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Why spinors do not form a vector space

Abstract : Spinors ψj of SU(2) represent group elements, i.e. three-dimensional rotations Rj , because they are shorthands of SU(2) representation matrices Rj obtained by taking their first columns. We explain that making linear combinations of spinors of SU(2) is feasible algebraically but geometrically meaningless. E.g. ψ1 + ψ2 ∈ C 2 would have to correspond to R1 + R2. But in SO(3), R1 + R2 would be a function that transforms r ∈ R 3 into R1(r) + R2(r), while in SU(2) it would correspond to a different function that is up to a multiplication constant equal to S : r → S(r) = R1(r) + R2(r) + 2 [ cos(ϕ1/2) cos(ϕ2/2) − s1•s2 ] r + 2 sin(ϕ1/2) sin(ϕ2/2)[ (s2•r) s1 + (s1•r) s2 ] + 2 sin(ϕ1/2) cos(ϕ2/2)(s1 ∧ r) + 2 cos(ϕ1/2) sin(ϕ2/2)(s2 ∧ r). The multiplication constant must be chosen such that |r| is preserved. Such extensions by linear combinations for spinors from a manifold to an embedding vector space are therefore mindless algebra with a spurious, conceptually impenetrable geometrical counterpart. This should not surprise anybody because the group axioms only define products of group elements, not linear combinations of them. Redefining spinors as vectors of a Hilbert space is therefore a purely formal, would-be scholar generalization.
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https://hal.archives-ouvertes.fr/hal-03289828
Contributor : Gerrit Coddens Connect in order to contact the contributor
Submitted on : Monday, July 19, 2021 - 5:33:50 AM
Last modification on : Thursday, July 22, 2021 - 3:35:36 AM

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  • HAL Id : hal-03289828, version 1

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Gerrit Coddens. Why spinors do not form a vector space. 2021. ⟨hal-03289828⟩

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