# Spinor Understanding: QFT vs pure Representation Theory

I have some questions on terminology used in QM & QFT and (pure mathematical) representation theory treating the concept of "spinor".

Let us focus on Dirac spinor as described in https://en.wikipedia.org/wiki/Dirac_spinor:

By the article it's a complex bispinor $$\psi =\left({\begin{array}{c}\psi _{L}\\\psi _{R}\end{array}}\right)$$ which is a solution of Dirac equation $$\left(i\gamma ^{\mu }\partial _{\mu }-m\right)\psi =0\;}$$

with $$\gamma^{\mu}$$ gamma matrices and $$\psi _{L}, \psi _{L}$$ Weyl spinors from $$(½,0)$$ and $$(0,½)$$ representations of the $$SO(1,3)$$ group (the Lorentz group without parity transformations).

From point of view of (pure mathematical) representation theory spinors are elements of fundamental representation of the Clifford-algebra.

Short review of representation theory: consider an algebra $$A$$ and one is looking for a vector space $$V$$ say of dimension $$n$$ and a group homomorphism $$\rho_A: A \to Mat_n(V)$$.

Using this language spinors are images under map $$\rho_{Cl}: Cl \to Mat_n(V)$$ for certain Clifford algebra $$Cl$$ and certain $$n$$-dimensional vector space $$V$$. In other words representant elements in the matrix algebra of elements of $$Cl$$.

The problem is that comparing these two viewpoints if we fixing a point x$$=(x_0,x_1,x_2,x_3)$$ what is $$\psi$$(x) $$=\left({\begin{array}{c}\psi _{L}\\\psi _{R}\end{array}}\right)$$(x) evaluated in x? An element of the image of an appropriate representation map $$\rho$$ (this would coinside with mathematical definition of a spinor) or an element of the vector space $$V$$ on which the images under $$\rho$$ act via $$\rho$$? But then calling $$\psi$$(x) a spinor would be misleading .

For sake of simplicity let us focus on the upper Weyl spinor $$\psi _{L}$$(x). By definition Weyl-Spinor-representation is the smallest (=fundamental) complex representation of $$\operatorname {Spin} (1,3)$$.

If this would be a "spinor" in usual sense there would be a representation map $$\rho: \operatorname {Spin} (1,3) \to Mat_n(V)$$ with certain vector space $$V$$ and $$\psi _{L}$$(x) would be contained in the image.

But why? $$\psi _{L}$$(x) is an element of $$\mathbb{C}^2$$ so intuitively it is an element of $$V$$ but thing violates the nomenclature.

Could anybody explain what I'm here confuse. Especially why the notation "spinor" for $$\psi _{L}$$(x) make here sense from mathematical viewpoint?

• You seem to be confusing the matrices of a representation with the elements of the vector space that the matrices operate on. Spinors are elements of the vector space. They aren’t “elements of the representation”. – G. Smith Sep 20 at 18:01

1. A Dirac spinor or bispinor transforms in the (only) irreducible representation of the Clifford algebra $$\mathrm{Cl}(1,3)$$. This representation is four-dimensional.
2. A Weyl spinor transforms in an irreducible complex representation of the Lorentz algebra $$\mathfrak{so}(1,3)$$ (and hence of $$\mathrm{Spin}(1,3)$$), of which there are two that are denoted by $$(1/2,0)$$ and $$(0,1/2)$$, the "left-handed" and "right-handed" representations. These representations are two-dimensional.
3. $$\mathfrak{so}(1,3)$$ is isomorphic as a Lie algebra to the even subalgebra of $$\mathrm{Cl}(1,3)$$, so the Dirac representation - irreducible as a representation of $$\mathrm{Cl}(1,3)$$ - is also a not necessarily irreducible representation of $$\mathfrak{so}(1,3)$$.
4. In fact, as a representation of $$\mathfrak{so}(1,3)$$ the Dirac representation is reducible and isomorphic to $$(1/2,0)\oplus (0,1/2)$$. This is what the physicist means when they write $$\psi = \begin{pmatrix} \psi_L \\ \psi_R\end{pmatrix}$$.