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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.

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  • $\begingroup$ Your question is by no means self-contained and very confusing. $\endgroup$
    – Tobias Fünke
    Dec 18, 2022 at 10:14
  • $\begingroup$ In short: Dirac needed to make his matrices four dimensional because his main goal was to come up with a Lorentz invariant theory. That's something the $2D$ qubits are not concerned with. $\endgroup$
    – Kurt G.
    Dec 18, 2022 at 10:43
  • $\begingroup$ The question was reworked & is now reopened due to 3 reopen votes. $\endgroup$
    – doublefelix
    Dec 19, 2022 at 11:23
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    $\begingroup$ @doublefelix The question is IMHO still not self-contained. Perhaps it is just me, but I really don't understand what the OP tries to explain here. $\endgroup$
    – Tobias Fünke
    Dec 19, 2022 at 11:37
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    $\begingroup$ @doublefelix If this really is the question, then OP should consider to rewrite the question as such. Given the 2 upvotes of my comment, I guess I am not the only one having trouble to understand the question. That being said, I have to say I still don't see how the question you pose can be related to most of the question. $\endgroup$
    – Tobias Fünke
    Dec 19, 2022 at 14:09

1 Answer 1

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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.

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    $\begingroup$ I don't think qubits are, in general, representations of anything. They are just two-dimensional Hilbert spaces, which have been picked out at will of a typically much larger space. (Of course, if the qubit it encoded in an actual spin-1/2, say, an electron spin, this is different, but this is rarely the case. -- Note, however, that a real spin-1/2 would have to be a fermion, which is not necessarily desirable for a qubit carrier.) $\endgroup$
    – Norbert Schuch
    Dec 19, 2022 at 12:14
  • $\begingroup$ Hm, I'd say this is still very different from the Dirac spinor. In that case, there is a natural physical action of the symmetry group on the representation. In the case of a qubit, there is none. (Of course, one can write down a natural SU(2) action on any 2dim Hilbert space, but there is no physical meaning to it, which is rather different from a spinor, or an actual spin-1/2 particle.) $\endgroup$
    – Norbert Schuch
    Dec 19, 2022 at 12:23
  • $\begingroup$ The purpose of this is to conclude that Dirac equation not only gives the position and momentum probabilities,also the spin for each probability. And because the spin can be entangled with another spin(qubit), and it modulates the initial wave function as spinor, then we can say that the electron position and momentum is not local, because a change in Bob spin makes not only a change in Alice spin, also in momentum and position, So there is the nexus between consciousness and physics, if the spin is the mechanism that produces the first one. $\endgroup$
    – Luis ALberto
    Dec 19, 2022 at 19:57
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    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic ♦
    Dec 20, 2022 at 9:46

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