I tried to google it but couldn't found an intuitive explanation (not existing in Wikipedia). I have also tried to read the Science paper by T. Senthil et al, but couldn't fully understand the derivations in the paper, neither the meaning of the statement:

This emergent global topological conservation law provides precise meaning to the notion of deconfinement.

Hope someone doing research in this area can give an intuitive explanation.


Antiferromagnetic spin systems can have "spinon" excitations. Conceptually, you can make these by starting from a VBS ground state made of singlets, then "breaking" one singlet into a tensor product of two spins, then "moving" the loose spins away from each other by rearranging the singlets, as illustrated in Fig. 4(c) of https://www.nature.com/nature/journal/v464/n7286/abs/nature08917.html. The important thing is that spinons carry spin-1/2, whereas flipping a single spin from up to down (or vice versa) creates a spin-1 "magnon" excitation. You can create one magnon with a local operation, but spinons can only be created in pairs.

Typically, spinons attract each other and are "confined" into bound pairs. (So basically you never see a single isolated spinon at zero temperature.) After you create two spinons, they always stay very near each other. A bound pair of two spin-1/2 spinons is a spin-0 or spin-1 excitation (kind of like a magnon), so it's not as interesting because it's "topologically trivial" because you can create it via local operations (like breaking one singlet, as mentioned above).

But if you tune the Hamiltonian parameters in a 2D antiferromagnet exactly to the critical point between VBS and antiferromagnetic ordering, something very unusual can happen: the attraction between spinons can vanish. You can therefore pull one of them very far away from the other one with little energy cost, even at zero temperature. You can even move one spinon so far away from its "twin" that you can ignore the twin completely, and study the spinon individually. The fundamental "charge" of this spin system due to the continuous global spin symmetry is $\hbar$, corresponding to flipping one spin, but this symmetry can "fractionalize" and you can get an effective local excitation with a "fractional charge" (really spin) of $(1/2) \hbar$.

Technical note: I pretended that particles are "deconfined" when their attractive potential energy function $V(r)$ vanishes. This isn't quite true: the actual definition of confinement is that the attractive potential energy $V(r)$ is bounded above, even as $r \to \infty$. Deconfined particles can still attract each other, but a finite amount of energy must be able to give them an "escape velocity" that can get them out to a separation $r = \infty$. E.g. a Coulomb-like attractive potential is not confining, but a harmonic-oscillator-like potential is: you need more and more energy to separate the spinons farther and farther (as if they were connected by a string with finite string tension), so moving them a macroscopic distance apart would require a huge amount of energy.

  • $\begingroup$ Thank you for your explanation, just one more question: when one spin-singlet pair breaks, do we know what would be the resulting tensor product state of two spins, i.e., if it's $| \uparrow \rangle \otimes | \uparrow \rangle$ or $| \uparrow \rangle \otimes | \downarrow \rangle$ or any other? (or maybe this doesn't really matter?) $\endgroup$ – Chuan Chen Jul 7 '17 at 2:09
  • $\begingroup$ @ChuanChen It depends on the exact nature of the local unitary operator that you apply to the singlet. If we are thinking of the state with one broken singlet as a low-lying excitation of the Hamiltonian, then the exact nature of the broken singlet will depend on the details of the Hamiltonian. (In fact, usually the unitary operator that creates an excitation won't act only on the one singlet. But it will still be "localized" near the singlet in the sense that it will decay exponentially quickly with distance from the singlet.) $\endgroup$ – tparker Jul 7 '17 at 2:25
  • $\begingroup$ @ChuanChen But the details don't matter much - the main point is that each spinon carries spin-1/2. $\endgroup$ – tparker Jul 7 '17 at 2:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.