Boundedness of quantum observables?

[A. Neumaier]

My complaint was about the fact the axioms for state description (#1 in the set I posted in the other thread) and state dynamics (#4 in the set I posted in the other thread) lead to the rebuttal of a free massive particle moving freely in R^3.

If the conflict is only with your axioms, it is much more likely that your axioms are faulty than that something is now nonexistent that has been considered to be respectable by everyone except you...

 

[dextercioby]

I'm not sure where this is going, but yes you can. For example:
$$\Psi\left( x, t \right) = \int{e^{-p^{2}}e^{-\iota \frac{p^{2}}{2m}t} e^{\iota p x} dp}$$

At any time this function is an element of $$\mathcal{L}^{2}\left(\mathbb{R}^{3}\right)$$ and satisfies Schrödinger's equation.

Alright, so you chose a function from the domain of selfadjointness. Alright, good start.

Question:
Do you have anything against the fact that for time-independent Hamiltonians, the time dependence of the solution of the SE can be factored out, it is nothing but a phase factor and can totally be removed from a discussion on whether the solution to the SE bears physical relevance or not ?

 

[Fra]

I apologize fur just jumping in, I admitt I didn't follw neither this entire discussion nor it's history in other threads.

I prefer to have the foundations free from allusion to measurement. The latter should be a derived many-particle process to be analyzed by the statistical mechanics of the equipment interacting with the observed system.

I personally doesn't understand why a foundations of mesurement theory, should avoid the measurement.

The idea to see the observer and it's environment, from a different perspective (and apply some complex system or stat mech) seems to me a way to not take the meaning of an intrinsic measurement theory seriously, it seems like you seek an "external view" w/o measurement, where the "intrinsic view" with measurement is explained.

The deep issue I have with that general approach is that your "explanation" in terms of this external view, is effectively introducting some sort of superobserver or alternatively, some level of realism that is IMO against the very spirit of seeking a "measurment theory".

I've always seen the conceptual heart of measurement theory in the context of science as the intent that all we shall do is try to describe what outcome we expect of nature in response our measurement. "We" and "our" are a bit unprecise here and the refined version I see is that "the observer can only have an EXPECTATION as to how a fellow subsystem will respond to measurement". Measurement theory merely should try to understand HOW this expectaiton is constructed/computed from the observers state, and how it's REVISED/UPDATED in the event that new unexpected information arrives. Ie. it does NOT explain WHAT new information that arrivees, it only explains the "logic of information update", the logic of making an optimal correction.

It seems to me that if you seek a foundation that doesn't take this process to be importance at the foundational level, then you are probably against what I call the spirit of intent behind measurement theory in the first place?

Ie. you seek som structural realism, or mathematical truth that has removed the observer notion from the fundamental picture?

Am I wrong in my impression? I'm curious to find elaboration as to what you mean by foundations of measurement hteory without allusion to measurement. It's sounds funny to me.

About bounded operators, it seems natural to me that if one takes the measurement and representation serious, then given any observer, the assumption that no observer can hold and store infinite information seems to me to equally suggest that although this bound is relative to the observer, given a definite observer, there is some bound?

Any other notiong of observable, one may question as to wether it's useful for physics?

My objection to QM as it stands, is that it consider the measurement as some silly projection. Ie. it considers only the communication channel. It ignores at the founding level the possibility of evaporating transmitters and saturated receivers, and how this must deform the effective communication. Ie. I think we need to take into acount not only the commmunication but also information processing and representation and coding at the observer. Ie. the "internal structure of the observer".

(Which in the case of QG, translates to matter, which is an open issue)

But to try to remove the observer and measurement, seems to me to be a step in the wrong direction?

/Fredrik

 

[A. Neumaier]

I personally don't understand why a foundations of mesurement theory, should avoid the measurement.

I am talking about the foundations of quantum mechanics, not that of measurement theory.

A foundation of measurement theory must assume quantum mechnaics and show how the complex quantum systems called measurement devices effect what is called a measurement of another quantum system.

But this is a difficult question of statistical mechanics, not a matter of philosophy.

Any other notion of observable, one may question as to wether it's useful for physics?

There is an established, very useful notion of observable as ''densely defined, self-adjoint operator on a Hilbert space''.The widespred use of it is proof of its usefulness, although it is not fully adequate to describe what experimentalist call an observable.

 

[A. Neumaier]

the fact that for time-independent Hamiltonians, the time dependence of the solution of the SE can be factored out, it is nothing but a phase factor and can totally be removed from a discussion on whether the solution to the SE bears physical relevance or not ?

This is not a fact but a severe misunderstanding.

Try to do it for the state mentioned in the post you answered.

 

[Fra]

I am talking about the foundations of quantum mechanics, not that of measurement theory.

A foundation of measurement theory must assume quantum mechnaics and show how the complex quantum systems called measurement devices effect what is called a measurement of another quantum system.

But this is a difficult question of statistical mechanics, not a matter of philosophy.

Ah ok. Perhaps I missed the context; I just jumped in.

So you are talking about finding a rigorous mathematical foundation of QM as it stands, right? Not talking about how to solve open issues that relates to unification and QG? (which may need REVISION of QM)

If so, I see your stance.

Yes that's a worthy goal on it's own. But it sounds like a pure mathematical project.

I know some mathematicians that work on fidning rigorous formalizations of the often "sloppy" mathematics that physicists has come up with, but without changing the physics. From a mathematicians views is often that physicists derivations is more like an informal argument than formal proof.

/Fredrik

 

[A. Neumaier]

So you are talking about finding a rigorous foundation of QM as it stands, right? Not talking about how to solve open issues that relates to unification and QG? (which may need REVISION of QM)

Yes. QM as far as it is based on a Hilbert space view.

 

[Fra]

Yes. QM as far as it is based on a Hilbert space view.

Ok, then I see. My confusion of context.

/Fredrik

 

[dextercioby]

This is not a fact but a severe misunderstanding.

Try to do it for the state mentioned in the post you answered.

Ok, then I'll present with my reasoning and kindly ask you to spot my flaws of any kind:

<Assume the the system is described by a function $\Psi (x,t)$ on a TVS call it $\mathcal{V}$.

We would like to solve the following PDE to find the general solution $\Psi (x,t)$ subect to the condition

$$\int \limits_{\mathbb{R}}\Psi^{*}(x,t) \Psi (x,t) \, dx \, =1 \, , \, \forall t\in\mathbb{R}$$ (1).

The PDE sounds like

$$\frac{\partial\Psi (x,t)}{\partial t} = \frac{1}{i\hbar} \frac{(-i\hbar)^2}{2m} \frac{\partial^2 \Psi (x,t)}{\partial x^2}$$.

subject to an initial condition $\Psi (x,t=0) = \psi (x)$.

The key assumption we're making is that $\Psi (x,t) = \psi (x) V(t)$ which when plugged in the PDE written above gives

$$\psi (x) \frac{d V(t)}{dt} = \frac{i\hbar}{2m} V(t) \frac{d^2 \psi (x)}{dx^2}$$

A manipulation of the above equation leads to

$$\frac{2m}{i\hbar} \frac{1}{V(t)} \frac{d V(t)}{dt} = \frac{1}{\psi (x)} \frac{d^2 \psi (x)}{dx^2} = K$$

K is a real constant with dimension of 1/area.

So

$$V(t) = \exp\left(\frac{i\hbar K}{2m}t\right)$$

, a phase factor (I assume the condition V(t=0)=1, so the integration constant is 0) and the other ODE is

$$\frac{d^2 \psi (x)}{dx^2} = K \psi (x)$$

which, compared to the spectral equation for the free Hamiltonian

$$-\frac{\hbar^2}{2m} \frac{d^2 \psi (x)}{dx^2} = E \psi (x)$$

gives

$$K = -\frac{2mE}{\hbar^2} \leq 0$$

for $E\geq 0$

So one finds complex exponential solutions for the x variable as well which are not normalizable as per (1).

What am I doing wrong?

 

[A. Neumaier]

The key assumption we're making is that $\Psi (x,t) = \psi (x) T(t)$

$$V(t) = \exp\left(\frac{i\hbar}{2m}t\right)$$, a phase factor. [/tex]

$$-\frac{\hbar^2}{2m} \frac{d^2 \psi (x)}{dx^2} = E \psi (t)$$

for $E\geq 0$

What am I doing wrong?

You have been sloppy in each of the four lines quoted above.

In the first line, you should have V(t) in place of T(t), in view of what follows.

In the second line, you lost both an integration constant and the constant K

In the third line, you write psi(t) in place of psi(x).

In the fourth line, you introduce an extra assumption without saying so.

After correcting these items (it would have paid off to go through all arguments after typing them, and have them checked for correctness), you get a correct derivation of _some_ solutions, namely all those that satisfy the assumption introduced in the first line above.

But since you didn't _prove_ this assumptions, your derivation tells nothing at all about solutions that do not satisfy it. There are lots of them, since by the superposition principle, any linear combination of the solutions you constructed is again a solution. You can easily convince yourself that these will usually not have the separable form you assumed.

 

[dextercioby]

You have been sloppy in each of the four lines quoted above.

I apologize for being sloppy. It's been corrected now.

In the fourth line, you introduce an extra assumption without saying so.[...] you get a correct derivation of _some_ solutions, namely all those that satisfy the assumption introduced in the first line above.

OK.

But since you didn't _prove_ this assumptions, your derivation tells nothing at all about solutions that do not satisfy it.

Is separability the assumption you need me to prove ? Or the E>0 one ?

There are lots of them, since by the superposition principle, any linear combination of the solutions you constructed is again a solution. You can easily convince yourself that these will usually not have the separable form you assumed.

I have obtained a basis on the space of all solutions. I claim that if separability of time dependence is assumed and E>0 (which can be proven), 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.

 

[A. Neumaier]

Is separability the assumption you need me to prove ? Or the E>0 one ?

You can prove neither, since they don't follow from the time-dependent Schroedinger equation. The fact that you had to assume the former but thought this could be done without losing any solution is your real mistake.

I have obtained a basis on the space of all solutions.

Well, how did you convince yourself of this fact? Please give the argument, so that I can correct you.

I claim that if separability of time dependence is assumed and E>0 (which can be proven), 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.

Even that is, with the arguments given so far, only a claim and not something proved.

But the space of the solutions is the space of all solutions, not only the separable ones
that your last statement claimed to classify. Thus even if this claim had been proved, you are still far away from having a proof that you know all solutions (and that none of them is normalizable - which you claimed but which the example of DarMM showed to be faulty).

 

[dextercioby]

You can prove neither, since they don't follow from the time-dependent Schroedinger equation. The fact that you had to assume the former but thought this could be done without losing any solution is your real mistake.

So you're challanging my separation of variables t and x. Why would this particular case be any different than the tons of cases in which PDE's in mathematical physics are solved
through this method ? In other words, help me understand why this 2 variable separation is problematic, and the one in the heat conduction equation for the 1d case is not.

Can you provide me with a solution of the SE for 1D case in which t and x are not separated ? (DarMM's solution is separable and accepting it would have to find other flaws in my argument than separability of variables).

Thank you

Daniel

p.s. if separability is assumed, E>0 follows.

 

[A. Neumaier]

So you're challenging my separation of variables t and x. Why would this particular case be any different than the tons of cases in which PDE's in mathematical physics are solved
through this method ? In other words, help me understand why this 2 variable separation is problematic, and the one in the heat conduction equation for the 1d case is not.

Heat conduction is not different. You only get separable solutions from the separable ansatz, and there are lots of nonseparable solutions (namely most linear combinations of separable ones).

What is missing in your argument is the discussion why you should get _all_ solutions as linear combinations of separable ones. So please try to prove this instead of just copying a textbook template without thinking yourself. If you fail, produce your attempt, so that I can see what you are having difficulties with.

Can you provide me with a solution of the SE for 1D case in which t and x are not separated ? (DarMM's solution is separable and accepting it would have to find other flaws in my argument than separability of variables).

DarMM's solution is normalizable but not separable. The time-dependent exponential depends also on p, so it can't be factored out.

if separability is assumed, E>0 follows.

How? It is impossible to correct you unless you explain the reasons why you think your claims are true.

 

[dextercioby]

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 ?

 

[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:

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

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:

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

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,

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