Spinors, vectors and quaternions

dimsun
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I am interested in using hypercomplex numbers and not using tensors.

Therefore a question about the difference between spinors and vectors.
I read that they both can be written as quaternions.

Vector:
Vq = ix + jy + kz

Spinor:
Sq = ix + jy + kz

So what is the difference between them?
Can anyone tell me?

Dimsun
 
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Can you give some references or link where you read this, for the people who are not so familiar with this stuff but would like to help? :)

I'm not into quaternions, but the short answer to your question is that spinors are representations of SU(2) and vectors are representations of SO(3), where SU(2) is the double cover of SO(3).
 
I think it goes something like this. If \sigma_x , \sigma_y , \sigma_z are the Pauli matrices, then \sigma_0 , i\sigma_x , i\sigma_y , i\sigma_z act like the unit quaternions, where \sigma_0 is the 2x2 identity matrix and i=\sqrt{-1}. A vector v can be written v = v_x\sigma_x + v_y\sigma_y + v_z\sigma_z and a spinor can be written \psi = \alpha\sigma_0 + \beta i\sigma_x + \gamma i\sigma_y + \delta i\sigma_z where \alpha, \beta, \gamma, \delta are all real. You get back to the idea of spinors as 2-component complex vectors by writing
|\psi\rangle = \left(\begin{array}{c}<br /> \alpha +i\delta \\<br /> -\gamma + i\beta\end{array}\right)

The set \sigma_0 , \sigma_x , \sigma_y , \sigma_z , i\sigma_x , i\sigma_y , i\sigma_z , i\sigma_0 forms a basis for the Clifford algebra.

This is in 3-d, no idea about 4-d.

At least I think some of that is right...
 
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