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Dirac Spinors is a 4 element vector, and a qubit state vector is two element vector. Two spinors are positive(1 and 4) and negative values (3 and 2), being the first value the spin up and the second the spin down.

This spinor vector is multiplied by the wave function at beginning, and then if evolutionated by operators (in split method) like in this simulations from dirac equation simulation.

So, after some time evolution, is correct to extract the qubit state vector from every point of the simulation? Suposse is 2D NxN discrete space, so I have 4 wave functions NxN. If I take the first psi at (x,y) as spin up, and the fourth psi at (x,y) as the spin down and normalize both, I have the qubit state vector?, taking into account that the initial 2º(negative down) and 3º(negative up) spinors were set to zero, so there will not be negative contribution.

So if this is correct, the qubit spin is just a point from the four Dirac's wave-function * spinnor.

Luis ALberto
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1 Answers1

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  1. The Dirac-spinor representation $(\frac{1}{2},0) \oplus (0,\frac{1}{2})$ is a direct sum of the left- and right-handed Weyl-spinor representation for the double cover $$G=SL(2,\mathbb{C})$$ of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. See also e.g. this related Phys.SE post.

  2. The qubit$^1$ can be viewed as the defining/fundamental/spinor representation for the double cover $$H=SU(2)$$ of the 3D rotation group $SO(3;\mathbb{R})$. See also e.g. my related Phys.SE answers here and here.

  3. Since $H\subseteq G$ is a subgroup, the qubit can be viewed as a restricted representation of the Weyl representation.

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$^1$ Since a qubit is an abstract quantum mechanical concept, the Lie group $H$ may not be canonically related to rotations in ordinary 3D position space and physical spin/angular momentum.

Qmechanic
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