Are fermions intrinsically non-local?

Question

Background:

When one studies quantum mechanics of more than one particle, one learns that all fundamental particles can be classified as either bosonic or fermionic. Fermions have a spinor structure, and pick up a phase of -1 when rotated by 2$\pi$ radians, while the (gauge) bosons have a vector nature and pick up a phase of 1. Mathematically, one says that spinors are a representation of the group SU(2), which is a double cover of the rotation group SO(3).

The standard way it is shown to undergraduates that something can rotate by 2$\pi$ and not end up in its initial configuration is the Dirac Belt trick, which essentially involves rotating one part of a belt while another point is fixed. Or, similarly, one can rotate a plate in one's hand while keeping one's body fixed (the Feynman Plate Trick).

One striking aspect of this demonstration, unlike engineering a physical representation of a system with 2$\pi$, $\pi$, or lower rotation symmetry, is that it depends essentially on the object's relation to its environment, in the sense that the connections to the environment (such as the body of the belt) are what prevent the system from having 2$\pi$ symmetry.

One known way of relating a system of (free) fermions to a system of (interacting) bosons is the Jordan-Wigner transformation. This transformation is, famously, highly nonlocal, in that a fermionic operator at one point is created by a long chain of the bosonic operators. This and related ideas have inspired Prof. Xiao-Gang Wen, in his textbook on many-body theory, to claim that fermions should be considered non-local excitations. However, I'm not aware of anyone else emphasizing this viewpoint.

Okay, with all that background established I have two questions:

1. Are fermions non-local objects, in a sense in which gauge bosons are not?

2. Is the fact that our physical representations of SU(2) symmetry, such as the Belt Trick, require some sort of connection to a fixed background a reflection of this fundamental non-locality, or a mere coincidence? Can this relationship be made more precise?

Edit: Thanks to all who contributed. It does not seem there is much consensus on what relation, if any, there is between SU(2) symmetry and non-local properties. I will give the bounty to the top-voted answer but consider the question not fully resolved and welcome any additional answers.

Answer 1

So what people mean by 'non-local' varies from context to context and person to person. Wen has a very particular meaning to this.

1) In fermionization in $D=1+1$ the Jordan-Wigner fermions are, in the bosonic language, operators supported over many sites. The emergent (mutual)-fermions in the toric code are also supported at the ends of strings.

2) Fermionic parity, $(-1)^F$, is a symmetry of all fermionic systems and cannot be broken explicitly by any local term.

However strings end at points and the Hamiltonians of fermionic systems are local. The Hilbert space (of emergent fermions) is also very nice and decomposable into a local tensor product.

So in this sense they are local in a way that gauge bosons are not! The (physical/gauge invariant) Hilbert space of a gauge theory does not factorize as a nice tensor product and gauge invariant operators are either Wilson/t'Hooft loops (non-local in definition) or when you have charged matter you need to include the configuration of the gauge field around it. This poses a lot of issues in making the entanglement entropy of a gauge theory well defined.

Wen and others then have this idea that maybe local bosons are fundamental because of their simplicity. Certainly gauge theories and fermions can emerge from them.

Answer 2

1. Are fermions non-local objects, in a sense in which gauge bosons are not?

As far as I understand, the answer is definitely NO. Fermionic particles are local objects as bosonic ones. Based merely on the non-local form of the bosonized Jordan-Wigner fermions, one cannot conclude that fermions are non-local. Jordan-Wigner transformation, like any other transformation, is merely an auxiliary tool which leads to an appropriate description of physical properties, esp. in 1-dimensional quantum systems. Notice that this is not the only way; one can obtain the physical properties by any other proper calculational method.

One can compare this issue with the different representations for spin density (a bosonic operator), as there are several representations for the spin density operator; namely, Schwinger boson, Holstein-Primakoff (boson), Dyson-Maleev (boson), Abrikosov pseudo-fermion or Majorana fermion representations. These are all exact representations and each one is useful for a certain class of problems – i.e., it reproduces the correct physical behaviour in a certain regime. So, one should not make a conclusion merely based on the representation used for a physical object/quantity, because the representation is merely a mathematical framework; only the observable physical properties which are obtained from that representation count as descriptions of the ‘reality’ or the ‘fact’.

In the case of Jordan-Wigner transformation applied to 1-dimensional interacting spin chains to map them to a fermionic model, there is no evidence in the resulting correlation functions for ‘non-locality’ of the fermions. Jordan-Wigner transformation merely provides a good description for low energy scales. In this case, if one takes literally the infinite bosonic-chain representation of a fermion (obtained after proper ‘bosonization’ of fermions), then one should be able to easily observe such a long chain (like a large heavy molecule)! Hence, such a literal interpretation is certainly absurd.

2. Is the fact that our physical representations of SU(2) symmetry, such as the Belt Trick, require some sort of connection to a fixed background a reflection of this fundamental non-locality, or a mere coincidence?

I believe such “physical representations” are merely cartoon pictures to make the problem intuitively clear and emphasize (and exaggerate) the differences between rotation of spin ($SU(2)$) and rotation in space ($SO(3)$). Beyond this intuitive illustration, the only guide is proper mathematical formalism.

Finally, remember that to have fermions with half-integer spin, there is no need to have any interaction or to use a certain representation. They appear naturally, for instance, as solutions to the non-interacting Dirac equation. The fundamental reason is the need to have a proper Lorentz-invariant quantum theory. Jordan-Wigner transformation is an elaborate tool applicable to interacting quantum systems in 1-dimension; an efficient representation for some physical systems at a certain (low-energy) regime. If the resulting fermions (or the elementary excitations) were ‘really’ non-local, this would be observed regardless of the representation used.

For a pedagogic description of bosonization and Jordan-Wigner transformation, see E. Miranda, “Introduction to bosonization”, Braz. J. Phys. 33:1 (2003) < http://dx.doi.org/10.1590/S0103-97332003000100002 >.

Comment Added: In a review of the question and comments, I realized that there is a need for clarification. Actually, Jordan-Wigner transformation maps spin operators to a fermion ‘chain’. Thereafter, the resulting fermionic Hamiltonian can be bosonized via a highly non-local transformation, the Mattis-Mandelstam formula. Consult the reference above.

Answer 3

Matthew Fisher once told me that he considers fermions intrinsically nonlocal for a very simple reason: a many-body bosonic creation operator can be written as a tensor product of single-body operators which only act nontrivially on the site where a particle is being created, i.e. $a_n^\dagger = I \otimes I \dots I \otimes a^\dagger \otimes I \dots$, where the $a^\dagger$ appears in the $n$th spot. But there's no such construction for a many-body fermionic creation operator, because they need to anticommute on different sites, which you can't get by tensoring together identities. So to figure out the action of a fermionic creation operator on a state, you need to consider the entire state (in order to figure out how many previous creation operators you need to anticommute it through), not just the single site at which the particle is being created. No need to consider spin, you can just think about the commutation relations.

Other people think that this is just a formal nonlocality of mathematical fermionic ladder operators, not actual physical fermions. It's largely a philosophical difference.

Answer 4

Are fermions intrinsically non-local?

Yes, definitely. It's quantum field theory, not quantum point-particle theory. An electron's field is what it is. And that field doesn't have a surface. From a great distance it will be swamped and undetectable, but there is no defineable place where that field stops.

When one studies quantum mechanics of more than one particle, one learns that all fundamental particles can be classified as either bosonic or fermionic. Fermions have a spinor structure, and pick up a phase of -1 when rotated by 2π radians, while the (gauge) bosons have a vector nature and pick up a phase of 1. Mathematically, one says that spinors are a representation of the group SU(2), which is a double cover of the rotation group SO(3).

Yes. But don't forget that you can make electrons and positrons out of photons in gamma-gamma pair production, so they aren't fundamental like energy is fundamental. And don't forget that photons have a definite E=hf wave nature, and take many-paths. For an analogy think about a seismic wave travelling from A to B across a flat gedanken plain. It isn't just the houses sitting on the AB line that shake. Houses 10km away from that line shake too. And 100km away, though much less. In this respect the seismic wave takes many paths. It isn't local. No wave is.

The standard way it is shown to undergraduates that something can rotate by 2π and not end up in its initial configuration is the Dirac Belt trick, which essentially involves rotating one part of a belt while another point is fixed. Or, similarly, one can rotate a plate in one's hand while keeping one's body fixed (the Feynman Plate Trick).

Yes, see this answer re a previous question on spinors. And note that the Einstein-de Haas effect demonstrates that "spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Also note that atomic orbitals electrons "exist as standing waves". And that the Dirac equation is a wave equation.

One striking aspect of this demonstration, unlike engineering a physical representation of a system with 2π , π , or lower rotation symmetry, is that it depends essentially on the object's relation to its environment, in the sense that the connections to the environment (such as the body of the belt) are what prevent the system from having 2π symmetry.

Sorry, I'm not clear what you mean here. Surely all we're talking about is a wave going 360° round a major axis whilst twisting 180° on an orthogonal minor axis, then round again for a total of 720° and 360°.

One known way of relating a system of (free) fermions to a system of (interacting) bosons is the Jordan-Wigner transformation. This transformation is, famously, highly nonlocal, in that a fermionic operator at one point is created by a long chain of the bosonic operators. This and related ideas have inspired Prof. Xiao-Gang Wen, in his textbook on many-body theory, to claim that fermions should be considered non-local excitations. However, I'm not aware of anyone else emphasizing this viewpoint.

More's the pity. IMHO point-particle thinking has been bad for physics.

Okay, with all that background established I have two questions: 1.Are fermions non-local objects, in a sense in which gauge bosons are not?

Yes and no. Fermions are non-local, and bosons are too.

2.Is the fact that our physical representations of SU(2) symmetry, such as the Belt Trick, require some sort of connection to a fixed background a reflection of this fundamental non-locality, or a mere coincidence? Can this relationship be made more precise?

Again apologies, I'm not sure what you mean. But if I can offer something that may be useful: IMHO the belt trick is merely how pair production "winds" waves into standing waves. Look up displacement current, electromagnetic geometry, and Wheeler's geons. Note though that the electromagnetic wave is not confined by gravitational attraction, but instead by its own displacement current. A wave ends up travelling through itself in a twisted double loop, forever displacing its own path into a closed path. Then instead of a wave propagating linearly at c, we've got a standing wave. Standing wave, standing field. This field has a centre, but there's nothing at that centre. Just as there's nothing at the centre of a hurricane. Scattering experiments have been used to claim that the electron is very small. But IMHO that's like hanging out of a helicopter probing a whirlpool with a bargepole. Then saying "I can't feel the billiard ball, it must be really really small".

Answer 5

Okay, I think I have a semi-convincing picture of this in my head. Both of the other answers contain at least part of the story I wanted; I will put the whole thing here in hopes of feedback and that it is useful to someone else.

As SM Kravec points out, fermionic parity is a non-local symmetry of a fermionic system. This suggests, as various people have proposed over the years, a topological nature of fermionic particles. To the ideas along these lines John Duffield has mentioned I would add, as far as I understand it, the new EP=EPR conjecture.

Comparing this with the physical demonstration of the Dirac belt trick, then, the obvious idea is that a minimal topological model for a fermion particle-antiparticle pair is to have them joined as two ends of a belt, whose windings tell us something about the phase of the total system. As in a real belt, rotating one of the ends (i.e., the particle or antiparticle by itself) has a 4$\pi$ symmetry. So the SU(2) behavior does indeed come from a type of non-locality in this view, specifically the non-locality associated with fermionic parity. Of course, a rotation of both particle and anti-particle by 2$\pi$, which in some sense is what should be considered a true rotation of the system, does bring you back to where you started.

This generalizes to several pairs of particles. At that point one must remember that, in 3D, exchange of two identical particles is topologically identical to rotation of one particle by 2$\pi$, so as a result exchange of two fermions must also give a -1 phase. Again, in some sense "full" exchange of the system would be switching two particles and also their corresponding antiparticles at the same time, which leaves the system invariant. Also, instead of having your Dirac belt connecting two particular particles, maybe it would be more in the spirit of indistinguishability to imagine a given particle as being in a superposition of being paired with every antiparticle, although it's not clear to me that there is any actual difference between these two choices.

No doubt this line of thinking has been explored by many people before, but I don't think I've ever seen it all quite laid out. If anyone does happen to know such a reference I would be grateful.