# Rigged Hilbert Spaces In Quantum Mechanics

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Mentor
In discussing stuff in another thread I used the standard Dirac notion expanding a state in position eigenvectors namely |u> = ∫f(x) |x>. By definition f(x) is the wave-function. I omitted the dx which is my bad but the following question was posed which I think deserved a complete answer. It was also off at a tangent to the main threads topic so really required a separate thread.

I'm confused. What is the state |x> supposed to be? Does ∫ f(x) |x> = ∫ f(x)dx•|x> or ∫ f(x) |x>dx ?

Its the Dirac bra-ket notation.

You are not the only one to be confused by it even though its in standard use in QM - so was the great Von-Neumann. He was scathing about it in his famous book on QM which with your mathematical background may be the presentation of QM you are most familiar with. Nowadays it can be made rigorous using the Rigged Hilbert Space (RHS) formalism:
https://en.wikipedia.org/wiki/Rigged_Hilbert_space

|x> is the eigenvector of the position operator. It does not exist in Hilbert Space - but in a RHS.

Heuristically here is what's happening. Suppose you have a very fine grid of positions with eigenvectors |xi> where each lies in some range Δx. We assume that positions lie in a large but finite range. These can be found in a finite dimensional Hilbert space and any element can be represented by ∑f(xi) |xi>. We then divide f(xi) and |xi> by √Δx so we have the new equation as ∑f(xi) |xi> Δx using these new f(xi) and |xi>. Then we let the range of positions go to infinity at the same time as Δx goes to zero. Then, heuristically |xi> goes to |x> where x is an exact position. f(xi) goes to a function f(x) and the sum goes to an integral ∫f(x) |x> dx Then we have this new continuous basis |x> - each is of infinite length and do not belong to a Hilbert space, but instead belong to this strange beast a RHS.

This is all simply heuristics. I have attached a document giving the full rigor - but its no easy read nor short.

Interestingly with your background in probability it also finds application in White Noise Theory (it called by its other name there Generalized Functionals and Schwartz Space - but it just is an example of a RHS):
http://www.asiapacific-mathnews.com/04/0404/0010_0013.pdf

Also interestingly since Von-Neumans scathing attack on it mathematicians were not lying down - it took the efforts of 3 great mathematicians - Gelfland, Schwartz and Grothendieck to sort it out.

Thanks
Bill

#### Attachments

• Quantum Mechanics In Rigged Hilbert Space Language.pdf
1.3 MB · Views: 514
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• Demystifier, Zafa Pi, eloheim and 2 others

Zafa Pi
In discussing stuff in another thread I used the standard Dirac notion expanding a state in position eigenvectors namely |u> = ∫f(x) |x>. By definition f(x) is the wave-function. I omitted the dx which is my bad but the following question was posed which I think deserved a complete answer. It was also off at a tangent to the main threads topic so really required a separate thread.
Thank you. I appreciate that.
However, I still don't know where the dx in ∫f(x) |x> is supposed to go. Since you say,
|x> is the eigenvector of the position operator. It does not exist in Hilbert Space - but in a RHS.
do you mean ∫f(x) |x> = ∫f(x)δ(x)dx? Though I am quite familiar with the Dirac bra-ket notation, as well as Schwartz distributions, Lighthill's generalized functions, and Mikusinski"s operational calculus, I admit to never seeing such an integral. Or do you mean (∫f(x)dx)⋅δ(x)? In which case I don't see any isomorphism with L2.

I vaguely recall seeing Rigged Hilbert Space, but remember no content. Thanks for the reference and I'll check it out.

Mentor
However, I still don't know where the dx in ∫f(x) |x> is supposed to go. Since you say,

It should be the usual notation for an integral ∫f(x) |x> dx,

But some like Zee abuse the notation and write at as ∫f(x) dx |x>

All RHS's are is, basically Hilbert spaces with distribution theory/generalized functions stitched on. Which is why the word Rigged is used - it not meant to mean rigged like a card game but like the scaffolding or rigging on a ship.

But doing that leads down some rather hairy roads such as Nuclear spaces worked on by Grothendieck, all leading to its jewel in the crown - the Gelfand-Maurin Theorem. Its proof is interesting because THE tome on this stuff is the three volume set Gelfand and Shilov - Generalized functions. Its proof is wrong.

However the attachment gives a correct proof that is valid for cases found in QM. I don't think its the full theorem which I have never seen a correct proof of.

Thanks
Bill

#### Attachments

• Quantum Generalized Eigenfunctions.pdf
116.5 KB · Views: 733
Even for an ordinary Riemann integral, $$\int f(x) \frac{dx}{x^2+1}=\int \frac{dx}{x^2+1}f(x)=\int dx \frac{f(x)}{x^2+1}=\int \frac{f(x)}{x^2+1}dx$$. The place where the ##dx## occurs doesn't matter as long as the formula making up the integrand is linear in ##dx##.

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• member 587159
Zafa Pi
Even for an ordinary Riemann integral, $$\int f(x) \frac{dx}{x^2+1}=\int \frac{dx}{x^2+1}f(x)=\int dx \frac{f(x)}{x^2+1}=\int \frac{f(x)}{x^2+1}dx$$. The place where the ##dx## occurs doesn't matter as lonmg as the formula making up the integrand is linear in ##dx##.
∫xf(x)dx ≠ ∫xdxf(x) = f(x)∫xdx. This was my problem. Check the beginning of post #3.

∫xf(x)dx ≠ ∫xdxf(x) = f(x)∫xdx. This was my problem. Check the beginning of post #3.
No. According to the customary rules, the first two expressions are equal, and the third expression is meaningless as ##x## appears both as a free and a bound variable.

Zafa Pi
It should be the usual notation for an integral ∫f(x) |x> dx,

But some like Zee abuse the notation and write at as ∫f(x) dx |x>

All RHS's are is, basically Hilbert spaces with distribution theory/generalized functions stitched on. Which is why the word Rigged is used - it not meant to mean rigged like a card game but like the scaffolding or rigging on a ship.

But doing that leads down some rather hairy roads such as Nuclear spaces worked on by Grothendieck, all leading to its jewel in the crown - the Gelfand-Maurin Theorem. Its proof is interesting because THE tome on this stuff is the three volume set Gelfand and Shilov - Generalized functions. Its proof is wrong.

However the attachment gives a correct proof that is valid for cases found in QM. I don't think its the full theorem which I have never seen a correct proof of.

Thanks
Bill
Reading this stuff would have been far easier for me 60 years ago. I'm also reading http://www.scottaaronson.com/democritus/lec9.html because of you, which is quite easy.

However, I take issue with you with you for saying that the L2 "wave" function isn't actually a vector in the (L2) Hilbert Space. It is pretty standard and the preference for the RHS in the "B wave vs vector" thread wasn't really necessary.

Zafa Pi
No. According to the customary rules, the first two expressions are equal, and the third expression is meaningless as ##x## appears both as a free and a bound variable.
Give me a break. If I asked anyone what ∫xdx is they would say ½x2 + c, not something like "the function whose rule is expressed by f(u) = ½u2 + an arbitrary constant". So if f(x) = 2 they would say f(x)∫xdx = x2 + c. Next you're going to tell me f(x) isn't a function, but rather a value in the range of f.. You're getting a bit too pedantic for me.

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Give me a break. If I asked anyone what ∫xdx is they would say ½x2 + c, not something like "the function whose rule is expressed by f(u) = ½u2 + an arbitrary constant". So if f(x) = 2 they would say f(x)∫xdx = x2 + c. Next you're going to tell me f(x) isn't a function, but rather a value in the range of f.. You're getting a bit too pedantic for me.

The integrals in quantum mechanics are all definite integrals!

Zafa Pi
The integrals in quantum mechanics are all definite integrals!
Aha, thanks. Now it's even clearer that ∫f(x)|x>dx ≠ ∫f(x)dx|x>.

Eye_in_the_Sky
do you mean ∫f(x) |x> = ∫f(x)δ(x)dx? Though I am quite familiar with the Dirac bra-ket notation, as well as Schwartz distributions, Lighthill's generalized functions, and Mikusinski"s operational calculus, I admit to never seeing such an integral. Or do you mean (∫f(x)dx)⋅δ(x)? In which case I don't see any isomorphism with L2.

Zafa, does this help?

Write,

|ψ> = ∫f(x)|x>dx .

Then,

<x'|ψ> = ∫f(x)<x'|x>dx

= ∫f(x)δ(x'-x)dx

= f(x') .

If f Є L2, then |ψ> is an element of the Hilbert Space (proper).

The converse also holds (modulo equivalence classes in L2 in terms of being equal "almost everywhere" – i.e. for any f,g Є L2

f ~ g

iff

∫ |f(x)-g(x)|2dx = 0 ).

Mentor
Reading this stuff would have been far easier for me 60 years ago. I'm also reading http://www.scottaaronson.com/democritus/lec9.html because of you, which is quite easy.

However, I take issue with you with you for saying that the L2 "wave" function isn't actually a vector in the (L2) Hilbert Space. It is pretty standard and the preference for the RHS in the "B wave vs vector" thread wasn't really necessary.

Is e^ipx in a Hilbert space? It is in the RHS. Its also the wave-function of a state with definite momentum p.

Thanks
Bill

Staff Emeritus
|x> is the eigenvector of the position operator. It does not exist in Hilbert Space - but in a RHS.

I remember reading about Rigged Hilbert space a while back, at your suggestion, and I was under the impression (or was it a mis-impression?) that there was a distinction between "bras" and "kets" in that formalism. I thought that a ket $|\psi\rangle$ meant an element of the Hilbert space, but that a bra $\langle \psi|$ meant a linear functional on Hilbert space. So $\langle x'|$ was a perfectly good "ket", since it's the function that takes a $\psi$ and returns $\psi(x')$. But that there was no corresponding "bra" $|x'\rangle$.

On the other hand, I suppose that the set of kets for one Hilbert space can be the bras for a different Hilbert space?

Now it's even clearer that ∫f(x)|x>dx ≠ ∫f(x)dx|x>.
These are equal, no matter what you think.

Staff Emeritus
These are equal, no matter what you think.

Is the issue about "variable capture"? There is certainly a distinction between

$(\int f(x) dx) |x\rangle$

and

$\int (f(x) |x\rangle dx)$

If that's the issue, then the meaning could be clarified by using different variables in the first case: It would mean the same thing as $\int f(x') dx' |x\rangle$.

The second is what is meant when people say that this is a representation of the "wave function".

• Zafa Pi and bhobba
Mentor
I remember reading about Rigged Hilbert space a while back, at your suggestion, and I was under the impression (or was it a mis-impression?) that there was a distinction between "bras" and "kets" in that formalism. I thought that a ket $|\psi\rangle$ meant an element of the Hilbert space, but that a bra $\langle \psi|$ meant a linear functional on Hilbert space. So $\langle x'|$ was a perfectly good "ket", since it's the function that takes a $\psi$ and returns $\psi(x')$. But that there was no corresponding "bra" $|x'\rangle$. On the other hand, I suppose that the set of kets for one Hilbert space can be the bras for a different Hilbert space?

You are basically correct. The kets are some subset of a Hilbert space - for example the space of continuously deferentialable functions of compact support. Its dual is the RHS and is larger than the Hilbert space and is a bra.

The functional's defined over the test space is the RHS and are expressed as bras - not ket's. To get around it you can define a RHS of ket's by taking the complex conjugate ie if |a> is a RHS test vector and <b| a member of the RHS of bras then you can define the RHS of kets by <a|b> = conjugate <b|a>. Sometimes you can even make sense of <a|b> where <a| and |b> are members of the RHS eg if they are members of the Hilbert space. This is part of the gelfland triple T⊂H⊂R where T is the test space and R is the RHS. Sometimes its possible to define <a|b> where a and b are from the RHS - in the Hilbert space H you can for sure.

Thanks
Bill

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• Greg Bernhardt
Eye_in_the_Sky
... gelfland triple R⊂H⊂T where T is the test space and R is the RHS.

You meant it, rather, the other way around: T⊂H⊂R.

• bhobba
Mentor
You meant it, rather, the other way around: T⊂H⊂R.

Yes - that is more usual. Not only that mine is incorrect - obviously.

Thanks
Bill

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Staff Emeritus
Yes - that is more usual. Not only that mine is incorrect - obviously.

Thanks
Bill

You spent an entire year working on that answer? • bhobba
Mentor
You spent an entire year working on that answer? When I worked for the government there was a saying - the wheels of government work slowly but surely. Old habits are hard to break - besides mistakes picked up by others help keep people on the ball. Penned while watching on TV the latest research that public servants believe they know best and the public are - well simply ignorant. From experience its more along the lines most public servants are process oriented, only the few that are results oriented could be accused of that          .

Thanks
Bill

• stevendaryl