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I've been studying the application of Clifford Algebra in quantum mechanics, more specifically in spin, but I'm stuck in a basic analogy that is part of the basics before starting to do things using Clifford Algebra. The point is that we have a spinor in spherical coordinates and then say that the equivalent representation of it in the way made below, but it looks like something taken out of nowhere, or simply established in a way that things could work. I am following three references about them:

  • Geometric Multiplication of Vectors An Introduction to Geometric Algebra in Physics (Compact Textbooks in Mathematics) by Miroslav Josipović;
  • Geometric Algebra for Physicists by Chris Doran, Anthony Lasenby;
  • William E. Baylis - Clifford (Geometric) Algebras With Applications to Physics, Mathematics, and Engineering - Birkhäuser Basel (1996)

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I would like some intuition or explanation about why we do this identification $$\psi = a^0 + a^k \sigma_k$$ where $I=\sigma_1 \sigma_2 \sigma_3$ but I'd like more than because it works in that way, if possible. I searched a lot but I couldn't find anything.

I really appreciate your attention and help.

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    $\begingroup$ Related: physics.stackexchange.com/q/656917/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 26 '21 at 5:05
  • $\begingroup$ Thanks @Qmechanic . I was also wondering about this relation between three dimensional and complex two dimensional representations. Thank you, it was very helpful! $\endgroup$
    – Bryant
    Aug 27 '21 at 12:33
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The capability of identifying a spinor with a rotor is a coincidence in 3D.

For a general dimension N, let's do an excise of counting

  • The degree of freedom of rotation (the rotor referenced in the OP): $\frac{N(N-1)}{2}$
  • the degree of freedom of a spinor (assuming it's the even part of the Clifford algebra):$2^{N-1}$

In 3D, both freedoms are 3, after you subtract the normalization factor freedom of a spinor. Therefore, there is a way to identify a spinor with a rotor.

In 4D, the degree of freedom of rotation is 6, while the degree of freedom of a spinor is 8. After subtracting the normalization factor freedom of a spinor, you still have one more degree of freedom ($6+1$) which can NOT be identified with a rotation. This fact has been noted early on by David Hestenes (please search his papers for more info).

In N>4 D, the situation is even worse. In summary, the gimmick cited in the OP only works in 3D.

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  • $\begingroup$ Oh, right. I can live with this (cool) coincidence. Thank you very much! I'm going to search for these papers to get more info. $\endgroup$
    – Bryant
    Aug 27 '21 at 12:30

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