Anticommutation relation different specie/type of fermion

Suppose we have two distinct fermions, say $X$ is Dirac, $Y$ is Majorana, part of different irreps of some Gauge group (e.g. SM). Alternatively, consider a lepton $l=l_L+l_R$ and a Majorana neutrino $N=N_L+N_R=N^c$. Assume $X \in \{X_1,\ldots,X_m\}$, $Y\in \{Y_1,\ldots,Y_n\}$ ie each fermion has a number of flavours.

Let $a(X)^\dagger |0 \rangle =|X \rangle$ denote the creation operator in the occupation number basis, $a(X,\mathbf{p})^\dagger |0\rangle =|X(\mathbf{p})\rangle$ be the creation operator in the momentum basis. Of course given arbitrary bases for Hilbert space, we can always relate them via

\begin{equation} a_{\tilde{\alpha}}^\dagger = \sum_\alpha \langle \alpha|\tilde{\alpha}\rangle \,a_\alpha^\dagger. \end{equation}

In our case, $\alpha \equiv n_\alpha$, the occupation number, whereas $\tilde{\alpha}\equiv \mathbf{p}$, the three momentum. We write then the related bases for Fock space $\mathcal{F} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus\ldots$ as

\begin{equation} |n_1,n_2,\ldots n_\alpha\rangle_{\text{occ. basis}} = \frac{(a(\phi_1)^\dagger)^n}{\sqrt{n_1!}} \ldots \frac{(a(\phi_\alpha)^\dagger)^n}{\sqrt{n_\alpha!}} |0\rangle \end{equation} Where $\phi$ represent the particle fields each operator creates, $n_i \in \mathbb{N_0}$ for bosons, $\{0,1\}$ for fermions.

Likewise, we write the momentum basis as

\begin{equation} |\phi_1(\mathbf{p}_1),\ldots, \phi_\alpha (\mathbf{p}_\alpha)\rangle_{\text{mom. basis}} = a(\phi_1,\mathbf{p}_1)^\dagger \ldots a(\phi_\alpha,\mathbf{p}_\alpha)^\dagger |0\rangle \end{equation}

Equivalently the relativistic momentum vector is \begin{equation} |\phi_1({p}_1),\ldots, \phi_\alpha ({p}_\alpha)\rangle_{\text{mom. basis}} = \sqrt{2E_{\mathbf{p}_1}\ldots E_{\mathbf{p}_\alpha}} \,a(\phi_1,\mathbf{p}_1)^\dagger \ldots a(\phi_\alpha,\mathbf{p}_\alpha)^\dagger |0\rangle \end{equation} I have several questions with this setup:

Suppose I consider the anti/commutations for $\xi = \pm 1$ and $[a(X_i),a(Y_j)]_{\xi} = a(X_i)a(Y_j) -\xi a(Y_j)a(X_i) \in \mathbb{C}$ \begin{equation} [a(X_i),a(Y_j)]_{\xi} \in \mathbb{C} \end{equation}

\begin{equation} [a(X_i),a(Y_j)^\dagger]_{\xi} \in \mathbb{C} \end{equation}

\begin{equation} [a(X_i)\dagger,a(Y_j)^\dagger]_{\xi} \in \mathbb{C} \end{equation}

Firstly, why do we need to take the antifermionic relation? By Pauli exclusion, if I take two of the same specie, they must vanish, for example

\begin{equation} |X_i,X_j\rangle = -|X_j,X_i\rangle \end{equation}

\begin{equation} |Y_i,Y_j\rangle = -|Y_j,Y_i\rangle \end{equation}

But if I take two different species, then nothing should stop us from implementing the commutation between them?, e.g. \begin{equation} |X_i,Y_j\rangle = +|Y_j,X_i\rangle \end{equation}

Of course one key feature of the fermion anticommutation relation is to prevent the unboundedness of the free Hamiltonian:

\begin{equation} [H^{(0)}_{X_i},a(X_i,\mathbf{p})]= - E_{X_i,\mathbf{p}}\, a(X_i,\mathbf{p}) \end{equation}

As well as causality outside the light cone. But this doesn't need to be protected between two different species? It is often quoted that we still impose the anticommutation relation in the relativistic limit, and can arbitrarily choose commutators/anticommutators in the non-relativsitic case, though this does not appear obvious to me. Also, the idea that we must impose a superalgebra does not match the fact creating $X_i$ then $Y_j$, should be order independent (as a counter example, creating an electron and muon field, the only distinguishing feature is mass, ie they belong to the same class of fermion, crucially with the same quantum numbers and thus we must anticommute them. Now create a Majorana neutrino state, with different quantum numbers to the electron; surely they should lie in a free part of Hilbert space away from the electron state, like a tensor product, so these operators should commute? )

Secondly, if we justify somehow the anticommutation relation between different species, then in the case of Majorana neutrinos, does this still also hold, that two different types of fermions must anticommute?

I think a more accurate statement would be that considering "abnormal" commutation relations is not necessary. It isn't always in conflict with the general principles of QFT (as long as the Hamiltonian still has a lower bound, etc), but we don't get any new models by considering such possibilities — at least not if two models are considered to be equivalent whenever they have the same net of local observables. A model that satisfies the general principles of QFT is supposed to be equivalent to one in which all fermion field operators anticommute with each other (including those for different species), and we can supposedly show this explicitly by using a Klein transformation.

I used the words "supposed" and "supposedly" because I haven't worked through all of the details myself, so there's a risk that I'm overlooking some important caveats. To compensate for my ignorance, I'll give some references.

Here's an excerpt from pages 146-147 in Streater and Wightman (1980), PCT, Spin and Statistics, and All That:

It turns out that `abnormal' commutation relations in which two integer spin fields or an integer spin and a half-odd integer spin field anti-commute, or two half-odd integer spin fields commute, can be realized but, in general, the resulting theories possess special symmetries. By virtue of these symmetries it turns out that there always exists another set of fields, satisfying normal commutation relations and related to the original fields by a so-called Klein transformation.

Streater and Wightman's book proves the spin-statistics theorem using a set of axioms that is expressed in terms of fields. The general principles of QFT can also be expressed using a different set of axioms, one that refers only to observables. In this case, the Doplicher-Roberts Reconstruction Theorem says that the net of observables can be (not that it must be) expressed in terms of a field system that has normal commutation relations. This theorem is reviewed on page 92 in "Algebraic Quantum Field Theory", https://arxiv.org/abs/math-ph/0602036. The theorem says that such a field system exists, but it does not say that such a field system is the only one that works. This is emphasized (in the context of the same system of axioms) on page 98 in Haag (1996), Local Quantum Physics:

We called this [connection between boson/fermion and commuting/anticommuting] a convention because it can be shown... that although different possibilities exist they can always be transformed to the "normal" case by a redefinition of the fields in a way which does not change the physical consequences.

Finally, here's the abstract of "How to commute", https://arxiv.org/abs/1312.0831:

A simple exposition of the rarely discussed fact that a set of free boson fields describing different... particle types can be quantized with mutual anticommutation relations is given by the explicit construction of the Klein transformations changing anticommutation relations into commutation relations. ... The analogous situation for two independent free fermion fields with mutual commutation or anticommutation relations is briefly investigated.

• This is a very insightful answer, I'll go through the references in detail and come up with a constructed hopefully well thought out response, thanks :) – MKF Oct 30 '18 at 1:35
• I came to similar conclusions a few weeks ago – MKF Oct 30 '18 at 1:38
• @MKF You may have already studied this more than I have. I'd be interested in knowing if you find any caveats that I overlooked. – Chiral Anomaly Oct 30 '18 at 1:40