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is it right to say that the Dirac spinor is a mathematical representation of a wave-function that satisfy the Dirac equation? or are there more requirements to it?

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    $\begingroup$ According to Wikipedia "spinors are the column vectors on which these matrices act." These matrices are Dirac's gamma matrices. Dirac's wave function is a spinor valued function of $x$ and $t$ that solves the Dirac equation. $\endgroup$
    – Kurt G.
    Commented Feb 17, 2022 at 18:22
  • $\begingroup$ Hi Kurt, thanks for your answer, but the fact that they satisfy the Dirac equation implies that the matrices can act upon them? $\endgroup$ Commented Feb 17, 2022 at 21:47
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    $\begingroup$ The Dirac equation is strictly speaking a system of four PDEs. $\psi$ has four components, hence, every $4\times 4$-matrix $\gamma^\mu$ can act on it and on $\partial_\mu\psi\,.$ Please go through this. $\endgroup$
    – Kurt G.
    Commented Feb 18, 2022 at 9:28
  • $\begingroup$ Hi Kurt, thanks a lot for the link, I checked it before, but what I'm really after is if my definition is correct, i.e. if it does ignore an important property of the Dirac spinors or not :) $\endgroup$ Commented Feb 18, 2022 at 14:38
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    $\begingroup$ Your definition of the Dirac spinor is correct. However it wasn't maths or theoretical physics if there were not much more general spinors. See this related post. $\endgroup$
    – Kurt G.
    Commented Feb 18, 2022 at 14:47

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Honestly, I somehow dislike this point of view for the simple reason that in order to state the differential equation you must first have a definition of the object which should solve it. Let me state this differently. A differential equation is an equation of the form ${\scr D}\Psi=0$, where $\scr D$ is one differential operator. But to properly define and make sense of $\scr D$ you must know its domain first. Without defining a spinor you cannot write the Dirac eqution.

That said, a Dirac spinor is one field $\Psi:\mathbb{R}^{1,3}\to \mathbb{C}^4$ on Minkowski spacetime which takes values in one vector space which carries one specific representation of the universal cover of the Lorentz group, ${\rm Spin}(1,3)\simeq {\rm SL}(2,\mathbb{C})$.

The spin group ${\rm Spin}(1,3)$ has irreducible representations labelled by $(A,B)$ where $A$ and $B$ are integers or half-integers greater or equal to zero. In particular, the objects living in the representations $\left(\frac{1}{2},0\right)$ and $\left(0,\frac{1}{2}\right)$ are respectively called left-handed Weyl spinors and right-handed Weyl spinors. Finally, the objects living in the direct sum $\left(\frac{1}{2},0\right)\oplus \left(0,\frac{1}{2}\right)$ are called Dirac spinors.

The representation space of the both left and right-handed Weyl spinors is $\mathbb{C}^2$, so the representation space of Dirac spinors is $\mathbb{C}^4$ as anticipated.

After you have a proper definition of a spinor field you make sense of the operator appearing in the Dirac equation ${\scr D} = \gamma^\mu \partial_\mu + m$.

To fully appreciate this story I like sections 5.4 and 5.6 of Weinberg's The Quantum Theory of Fields.

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  • $\begingroup$ Hi Gold, thank you very much for your clarification, but isn't this answer more directed to the definition of a spinor? My question was more focussed on what make a spinor a "Dirac" spinor (after defining what a spinor is). $\endgroup$ Commented Feb 20, 2022 at 15:21
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    $\begingroup$ I guess this is an issue with terminology. I'm used to the term "Dirac spinor" being used to mean one specific representation of the universal cover of the Lorentz group, namely the $(\frac{1}{2},0)\oplus (0,\frac{1}{2})$ representation as I explained. $\endgroup$
    – Gold
    Commented Feb 20, 2022 at 17:40

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