at least an operator in the CSCO with (partially/totally) continuos spectrum.
4-5. The CSCO has Ω for all the operators as a common (dense everwhere in the Hilbert space topology) domain. However, bringing in the commutation condition for these operators, we are 'thown' again in the Ωx, because the commutativity is realized through generalized projectors onto spectral subspaces of Ωx. Again, the need to go to topological (anti)duals stems from the 'continuous spectrum' condition.

4-5. Example: For the 3D free Galilean particle, x, y, z actually form a set/algebra of operators for which any commutator of any 2 operators vanishes strongly. However, x,y,z have generalized eigenvectors (Dirac distributions from Sx(R3)), so they achive the CSCO status only when 'promoted' as mappings in the distributional space, as per the definition of a CSCO.

7. Jan 30, 2013

Petro z sela

3. Ok, I am aware that eigenvectors of unbounded (continuous, continuous+discrete spectrum) lie in Ω$\times$. But I think that only vectors from Ω can be decomposed over the CSCO basis.

Thanks, I'm waiting for 1, 2 )

8. Jan 30, 2013

strangerep

A "Gilbert" space, huh? :-)
From Wikipedia, i.e., http://en.wikipedia.org/wiki/Separable_space ....
So: 1: Yes.

As for 2: I can think of cases in which $\Omega^\times$ has a countable dense set. Not sure if that's true in general, though.
I believe that vectors from $\Omega^\times$ can also be decomposed, else Fourier transforms wouldn't be anywhere near as useful as they are...

9. Jan 31, 2013

Petro z sela

Sorry for "Gilbert". I am not so good in English ).

1. I agree. Thanks. I must be considerate towards "wikipedia" ).

2 and 3 are still actual.

10. Jan 31, 2013

strangerep

As for (2), you might get answers from better-informed people if you ask in the mathematics forums (i.e., functional analysis & topology).

And BTW, it would help if you tell us which references on rigged Hilbert space you have already studied... (?)

But let's go back... Question 3 was:
Summarizing Antoine in "Quantum Mechanics beyond Hilbert space" ...

We assume $\Omega$ is complete with respect to a locally convex topology $\tau_\Omega$ on $\Omega$ (i.e., a finer topology than the norm topology inherited from $H$).

$\Omega^\times$ is the space of $\tau_\Omega$--continuous antilinear functionals on $\Omega$.

We also assume that $\Omega$ is reflexive (with some technical conditions).

In most cases, $\Omega$ is the intersection of a countable family of Hilbert spaces. That makes it a Frechet space, according to Antoine.

We also assume that $\Omega$ is "nuclear" which is a technical condition about the respective embeddings of members of the countable family of Hilbert spaces mentioned above.

Then if $A$ is a closed operator in $H$, which maps $\Omega$ into itself (continuously,... blah blah blah),.... then $A$ may be extended by duality to a linear operator $A^\times : \Omega^\times \to \Omega^\times$ which is an extension of the usual adjoint $A^*$ ...

If $A$ is self-adjoint and $\Omega$ is nuclear and complete, the nuclear spectral theorem asserts that $A$ possesses a complete orthonormal set of generalized eigenvectors $\xi_\lambda \in \Omega^\times$, which means that for any $\phi,\psi \in \Omega$, we have
$$\def\<{\langle} \def\>{\rangle} \<\phi|\psi\> ~=~ \int_R \<\phi|\xi_\lambda\> \; \<\xi_\lambda|\psi\> \; d\mu(\lambda)$$
So... I can see why you ask whether only vectors from $\Omega$ can be decomposed in terms of the $\xi_\lambda$. But every element (functional) of $\Omega^\times$ is only defined in terms of its action on elements of $\Omega$. $\Omega^\times$ was constructed in terms of a particular operator (or operators), and it's the spectrum of those operator(s) and the associated generalized eigenvectors which matter.

Hence I believe the answer to 3 is that elements of $\Omega^\times$ can also be decomposed in terms of the generalized eigenvectors of such an operator $A$.

[WARNING: Although I have studied these subjects, I am not an expert. ]

11. Jan 31, 2013

micromass

OK, but $\Omega$ is not a subspace of H. The point is that $\Omega$ has a finer topology than the subspace topology. So all we demand is that $\Omega\rightarrow H$ is a injective bounded operator. We don't demand subspace topology.

I think the answer to 1 is "no", but I can't seem to find a good counterexample. Every example of rigged Hilbert spaces seem to be separable.

12. Jan 31, 2013

micromass

I don't think that reflexive is part of the definition, is it?? Anyway, if it is reflexive, then $\Omega$ is separable if and only if $\Omega^\times$ is separable. So the answer to (1) and (2) are equivalent in that case.

13. Jan 31, 2013

micromass

The triple $\ell^1\subset \ell^2 \subset \ell^\infty$ is a counterexample to (2).

14. Jan 31, 2013

micromass

And it seems to me that $L^\infty([0,1])\subset L^2([0,1])\subset (L^\infty([0,1]))^\times$ is a counterexample to (1).

15. Jan 31, 2013

strangerep

Ah, excellent! Micromass is getting involved. I might actually learn something...

OK, thanks. I meant "subspace" in the "sub vector space" sense, which was sloppy terminology.

Yes, QM people don't like nonseparable spaces since they're more difficult to work with.

Well, it's certainly part of Antoine's definition, but perhaps different authors use other conventions? Which author(s) did you have in mind?

Certainly, these spaces in physics are usually (always?) assumed to be reflexive.

Ah, of course. Well, that certainly answers those questions if one restricts to the RHS's typically used in QM.

Could you expand on both of those points, pls? I've always had difficulty coming to grips properly with stuff involving $\ell^\infty$ and $L^\infty$ . :uhh:

16. Jan 31, 2013

micromass

Ah ok, I just looked "rigged Hilbert space" up in wikipedia. There they don't ask anything about being reflexive.

I'll try to expand on the L-spaces in my next post.

17. Jan 31, 2013

micromass

So, to my understanding (from wikipedia), we are looking for a couple $(\Phi,H)$, where H is a Hilbert space and $\Phi$ is a dense subset of H. Furthermore, we equip $\Phi$ with a topology (which is unrelated to the subspace topology of H).

We must have a map $i:\Phi\rightarrow H$ which is injective and a bounded linear operator. This has to a map $i^*=H^*\rightarrow \Phi^*$, with $(i^*|v>)(|w>)=<v|w>$.

So, let's define the real (complex works too) vector space

$$\ell^2=\{(x_n)_n~\vert~\sum|x_n|^2<+\infty\}$$

We can equip this space with a natural inner product

$$<(x_n)_n,(y_n)_n>=\sum x_ny_n$$

This is well-defined. Indeed, recall the standard inequality $ab\leq \frac{1}{2}(a^2+b^2)$, this implies

$$|<(x_n)_n,(y_n)_n>|=|\sum x_ny_n| \leq \frac{1}{2}(\sum x_n^2 + \sum y_n^2)$$

the last two series converge by definition of $\ell^2$. So the inner product is finite. Checking the properties of the inner product is trivial.

It is well-known that any inner product defines a norm. In this case, the norm on $\ell^2$ is

$$\|(x_n)_n\|_2=\sqrt{\sum |x_n|^2}$$

The space $\ell^2$ is complete and thus defines a Hilbert space.

We can also define

$$\ell^1=\{(x_n)_n~\vert~\sum |x_n|<+\infty\}$$

This will not turn out to be a Hilbert space, as there is no natural inner product. But we can make a norm by

$$\|(x_n)_n\|_1=\sum |x_n|$$

This turns out to be a complete space.

It is true that $\ell^1\subset \ell^2$ (as subsets). And we can prove that $\ell^1$ is dense in $\ell^2$. Furthermore, we can prove that

$$\|(x_n)_n\|_2\leq \|(x_n)_n\|_1$$

So the injection $i:\ell^1\rightarrow \ell^2$ is bounded with norm (at most) 1.

Now, what is the dual of $\ell^1$. It turns out that it is $\ell^\infty$. This is defined by

$$\ell^\infty=\{(x_n)_n~\vert~\sup |x_n|<+\infty\}$$

This is again not a Hilbert space, but we do have a canonical norm

$$\|(x_n)_n\|_\infty=\sup |x_n|$$

It turns out that $\ell^\infty$ is complete. However, we can prove that $\ell^\infty$ is not separable. The proof of this is essentially some kind of diagonal argument.

Every element of $\ell^\infty$ induces some bounded functional in $(\ell^1)^\times$. Indeed, if $(x_n)_n\in \ell^\infty$, then we define

$$T_{(x_n)_n}:\ell^1\rightarrow \mathbb{R}:(y_n)_n\rightarrow \sum x_ny_n$$

The map $(x_n)_n\rightarrow t_{(x_n)_n}$ defines an isometric bijection between $\ell^\infty$ and $(\ell^1)^\times$.

It is clear that $\ell^2\subset \ell^\infty$. We even have

$$\|(x_n)_n\|_\infty\leq \|(x_n)_n\|_2$$

So the injection $\ell^2\rightarrow \ell^\infty$ is bounded with norm (at most) 1.

Now, if $(x_n)_n\in \ell^2$, then $T_{(x_n)_n}$ is in $(\ell^1)^\times$. Furthermore, we have $T_{(x_n)_n}((y_n)_n)=\sum x_ny_n = <(x_n)_n,(y_n)_n>$. So we are indeed dealing with a rigged Hilbert space.

18. Jan 31, 2013

strangerep

The Wiki page on "rigged Hilbert space" is sadly minimal. I've been wishing for several years for an expert to flesh it out some more.

And the "best" textbook I know of, i.e., Gel'fand & Vilenkin vol 4, is quite difficult for the average physics type, including me. It's also growing a bit old.

Cheers.

19. Jan 31, 2013

micromass

Hmm, I see. There is indeed more to a rigged Hilbert space than the wiki page shows. In particular, we want $\Phi$ to be reflexive. This implies that my example of the $\ell^2$-space would be wrong, since $\ell^1$ is not reflexive.

I don't have access to Antoine, sadly, but I'll look on the internet for a more complete definition.

20. Jan 31, 2013

strangerep

Are you sure? Try Google Scholar for Antoine "beyond hilbert space". You should be able to get a pdf from citeseer, or a ps from some other places.

I'll also dig out some other references and post them shortly.

Dunno why you say that. You always make me happier when you drop by...