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  What really are superselection sectors and what are they used for?

+ 12 like - 0 dislike
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When reading the term superselection sector, I always wrongly thought this must have something to do with supersymmetry ... DON'T laugh at me ... ;-)

But now I have read in this answer, that for example for a free QFT highly excited states , that would need infinite occupation numbers to build them up, and that lie therefore outside the Fock space are said to lie in a (different?) superselection sector. If a state has finite or infinite energy depends on the Hamiltonian, and a finite energy and physically relevant Hilbert space can be obtainend from the inacessible infinite energy states of another Hamiltonian.

This makes me now want to really know what a superselection sector is. What are the key ideas behind the definition of a superselection sector? Are they an underlaying concept to derive quantum field theories with a physical hilbert space that has only finite energy states, or what is their common use in physics?

asked Mar 11, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Most voted comments show all comments
@MarkMitchison Thanks for the references. I'll read them when I can. :) So to construct a silly example consider a Minkowski universe. Would the total momentum $P$ of the universe be considered a superselection variable? (You can boost it away, but we won't.) Since you lack a reference for absolute position, and it is conserved so you can't interfere states with different values of $P$.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Michael Brown
@MichaelBrown as far as I know rigorous results exist for compact transformation groups, I don't know what happens for the non-compact Poincaré group. The question also arises: what does writing down a pure state of the universe even mean? A better example is angular momentum: if I prepare a single spin-1/2 aligned along the x-axis in one of the states $|\pm\rangle = \frac{1}{\sqrt{2}}(|\uparrow_z\rangle \pm |\downarrow_z\rangle)$, you can observe the relative phase by a Stern-Gerlach measurement along the x-axis.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Mark Mitchison
(contd.) But if you don't know how I've defined my Cartesian axes, then you don't know how to align your Stern-Gerlach magnets and you get a random result, consistent with the classical mixture $\frac{1}{2}(|\uparrow_z\rangle\langle\uparrow_z| + |\downarrow_z\rangle\langle\downarrow_z|)$. From your perspective, this is equivalent to a superselection rule that prohibits coherence between different eigenstates of $\sigma^z$, due to the lack of a shared Cartesian reference frame.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Mark Mitchison
I have just noticed I mistakenly down-voted this question when I tried to up-vote it. Sorry about this, Dilaton. I would need you to edit your question to undo the down-vote.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user drake
@drake thanks a lot for this explanation, I'll add the quantum mechanics tag an edit which applies too to my question. Not sure if editing the tags is enough for this purpose.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Dilaton
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@Michael Brown: you figured out the answer when you said "Obviously, if you start in a charge eigenstate you will stay in one because charge is conserved." Because a system in a charge eigenstate never changes its charge, there is no way to tell whether it's in a superposition of states with different charges, so you might as well assume that it's not.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Peter Shor
@MichaelBrown you should check out this paper by Aharonov and Susskind, where they explain how to prepare a superposition of a neutron and a proton. I'm sorry it's behind a paywall :(. The point is that superselection rules are equivalent to lacking a reference frame for the conjugate variable. Charge (number) superselection is equivalent to a lack of a phase reference. Of course, constructing such a reference frame is not necessarily practical. This review has a good list of references.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Mark Mitchison

1 Answer

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A superposition sector is a subspace of the Hilbert space ${\mathcal H}_i$ such that the total Hilbert space of the physical system may be described as the direct sum $$ {\mathcal H} = {\mathcal H}_1 \oplus {\mathcal H}_2 \oplus\cdots \oplus {\mathcal H}_N$$ where $N$ may be finite or infinite such that if the state vector belongs to one of these superselection sectors $$|\psi(t)\rangle\in{\mathcal H}_I,$$ then this property will hold for all times $t$: it is impossible to change the superselection sectors by any local operations or excitations.

An example in the initial comments involved the decomposition of the Hilbert space to superselection sectors ${\mathcal H}_Q$ corresponding to states with different electric charges $Q$. They don't talk to each other. A state with $Q=-7e$ may evolve to states with $Q=-7e$ only. In general, these conservation laws must be generalized to a broader concept, "superselection rules". Each superselection rule may decompose the Hilbert space into finer sectors.

It doesn't mean that one can't write down complex superpositions of states from different sectors. Indeed, the superposition postulate of quantum mechanics guarantees that they're allowed states. In practice, we don't encounter them because the measurement of total $Q$ – the identification of the precise superselection sectors – is something we can always do as parts of our analysis of a system. It means that in practice, we know this information and we may consider $|\psi\rangle$ to be an element of one particular superselection sector. It will stay in the same sector forever.

In quantum field theory and string theory, the term "superselection sector" has still the same general meaning but it is usually used for different parts of the Hilbert space of the theory – that describes the whole spacetime – which can't be reached from each other because one would need an infinite energy to do so, an infinite work to "rebuild" the spacetime. Typically, different superselection sectors are defined by different conditions of spacetime fields at infinity, in the asymptotic region.

For example, the vacuum that looks like $AdS_5\times S^5$ ground state of type IIB string theory is a state in the string theory's Hilbert space. One may add local excitations to it, gravitons, dilatons ;-), and so on, but that will keep us in the same superselection sector. The flat vacuum $M^{11}$ of M-theory is a state in string theory's Hilbert space, too. There are processes and dualities that relate the vacua, and so on. However, it is not possible to rebuild the spacetime of the $AdS$ type to the spacetime of the $M^{11}$ time by any local excitations. So if you live in one of the worlds, you may assume that you will never live in the other.

Different asymptotic values of the dilaton ;-) or any other scalar field (moduli...) or any other field that is meaningful to be given a vev define different superselection sectors. This notion applies to quantum field theories and string theory, too. In particular, when we discuss string theory and its landscape, each element of the landscape (a minimum of the potential in the complicated landscape) defines a background, a vacuum, and the whole (small) Hilbert space including this vacuum state and all the local, finite-energy excitations is a superselection sector of string theory. So using the notorious example, the F-theory flux vacua contain $10^{500}$ superselection sectors of string theory.

In the case of quantum field theory, we usually have a definition of the theory that applies to all superselection sectors. A special feature of string theory is that some of its definitions are only good for one superselection sector or a subset of superselection sectors. This is the statement that is sometimes misleadingly formulated by saying that "string theory isn't background-independent". Physics of string theory is demonstrably background-independent, there is only one string theory and the different backgrounds (and therefore the associated superselection sectors – the empty background with all allowed local, finite-energy excitations upon it) are clearly solutions to the same equations of the whole string theory. We just don't have a definition that would make this feature of string theory manifest and it is not known whether it exists.

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Luboš Motl
answered Mar 11, 2013 by Luboš Motl (10,278 points) [ no revision ]
Thanks a lot Lumo for these very nice explanations, reading this answer saves my (otherwise not so stellar) day :-)

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Dilaton
... and I feel really touched by some of your cool explanations about the superselection sectors in string theory, ha ha ... :-D ;-P

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Dilaton
"It doesn't mean that one can't write down complex superpositions of states from different sectors. Indeed, the superposition postulate of quantum mechanics guarantees that they're allowed states. In practice, we don't encounter them" - That's what I needed. I've always heard of superselection as a rule on what superpositions you are allowed to make, which never sat well with me. This helps. Thanks. :)

This post imported from StackExchange Physics at 2014-03-12 15:21 (UCT), posted by SE-user Michael Brown
It was a pleasure, Michael and Dilaton. We don't encounter the mixture because it's always the easiest measurement to determine in which superselection sector the particle is. So it's an eigenstate of the "which sector" operator, e.g. $Q$, and it stays an eigenstate i.e. in the sector at all times.

This post imported from StackExchange Physics at 2014-03-12 15:22 (UCT), posted by SE-user Luboš Motl

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