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In Peskin-Schroeder Chapter 3.3: Free Particle Solutions of the Dirac Equation the form of the general Dirac Spinor $u(p)$ along 3-direction is derived from $u(p_0)$ in the system of rest by applying a boost. I understand how the following expression is derived from the already known transformation of spinors, which gives:

u(p) = \begin{bmatrix} (\sqrt{E+p^3} \, \frac{1-\sigma^3}{2} + \sqrt{E-p^3} \, \frac{1+\sigma^3}{2}) \,\xi \\ (\sqrt{E+p^3} \, \frac{1+\sigma^3}{2} + \sqrt{E-p^3} \, \frac{1-\sigma^3}{2}) \,\xi \end{bmatrix}

Here $\xi$ denotes any two-dimensional spinor, $\sigma^3$ the Pauli matrix for 3-direction However, as a next step, they say: "the last line can be simplified to give"

u(p) = \begin{bmatrix} \sqrt{p \cdot \sigma} \, \xi \\ \sqrt{p \cdot \overline{\sigma}} \, \xi \end{bmatrix}

This is the point where I cannot follow...Even in my greatest imagination I cannot explain how $\sigma$ changes its place under the squere root. I'm sure to have a total blackout, but cannot find the problem I have...

By the way: One page before they introduced

$\sigma^\mu = (1, \vec \sigma)$

$\overline{\sigma}^\mu = (1, -\vec \sigma)$

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You may wish to compare the squares of the coefficients at $\xi$ in the two lines.

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  • $\begingroup$ Yes much easier, as I thought $\endgroup$ – michael Feb 7 '17 at 22:19

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