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I was reading through my particle physics textbook and saw a property of the Dirac spinors that I did not understand. The spinor is defined by $u^s(p)=\sqrt{\frac{E+m}{2m}} \begin{bmatrix}\phi^s \\ \frac{\sigma p}{E+m}\phi^s \end{bmatrix}$. The normalisation of this means that the scalar invariant $\bar{\phi}\phi$ (where the conjugate spinor has its usual meaning) means the $\bar{u^s}(p)u^{s'}(p)=\delta_{s s'}$.

What I don't understand is how the normalisation yields a Kronecker delta? I have been trying to work through the maths to prove this but I end up with a 2x2 matrix and I don't know where to go from there?

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  • $\begingroup$ what is the relation of the $s$ and $s'$ labels to the $a$ and $b$ labels? Are they supposed to be the same? $\endgroup$
    – mike stone
    Commented Oct 27, 2023 at 12:27

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Its the two Weyl eigen-spinors $\phi^s$, that are mutually orthogonal that give the Kronecker delta.

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