Continuous dual space and conjugate space

[Rasalhague]

I've been reading Ballentine, Chapter 1. Have I got this the right way around? Taking our inner product to be linear in its second argument and conjugate linear in its first, the (continuous?) conjugate space of a Hilbert space \cal{H} is the following set of linear functionals, each identified with an element of \cal{H} via the isomorphism defined by the inner product:

\cal{H}^{\times}=\left \{ F_\alpha \text{ continuous } \; | \; F_\alpha(\beta) \equiv (\alpha,\beta) \right \}.

The continuous dual space is the following set of conjugate linear functionals, each identified with an element of \cal{H} via the conjugate linear analogue of an isomorphism (anti-isomorphism?), defined by the inner product:

\cal{H}'=\left \{ F_\alpha \text{ continuous } \; | \; F_\alpha(\beta) \equiv (\beta,\alpha) \right \}.

Is Ballentine's terminology exceptional? Other sources use the name "(continuous) dual space" together with the symbol \cal{H}^{\times} or \cal{H}^* for what Ballentine calls "the conjugate space".

 

[Fredrik]

Taking our inner product to be linear in its second argument and conjugate linear in its first

That's how physicists usually do it. Mathematicians do it the other way for some reason.

\cal{H}'=\left \{ F_\alpha \text{ continuous } \; | \; F_\alpha(\beta) \equiv (\beta,\alpha) \right \}.

If you define F_\alpha that way, it's automatically continuous (because the CBS inequality implies that it's bounded), so you wouldn't have to explicitly mention that it's continuous, but you have made it antilinear (=conjugate linear). I would just define H* to be the set of bounded linear functions from H into ℂ and then prove that f is in H* if and only if f=\langle x,\cdot\rangle for some x\in H. (Posts 13-14 here).

I don't know what the conjugate space is and I'm feeling too lazy to go get my Ballentine right now. Maybe tomorrow, but you'll probably have figured it out by then.

 

[Rasalhague]

That's how physicists usually do it. Mathematicians do it the other way for some reason.

Yeah, I knew about the two conventions. I actually met the mathematicians' inner product first. I guess they don't want to make it too easy for us : )

If you define F_\alpha that way, it's automatically continuous (because the CBS inequality implies that it's bounded), so you wouldn't have to explicitly mention that it's continuous, but you have made it antilinear (=conjugate linear). I would just define H* to be the set of bounded linear functions from H into ℂ and then prove that f is in H* if and only if f=\langle x,\cdot\rangle for some [/itex]x\in H[/itex]. (Posts 13-14 here).

Oh, of course, yes. It's bounded, in the bounded operator sense, with the nonnegative constant M = ||\alpha ||, and that's equivalent to being continuous.

Thanks for the link! Ballantine doesn't go into the full proof.

I don't know what the conjugate space is and I'm feeling too lazy to go get my Ballentine right now. Maybe tomorrow, but you'll probably have figured it out by then.

I think Ballentine's "conjugate space" is what most people call the "continuous dual space", often omitting the qualifier continuous. (Is that omission because the algebraic dual space tends not to play a role in applications?) In fact, it seems more natural to reverse his terminology and call the space of conjugate linear functionals the conjugate space, which is what made me think I might have mixed them up somehow, but he does say:

We now define its conjugate space, \cal{H}^{\times}, as consisting of all vectors f=\sum_n b_n \phi_n for which the inner product (f,h)=\sum_n b_n^* c_n is convergent for all h = \sum_n c_n \phi_n in \cal{H}.

The conjugate space \cal{H}^{\times} is closely related to the dual space \cal{H}'. The only important difference is that the one-to-one correspondence between vectors in \cal{H} and vectors in \cal{H}' is antilinear, whereas \cal{H} and \cal{H}^{\times} are strictly isomorphic. So one may regard \cal{H}' as the complex conjugate of \cal{H}^{\times}.

- Ballentine (1998): Quantum Mechanics: A Modern Development, Ch. 1, p. 27, footnote b.

 

[Rasalhague]

Hang on... The inner product implicitly defines a conjugate linear bijection (anti-isomorphism) between \cal{H} and the space of continuous linear functionals, which most people call the continuous dual space of \cal{H}:

z \alpha \mapsto (z \alpha, \cdot) = \overline{z}(\alpha,\cdot)=\overline{z}F_\alpha.

It also implicitly defines a linear bijection, i.e. an isomorphism, between \cal{H} and the space of continuous conjugate-linear functionals:

z \alpha \mapsto (\cdot,z \alpha) = z(\cdot,\alpha)=zG_\alpha.

The first Ballentine quote suggests to me that his conjugate space is that of the continuous linear functionals, for which the bijection implicitly defined by the inner product is conjugate linear:

\left \{ F_\alpha \; | \; F_\alpha(\beta) \equiv (\alpha,\beta) \right \}.

But the second quote seems to be saying that the conjugate space is the space of continuous conjugate-linear functionals, for which the bijection implicitly defined by the inner product is linear:

\left \{ G_\alpha \; | \; F_\alpha(\beta) \equiv (\beta,\alpha) \right \}.

 

[Rasalhague]

Other naming and notational conventions:

Then we can construct the dual space \Phi^{\times} (space of antilinear functionals over a vector space) of \Phi. There is also an antidual space corresponding to \Phi, the linear functionals correspond to bras in the Dirac bra-ket notation and the antilinear functionals correspond to kets.

- http://www.abhidg.net/RHSclassreport.pdf (PDF)

The dual space \Phi' and the antidual space \Phi^{\times} contain respectively the bras and the kets associated with a continuous spectrum of observables.

- R. de la Madrid (2005): "The role of the rigged Hilbert space in Quantum Mechanics," Eur. J. Phys. 26, 287

 

[dextercioby]

Yes, you have to be careful with the various notations and conventions. Typically you should choose a book (for example Ballentine, or Gallindo & Pascual which is more mathematical) and stick to it.

Reading from various sources can definitely throw into confusions. It's actually quite typical for mathematics to use different terminology for the same thing.

 

[Rasalhague]

Reading from various sources can definitely throw into confusions. It's actually quite typical for mathematics to use different terminology for the same thing.

Hi, Dexter! So I've noticed : )

Yes, you have to be careful with the various notations and conventions. Typically you should choose a book (for example Ballentine, or Gallindo & Pascual which is more mathematical) and stick to it.

On the whole, I agree with the strategy of concentrating on one good book at a time, when starting out on a new subject. I've just been doing that with Gillespie. But actually, in this case, I think reading around may have helped. Even the best authors will occasionally use some turn of phrase that's confusing, for whatever reason. So there are different names, different conventions for how to apply the symbols, but the basic idea seems clearer after seeing 3 or 4 descriptions of it. If it's not too overwhelming, it can also help to show up which aspects are fundamental to an idea, and which are arbitrary conventions for how to express it.

Do the Ballentine passages I quoted look contradictory to you?

 

[dextercioby]

I don't find it contradictory, there are indeed 2 different spaces, the topological dual wrt the H-space strong topology (H') and the conjugate space which is customarily denoted by H^x which appears in most sources on rigged Hilbert spaces. These 2 spaces are related, they are anti-isomorphic, the bijection is antilinear wrt the field of complex nrs.

H^x is really important because of the RHS utility. But from an analytical perspective, one could have built RHS with H' without any issue.

 

[Rasalhague]

I don't find it contradictory

What about the issue I raised in #4? If the continuous dual space, \cal{H}', comprises the continuous linear functionals on \cal{H}, and the conjugate space, \cal{H}^{\times}, comprises the conjugate-linear functionals, then the inner product implicitly defines a conjugate-linear bijection between \cal{H} and its continuous dual space, \cal{H}', and a linear bijection (isomorphism) between \cal{H} and \cal{H}^{\times}. Is that right?

This seems to be what Ballentine is referring to in the second quote. But in the first quote, he seems to define the conjugate space as that of the linear functionals (f,\cdot):\cal{H}\rightarrow \mathbb{C}, where f \in \cal{H}, rather than that of the conjugate-linear functionals (\cdot,f). Or does he? Perhaps it's just the way he puts the vector in the first argument slot of the inner product; I guess if the integral converges for the arguments one way around, it should converge with the arguments interchanged, shouldn't it? So, in principle, it doesn't really matter, in defining the set of vectors which are identified with \cal{H}^{\times}, which slot he uses.

 

[Fredrik]

If the continuous dual space, \cal{H}', comprises the continuous linear functionals on \cal{H}, and the conjugate space, \cal{H}^{\times}, comprises the conjugate-linear functionals, then the inner product implicitly defines a conjugate-linear bijection between \cal{H} and its continuous dual space, \cal{H}', and a linear bijection (isomorphism) between \cal{H} and \cal{H}^{\times}. Is that right?

Yes.

This seems to be what Ballentine is referring to in the second quote. But in the first quote,

Are those quotes? On page 27, he defines things differently. His definitions are very strange in my opinion. They go roughly like this. Let B be an orthonormal set in an inner product space. Let F(B) be the free vector space over B (a vector space that has a basis that's in bijective correspondence with B). Define H as the set of all x in F(B) such that [condition 1], and say that such an H is called a Hilbert space. (This appears to be his definition of Hilbert space!) Define H^× as the set of all x in F(B) such that [condition 2]. Then he argues that conditions 1 and 2 are equivalent, so that H and H^× are actually the same subset of F(B). H=H^×.

 

[Rasalhague]

The quotes I meant are those in #3, the first one containing "condition 2". I guess he could just as well have said (h,f) as (f,h), so I don't think it is actually contradictory. If one exists, so should the other.