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 $${\displaystyle \left(i\gamma ^{\mu }\partial _{\mu }-m\right)\psi =0\;}$$

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?

  • 3
    $\begingroup$ 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”. $\endgroup$ – G. Smith Sep 20 '19 at 18:01

Spinors are vectors in the representation vector space, not matrices in the image of the representation map.

  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 degree 2 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 $\psi = \begin{pmatrix} \psi_L \\ \psi_R\end{pmatrix}$.


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