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Consider a quantum system described by the $\ast$-algebra $\mathscr{A}$. Let $\omega : \mathscr{A}\to \mathbb{C}$ be one state. By the discussion on this question a choice of algebraic state is a choice of $\ast$-representation of $\mathscr{A}$ on a Hilbert space by means of the GNS construction which generates a triple $(\mathscr{H}_\omega,\pi_\omega,\Omega_\omega)$.

It is a fact that all density matrices on $\mathscr{H}_\omega$ give rise to algebraic states. We then define $\mathfrak{F}(\omega)$ the folium of $\omega$ to be the set of all algebraic states which can be represneted as density matrices on $\mathscr{H}_\omega$. A state is then said normal with respect to $\omega$ if it belongs to $\mathfrak{F}(\omega)$.

So when we perform the GNS construction and work with the Hilbert space $\mathscr{H}_\omega$ we look just to the normal states with respect to $\omega$ and forget for a moment the others.

This invites two natural questions:

  1. It seems that in contrast with usual QM, on the algebraic approach we choose one state by mathematical reasons.

    In QM we pick one observable, it has a basis. We write one state on this basis, and it has the meaning of a probability distribution for the measurements of the observable.

    We prepare the system on a state by measurement. So there, in a sense, we don't choose the state, by the way the system was prepared and by the physical meaning of the eigenstates of observables we know the initial state.

    This is in contrast with the algebraic approach, where we choose one state to generate the GNS triple. I mean, there doesn't seem to be a physical motivation to choose one instead of another, and this seems highly different than QM where the act of preparing the system forces one state initial state on us.

    So why on the algebraic approach we seem to choose one state by mathematical reasons, while in QM it seems that the state actually comes from the physics?

    How the actual physics of the problem dictates the state in the algebraic approach, as in usual QM?

  2. More than that, given one $\omega$ it generates the GNS triple $(\mathscr{H}_\omega,\pi_\omega,\Omega_\omega)$. It has a whole collection of other states $\mathfrak{F}(\omega)$ on the same Hilbert space. So why would one choose $\omega$ instad of any other $\omega'\in \mathfrak{F}(\omega)$? I can't see a reason connected to physics here.

In both questions the issue is that I'm failing to see the connection between the math and the physics.

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    $\begingroup$ I totally agree with you that the physics is hard to see in this highly abstract math, and I would also see this as a weakness of the algebraic approach. For an example where you see the choice of state clearly, consider the vacuum (="Dirac sea") in a QFT. This is a state of reference called "no particles", but depending on external fields, this choice is not unique. But what you see here is that the problem how to choose the state is not only present in the algebraic approach, also in the usual one you have to ask: "So what is now the right vacuum?" $\endgroup$ – Luke Feb 12 '18 at 14:56
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The GNS construction generates a representation of the $C^*$ algebra over a Hilbert space. We know that if we start from a pure state, we reach an irreducible representation, which describes an elementary quantum system. If we start from another state, the representation describes another quantum system. Some of these representations are unitarily equivalent, and we view them as equivalent in the physical sense. For example, two spin-$\frac{1}{2}$ system corresponding to rotated axes belong to the spin-$\frac{1}{2}$ type of systems.

It is true that when the operator algebra is the Heisenberg-Weyl algebra, then all the GNS representations are unitarily equivalent (The Stone-von Neumann theorem), but this is really an exception. In general we have many inequivalent representations, each parametrized by the state we started from. This can happen even in quantum mechanics when our algebra is $\mathrm{Mat}(N)$.

About the physical significance of the inequivalent representations

I know of two approaches which can give a physical meaning to the above inequivalent representations

  1. Rieffel's approach: As stated above, the different representations represent different quantum systems. Rieffel investigated the possibility that these representations arise from the quantization of a classical phase space. We know that when we quantize a system, we can get plenty of inequivalent quantum systems. Thus in this approach the inequivalent representations are actually inequivalent quantizations of some system defined on a classical phase space (generally a Poison manifold).

  2. A second way that we may interpret the inequivalent representations, is that we can imagine the space of (distinct) functionals in the GNS construction (or a subspace of which) as a classical (phase) space by itself. In this way we reach a parametrized quantum system. This approach is particularly transparent when our representations are coherent space representation, then it is actually parametrized by a phase space. Parametrized quantum systems have many modern applications from quantum control, to topological insulators.

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  • $\begingroup$ there's one point that is confusing me. I thought one $\ast$-algebra $\mathscr{A}$ was supposed to describe one specific system (like one harmonic oscillator, one free KG field, etc). By your answer it seems that the repsentations describe specific systems, and that given one $\ast$-algebra we can have representations describing different systems. So isn't $\mathscr{A}$ in the end meant to represent one single quantum system? Why it describes more than one? $\endgroup$ – user1620696 Feb 12 '18 at 20:31
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    $\begingroup$ Take for example the (universal enveloping) algebra of spin ($\frak{su}(2)$). Clever choices of the GNS states can generate all of its representations corresponding to spin $S=0, \frac{1}{2}, 1, …$ etc. We can call all these representations spinning systems; while each representation is an elementary spinning system as it is irreducible. $\endgroup$ – David Bar Moshe Feb 13 '18 at 10:47
  • $\begingroup$ Please give some references showing the use of parametrized quantum systems featuring inequivalent representations! $\endgroup$ – Arnold Neumaier Mar 5 '18 at 18:02
  • $\begingroup$ @ArnoldNeumaier The most known example is the case of spontaneous symmetry breaking. In this case, the inequivalent vacua can be labeled by Goldstone field configurations. Please see for example arxiv.org/abs/1111.3228v2 section 4. Another example (which is not completely orthogonal) is a Bogoliubov transformation with non-Hilbert-Schmidt off diagonal elements, please see www1.maths.leeds.ac.uk/~siru/papers/p10.pdf . In this case, the "vacuum manifold" can be parametrized by an infinite dimensional symmetric space. $\endgroup$ – David Bar Moshe Mar 6 '18 at 11:32
  • $\begingroup$ cont. A special case of the above is for example a complex scalar field with a modified complex structure connected to the original one by a Bogoliubov transformation. This is also related to fermions external field problems. Here the interaction Hamiltonian is required to have Hilbert-Schmidt off diagonal elements in order for interaction Hamiltonian be implementable in second quantization over the free vacuum www1.maths.leeds.ac.uk/~siru/papers/p8.pdf. $\endgroup$ – David Bar Moshe Mar 6 '18 at 11:33
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The usual physical choice for the preferred state in relativistic quantum mechanics, at least in flat spacetime, is the vacuum (or ground state). The idea is that such state has some very special physical properties that make it uniquely defined for a given theory. Of course, this is not always the case. First of all, it should be a state that is pure and invariant with respect to the action on the algebra of canonical (anti)commutation relations of the orthochronous, proper Poincaré group. In addition, it should be a ground state with respect to the dynamical system given by time translations (i.e. it has to be a ground state for the Hamiltonian). To "find" the vacuum state for a given (interacting) quantum field theory is very challenging, and there are very few results of that type.

In curved spacetimes there is usually no notion of a vacuum, and other type of distinguished states are used, such as the (quasifree) Hadamard states, also defined using special physical properties (I not at all an expert on quantum fields on curved spacetimes, so I will not comment further on this).

Let me elaborate a bit more on the idea of vacua in the algebraic approach, and how distinguishing among them yield inequivalent and physically distinct representations.

Given a classical symplectic space $(X,\sigma)$ (of test functions for the classical fields that are seen as distributions), we can apply to it the "quantization" (more precisely, a suitably defined quantization functor $\mathbb{W}_h$) to obtain the algebra of canonical commutation relations $\mathbb{W}_h(X,\sigma)$. Suppose that, as it is usually the case, $(X,\sigma)$ carry a symplectic representation of the Poincaré group, or of any other group $G$ (let us call such representation $(s_g)_{g\in G}$); then $\bigl(\mathbb{W}_h(s_g)\bigr)_{g\in G}$ is a representation of the Poincaré group on the quantum observables $\mathbb{W}_h(X,\sigma)$, mapping quantum fields in quantum fields (as required by the usual axioms of quantum field theory, e.g. in the Gårding-Wightman formulation).

Now, it is clear that both free and interacting theories of the same kind (e.g. theories describing scalar, non-charged, fields) have the same space of test functions, and therefore the same abstract algebra of canonical commutation relations. How can we therefore distinguish between a free and an interacting theory?

By choosing the appropriate vacuum state.

Any vacuum state is $G$-invariant, and Haag's theorem tells us that given two $G$-invariant states (there is an additional technical condition that is usually satisfied by physical states), either they are equal or disjoint (in the sense that one is not normal with respect to the other). Therefore, to different vacua (actually $G$-invariant states) there correspond inequivalent representations of the canonical commutation relations, corresponding in turn to distinct physical systems. This is on one hand comforting, and on the other a bit troublesome, for it implies that the free and interacting systems corresponding to the same kind of field (e.g. a scalar field) have to be described by inequivalent representations of the same canonical commutation relations.

The fact is that while the representations of free theories are very well understood from the mathematical standpoint, and are the so-called Fock representations, the representations of interacting theories (in other words, the interacting $G$-invariant (vacuum) states) are unknown for all "interesting" relativistic interacting theories in $3+1$ dimensions. This is a notoriously very difficult open problem in mathematical physics (there is a Clay prize of $\$ 1,000,000$ for solving this problem for a Yang-Mills theory), and cannot be investigated (as far as I know) using the algebraic approach, for the latter is too general. The few cases (in dimension $1+1$ and $2+1$) where the interacting vacuum is known, have been studied long time ago (sixties and seventies) by famous mathematical physicists such as Glimm, Guerra, Jaffe, Rosen and others (and more recently by the Fields medal Martin Hairer) using operator and Feynman-integral techniques. Their results can of course be translated in the algebraic formalism, and the so defined interacting vacuum can be proved to be disjoint from the free Fock vacuum, but the former cannot be characterized directly by algebraic considerations.

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