I like to think of QFT Weinberg-style: particles come first, and fields come later; and the latter are constructed so as to describe the former. Fields are not to be thought of as fundamental, but only as convenient tools to study particles.

In this setting, what justifies the postulate that a fermion particle must be described by a field transforming according to a representation of the Clifford algebra?

More details:

Particles are classified according to the irreducible representations of the orthogonal group, $\mathrm{SO}(d-1)$ (where $d$ is the number of spacetime dimensions; I'm taking all particles to be massive for simplicity).

Once you have a particle described by a representation $R$ of $\mathrm{SO}(d-1)$, you introduce a field $\psi$ which, by definition, lives in a representation $R'$ of the Lorentz group, $\mathrm{SO}(1,d-1)$. The non-trivial question is which representation $R'$ corresponds to a representation $R$; that is, what field is to be used to describe a certain particle. In general, the answer is non-unique, so we should actually look for "the simplest" $R'$ for a given $R$.

In his book, Weinberg addresses this question in $d=4$. For example, the trivial representation of the orthogonal group corresponds to the trivial representation of the Lorentz group (i.e., scalar particles correspond to scalar fields). Similarly, the spin $s=1/2$ representation of the orthogonal group corresponds to the $(0,\frac12),(\frac12,0)$ representations of the Lorentz group. To conserve parity, these are to appear together, so $$ \text{spin $s=1/2$}\quad \Longleftrightarrow \quad (0,\tfrac12)\oplus(\tfrac12,0) $$

Now comes the key point: as it turns out, the right hand side of this equation actually corresponds to the an irreducible representation of the Clifford algebra $$ \gamma^{(\mu}\gamma^{\nu)}=\eta^{\mu\nu} $$ but this is only an a-posteriori realisation. There was no reason to expect Clifford to be relevant from the beginning. It just happens to be so.

In higher dimensions, the logic is inverted. One declares that in higher dimensions the Clifford algebra is fundamental, and fermions are whatever particle is described by such fields. In the spirit of Weinberg, this is pretty unconvincing: particles should come first. Given higher dimensional fermions, one must ask which fields are to be used to describe them. And it may very well be the case that the answer is again Clifford, but this must be an a-posteriori conclusion, not a postulate.

Thus, my question: what justifies the use of the Clifford algebra to describe higher dimensional fermions? How can we prove that such a representation of the Lorentz group is indeed the simplest representation that is able to describe fermions, when we take the latter as fundamental instead of the former?

To be specific, let us define higher dimensional fermions as the first non-trivial projective representation $R$ of $\mathrm{SO}(d-1)$ (or a direct sum thereof, if necessary, in order to conserve parity).

  • 2
    $\begingroup$ This is probably unrelated, but would you mind me asking how this "particles – fundamental, fields – auxiliary" belief works with QFT in curved spacetime? $\endgroup$ Feb 14, 2018 at 16:07
  • $\begingroup$ @SolenodonParadoxus I don't believe in spacetime curvature ;-) . $\endgroup$ Feb 14, 2018 at 16:10
  • $\begingroup$ Are you asking about why we use Clifford algebra representations at all, or about why we use representations of $Cliff(D-1,1)$ instead of $Cliff(D-1)$. $\endgroup$ Feb 15, 2018 at 9:41
  • $\begingroup$ Actually after reading your comment to Qmechanic, I see that you are asking about the first possibility. $\endgroup$ Feb 15, 2018 at 10:00
  • $\begingroup$ I'd just like to point out that all representations of semisimple Lie groups (in particular, $Spin(d-1)$), have been classified. This classification (together with spin-statistics theorem) says that the simplest representations which can satisfy Fermi statistic are the spinor ones. The OP would probably benefit from learning representation theory of semisimple Lie groups. $\endgroup$ Feb 26, 2018 at 16:11

2 Answers 2


OP seems to asks about the logic behind "Little group $\to$ Lorentz group $\to$ Clifford algebra". In this answer we discuss the last leg.

Given an (indefinite) orthogonal / a (restricted) Lorentz group $SO^{+}(V,Q)$ [and its double cover, the spin group $Spin^{+}(V,Q)$] over a $\mathbb{F}$-vector space $V$ with a quadratic form $Q$, there always exists functorially a corresponding Clifford algebra $Cl(V,Q)$.

Here we assume that the pertinent finite-dimensional representations are derived from the spinor representations [i.e. the fundamental representations] of $Spin^{+}(V,Q)$ and tensor product representations thereof.

In a nutshell, spinor representations are realized as exterior algebras $\bigwedge^{\bullet}(V)$. The Clifford quadratic form $Q$ and gamma matrices can be build out of interior and exterior multiplication, and the Clifford algebra $Cl(Q,V)$ is a pertinent "quantization" of the exterior algebras $\bigwedge^{\bullet}(V)$. So the Clifford algebra $Cl(Q,V)$ comes for free. The (indefinite) orthogonal / Lorentz algebra $so(V,Q)$ itself becomes realized as anti-commutators of gamma matrices.

This does not imply that the Clifford algebra [e.g. the $Pin(V,Q)$ group] enhance the symmetry of the theory.

  • $\begingroup$ I'm not sure I understand your point. Are you arguing that the field that describes a fermion must be Clifford because Clifford is a natural construction of Lorentz fields? That doesn't rule out other representations of Lorentz, so it leaves open the possibility of using non-Clifford fields. In other words, just because Clifford is a valid possibility that doesn't mean that there are no other possibilities, right? $\endgroup$ Feb 14, 2018 at 20:25
  • $\begingroup$ Thanks. By assuming that you are going to work with spin representations, haven't you actually assumed what you wanted to prove? i.e., to my question "why Clifford?" you answer "because spin", but then we're back to square one: "why spin?". What justifies the assumption that fermion particles are described by fields that transform according to a spin representation or the Lorentz group? $\endgroup$ Feb 14, 2018 at 21:40
  • $\begingroup$ That is essentially the first leg. $\endgroup$
    – Qmechanic
    Feb 14, 2018 at 22:19

I am answering the question, why we use Clifford algebra representations in quantum mechanics and quantum field theory to describe fermions or spin.

The short answer, which may seem rather surprising, is that we don't have to. Clifford algebras have so many advantages in describing spin but they are not the unique way to do so.

The answer to the question why spin is empirical: When we quantize a classical system having a symmetry group $G$, the corresponding quantum mechanical systems transform according to the irreducible representations of the universal covering group $\tilde{G}$ rather than $G$ itself. This is an experimental fact. This is why half integral spinor representations emerge in quantum systems with rotation symmetry.

The answer to the question why spin is needed to describe fermions is given by the spin-statistics theorem which I'll not elaborate here on.

I'll concentrate on the question: are Clifford algebras necessary for the description of spin.

Before giving the answer, let me recount a historical anecdote: Before Berezin introduced his Grassmann algebra description of the path integral, people used to invert the fermionic determinant by hand. They did so during the whole era of great achievements of QED. (Grassmann algebras can be regarded as the classical counterparts of Clifford algebras).

It is not by chance that I mentioned Berezin. Berezin (together with Marinov) was the first one to realize that Grassmann algebras describe spin classically and they can be quantized by a process which is the fermionic counterpart of canonical quantization to from Clifford algebras: Ref 1, Ref2. But what is really interesting is that he was also the first one who described a spinor representation not by means of a Clifford algebra Ref3. In fact the Clifford algebra is so very well hidden in Berezin's realization that it is very difficult to construct the representatives of its generators in this realization.

I'll elaborate a little about this realization in the case of the group $SO(2N)$

It is known that representations of compact Lie groups are in a 1-1 correspondence to (integral) orbits of the coadjoint representation. This is the Borel-Weil-Bott theorem.

In the case of the spinor representation of $SO(2N)$ the corresponding coadjoint orbit is $SO(2N)/U(N)$ this is a complex manifold of (complex dimension) $\frac{N^2-N}{2}$. It is a compact symplectic manifold that can serve as phase space.

What the Borel-Weil theorem really means is that we can formulate a classical mechanical system on $SO(2N)/U(N)$ and quantize it by means of geometric quantization (which is just a small generalization of the well-known canonical quantization) and obtain the spinor representation of $SO(2N)$ on the quantum Hilbert space. In practice this representation will be given by means of coherent states built as holomorphic functions of the $\frac{N^2-N}{2}$ (bosonic) coordinates. No Grassmann or Clifford algebras are involved in the construction.

Now, one reason that this realization is not used in quantum field theory is that the Grassmann $\rightarrow$ Clifford has the spin statistics connection encoded within because of its anti-symmetry properties. If we had used the Berezin's quantization realization in computing some quantum field theoretical amplitude would have needed to introduce the anti-symmetry by hand in addition to all other complications.

Having said that; still one of my dreams, is to compute some simple tree level QED amplitude using this representation. I know that I will need a very powerful computer algebra package for that.

  • 1
    $\begingroup$ "Grassmann algebras can be regarded as the classical counterparts of Clifford algebras" - I hear this a lot, but how is it true? Clifford algebras can be understood as algebras of reflections; a rotation in a plane is a product of two reflections; when an object is continuously rotated through 360° the angle between the reflection planes rotates by 180° and picks up a factor of -1; etc. That's all classical geometry. Grassmann algebras seem much less related to spin, if at all. Grassmann numbers only appear in the path integral, and seem to be connected to quantum statistics, not spin as such. $\endgroup$
    – benrg
    Mar 4, 2023 at 23:46

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