- #1
leo.
- 96
- 5
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?
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?