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In many treatments of the Dirac equation (e.g. Peskin and Schroder, pages 45-46) after subbing in $\psi(x) = e^{-ix_\mu p^\mu}u(\vec p)$, with $u$ a constant spinor, into the Dirac equation, we consider the rest frame and, solving for $u$, find: $$u(0)=\sqrt{m}\begin{bmatrix} \xi \\ \xi \end{bmatrix},$$ with $\xi$ an arbitrary two component spinor, and a convenient normalisation. The general $\vec p$ case is then obtained, for example by boosting the above spinor. The result is quoted as $$u(\vec p)=\begin{bmatrix} \sqrt{p_\mu \sigma^\mu} \xi \\ \sqrt{p_\mu \bar{\sigma}^\mu} \xi \end{bmatrix},$$ with $$\sigma^\mu = (\mathbb{1},\sigma^i) \quad \quad \bar{\sigma}^\mu = (\mathbb{1},-\sigma^i)$$ where "it is understood that in taking the square root of a matrix, we take the positive root of each eigenvalue". Now,

$$ p_\mu \sigma^\mu = \begin{bmatrix} p_0+p_3 & p_1+i p_2 \\ p_1 - i p_2 & p_0 -p_3 \end{bmatrix},$$ and the eigenvalues of this matrix are $p_0 \pm |\vec p|$. So am I to understand that $$\sqrt{p_\mu \sigma^\mu}=\begin{bmatrix} \sqrt{p_0 +|\vec p|} & 0 \\ 0 & \sqrt{p_0 -|\vec p|} \end{bmatrix}\quad ?$$

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