Boundedness of quantum observables?

[strangerep]

DarMM's solution is a valid solution of the SE. I used Fourier transformations
(which are allowable on L^(RxR,dx)) and got the general solution, to which the
function written by DarMM is only a particular case.

Now I'm bothered by the fact that my separation of variables leads me to
non-normalizable solutions and how these are related to my Fourier analysis
which apparently stays in the Hilbert space.

Aren't you really working in rigged Hilbert space here? I.e.,

$$\Omega ~\subset~ \mathbb{H} ~\subset~ \Omega'$$

where $$\Omega$$ is a nuclear space, $$\Omega'$$ its dual,
and $$\mathbb{H}$$ the Hilbert space.

We rely on the nuclear spectral theorem to ensure that the
(non-normalizable) eigenstate solutions span $$\Omega'$$. Then, since
$$\mathbb{H}$$ is a (dense) subspace thereof, any element of
$$\mathbb{H}$$ can be expressed as a linear combination of them,
as in DarMM's solution:

$$\Psi\left( x, t \right) ~=~ <br /> \int{e^{-p^{2}}e^{-\iota \frac{p^{2}}{2m}t} e^{\iota p x} dp}$$

The Fourier transform is a particular case of the more general Gel'fand
transform, which still works in the larger context of rigged Hilbert
space. If a vector is in $$\mathbb{H}$$, then Fourier-transforming it
won't move it out of $$\mathbb{H}$$ .

(Or did I misunderstand your point?)

 

[Hurkyl]

a basis of solutions is given by the e^(iAx)e^(iBt) , A,B in R, nonzero. From these you can form linear combinations which would again be solutions of the initial equation.

The statement I have emphasized in your quote is the key one. Your basis states separate into a product of a function in x with a function in t -- but you are asserting that all linear combinations separate as well.

This is actually a special case of tensor products, and the familiar phenomenon that very few tensors are pure, despite the fact you can always find a basis of pure tensors.

 

[dextercioby]

Aren't you really working in rigged Hilbert space here? I.e.,

$$\Omega ~\subset~ \mathbb{H} ~\subset~ \Omega'$$

where $$\Omega$$ is a nuclear space, $$\Omega'$$ its dual,
and $$\mathbb{H}$$ the Hilbert space.

We rely on the nuclear spectral theorem to ensure that the
(non-normalizable) eigenstate solutions span $$\Omega'$$. Then, since
$$\mathbb{H}$$ is a (dense) subspace thereof, any element of
$$\mathbb{H}$$ can be expressed as a linear combination of them,
as in DarMM's solution:

$$\Psi\left( x, t \right) ~=~ <br /> \int{e^{-p^{2}}e^{-\iota \frac{p^{2}}{2m}t} e^{\iota p x} dp}$$

The Fourier transform is a particular case of the more general Gel'fand
transform, which still works in the larger context of rigged Hilbert
space. If a vector is in $$\mathbb{H}$$, then Fourier-transforming it
won't move it out of $$\mathbb{H}$$ .

(Or did I misunderstand your point?)

DUHHHHHH! I feel so ashamed that I didn't see this.

Thanks, Mike.

 

[strangerep]

DUHHHHHH! I feel so ashamed that I didn't see this.

Join the club. I regularly feel ashamed around here. But I learned a little more from
your exchange with the others, so I thank you for pursuing it.

BTW, my first name is "Strange"! :-)

 

[strangerep]

Getting back to the main topic of this thread,...

I don't like the C^*-algebraic foundations of quantum mechanics since
it assumes that every observable must be bounded and self-adjoint.

[...]

It's curious to me that this seems not to coincide exactly with Segal's
vision of his algebraic approach to QM...

During 1947, Segal published these two papers:

[1] I.E.Segal, "Irreducible Representations of Operator Algebras",
Bul. Am. Math. Soc, vol 53, no 2, (1947), p73.

in which he introduces C*-algebras in the specifc context of bounded
operators on Hilbert space. I think this is the one that people
mean when they associate C*-algebras inextricably with bounded
self-adjoint operators on Hilbert space.

But then there's also this (subsequent) paper:

[2] I.E.Segal, "Postulates for General Quantum Mechanics",
Ann. Math, 2nd series, vol 48, no 4, (Oct 1947), p 930.

in which he proposes axioms for observable algebras and associated
states over these algebras. In this paper, he doesn't call them
"C*-algebras", afaict. Bounded operators on Hilbert space are only
given as an example, but his algebraic framework is clearly more
general that this.

He calls such an algebra a (closed) "system", and defines states as
linear functionals w on the algebra such that, for A in the algebra,
$w(A^2) \ge 0$, and $w(I) = 1$. A "pure state" is
one which is "not a linear combination with positive coefficients, of
two other states". w(A) is called the expectation value of A in the
state w. A collection of states on the algebra is called "full" if
for every two observables there is a state in the collection in which
the observables have different expectation values.

Later in the paper, Segal proves a theorem which I find quite
nontrivial: A system has a full collection of pure states.

I took partially inconsistent comments from DarMM about unbounded
observables in the C^* algebra approach to rigorous field theory as my
starting point.

The intended goal was to discuss the limitations of C^* algebras in
this regard, and what the possible alternatives are.

Let me try to open a line of discussion in pursuit of that goal...

In Segal's 2nd paper, Hilbert space plays no role in the theory (see his
introduction), although we can of course construct one by choosing a
fiducial vector. The algebraic framework then encompasses all those
pesky inequivalent representations via different choices of fiducial
vector from which to construct each representation. Passing between
such representations generally involves unbounded operators
(Bogoliubov transformations, etc).

But it's unclear to me how one should regard the postulated norm on the
algebra. Usually one relates it to a supremum norm over the vectors in
a representation, but this seems inappropriate in the context of
infinitely many inequivalent representations. So what is the algebraic
norm in this more general case? Is it merely abstract and nonconstructive,
or can a representation-independent construction be given?

[Edit:] BTW, maybe if someone would explain the "type I/II/III" business
more clearly to me I could get a better handle on all this?

 

[Hurkyl]

IIRC, in a C*-algebra, the spectrum and norm are fully determined by the algebra.

A complex number x is in the spectrum of T if and only if (T-x) is not invertible. The norm is the supremum of the absolute values of the numbers in the spectrum.

 

[strangerep]

IIRC, in a C*-algebra, the spectrum and norm are fully determined by the algebra.

A complex number x is in the spectrum of T if and only if (T-x) is not invertible.
The norm is the supremum of the absolute values of the numbers in the spectrum.

Ah, thanks. I actually remember this now -- after you've reminded me. :-)

So one can indeed have "unbounded" elements in the algebra even if there's
no representation anywhere in sight.

 

[Hurkyl]

Ah, thanks. I actually remember this now -- after you've reminded me. :-)

So one can indeed have "unbounded" elements in the algebra even if there's
no representation anywhere in sight.

I imagine similar sorts of definitions would make sense in the more general context and some elements could have infinite "norm". They're all bounded in a C*-algebra.

(I've only studied C*-algebra a little, and the more general context much less, so I can't really say much about more general but similar sorts of algebras)

 

[A. Neumaier]

Hi, Arnold, you're right. DarMM's solution is a valid solution of the SE. I used Fourier transformations (which are allowable on L^(RxR,dx)) and got the general solution, to which the function written by DarMM is only a particular case.

Now I'm bothered by the fact that my separation of variables leads me to non-normalizable solutions and how these are related to my Fourier analysis which apparently stays in the Hilbert space.

Any clue ?

Yes. The most general superpositions of solutions psi_s indexed by a parameter s (which may be an integer, a number, a function, or more composite things) is an integral
$$\int d\mu(s)\psi_s$$
in terms of an arbitrary measure mu. This is the way you can get normalizable solutions from unnormalizable ones. Indeed DarMM's solution is an integral over separable solutions.

On the other hand, you need to do more work to actually show that _every_ solution is in fact an integral over separable solutions.

In general, an ansatz _always_ gives only particular solutions, and the question of finding all solutions is a completely separate one.

 

[A. Neumaier]

So one can indeed have "unbounded" elements in the algebra even if there's
no representation anywhere in sight.

The key is the norm, not the representation. But if the norm is real-valued and defined for all elements of the algebra, all elements of the algebra are bounded!

In particular, the paper ''Postulates for General Quantum Mechanics'' by Segal that you referenced, doesn't admit unbounded operators since it doesn't allow the norm to be infinite.

 

[A. Neumaier]

But it's unclear to me how one should regard the postulated norm on the
algebra. Usually one relates it to a supremum norm over the vectors in
a representation, but this seems inappropriate in the context of
infinitely many inequivalent representations. So what is the algebraic
norm in this more general case?

In a C^* algebra (and so presumably also in Segal's paper - although I lost interest when I saw the norm postulated), the norm of an operator A is the supremum of ||J(A)psi|| over all unitary representations J of the algebra and all unit vectors psi in the representation space of J.

 

[strangerep]

the paper ''Postulates for General Quantum Mechanics'' by Segal that you
referenced, doesn't admit unbounded operators since it doesn't allow the norm
to be infinite.

Yes. The point still puzzling me is that Segal intended (IIUC) his framework
to be capable of handling unitarily inequivalent reps under
a common framework, but I don't see how that sits consistently with
banishing unbounded operators.

In a C^* algebra (and so presumably also in Segal's paper - although I lost
interest when I saw the norm postulated), the norm of an operator A is the
supremum of ||J(A)psi|| over all unitary representations J of the algebra and
all unit vectors psi in the representation space of J.

I don't recall seeing that in Segal's paper. I got the impression the norm
is merely postulated, as you said.

But is your definition above equivalent to Hurkyl's? I.e.,

A complex number x is in the spectrum of T if and only if (T-x) is not
invertible. The norm is the supremum of the absolute values of the
numbers in the spectrum.

(My intuition says "yes", but I've learned not to trust it.)

 

[strangerep]

Thinking more about (un)boundedness of observables in a strictly
algebraic context (i.e., without Hilbert space), I'm wondering
how one could derive the usual angular momentum spectrum
in this context (if we didn't already know it).

In standard QM with Hilbert space, etc, one takes the usual two
commuting observables $$J^2 , J_z$$ and with the help of the
su(2) commutation relations and the Hermitian inner product, one
derives the spectrum in a page or two.

But the spectrum of $$J^2$$ is unbounded, right?
I.e., 0, 1/2, 1, 3/2, 2, ...

Does this mean that even this elementary case can't be handled
in the context of C* algebras alone?

 

[Fredrik]

Thinking more about (un)boundedness of observables in a strictly
algebraic context (i.e., without Hilbert space), I'm wondering
how one could derive the usual angular momentum spectrum
in this context (if we didn't already know it).

In standard QM with Hilbert space, etc, one takes the usual two
commuting observables $$J^2 , J_z$$ and with the help of the
su(2) commutation relations and the Hermitian inner product, one
derives the spectrum in a page or two.

But the spectrum of $$J^2$$ is unbounded, right?
I.e., 0, 1/2, 1, 3/2, 2, ...

Does this mean that even this elementary case can't be handled
in the context of C* algebras alone?

This confuses me as well. An irreducible representation of the algebra will take the J_i and therefore J² to operators on a finite-dimensional vector space. The representative of J² is even proportional to the identity operator on that space.

But a linear operator on a Hilbert space, with the spectrum {0,1/2,1,3/2,...}, must be unbounded, because:

1. The set of bounded linear operators on a Hilbert space has the structure of a Banach algebra.
2. The spectrum of a member of Banach algebra is a non-empty compact subset of ℂ.
3. The set of eigenvalues is a subset of the spectrum.
4. Any compact subset of a metric space is bounded.

If J_i is an unbounded linear operator on a Hilbert space for i=1,2,3, and J²=J_iJ_i, then the J_i must be unbounded too. (The sum and the composition of two bounded linear operators is a bounded linear operator).

If we define the J_i as linear operators on an infinite-dimensional Hilbert space, satisfying the usual commutation relations, they are definitely unbounded (Edit: Not really. See the comment I added at the end). Does it perhaps still make sense to define them as the generators of a 3-dimensional C*-algebra such that the commutation relations are satisfied? I mean, we're going to represent them as bounded operators anyway (since linear operators on finite-dimensional normed vector spaces are always bounded).

Edit: I think the first sentence of the preceding paragraph is actually false. What if the Hilbert space is the direct sum of infinitely many copies of the spin-1/2 spaces? Then we seem to end up with bounded operators. This means that the commutation relations are insufficient to single out 3 specific operators, or even a specific subalgebra of operators, on L²(ℝ³). That's actually pretty obvious if we think about it in terms of representations.

I think this means that we can deal with the spin operators in a C*-algebra framework, but that {0,1/2,1,3/2,...} will not appear as the spectrum of a single operator, but as numbers labeling the different irreducible representations. Once we have all the irreducible representations, we can construct an operator with that spectrum if we want to.

 

[A. Neumaier]

T
But the spectrum of $$J^2$$ is unbounded, right?
I.e., 0, 1/2, 1, 3/2, 2, ...

Does this mean that even this elementary case can't be handled
in the context of C* algebras alone?

Yes. No.

One handles unbounded self-adjoint operators A by considering instead the bounded family of operators exp(isA), s real. In each representation, their spectrum is in 1-to-1 correspondence with each other. Thus, in _some_ sense, one doesn't need unbounded operators. But many things become quite awkward when expressed in terms of the exponentials rather than their generators.

To answer your other question; C^* algebras handle the problem of inequivalent representations. The CCR for unbounded operators translate into Weyl relations for the corresponding bounded exponentials, and these still have lots of inequivalent representations for systems with infinitely many degrees of freedom.

 

[A. Neumaier]

But is your definition above equivalent to Hurkyl's?

He defines the spectrum, I defined the norm. In C^*-algebras, both statements are theorems.

 

[A. Neumaier]

What if the Hilbert space is the direct sum of infinitely many copies of the spin-1/2 spaces? Then we seem to end up with bounded operators. [...]

I think this means that we can deal with the spin operators in a C*-algebra framework, but that {0,1/2,1,3/2,...} will not appear as the spectrum of a single operator, but as numbers labeling the different irreducible representations. Once we have all the irreducible representations, we can construct an operator with that spectrum if we want to.

An operator with unbounded spectrum cannot be bounded. But an element in an algebra whose spectrum (as defined in the algebra) is unbounded can be bounded in some unitary representation, since these do not need to reproduce the whose spectrum. This is what happens with J^2. All irreducible unitary representations of SO(3) are finite-dimensional and hence have bounded J^2, but nevertheless, J^2 is unbounded in the standard single-particle bosonic representation of SO(3).

 

[Fredrik]

...J^2 is unbounded in the standard single-particle bosonic representation of SO(3).

I think this detail is incorrect. Wouldn't we be dealing with a Hilbert space that's the tensor product of $L^2(\mathbb R^3)$ and the 2j+1-dimensional space of spin states? Wouldn't our "$J^2$ " be of the form $1\otimes J^2$, where this other $J^2$ acts only on the spin states, and is in fact proportional to the identity operator on that space?

 

[dextercioby]

Well, the general uniparticle Hilbert space is actually a tensor product

$$L^2(\mathbb{R}^3) \otimes \mathbb{C}^{2s+1}$$,

where s is the value for spin. On this space J^2 is defined by

$$J^2 := L^2\otimes 1_{\mathbb{C}^{2s+1}} + 1_{L^2(\mathbb{R}^3)} \otimes S^2$$

and is neither bounded, nor unbounded, unless one comes up with a definition of norm of an operator on a tensor product of Hilbert spaces.

 

[Fredrik]

You can define an inner product on a tensor product space $\mathcal H=\mathcal H_1\otimes\mathcal H_2$ by

$$\langle x\otimes y,x'\otimes y'\rangle_{\mathcal H}=\langle x,x'\rangle_{\mathcal H_1} \langle y,y'\rangle_{\mathcal H_2}$$

and use that to define the norm on $\mathcal H$. Then you can define an operator norm the usual way,

$$\|A\|=\sup_{x\in\mathcal H}\frac{\|Ax\|}{\|x\|}$$

 

[dextercioby]

So J is unbounded because of L, irrespective of spin.

 

[Fredrik]

*** Deleted *** I see that I overlooked an essential part of what you said. I need to think for a minute...

OK, I've thought about it. Your $J^2$ is the total (squared) angular momentum operator. I agree that it's unbounded, because $\vec L=\vec x\times \vec p$, and both x and p are unbounded. The operator $1\otimes S^2$ is however bounded, with $\|1\otimes S^2\|\leq s(s+1)$.

 

[dextercioby]

Yes, but that's the spin operator; it's bounded, because the spin space is finite-dimensional.

 

[A. Neumaier]

Your $J^2$ is the total (squared) angular momentum operator. I agree that it's unbounded, because $\vec L=\vec x\times \vec p$, and both x and p are unbounded.

No. It is unbounded because its spectrum consists of all j(j+1) with integral j (as can be seen by decomposing the wave functions into spherical harmonics) - other functions of x and p may well be bounded!

And for a scalar particle (which I had meant when I wrote ''bosonic particle'' - sorry for introducing ambiguity), J=L, so that spin need not be discussed to draw the conclusion.

 

[strangerep]

But is your definition above equivalent to Hurkyl's?

He defines the spectrum, I defined the norm.

Actually, Hurkyl defined a norm as well:

A complex number x is in the spectrum of T if and only if (T-x) is not
invertible. The norm is the supremum of the absolute values of the
numbers in the spectrum.

I was really just checking whether both his and your definitions
of norms are essentially the same norm, i.e., just different ways
of defining the same norm (meaning that both norms always give the
same result when applied to any arbitrary element of the algebra).

In C^*-algebras, both statements are theorems.

Which statements precisely? (Sorry, I just need this to be crystal-clear...)

 

[A. Neumaier]

Actually, Hurkyl defined a norm as well:

I was really just checking whether both his and your definitions
of norms are essentially the same norm, i.e., just different ways
of defining the same norm (meaning that both norms always give the
same result when applied to any arbitrary element of the algebra).

Yes. I didn't ''notice'' this although I read it. Yes, I have seen both versions in print, so that, at least, both definitions are provable from the axioms of a C^* algebra. Though it might not be trivial to prove this.

 

[strangerep]

But the spectrum of $$J^2$$ is unbounded, [...]
I.e., 0, 1/2, 1, 3/2, 2, ...

Does this mean that even this elementary case can't be handled
in the context of C* algebras?

One handles unbounded self-adjoint operators A by considering instead
the bounded family of operators exp(isA), s real. [...]

I sense from your diminishing answers in this thread that you don't
have much interest in this subtopic, but I'll try to pursue it a
bit further anyway...

[To other readers: I want to concentrate just on the spectrum arising
from so(3), without the extra complication of position, linear momentum,
Hamiltonian, etc, i.e., without orbital angular momentum.]

I'll start by re-stating part of my earlier post:

Thinking more about (un)boundedness of observables in a strictly
algebraic context (i.e., without Hilbert space), I'm wondering
how one would derive the usual angular momentum spectrum
in this context (if we didn't already know it).

In standard QM with Hilbert space, etc, one takes the usual two
commuting observables $$J^2 , J_z$$ and with the help of the
su(2) commutation relations and the Hermitian inner product, one
derives the spectrum in a page or two.

One does the latter by assuming common eigenstates $$|j,m\rangle$$
of $$J^2 , J_z$$, and using the positive-definite inner
product, as well as properties of the so(3) algebra to show that the
spectrum of $$J^2$$ is given by j = 0, 1/2, 1, 3/2, 2, ...
and if a particular such j is specified, then m runs from
-j to +j in integer amounts.

So I pose this problem: if one didn't know these results, and wanted to
derive them using strictly C*-algebraic means, how does one proceed to
do so? Arnold suggested working with the exponentiated operators, but I
don't have a clue how to get started. Indeed, what does "joint spectrum
of $J^2$ and $J_z$" mean in the algebraic context?
In the Hilbert space context we look for common eigenstates, but what does
one do in the algebraic context? If we introduce representations over the
algebra, that's essentially a retreat to Hilbert-space-like techniques, afaict.

Maybe my problem here is that in the Hilbert space context we're really
attacking a different physical question, i.e., finding the unirreps -
which means finding all subspaces which are invariant under SO(3)
rotations and determining their respective dimensions and other
properties. In contrast, finding all values u,v such that neither
$$(J^2 - u)$$ nor $$(J_z - v)$$ have inverses in the
so(3) universal enveloping algebra seems like a distinctly different
problem. So maybe I should be thinking in terms of automorphisms of
the algebra, finding operators A(u,v) such that

$$e^{i(aJ^2 + bJ_z)} \; A \; e^{-i(aJ^2 + bJ_z)} ~=~ A ~,~~~~~~~~~ (1)$$

where A is a polynomial expression in the so(3) U.E.A. and
a,b are real parameters. Is that how one should start?
Presumably one should reach the outcome that the set of
such polynomials A is characterized by two values that
correspond to the usual j,m of the standard treatment?

[Edit: I just realized (1) can't be the right approach -- because $J^2$
commutes with everything. Therefore the $J^2$ parts of the exponentials
cancel without effect.]

I hope someone can help disentangle my obvious confusion about all this.

 

[Fredrik]

What would be the problem with this approach other than that it mentions Hilbert spaces?

Let's say that we have a C*-algebra $\mathcal A$ that's equal to the subalgebra generated by three of its members, $E_1,E_2,E_3$, which satisfy $[E_i,E_j]=i\varepsilon_{ijk}E_k$. Suppose that $\pi:\mathcal A\rightarrow\mathcal B(\mathcal H)$ is a representation. (B(H)=the set of bounded linear operators on a Hilibert space H). Define [itex]J_i=\pi(E_i)[/tex]. Proceed the way you're familiar with. Now you can conclude a bunch of things about what the irreducible representations must look like.[/itex]

 

[strangerep]

What would be the problem with this approach other than that it mentions Hilbert spaces?

There's no problem if you bring in Hilbert spaces. That's (equivalent to) the standard
approach. But I'd always understood that part of the "benefit" of the algebraic approach
was that one could do without Hilbert spaces (i.e., find spectra, etc). I'm trying to discover
whether that's really correct or not, and precisely why.

 

[DarMM]

There's no problem if you bring in Hilbert spaces. That's (equivalent to) the standard
approach. But I'd always understood that part of the "benefit" of the algebraic approach
was that one could do without Hilbert spaces (i.e., find spectra, etc). I'm trying to discover
whether that's really correct or not, and precisely why.

The C*-algebra approach is not really focused on doing without Hilbert spaces, but really it is about being able to move between them.
For example if I create the abstract C*-algebra of a scalar quantum field theory, states become linear functionals on that algebra. By the GNS theorem, each state is equivalent to a vector in a Hilbert space and the C*-algebra the set of bounded operators on that Hilbert space. The Hilbert space of the state is called its GNS space.
Commonly, many states give the same GNS space, they just turn out to be different vectors in it. This is what is treated as a Hilbert space in the standard approach and is called a folium in the algebraic approach.

The advantage of the algebraic approach is that you can easily use a state from any folium. For example I could use a thermal state for a free QFT, which isn't a Fock state, I simple choose a state from a thermal folium of the right temperature.

So all states, even if they are in different Hilbert spaces, belong to the same space of linear functionals on the algebra. This allows you to analyse difficult problems. For instance free scalar field theory and $$\phi^{4}$$ have the same C*-algebra in two dimensions. However the automorphisms on that algebra are different (different time evolution or different Hamiltonian in conventional language). It turns out that these two automorphisms never have a well-defined representation in one folium simultaneously. Hence the folium the free theory is defined in (the Fock folium) is different from the folium of $$\phi^{4}$$. This is the algebraic version of Haag's theorem.

In summary, you do not dispose of Hilbert space, you just become detached from it and you are able to treat all states in a unified way.