# Why can we choose spin-1/2 degrees of freedom to commute?

Edit 2: The previous title of this question was "Why are qubits bosonic?" Thanks to the answers that have been provided so far, I now realize I asked my question in a sloppy way. The question I would actually like answered is something more like "Why can local spin-1/2 degrees of freedom in condensed matter systems be chosen to commute, in light of the spin-statistics connection?" What follows is the original body of the question.

I have a very simple, probably even trivial question. In the condensed matter literature, I see local qubits referred to as bosonic degrees of freedom. Now, a qubit is a spin-1/2 system (or at least can be realized by or mapped to such a system), and at least in relativistic QM, spin-1/2 particles are fermions. So why are qubits bosonic degrees of freedom?

One thought I have is that in condensed matter, there's no Lorentz symmetry (at least microscopically), so there's no spin-statistics connection and hence no requirement that spin-1/2 particles be fermions. Perhaps we just get to choose their exchange statistics as we wish.

Another thought I had is that since we are considering qubits on different sites of a lattice, the qubits are distinguishable, hence they commute and we treat them as bosons. But since the boson/fermion classification applies to indistinguishable particles, I think this is probably not it.

Edit: Here are a couple examples of what I'm talking about.

In Ref. 1:

An example of local bosonic model is given by a spin 1/2 system on a lattice.

In Ref. 2:

The first bosonic model is a spin-1/2 model on a d-dimensional cubic lattice.

[1] A. Hamma, F. Markopoulou, I. Prémont-Schwarz, and S. Severini, Phys. Rev. Lett. 102, 017204 (2009), arXiv:0808.2495

[2] M. B. Hastings and X.-G. Wen, Phys. Rev. B 72, 045141, arXiv:cond-mat/0503554

• The existing answers answer your main question regarding qubits. However, what you mention about condensed matter systems not needing to obey the spin-statistics theorem is correct. There can be bosons (i.e., whose statistics is bosonic) with half-integer spins, check out Schwinger bosons. Sep 18 at 18:33
• Sep 18 at 19:23

Qubits are neither fermionic nor bosonic, but you can use either fermionic or bosonic degrees of freedom to store qubits.

The notion of a qubit has nothing to do with exchange symmetry nor with the commutators or anticommutators that apply to raising and lowering operators for quantum fields. That's why a qubit, qua qubit, is neither fermionic nor bosonic.

The term "qubit", strictly defined, is an information measure. It indicates the size of a Hilbert space. To be precise, a number of qubits is the number of 2-state systems which would be required to faithfully store or transmit the quantum states of some given quantum system.

The question changed somewhat in an edit, so now I will answer the new version of the question. It concerns why spin-$$1/2$$ systems can be treated as if we don't need to care about exchange symmetry when they are localized. The reason is as follows.

Take two fermions such as two electrons, and let them be localized at locations $$A$$ and $$B$$. On a particle (as opposed to field theory) formulation, their joint state has to be antisymmetric with respect to exchange of labels. So for example this joint state is impossible: $$|\psi\rangle = |\uparrow,A\rangle_1 \otimes |\downarrow,B\rangle_2$$ but this joint state is possible: $$|\psi\rangle = \frac{1}{\sqrt{2}}\left( |\uparrow,A\rangle_1 \otimes |\downarrow,B\rangle_2 - |\uparrow,A\rangle_2 \otimes |\downarrow,B\rangle_1 \right). \tag{1}$$ The notation is such that the arrows indicate spin state, the letters A and B indicate spatial states, and the subscripts 1 and 2 indicate labels on electrons.

The first state above (the impossible one) indicates, among other things, that electron 1 is up and electron 2 is down, but since electrons are indistinguishable it is not possible to talk about them like that. The second state above (the possible one) does not say whether electron 1 or 2 is up or down, but it does say that there is a correlation between spin state and location, such that a spin state observed at A will be found to be up, and a spin state observed at B will be found to be down. Now as long as the electrons are confined to non-overlapping locations, we will always thus find that the locations can be used to say which spin we are referring to. So now instead of saying "electron 1, electron 2" (which would be a mistake) we can say "electron A, electron B" and use the shorthand $$|\psi\rangle = |\uparrow \rangle_A \otimes |\downarrow \rangle_B . \tag{2}$$ The point of this way of proceeding is if you regard equation (2) as a shorthand for equation (1), then you will find that when you calculate inner products all the cross terms involving $$\langle A | B \rangle$$ vanish so the maths turns out to work just as if you were really using product states, not antisymmetrised states. One way of understanding it is to say that the locations are providing a physical property which effectively labels the electrons, no matter what other labels such as 1 and 2 we may provide. To get used to this I recommend working with a fully written antisymmetrised state in one or two simple examples, and note how it all comes out the same as if you had adopted a notation such as equation (2).

• Thanks for your answer. I've edited my question with explicit references to the type of thing I'm talking about. It was probably misleading of me to even use the word "qubit" in my question, since my question is really about references to spin-1/2 degrees of freedom as bosonic. I used "qubit" because in my head I tend to think of "local spin-1/2 model" and "local qubit model" as meaning the same thing. (In my experience they typically do mean the same thing in the condensed matter theory literature.)
– d_b
Sep 17 at 1:18
• +1 for introducing the word "qua" in physics writing ;) Sep 18 at 18:30

Qubits in the quantum-circuit model are assumed to be distinguishable, so it makes no difference whether they're implemented using bosons or fermions.

"A qubit is a spin-1/2 system" is true only in the sense that an abstract qubit is isomorphic to an abstract spin-1/2 system, such as an electron confined so that it has no spatial degrees of freedom. But an abstract qubit is also isomorphic to the polarization states of a photon, or to any two-state subsystem of any quantum system whatsoever. Finding a good representation of qubits in quantum computers is an engineering problem. There are no fundamental theoretical limitations on what can be used as a storage medium.

I'm not familiar with the term "bosonic qubit", but a search of the arXiv suggests that it refers to various methods of representing qubits in bosonic systems which have good fault tolerance properties. Qubits stored this way are distinguishable, as required by the abstract computational model. They may be distinguished by position, but not necessarily; for example, in principle, you could store three qubits in the occupation number ($$0$$ to $$7$$) of a single quantum state. Indistinguishability of the bosons implies an $$n!$$-fold permutation symmetry of the wavefunction in state $$n$$, but the qubits are distinguishable because there is no permutation symmetry of the binary digits of $$n$$.

• Thanks for taking the time to answer. I should say that I'm really not concerned at all with how the qubits/spins are implemented. This is something that shows up in the theory literature without any discussion of practical implementation. I took a quick look at some search results for "bosonic qubit," and this term seems to show up in literature much more oriented toward practical implementation of qubits in the lab. It doesn't seem related to what I am talking about.
– d_b
Sep 17 at 1:11

I think you need to be careful about the terminology. None of the papers or quotes you posted uses the word "qubit".

"Qubit" is a term you often find in Quantum Information Theory literature and refers to any two-state quantum system. These could be bosonic, fermionic or even (in the case of topological quantum computation) anyonic in nature.

I think what you are talking about is "spin systems", in particular spin models with local spin-1/2 degrees of freedom. A spin-1/2 is a qubit, since it is a two-level quantum system. But not all qubits are spin systems.

The reason these models are called "bosonic", is mainly because the local operators associated to spin systems commute at long distances instead of anti-commuting. Just like bosons would. In some fields people talk about systems with "bosonic locality" vs "fermionic locality". For example, there is a subtle difference between bosonic and fermionic topological orders. (In fancy field theory language: difference between a TQFT vs spin TQFT).

Note that bosonic systems can emerge from degrees of freedom that are fundamentally fermionic. For example, the spin-1/2 models people usually study in real life emerge in systems of strongly interacting electron systems (which are fermionic). For example, the large $$U$$ limit of the Hubbard model.

This is a subtle issue that is hard to explain in all details. But generally in condensed matter physics the difference between bosonic vs fermionic system is essentially about whether the operators associated to the fundamental degrees of freedom of the system commute or anti-commute at long distances. Even if those degrees of freedom are not literal particles moving around.

• Thanks for this. You are right that I should have been more careful. I used the word "qubit" when I was really thinking of spin-1/2, since I tend to think of those as synonyms (but I realize they don't always mean the same thing). I understand that bosonic and fermionic refer to the commutativity or anticommutativity of local operators at long distances. I was under the impression that this was connected to exchange statistics, although maybe I am mistaken about that.
– d_b
Sep 17 at 3:52
• I guess a more careful version of my question is "Why are spin-1/2 degrees of freedom allowed to commute instead of anticommute?" And I guess maybe the answer just comes down to the fact that these degrees of freedom are not particles in a relativistic quantum field theory, but maybe there is something more subtle there.
– d_b
Sep 17 at 3:54