Mathematical Quantum Field Theory - Fields - Comments

[Urs Schreiber]

derive the usual quantum angular momentum spectrum without using a Hilbert space (directly or indirectly). Cf. Ballentine section 7.1 where he derives that spectrum from little more than the requirement to represent SO(3) unitarily on an abstract Hilbert space.

You'll notice that what this computation (in Ballentine or elsewhere) really uses is the commutation relations of the angular momentum observables together with one ground state they act on. For amplification see for instance also the quantization of the hydrogen atom in the Heisenberg picture, e.g. section 12 of Nanni 15.

In passing from quantum mechanics to quantum field theory, it is important to notice that it is the Heisenberg picture (or interaction picture for perturbation theory) that generalizes well, not the Schrödinger picture (see also Torre-Varadarajan 98 for issues with the Schrödinger picture in QFT). Those quantum field observables $\mathbf{\Psi}(x)$ that we have been discussing elsewhere, they are the hallmark of the Heisenberg picture. And in the Heisenberg picture the Hilbert space of states is secondary. What is primary is the algebra of observables together with one state on them (the vacuum or ground state). Of course, as Arnold recalled above, under good circumstances a Hilbert space of states may be reconstructed from this data, but this is usally an afterthought, not key for the core computations.

If one looks at old articles such as the original Epstein-Glaser article on causal perturbation theory, they always carry around phrases like "assuming that all operators involved can be chosen to have joint dense domain of definition" etc., which comes from insisting that the quantum observables be represented on a Hilbert space. But in fact this is an unnecessary asumption for the results of causal perturbation theory, everything goes through directly with considering just the "abstract" algebra of observables. It's easier, less conceptual baggage.

That is not to say that having a Hilbert space representation is not useful. But it's not conceptually primary for the development of QFT.

 

[vanhees71]

Well, then I've big trouble making any sense of QFT at all since then I don't know, how to define states and how to interpret the S-matrix physically, or how do you define states and probabilities in a theory without an underlying Hilbert space. Of course, the general state is uniquely represented by a Statistical Operator rather than a Hilbert-space vector (or ray), but to make sense of the Statistical Operator you need at least a trace operation. Can this be defined without relying on a Hilbert-space concept?

I always thought the success of perturbative QFT, despite being mathematically quite undefined, is that the physicists intuitively take the limits in their calculations in a correct way, even although they are not able to rigorously define them in a mathematically satisfactory strict sense. Of course, I have in mind things like using a finite quantization volume (with periodic boundary conditions), adiabatic switching a la Gell-Mann and Low and various regularization procedures for divergent perturbative integrals (loops and Feynman diagrams), or appropriate counterterm-subtraction techniques without regularization (a la BPHZ) to make sense of the ill-defined divergent integrals.

I thought, particularly the Epstein-Glaser causal approach, is just another particularly physical way to take these limits in using "smeared" field operators, which makes a lot of physical sense, particularly in view of the Wilson approach to renormalization and his physical interpretation of the renormalization-group equations, and I also thought that this is at least some step towards a mathematically more satisfying foundation in providing the correct rules of mutliplying distribution-like operators or Green's and vertex functions.

 

[Urs Schreiber]

Well, then I've big trouble making any sense of QFT at all since then I don't know, how to define states and how to interpret the S-matrix physically, or how do you define states and probabilities in a theory without an underlying Hilbert space.

This is the old concept of state in AQFT, I recommend the lecture notes Fredenhagen 03 (section 2) for the non-perturbative ($C^\ast$-algebraic) version and Fredenhagen-Rejzner 12 (around def. 2.4) as well as Rejzner 16 (starting with section 2.1.2) for the perturbative QFT version (just dropping the $C^\ast$-condition).

But you are secretly very familiar with this Heisenberg-picture perspective, because this is precisely what traditional perturbative QFT is based on. There we have the $n$-point functions or scattering amplitudes defined as linear functionals on the collection of products of field observables, with respect to a "vacuum state" which is implicitly defined thereby, and no Hilbert space needs to be mentioned to speak about these. In fact for pQFT in curved spacetime no such Hilbert space needs to exist, and still the standard machine applies.

We will come to this in the series in chapters 14. Free quantum fields and 16. Quantum observables.

Regarding Epstein-Glaser 73, what they achieved is to axiomatiize the properties of the perturbative S-matrix, thereby making sense of the would-be path integral, and proving from these (very simple) axioms the construction of pQFT via renormalization by splitting/extension of the distributions given by the Feynman diagrams (here). Hence they managed to make good sense of (Lorentzian) pQFT.

We come to this in the series in chapter 15. Scattering.

 

[strangerep]

You'll notice that what this computation (in Ballentine or elsewhere) really uses is the commutation relations of the angular momentum observables together with one ground state they act on.

Well,... no, I don't notice that. (Do you have a copy of Ballentine in front of you?)

I see (on p160) that he starts by recognizing $J^2$ and $J_z$ as commuting self-adjoint operators. Hence if they act on a Hilbert space, that Hilbert space is spanned by common eigenvectors parameterized by the eigenvalues of these operators ($\beta,m$). Then he uses positivity of the Hilbert space inner product to derive an inequality $\beta \ge m^2$ in eq(7.4). Then he use the ladder operators $J_\pm$ and the inequality (7.4) to show that repeated application of either ladder operator eventually gives 0, and for any given value of $\beta$ there is a maximum value "$j$" of $m$. Thus he deduces the dimension of the Hilbert space for any given value of $\beta$, and also that $\beta = j(j+1)$ -- eq(7.10).

For amplification see for instance also the quantization of the hydrogen atom in the Heisenberg picture, e.g. section 12 of Nanni 15.

Nanni's section 12 (wherein he searches for a suitable factorization of the Hamiltonian) relies on results back in sections 6,7,8 concerning angular momentum properties of wave functions found from the usual Schrodinger equation.

 

[vanhees71]

But you are secretly very familiar with this Heisenberg-picture perspective, because this is precisely what traditional perturbative QFT is based on. There we have the $n$-point functions or scattering amplitudes defined as linear functionals on the collection of products of field observables, with respect to a "vacuum state" which is implicitly defined thereby, and no Hilbert space needs to be mentioned to speak about these. In fact for pQFT in curved spacetime no such Hilbert space needs to exist, and still the standard machine applies.

Well, I guess that refers to free particles in the Heisenberg picture (which is anyway the most natural picture to use; it's only due to overemphasizing wave mechanics rather than the canonical approach, why in QM 1 we usually start with the Schroedinger picture). Then we construct the Fock space as the occupation-number eigenstates (usually in the single-particle momentum representation). Now, isn't his a Hilbert space (at least for a finite quantization volume imposing periodic boundary conditions)? I'll have a look at the mentioned lecture notes by Fredenhagen et al.

 

[Urs Schreiber]

Then he uses positivity of the Hilbert space inner product

Right, one uses the positivity of a given state (one of the axioms on a state on an algebra of observables).

Nanni's section 12 (wherein he searches for a suitable factorization of the Hamiltonian) relies on results back in sections 6,7,8 concerning angular momentum

Section 8 that the discussion relies on is aptly titled "Commutation relations". That means: The algebra structure of the algebra of observables. That's what enters.

You can tell that most constructions in quantum mechanics don't necessarily need the Hilbert space concept by the simple fact that standard textbooks and most students don't even know the functional analysis involved when really considering Hilbert spaces, and not just commutation relations. There are issues of dense sub-domains of unbounded operators, of self-adjoint extensions of symmetric operators on a Hilbert space etc. that need to be discussed if one is really speaking Hilbert spaces. The reason that students and textbooks get away without even mentioning this (such as Ballentine's book) is due to the fact that from just the algebra structure (commutators) of the observables and the basic properties of states (linearity, normalization, positivity) as functions from observables to complex numbers, one can obtain most results.

 

[Urs Schreiber]

Well, I guess that refers to free particles

No, this applies generally, also to interacting theory. In pQFT what changes as one turns on the interaction is that the Wick normal ordred product on observables gets deformed into the "retarded products" (the infinitesimal version of Bogoliubov's formula) and then it's still the vacuum state (or more generally Hadamard state) which serves to send such products of observables to their actual expectation value, which is the corresponding correlator.

in the Heisenberg picture (which is anyway the most natural picture to use; it's only due to overemphasizing wave mechanics rather than the canonical approach, why in QM 1 we usually start with the Schroedinger picture).

Absolutely. And in the Schrödinger picture a Hilbert space of states needs to be assumed from the outset, while in the Heisenberg picture it is an afterthought, if it exsists at all. What matters in the Heisenberg picture is that we know one state, the vacuum state, given as a positive linear non-degenerate function which sends observables to their expectation value. In optimal situation the GNS construction allows to reconstruct a Hilbert space from this, but not generally (notably not in interacting pQFT).

Then we construct the Fock space as the occupation-number eigenstates (usually in the single-particle momentum representation).

And that's where it begins to be misleading. First of all the representation of observables as operators on a Fock space is not necessary for the construction of the Wick algebra and second it does not generally exist for QFT on curved backgrounds. Instead what one needs is the Hadamard propagator which allows to construct the Wick algebra of observables and at the same time defines a single state (positive linear non-degenrate functional on observables) on this.

Now, isn't his a Hilbert space

The Fock space is a Hilbert space, yes. In good cases it happens to exist. In general it does not, and even if it does, it is not actually necessary to do any and all of pQFT.

 

[bhobba]

There are issues of dense sub-domains of unbounded operators, of self-adjoint extensions of symmetric operators on a Hilbert space etc. that need to be dscussed if one is really speaking Hilbert spaces.

Too true.

Whats your view of resolving it with Rigged Hilbert Spaces and the Generalized Spectral Theorem?

Thanks
Bill

 

[Urs Schreiber]

Whats your view of resolving it with Rigged Hilbert Spaces and the Generalized Spectral Theorem?

These are good tools if one wants to be serious about Schrödinger-picture quantum mechanics.

There is good stuff to be found in the Hilbert space Schrödinger picture, if done with due care (as amplified in texts like "Self-adjoint extensions of operators and the teaching of quantum mechanics").

One just has to exercise care with regarding this good QM stuff as indication that the Schrödinger picture is a robust foundation for quantum physics in a generality that includes QFT. It turns out that it is not. On the one hand there are intrinsic problems with applying the Hilbert space Schrödinger picture to QFT even in principle (Torre-Varadarajan 98). On the other hand the Heisenberg/interaction picture without a choice of Hilbert space (just with the option to find one, if possible) works wonders and is in fact, more or less secretly, precisely what everyone uses in practice anyway, even if it superficially seems as if Hilbert spaces are being used.

 

[bhobba]

On the one hand there are intrinsic problems with applying the Hilbert space Schrödinger picture to QFT even in principle (Torre-Varadarajan 98).

Dirac always disliked re-normalization. But I did read somewhere by using the Heisenberg picture he did obtain the same results without it. Evidently they were long but it did work.

Thanks
Bill

 

[vanhees71]

No, this applies generally, also to interacting theory. In pQFT what changes as one turns on the interaction is that the Wick normal ordred product on observables gets deformed into the "retarded products" (the infinitesimal version of Bogoliubov's formula) and then it's still the vacuum state (or more generally Hadamard state) which serves to send such products of observables to their actual expectation value, which is the corresponding correlator.

Well, the physical interpretation of Heisenberg field operators is highly non-trivial, if not even one can say it basically doesn't exist. That's the reason why one finally only discusses S-matrix elements, which rely on asymptotic free in and out states which have a proper particle interpretation. To make sense of transit states is usually not even considered!

Absolutely. And in the Schrödinger picture a Hilbert space of states needs to be assumed from the outset, while in the Heisenberg picture it is an afterthought, if it exsists at all. What matters in the Heisenberg picture is that we know one state, the vacuum state, given as a positive linear non-degenerate function which sends observables to their expectation value. In optimal situation the GNS construction allows to reconstruct a Hilbert space from this, but not generally (notably not in interacting pQFT).

And that's where it begins to be misleading. First of all the representation of observables as operators on a Fock space is not necessary for the construction of the Wick algebra and second it does not generally exist for QFT on curved backgrounds. Instead what one needs is the Hadamard propagator which allows to construct the Wick algebra of observables and at the same time defines a single state (positive linear non-degenrate functional on observables) on this.

So the trick is that you only need the vacuum state and then reconstruct everything through the N-point functions, defined as "vacuum expectation values"? That's very interesting since it sounds intuitively to be sufficient to define S-matrix elements for definite scattering processes since for asymptotic free states you have a particle interpretation.The Fock space is a Hilbert space, yes. In good cases it happens to exist. In general it does not, and even if it does, it is not actually necessary to do any and all of pQFT.[/QUOTE]

 

[Urs Schreiber]

Dirac always disliked re-normalization.

Clearly the old Schwinger-Tomonaga-Feynman-Dyson renormalization is to be disliked. But I think this is unrelated to the issue of the Schrödinger picture that I just mentioned. Essentially nobody ever works or worked in the Schrödinger picture in QFT, it's only that people fall back to it when trying to conceptualize what they are doing

 

[Urs Schreiber]

the physical interpretation of Heisenberg field operators is highly non-trivial

Here I am not fully certain what this is arguing about. If you mean Heisenberg picture as opposed to interaction picture, hence non-perturbative as opposed to perrturbative, then there is a dearth of examples, sure, but no conceptual issue. On the contrary, the Haag-Kastler AQFT axtioms are all about this: axiomatizing the Heisenberg picture observables in QFT, and that's just where the algebraic definition of quantum state that I have been highlighting originates. The point of "perturbative AQFT" is to notice that if one keeps everything about Haag-Kastler except the demand that the star-algebras of observables have $C^\ast$-algebra structure, then one gets a precise conceptualization of traditional perturbative quantum field theory.

So the trick is that you only need the vacuum state and then reconstruct everything through the N-point functions, defined as "vacuum expectation values"? That's very interesting since it sounds intuitively to be sufficient to define S-matrix elements for definite scattering processes since for asymptotic free states you have a particle interpretation.

I would say that's just how pQFT works: We fix a vacuum state, given by a linear map

$$\langle -\rangle \;:\; \mathrm{NiceEnoughObservables} \longrightarrow \mathbb{C}$$

(on curved spacetimes a Hadamard state) and then for incoming field excitations at $x_{in,i}$ in state $a_{in,i}$ and outgoing field excitations at $x_{out,j}$ in state $b_{out,j}$ we form the observable

$$ \left( \mathbf{\Phi}^{a_{out,1}}(x_{out,1}) \cdots \mathbf{\Phi}^{a_{out,n_{out}}}(x_{in,n_{out}}) \right)^\ast \, S_g \, \mathbf{\Phi}^{a_{in,1}}(x_{in,1}) \cdots \mathbf{\Phi}^{a_{in,n_{in}}}(x_{in,n_{in}}) \;\in\; \mathrm{Observables} $$

and then apply the above vacuum state to it to produce a function (in fact a generalized function) in the positions $x$

$$\left\langle \left( \mathbf{\Phi}^{a_{out,1}}(x_{out,1}) \cdots \mathbf{\Phi}^{a_{out,n_{out}}}(x_{in,n_{out}}) \right)^\ast \, S_g \, \mathbf{\Phi}^{a_{in,1}}(x_{in,1}) \cdots \mathbf{\Phi}^{a_{in,n_{in}}}(x_{in,n_{in}}) \right\rangle $$

If we are attached to the idea of Hilbert spaces, then we write this as

$$ \left\langle \mathbf{\Phi}^{a_{out,1}}(x_{out,1}) \cdots \mathbf{\Phi}^{a_{out,n_{out}}}(x_{in,n_{in}}) \right\vert \, S_g \, \left\vert \mathbf{\Phi}^{a_{in,1}}(x_{in,1}) \cdots \mathbf{\Phi}^{a_{in,n_{in}}}(x_{in,n_{in}}) \right\rangle $$

and feel that we have justified the term "matrix" in "S-matrix". But the previous notation is better for reminding us that all we actually need to use is a single state: the vacuum state (generally: Hadamard state).

 

[strangerep]

Section 8 that the discussion relies on is aptly titled "Commutation relations". That means: The algebra structure of the algebra of observables. That's what enters.

Yes, that's what "enters". But section 8 doesn't derive the quantum angular momentum spectrum. The output is important here, not just what "enters". The closest it gets is a passing mention of an eigenvalue equation for $L^2$ in terms of ordinary wave functions.

You can tell that most constructions in quantum mechanics don't necessarily need the Hilbert space concept [...]

"Most" being the key word here. The point of my challenge to try and establish whether derivation of the quantum angular momentum spectrum is one of the cases in QM for which a Hilbert space is essential.

You're not the first person to whom I have offered this challenge. But so far, no one has actually provided a satisfactory response (nor reference) to the point of the challenge, instead evading that point by giving references that don't actually address the point, and (eventually) by unhelpful denigration of other authors. I grow concerned that you seem to be sliding into the latter category.

I also notice that you ignored my question about whether you have a copy of Ballentine there to refer to. I guess your non-response means "no"?

 

[vanhees71]

Here I am not fully certain what this is arguing about. If you mean Heisenberg picture as opposed to interaction picture, hence non-perturbative as opposed to perrturbative, then there is a dearth of examples, sure, but no conceptual issue.

It's about physics. You have no particle interpretation of transient states. For practical purposes, it's a delicate issue. One example is the off-equilibrium production of photons in heavy-ion collisions. There was quite some debate due to these problems. We have investigated it for a simple toy model (photon production due to a time-dependent scalar background field):

F. Michler, H. van Hees, D. D. Dietrich, C. Greiner, Asymptotic description of finite lifetime effects on the photon emission from a quark-gluon plasma
Phys. Rev. D 89, 116018 (2014)
arXiv: 1310.5019 [hep-ph]

F. Michler, H. van Hees, D. D. Dietrich, S. Leupold, C. Greiner, Off-equilibrium photon production during the chiral phase transition
Contribution to the proceedings of the 51st International Winter Meeting on Nuclear Physics, 21-25 January 2013, Bormio (Italy)
PoS Bormio2013, 055 (2013)
arXiv: 1304.4093 [nucl-th]

F. Michler, H. van Hees, D. D. Dietrich, S. Leupold, C. Greiner, Non-equilibrium photon production arising from the chiral mass shift
Ann. Phys. 336, 331 (2013)
arXiv: 1208.6565 [nucl-th]

It's not about the mathematics, and I'm quite sure that the formalism gets the standard perturbation theory right, but it's about the physics interpretation, and there also the axiomatic approach deals with ther properly defined in physically interpretable S-matrix elements, i.e., the transition amplitudes from asymptotic free into asymptotic free out states. Only the asymptotic free states have a clear particle interpretation, not any quantities in "transient states".

 

[vanhees71]

Yes, that's what "enters". But section 8 doesn't derive the quantum angular momentum spectrum. The output is important here, not just what "enters". The closest it gets is a passing mention of an eigenvalue equation for $L^2$ in terms of ordinary wave functions.

"Most" being the key word here. The point of my challenge to try and establish whether derivation of the quantum angular momentum spectrum is one of the cases in QM for which a Hilbert space is essential.

You're not the first person to whom I have offered this challenge. But so far, no one has actually provided a satisfactory response (nor reference) to the point of the challenge, instead evading that point by giving references that don't actually address the point, and (eventually) by unhelpful denigration of other authors. I grow concerned that you seem to be sliding into the latter category.

I also notice that you ignored my question about whether you have a copy of Ballentine there to refer to. I guess your non-response means "no"?

I don't understand what's the issue with angular momentum. It's a nice operator algebra of a compact semisimple Lie algebra and as such doesn't make any trouble at all in the standard Hilbert-space theory. You construct them algebraically via raising- and lowering operators or, even more convenient, using the fact that the 2D harmonic oscillator has SU(2) symmetry and construct everything with annihilation and creation phonon operators.

For orbital angular momentum you also get a very elegant derivation of the spherical harmonics by just writing the algebraic findings in position representation. I think it's very easy to make this also mathematically rigorous in the standard Hilbert-space representation. It's of course the same in relativistic and non-relativistic physics, because SO(3) is a subgroup of both the proper orthochronous Poincare as well as the Galileo group.

In other words: What do you use instead of the standard Hilbert space concept to define states (represented by statistical operators, where pure states are special cases being represented by projection operators) and why should one do so?

 

[A. Neumaier]

in the Heisenberg picture the Hilbert space of states is secondary. What is primary is the algebra of observables together with one state on them (the vacuum or ground state). Of course, as Arnold recalled above, under good circumstances a Hilbert space of states may be reconstructed from this data, but this is usually an afterthought, not key for the core computations.

The GNS construction always associates to the vacuum state a Hilbert space representing the algebra. The only requirement is that the vacuum state is a positive linear functional of the associated *-algebra; this basic property is a necessary physical requirement. Thus the Hilbert space is as relevant to the Heisenberg picture as it is to the Schrödinger picture. The only difference is that the Heisenberg picture is manifestly covariant, while the Schroedinger picture assumes a preferred frame (or foliation) .

Note that perturbative QFT neither constructs the observable algebra nor a positive vacuum state. (Constructing both would imply having constructed a model of the Wightman axioms.) Instead it constructs an approximate algebra in a Fock space corresponding to an asymptotic state space. This asymptotic subspace is unphysical, as it treats both infraparticles such as the electron and confined quarks as asymptotic particles. This is the deeper origin of the infrared problems.
Thus taking the Hilbert space as only an afterthought of QFT is one of the roots of the main unsolved problems in the area.

 

[vanhees71]

Clearly the old Schwinger-Tomonaga-Feynman-Dyson renormalization is to be disliked. But I think this is unrelated to the issue of the Schrödinger picture that I just mentioned. Essentially nobody ever works or worked in the Schrödinger picture in QFT, it's only that people fall back to it when trying to conceptualize what they are doing

Well, there's one book, treating the Schrödinger picture in relativistic QFT (although I never understood, why I should use it for this purpose anyway):

B. Hatfield, Quantum Field Theory of Point Particles and Strings, Addison-Wesley, Reading, Massachusetts, 10 ed., 1992.

 

[A. Neumaier]

Well, the physical interpretation of Heisenberg field operators is highly non-trivial, if not even one can say it basically doesn't exist. That's the reason why one finally only discusses S-matrix elements

How can you say that? In your work you don't only discuss S-matrix elements but all the correlation functions, related by Kadanoff-Baym equations!

 

[A. Neumaier]

The point of "perturbative AQFT" is to notice that if one keeps everything about Haag-Kastler except the demand that the star-algebras of observables have a $C^*$-algebra structure,

And one has to give up the idea that $\hbar$ is a number - instead it is only a formal parameter! And one has to give up the idea that operators act on more than a compact part of space-time - to avoid all the infrared problems.

 

[vanhees71]

How can you say that? In your work you don't only discuss S-matrix elements but all the correlation functions, related by Kadanoff-Baym equations!

Yes, but at the end we use these correlation functions to measure spectra of "particles", and these are defined as asymptotic free space. Of course, we do this in the naive mathematically non-rigorous way, using the usual recipies like adiabatic switching and all that. Our conclusion at the time of writing these articles (see, particularly the Annals of Physics one) that one has to do the good old Gell-Mann-Low switching for both "switching on and off the interactions" to make physical sense of the photon spectra. The considered quantities at "finite times" ("transient states") are off by orders of magnitude and, as far as we could figure out, don't have a clear physical interpretation but are calculational tools only.

 

[vanhees71]

And one has to give up the idea that $\hbar$ is a number - instead it is only a formal parameter! And one has to give up the idea that operators act on more than a compact part of space-time - to avoid all the infrared problems.

Well, $\hbar$ is still a number, which is empirically defined. In fact it's a unit-conversion factor and I'm pretty sure that it will be defined officially next year to update the SI for the 21st century and to take legal effect in 2019.

 

[A. Neumaier]

Well, $\hbar$ is still a number, which is empirically defined. In fact it's a unit-conversion factor and I'm pretty sure that it will be defined officially next year to update the SI for the 21st century and to take legal effect in 2019.

But in the theoretical exposition of Urs Schreiber (and implicitly in perturbative QFT in general) it is a parameter in a power series with zero convergence radius. Thus inserting a finite positive value gives results depending on the order of calculation, and diverging if the order is taken too high. Haag and Kastler, to whom he had referred, were using true operators, not formal power series operators.

The considered quantities at "finite times" ("transient states") are off by orders of magnitude and, as far as we could figure out, don't have a clear physical interpretation but are calculational tools only.

Well, at least in the equilibrium case (and in fact more generally in the hydrodynamic limit), they have a very tangible measurable meaning at finite times.

 

[vanhees71]

In the equilibrium case you also have a kind of "adiabatic switching" by pushing the initial time to $-\infty$ to get rid in a subtle way of the necessity to consider the vertical pieces in the extended Schwinger-Keldysh contour. This is another often discussed subtlety in the real-time community. In my opinion it's completely settled in F. Gelis's papers, where it is shown that in fact you can take the initial time finite (but "earlier" than any time argument in the to be evaluted Green's functions), as to be expected from the fact that one deals with equilibrium which is by definition stationary and thus time-translation invariant. From another point of view, it's only important to keep track of the correct "causal regularization" of the on-shell $\delta$ distributions in the free Schwinger-Keldysh-contour propgators, used in perturbation theory.

F. Gelis, The Effect of the vertical part of the path on the real time Feynman rules in finite temperature field theory, Z. Phys. C, 70 (1996), p. 321–331.
http://dx.doi.org/10.1007/s002880050109

F. Gelis, A new approach for the vertical part of the contour in thermal field theories, Phys. Lett. B, 455 (1999), p. 205–212.
http://dx.doi.org/10.1016/S0370-2693(99)00460-8

 

[A. Neumaier]

Right, one uses the positivity of a given state (one of the axioms on a state on an algebra of observables).

In the quoted Nlab article you write (and implicitly you usie this in the present discussion, too):

the definition [of a state] makes sense generally for plain star-algebras, such as for instance for the formal power series algebras that appear in [URL='https://www.physicsforums.com/insights/paqft-idea-references/']perturbative quantum field theory[/URL]

But the notion of positivity is questionable in algebras over rings of formal power series since the latter have no total linear order. Using a partial order instead provides some notion of positivity but not the physical one.In the physical setting, $\hbar_{phys}-\hbar_{formal}\ge 0$, while in the formal setting, this is not the case.

 

[Urs Schreiber]

But the notion of positivity is questionable in algebras over rings of formal power series since the latter have no total linear order.

It's formal power series algebras equipped with star-algebra structure and positivity is defined in terms of the star algebra structure.

 

[strangerep]

I don't understand what's the issue with angular momentum. It's a nice operator algebra of a compact semisimple Lie algebra and as such doesn't make any trouble at all in the standard Hilbert-space theory.

The issue is that I've heard various purist advocates of the algebraic approach suggesting that Hilbert space is not essential to quantum physics, but only an afterthought. I claim Hilbert space is essential, and no one has yet satisfactorily refuted this by deriving the quantum angular momentum spectrum without reliance on Hilbert space.

You construct them algebraically via raising- and lowering operators or, even more convenient, using the fact that the 2D harmonic oscillator has SU(2) symmetry and construct everything with annihilation and creation phonon operators.

Well, the ladder operators come later in a treatment that relies on nothing more than the algebra and abstract Hilbert space. Cf. Ballentine section 7.1. The extra baggage of a harmonic oscillator is unnecessary for deriving the spectrum.

 

[atyy]

One just has to exercise care with regarding this good QM stuff as indication that the Schrödinger picture is a robust foundation for quantum physics in a generality that includes QFT. It turns out that it is not. On the one hand there are intrinsic problems with applying the Hilbert space Schrödinger picture to QFT even in principle (Torre-Varadarajan 98). On the other hand the Heisenberg/interaction picture without a choice of Hilbert space (just with the option to find one, if possible) works wonders and is in fact, more or less secretly, precisely what everyone uses in practice anyway, even if it superficially seems as if Hilbert spaces are being used.

In the Heisenberg picture, there is an initial state which does not evolve with time. The initial state can be any state in the Hilbert space. How can one do away with this arbitrary initial state?

 

[A. Neumaier]

It's formal power series algebras equipped with star-algebra structure and positivity is defined in terms of the star algebra structure.

But as I had mentioned, the positivity obtained is not the physical one, as for formal power series in a variable $x$, the rule $\xi-x\ge 0$ holds for no real $\xi$ while after picking the physical value of $x$ (in a nonperturbative theory) one has $\xi-x\ge 0$ for every real $\xi$ exceeding the physical value.

 

[vanhees71]

Well, the ladder operators come later in a treatment that relies on nothing more than the algebra and abstract Hilbert space. Cf. Ballentine section 7.1. The extra baggage of a harmonic oscillator is unnecessary for deriving the spectrum.

Well, I also think, the Hilbert-space structure is an essential element in teaching at least QM. Since relativistic QFT in (1+3) dimensions is not rigorously defined, I understand that mathematicians try a different approach to define states. Of course, in QT it is of utmost importance to distinguish between observables and states. It's the very point dinstinguishing QT from classical theories that observables and states are disinct entities of the theory.

Concerning the treatment of angular momentum, I never understood, why one should bother students with the wave-mechanical derivation of the angular-momentum eigenvectors, i.e., an old-fashioned treatment of the spherical harmonics. It's so much more transparent to treat the algebra su(2) and its representations. The only cumbersome point is to show that the special case of orbital angular momentum has no half-integer representations, and for that you need the "harmonic-oscillator approach". See, e.g.,

D. M. Kaplan, F. Y. Wu, On the Eigenvalues of Orbital Angular Momentum, Chin. Jour. Phys. 9, 31 (1971).
http://psroc.phys.ntu.edu.tw/cjp/issues.php?vol=9&num=1
http://psroc.phys.ntu.edu.tw/cjp/issues.php?vol=9&num=1
Of course, with that analysis at hand, you can very easily derive all properties of the spherical harmonics by using the position representation (in spherical coordinates).

 

[samalkhaiat]

The issue is that I've heard various purist advocates of the algebraic approach suggesting that Hilbert space is not essential to quantum physics,

This is not correct. Notions such as completeness (by a norm) and continuity (i.e., boundedness) of any element of an operator algebra need to be defined with respect to some vector space topology. Hermitian adjoint can only be defined on a vector space with scalar product. Moreover, every (abstract) non-commutative C^{*}-algebra can be realized as (i.e., isomorphic to) a norm-closed , *-closed subalgebra of \mathcal{L}(\mathcal{H}), the algebra of bounded operators on some Hilbert space \mathcal{H}. Precisely speaking, for every abstract C^{*}-algebra \mathcal{A}, there exists a Hilbert space \mathcal{H} and injective *-homomorphism \rho : \mathcal{A} \to \mathcal{L}(\mathcal{H}). That is \mathcal{A} \cong \rho (\mathcal{A}) \subset \mathcal{L}(\mathcal{H}), as every *-homomorphism is continuous (i.e., norm-decreasing).

In general, one can say the following about quantization: Given a locally compact group G, its (projective) unitary representation on some Hilbert space \mbox{(p)Urep}_{\mathcal{H}}(G) and the group (Banach) *-algebra \mathcal{A}(G), then you have the following bijective correspondence \mbox{(p)URep}_{\mathcal{H}}(G) \leftrightarrow \mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right) \ , \ \ \ \ (1) where \mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right) is the representation of the (Banach) *-algebra \mathcal{A}(G) on the same Hilbert space \mathcal{H}, i.e., *-homomorphism from \mathcal{A}(G) into the algebra of bounded operators \mathcal{L}(\mathcal{H}) on \mathcal{H}. Similar bijective correspondence exists when \mathcal{A} is a C*-algebra. And both ends of the correspondence lead to quantization. When G = \mathbb{R}^{2n} is the Abelian group of translations on the phase-space S = T^{*}\left(\mathbb{R}^{n}\right) \cong \mathbb{R}^{2n} (or its central extension H^{(2n+1)}, the Weyl-Heisenberg group) then (a) the left-hand-side of the correspondence leads (via the Stone-von Neumann theorem) to the so-called Schrodinger representation on \mathcal{H} = L^{2}(\mathbb{R}^{n}) [Side remark: of course Weyl did all the work, but mathematicians decided (unjustly) to associate Heisenberg’s name with the group H^{2n+1}], while (b) the right-hand-side of the correspondence leads to the Weyl quantization which one can interpret as deformation quantization (in effect, Weyl quantization induces a non-commutative product (star product) on the classical observable algebra, thus deforming the commutative associative algebra of functions C^{\infty}(\mathbb{R}^{2n})).

 

[bhobba]

I have said it before and will say it again - I wish Samalkhaiat had the time to post more. His answers cut straight though.

The c*Algebra approach is, as it must be, equivalent to the normal Hilbert-Space approach - but can be formulated in a way where its association with classical mechanics is clearer:
http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Gleason.pdf

Thanks
Bill

 

[A. Neumaier]

The c*Algebra approach is, as it must be, equivalent to the normal Hilbert-Space approach

Actually it is more general, as the same algebra may have states corresponding to different Hilbert spaces (more precisely, unitarily inequivalent representations).

Thus it is able to account for superselection rules (restrictions of the superposition principle), which have no natural place in a pure Hilbert space approach.

Also it accounts for quantum systems having no pure states (such as those required in interacting relativistic quantum field theory).