States in usual QM/QFT and in the algebraic approach

[leo.]

Studying QFT on curved spacetimes I've found the algebraic approach, based on $\ast$-algebras. In that setting, a quantum system has one associated $\ast$-algebra $\mathscr{A}$ generated by its observables.

Here we have the algebraic states. These are defined as linear functionals $\omega : \mathscr{A}\to \mathbb{C}$ such that $\omega(a^\ast a)\geq 0$ and $\omega(1)=1$. The suggested way to view these is that given one observable $a\in \mathscr{A}$ and one state $\omega$ the evaluation $\omega(a)$ is the mean value of $a$ on that state. So at first, one algebraic state encodes all mean values of all observables if the system is physically on that state.

If $\omega$ cannot be written as a non-trivial convex linear combination of other states, it is called pure. Otherwise it is a mixed state.

On the other hand, we have the usual QM approach. A system is described by a Hilbert space $\mathscr{H}$ with algebra of operators $\mathscr{A}(\mathscr{H})$. A state on $\mathscr{H}$ is a density operator $\rho$ acting on it, this encompasses both mixed and pure states. The pure states reduces to state vectors, which are unit vectors on $\mathscr{H}$.

These state vectors have a clear and immediate way write down. From the postulates of quantum mechanics, they encode the probabilities for measurements of the observables. More than that, if we pick one observable $A$ with basis $|a_i\rangle$ assumed non-degenerate for simplicity, we immediately can write any state as $|\psi\rangle = \sum c_i |a_i\rangle$ and those $c_i$ have the immediate meaning as the probability amplitude for the value $a_i$ of the observable.

Another aspect: we don't choose one state vector. What we can is prepare the system on a state $|\psi\rangle$ by measurement and by the state collapse postulate, the state will then evolve in time according to the dynamics. The fact is that at each moment there is one state that describes the system. We don't choose the state, the state is forced on us by physics: it encodes the information about the system.

Now algebraic states have something strange to them. At first, they seem to be the same thing as vector/density operator states, on a more abstract level. But there's more to it and this is the discussion I want to make, because there are mainly two points which make me uncomfortable with this yet.

First point: we have the GNS theorem. The GNS theorem tells us that for any state $\omega$ on $\mathscr{A}$ we can build the so-called GNS triple $(\mathscr{H}_\omega,\pi_\omega,\Omega_\omega)$ where $\mathscr{H}_\omega$ is a Hilbert space, $\pi_\omega$ a $\ast$-representation of $\mathscr{A}$ on $\mathscr{H}_\omega$ and $\Omega_\omega$ a ciclyc vector, i.e, $\pi_\omega(\mathscr{A})\Omega_\omega$ is dense in $\mathscr{H}_\omega$. It turns out that if $a\in \mathscr{A}$ is one observable $$\omega(a)=\langle \Omega_\omega, \pi_\omega(a)\Omega_\omega\rangle$$

More than that, any density operator $\rho$ on $\mathscr{H}_\omega$ yields one algebraic state by means of $\omega_\rho(a)=\operatorname{Tr}(\rho \pi_{\omega}(a))$. All the states which can be realized as density operators on this Hilbert space are said normal with respect to $\omega$ and comprise its folium $\mathfrak{F}(\omega)$. But there are states left out. More than that, Haag says on his book that if we pick one mixed state from $\mathfrak{F}(\omega)$ and generate another GNS triple, the state will be a vector state there instead of a density operator. So I imagine that given $\omega'\in \mathfrak{F}(\omega)$ the representation it generates is different from the one coming from $\omega$.

Thus: one algebraic state yields one whole collection of "usual QM states" and not all algebraic states are there. It even looks like another usual QM state on this collection if viewed as algebraic, would yield a different representation.

Second point: by reading Wald's review paper on QFT on curved spacetime with the algebraic approach, it seems that one chooses the algebraic states. So there are lots of discussion about what states can be chosen, and how can we choose one.

Contrasting to usual QM this is strange. As I said, there all vector states are valid, all density matrices constructed from them are valid as mixtures, and we don't choose one state. The state is what it is: it encodes the information about the system and this is something we don't choose.

Thus: choosing one state seems like taking one spin 1/2 particle and saying: I want the state of the system to be $\frac{1}{\sqrt{2}}(|-\rangle + |+\rangle)$. It simply doesn't make sense. The system could be prepared with one initial state, but we need to work with all the others, as it will evolve.

So how do we reconcille the algebraic states and the Hilbert space states? Why the algebraic states seems to have the only purpose of generating a Hilbert space, instead of encoding the information about the system? How can we deal with this thing of choosing a state, since it doesn't make sense at all, when we contrast with the usual QM picture?

 

[A. Neumaier]

Usually a state is chosen by preparation, and the prepared state is typically mixed except in idealized or very simple experiments. The particular choices made in QFT in curved spacetime are dictated by tractability (generalizing the vacuum state in flat space), and many other states can be derived from these by operating on the preferred class by operators form the algebra.

It usually does not matter whether that state is described in the algebraic sense or in the sense of a density matrix. For standard quantum mechanical problems and in perturbative quantum field theory, they are completely equivalent.

In nonperturbative quantum field theory, the algebraic approach is more general as it allows for the (needed) presence of superselection sectors.

 

[leo.]

Thanks for the response A. Neumaier ! What still somehow bothers me is that on the algebraic setting we pick one distinguished state to generate the whole representation, so that this state becomes distinguished somehow.

I can see one parallel with usual perturbative QFT, because there the vacuum is one distinguished state which generates all the multi-particle momentum/spin eigenstates through creation operators. On the other hand this seems quite strange in usual QM since in general it seems there is no distinguished state there.

If I understood the idea of the algebraic approach, we must pick one state and then work in its GNS representation. But couldn't we just have taken any other state in its folium? More precisely, given the state $\omega$ what makes one use it instead of any other state on $\mathfrak{F}(\omega)$? I could be completely wrong, but it seems there is a difference: Haag mentions that if $\eta\in \mathfrak{F}(\omega)$ is represented by a density matrix on $\mathscr{H}_\omega$ once we build the GNS triple $(\mathscr{H}_\eta,\pi_\eta,\Omega_\eta)$, it will be one usual state vector in $\mathscr{H}_\eta$.

 

[A. Neumaier]

this seems quite strange in usual QM since in general it seems there is no distinguished state there.

Here the distinguished state is the ground state of a reference Hamiltonian. If you take as reference Hamiltonian a sum of independent harmonic oscillators, you get the usual creation/annihilation picture. This is the starting point of perturbation theory for anharmonic oscillators.

Note that a Hilbert space is never given alone but always together with a representation of some key observables.

couldn't we just have taken any other state in its folium?

Pure states in the same folium generate via GNS the same Hilbert space.in the sense of equivalent representations. Mixed states become pure under GNS, making the Hilbert space unnecessarily large.

 

[leo.]

Here the distinguished state is the ground state of a reference Hamiltonian. If you take as reference Hamiltonian a sum of independent harmonic oscillators, you get the usual creation/annihilation picture.

So let's see if I got this straight: the $\ast$-algebra is generated by the observables of the theory, hence we can define the necessary Hamiltonian in terms of them. In one usual one-particle non-relativistic QM problem, we would necessarily have $X,P\in \mathscr{A}$ such that $[X,P]=i\hbar$ and then we can define $$H=\dfrac{P^2}{2m}+V(X)$$ as usual.

Depending on the problem this $H$ itself could serve as a reference Hamiltonian, but we could also take the free part, or any other $H = H_0 + W$ splitting as done in perturbation theory. What would be a good reference Hamiltonian would depend on the actual system?

Then, we assume that the energy is bounded from below, so there should be one ground state. This is certainly one state which we expect to find the system in, so it is one state we would need to work with. Then we build its GNS representation, and work on its folium. Is that, roughly, the idea?

In other words: we have one state which is reasonable that the system be found on it, or that the system could be prepared on it and once we have it, in a sense we have restricted attention to a single folium, and hence to the usual Hilbert space picture?

Furthermore, if we restrict attention to a certain folium, and consider for simplicity the pure states, which one we use to generate the GNS representation doesn't matter, since the representations are unitarily equivalent, so in the end there is nothing "special" with the state we started with, it is just that it was reasonable to be one state attained by the system. So if we have some condition singling out one folium, the algebraic approach and the usual QM approach based on Hilbert spaces become the same thing right?

 

[A. Neumaier]

Depending on the problem this H itself could serve as a reference Hamiltonian

A reference Hamiltonian should be explicitly diagonalizable, otherwise it is useless for perturbation theory.

Given a Hamiltonian of interest, one chooses a nearby explicitly diagonalizable Hamiltonian as reference for perturbation. The naive way is just to delete the obvious interaction terms, but this is suboptimal. Better choices are found by variational perturbation theory; see, e.g., my notes here.

we have one state which is reasonable that the system be found on it, or that the system could be prepared on it and once we have it, in a sense we have restricted attention to a single folium, and hence to the usual Hilbert space picture?

Yes, indeed.

So if we have some condition singling out one folium, the algebraic approach and the usual QM approach based on Hilbert spaces become the same thing right?

Yes.