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Why do we call a 1/2 spin particle satisfying the Dirac equation a spinor, and not a vector or a tensor?

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up vote 4 down vote accepted

It can be instructive to see the applications of Clifford algebra to areas outside of quantum mechanics to get a more geometric understanding of what spinors really are.

I submit to you I can rotate a vector $a = a^1 \sigma_1 + a^2 \sigma_2 + a^3 \sigma_3$ in the xy plane using an expression of the following form:

$$a' = \psi a \psi^{-1}$$

where $\psi = \exp(-\sigma_1 \sigma_2 \theta/2)= \cos \theta/2 - \sigma_1 \sigma_2 \sin \theta/2$.

It's typical in QM to assign matrix representations to $\sigma_i$ (and hence, $a$ would be a matrix--a matrix that nonetheless represents a vector), but it is not necessary to do so. There are many such matrix representations that obey the basic requirements of the algebra, and we can talk about the results without choosing a representation.

The object $\psi$ is a spinor. If I want to rotate $a'$ to $a''$ by another spinor $\phi$, then it would be

$$a'' = \phi a' \phi^{-1} = \phi \psi a \psi^{-1} \phi^{-1}$$

I can equivalently say that $\psi \mapsto \psi' = \phi \psi$. This is the difference between spinors and vectors (and hence other tensors). Spinors transform in this one-sided way, while vectors transform in a two-sided way.

This answers the difference between what spinors are and what tensors are; the question of why the solutions to the Dirac equation for the electron are spinors is probably best for someone better versed in QM than I.

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Actually, Dirac equation is some what a "square root" of Klein-Gordon equation, so intuitively it can't represent a Vector or Tensor, as "symbolically" spinor corresponds to a square root of "differential", so the transformation rules had to differ from Tensors (actually one is in some vague sense taking "square root" of Tensor transformation rules, spinors actually come from "HALF" of Gram density for tensor) , hence for vectors. The above discussion can be rigorized in "Principal Bundle" or rather say "Vector bundle" Settings, one very good book is "Spin Geometry (PMS-38) - Princeton University Press by HB Lawson" here one can find why "The Dirac Equation" only describes 1/2 - spin particles.

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A four-vector transforms under the Lorentz-Group $SO(3,1)$, i.e. a "standard Lorentz transformation". The Lorentz transformation for a spinor is under $SU(2)\times SU(2)$ (to be exact the representation $2 \times \bar 2$) which is locally isomorph to $SO(3,1)$ but not the same. To get a better understanding, you could read chapter two here and meditate a bit about it. It took me a while to get everything sorted in my head. (In fact, I doubt everything is sorted yet, but at least I start to get the grasp of it.)

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Spinor is a vector in the basis of not space-time, but its spin states; in on sense, spinor is not a vector, since it will not transform as you transform the space (rotation, etc) .

Generally speaking, tensors (including scalar, vector, tensor of rank 2,3,4...etc) are just mathematical objects (you lumped together) that transformed as the space-time coordinate transforms as an entirety. A scalar is so that: $$ A'=A $$ after arbitrary coordinate transform; a Vector is so that: $$ A'=MA, or \ A'_i=M_i^{\ j}A_j $$ where M is the unitary (orthogonal) transformation matrix. when it comes to a tensor of rank two: $$ A'=MAM^{-1} or \ A'_{\nu\mu}=M_{\mu}^{\ \rho}M_{\nu}^{\ \theta}A_{\rho\theta } $$

Higher orders of tensors just transform accordingly (by adding more terms of M) .

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What makes you say a spinor doesn't change under space-time transformations? It does, and it's precisely because it doesn't transform as a tensor that it merits the attention it gets. – SamRoelants Mar 26 '15 at 8:26

protected by Qmechanic Dec 26 '13 at 14:09

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