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