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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?

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    $\begingroup$ 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
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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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The answers given above by KarlPeter is WRONG. The answer given by ACuriousMind is correct but it might be difficult to make sense of it for physicists. Spinors of representation theory are not vectors. They can a priori not be summed. The answers given by KarlPeter confuses these pure mathematical spinors and entities that intervene in the application of spinors to quantum mechanics where sums of spinors are introduced despite the representation theory. Quantum mechanics treats thus spinors a though they were vectors which according to the representaion theory they are not.

Because people do not know that the quantities used in quantum mechanics are sums they call them also spinors, but they are no longer the same thing. We must thus understand why we can sum spinors in quantum mechanics despite the representation theory. The answer is unfortunately very long.

For understanding mathematical spinors you may consult: https://hal.archives-ouvertes.fr/cea-01572342v1 especially its section 2 (one can skip 2.6 nand 2.8). That gives the mathematical meaning of spinors.

To understand how they are used in the Dirac equation you should next consult: https://hal.archives-ouvertes.fr/hal-03084916v2 A part of this document derives the Dirac equation from scratch. It will allow you to understand the meaning of the Dirac equation, while the traditional treatment cannot offer you this understanding because Dirac has guessed his equation. You will see then that the bi-spinor is a super position state, i.e. the sum of two spinors of the 4x4 Weyl representation, which has a block structure containing the two different 2x2 Weyl representations. Sums of spinors are not defined in the representation theory because sums of group elements are a priori not defined. As quantum mechanics uses sums of spinors, a meaning for such sums must be found. It turns out that such sums represent sets of group elements. Hence the bi-spinor is not a pure spinor but represents a set of spinors, half of them being left-handed and half of them being right-handed. These sets now describe statistical ensembles of electrons. Each electron has its own pure spinor. Without introducing such sets the Dirac equation cannot be derived.

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  • $\begingroup$ You said "Spinors of representation theory are not vectors." and @ACuriousMind said "Spinors are vectors in the representation vector space". Can you please explain why both of you are correct? $\endgroup$ – MadMax Feb 4 at 17:03

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