Recently, I encounter a problem about the definition of Fermion operator. In our standard textbooks, the Fermions are defined by their exchange/braiding property, that is, if a minus sign appears by exchanging two Fermions, we say that they are Fermions. Bosonic particles do not have this sign. However, mathematically in the textbooks, the Fermion operators are defined in the following way.

\begin{equation} \{c_i, c_j\} = \{c_i^\dagger, c_j^\dagger\} =0, \quad \{c_i, c_j^\dagger\} = \delta_{ij} \end{equation}

The first two equations follow exactly the property of braiding. So my question is, why we still need the second equation to fully define a Fermion? OF course, the exact the same condition happens to the definition of boson operator. This is a basic question in quantum mechanics, but it seems that the textbooks do not give a detailed discussion for this issue. In the Fock space, of course, the second equation seems to be redundant.

I have this question because of the following paper, http://dao.mit.edu/~wen/pub/edgere.pdf by prof. xiaogang wen. In Eq. 2.10, Prof. Wen said that the wavefunction in Eq. 2.9 (in the above link) is fermionic only when $1/\nu = m$ is an odd number. In his discussion, we do not need to discuss the second equation.

I fact in the above paper,Prof. Wen did not check that his defined wave function respect $\{\psi(x), \psi^\dagger(x')\} = \delta(x-x')$. OF course, I can find this discussion in other refs.

I think my question maybe formally asked in a straightforward way. If I can define a operator satisfying the following condition

\begin{equation} \{c_i, c_j\} = \{c_i^\dagger, c_j^\dagger\} =0, \end{equation} then are $c_i$ Fermion particles? The potential controversial in this new definition is that the creation and destruction operators are not well-defined.


closed as unclear what you're asking by ACuriousMind, Kyle Kanos, Brandon Enright, Danu, JamalS Feb 15 '15 at 12:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ I do not follow your question at all. First, bosons do not have "the same" definition, because bosonic operators have commutation, not antic. relations. Second, the $c_i$ are not the particles, they create the particle states from a vacuum. Third, a fermion is not, in general, defined by its braiding properties because you need the notion of "switching particles" for that, and we talk of fermionic fields and their modes even without them being particles. To speak of switching, you need the Fock space, and for that, you need $c_i$ with all the properties in your first equation line. $\endgroup$ – ACuriousMind Feb 15 '15 at 1:09
  • $\begingroup$ I think a decent source for what @ACuriousMind stated is Weinberg's first field theory book. $\endgroup$ – Ryan Unger Feb 15 '15 at 1:35
  • $\begingroup$ Why is this "unclear what you're asking"? The last paragraph seems quite clear, in fact, the second but last sentence all by itself seems an entirely well-posed question. $\endgroup$ – Norbert Schuch Feb 15 '15 at 19:22

You may wish to look at http://en.wikipedia.org/wiki/Fermion#Composite_fermions , in particular: "The number of bosons within a composite particle made up of simple particles bound with a potential has no effect on whether it is a boson or a fermion.

Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared to size of the system) distances. At proximity, where spatial structure begins to be important, a composite particle (or system) behaves according to its constituent makeup."


"The quasiparticles of the fractional quantum Hall effect are also known as composite fermions, which are electrons with an even number of quantized vortices attached to them."

The commutation relations for composite particles only approximately coincide with the "ideal" commutation relations that you use (see, e.g., the book "Quantum Mechanics" by Lipkin, where commutation relations for composite particles containing two fermions are derived).


I think these equations are not the definition of fermionic operators but property we can derive from the definition of boson and fermion.

we can define the creation and annihilation operators of a particle first, and according to the exchange symmetry we can derive the commutation relations of bosonic and fermionic operators. I guess this is the way to understand this.


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