# Do bras and inner products relate in a Rigged Hilbert Space?

1. Aug 18, 2013

### lugita15

One of the most important results of functional analysis is that for every bounded linear functional f: H → ℂ on a Hilbert space H, there exists a fixed |v> in H such that f(|u>) is equal to the inner product of |v> with |u> for all |u> in H. This justifies the labeling of f as <v| in the bra-ket notation used in quantum mechanics.

I want to know the relationship of bra's to inner products carries over to Rigged Hilbert Space. To construct the Rigged Hilbert Space, we start with a Hilbert space H and pick out a nuclear subspace $\Phi$, a dense subspace of H on which certain unbounded linear operators on H are defined. The space $\Phi^{\times}$ of continuous anti-linear functionals will be our ket space, and the space $\Phi'$ of continuous linear functionals will be our bra space. We can then define a distribution-valued inner product on $\Phi^{\times}$. Also, we can define a conjugate map between $\Phi'$ and $\Phi^{\times}$, so that for every |ψ> in the ket space there corresponds a bra <ψ| and vice versa.

Now my question is, can the action of an arbitrary bra <ψ| on the ket space be equated with the action of the distribution-valued inner product with |ψ>?

Any help would be greatly appreciated.

2. Aug 19, 2013

It would be helpful if you could provide a reference to a book or a paper where your terms are being used. The term "distribution-valued inner product" is not clear from your short description. Are you thinking about a general theory or about some particular example?

3. Aug 19, 2013

### micromass

Staff Emeritus
You will need to clarify this. First of all, how does the bra <ψ| act on the ket-space exactly?
And how do you define your distribution-valued inner product?

4. Aug 19, 2013

### dextercioby

A Hilbert space is reflexive and for topological duals for subsets of it you can reach back to 'vectors' by taking a second dual.

5. Aug 19, 2013

### strangerep

I believe that last sentence is not correct. Although I myself have used the term "distribution-valued inner product" in the past, this can be misleading, so (in the spirit of the PF values) I must try to clarify (and correct) my earlier usage.
Further corrections/elaborations welcome!

$\def\<{\langle} \def\>{\rangle}$Reviewing the nuclear spectral theorem, a symmetric linear operator $A$ defined everywhere on the space $\Phi$ and admitting a self-adjoint extension to the Hilbert space $H$, can be extended by duality to $\Phi^\times$, is continuous on $\Phi^\times$ (in the operator topology in $\Phi^\times$) and its extension (still denoted $A$ by a slight abuse of terminology) possesses a complete system of generalized eigenvalues $a$ and associated eigenfunctionals $|a\>$ belonging to the dual space $\Phi^\times$. By "complete", we mean that any element of the small space $\Phi$ can be decomposed in terms of the $|a\>$.

Moreover, (by duality), any operator on $\Phi^\times$ can be decomposed in terms of the $|a\>$. In particular, the identity operator can be thus decomposed, leading to the expression:
$$I ~=~ \int da\, |a\>\<a|$$
Let $|\phi\> \in \Phi$, with spectral decomposition
$$|\phi\> ~=~ \int da\, \phi(a) \, |a\> ~.$$
Here, $\phi$ must typically be a Schwartz function (for the case of canonical position/momentum operators in QM). Acting on $|\phi\>$ with the identity operator above, we have
$$|\phi\> ~=~ \int da\, |a\>\<a| \; \int da'\, \phi(a') \, |a'\> ~=~ \iint da\, da'\, \phi(a') \, |a\>\<a| \,|a'\>$$
This must hold for arbitrary $\phi\in\Phi$, hence we conclude
$$\<a|a'\> ~=~ \delta(a-a') ~.$$
So far, so good. But suppose we now try to extend such spectral decomposition of vectors to any (arbitrary) element of $\Phi^\times\setminus H$ similarly. One might write down a (formal!) expression like this:
$$|F\> ~=~ \int da\, F(a) \, |a\> ~,$$
where $|F\>\in\Phi^\times\setminus H$ but the quantity $F(a)$ is merely an undefined formal expression at this point. Enquiring what type of quantity $F$ could be, we act with $|F\>$ on an arbitrary element $\phi\in\Phi$, as follows:
$$\<\phi|F\> ~=~ \iint da\,da'\, \bar\phi(a) \, F(a) \, \<a|a'\> ~=~ \int da\, \bar\phi(a) \, F(a) ~.$$
This could make sense if ($\phi$ being a Schwartz function), we insist that $F$ be a tempered distribution.

But what if we now try to make sense of an "inner product" of the form $\<F'|F\>$?
Repeating the above formal manipulation, we'd get
$$\<F'|F\> ~=~\int da\, \bar F'(a) \, F(a) ~,$$and (afaict) this does not make sense in general, since we're essentially trying to perform pointwise multiplication of distributions.

Summary: although a "distribution-valued inner product" can be defined between certain non-normalizable vectors (i.e., our basis of eigenfunctionals), as $\<a'|a\> = \delta(a-a')$, this does not extend sensibly to arbitrary elements of the large space $\Phi^\times$, and one must always be careful to interpret all such formal expressions only in terms of their action on the small space $\Phi$.

Last edited: Aug 19, 2013
6. Aug 19, 2013

### bhobba

Getting to grips with RHS's is far from trivial and will require a good deal of effort - I know I have spent more time mucking around with it than I care too admit.

My suggestion is start out simple with the overview treatment in Ballentine, then get a good background in analysis. I like the following (although I have a number of textbooks dating back from my undergrad days)
https://www.math.ucdavis.edu/~hunter/book/pdfbook.html

Then read the following PhD Thesis which really is the best account of it and its relation to QM I have come across:
http://physics.lamar.edu/rafa/dissertation.htm

Then you can come to grips with the key theorem on generalized eigenfunctions that makes this all worthwhile - see attached document.

Have fun.

Thanks
Bill

#### Attached Files:

• ###### Quantum Generalized Eigenfunctions.pdf
File size:
116.5 KB
Views:
189
7. Aug 20, 2013

8. Aug 20, 2013

### strangerep

9. Aug 27, 2013

### lugita15

Thanks strangerep, this was a really helpful post.
Does that mean that the standard use of bra-ket in quantum mechanics, where we freely take inner products of any kets we please, is illegitimate? There are various theories of multiplication of distributions, like the Colombeau algebra for instance. Can any of those be used to legitimize how bra-ket is used?
Instead of treating the inner product as a function on $\Phi^\times \times \Phi^\times$, could we consider it to be a (two-dimensional) distribution? δ(x-y) is undefined if x=y, but that doesn't stop us from defining the distribution δ(x-y). Could we do something similar with this inner product?

10. Aug 27, 2013

### strangerep

Well, strictly speaking, we don't freely take inner products of any kets we please. Consider $\<a|a\>$ which is ill-defined...

I looked at Colombeau algebras some time ago, but I don't know enough about them to offer an opinion. I once asked Arnold Neumaier about Colombeau algebras (since its proponents propagated a fair bit of hype). He said (iirc) that they're basically equivalent to Young measures and he ended up disappointed and disinclined to spend any more time on that stuff.

Jenny Harrison also claimed to have developed a generalized framework for this stuff, but her papers are over my head.

Hmmm. You've got to be more careful than that if you want to define the delta distribution rigorously. And whatever you come up with would have to be compatible with the ordinary 1D distribution $\<a'|a\> = \delta(a-a')$. Off the top of my head I have no idea how one might do that.