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In Peskin and Schroeder the Dirac equation is solved in the rest frame for solutions with positive frequency: $$\psi(x) = u(p) e^{-ip\cdot x}$$ $$u(p_0) = \sqrt{m} \begin{pmatrix} \xi \\ \xi \end{pmatrix},$$ for any numerical two-component spinor $\xi.$ Boosting to any other frame yields the solution: $$u(p) = \begin{pmatrix} \sqrt{p \cdot \sigma}\ \xi \\ \sqrt{p\cdot \bar{\sigma}}\ \xi \end{pmatrix},$$ where in taking the square root of a matrix, we take the positive root of each eigenvalue.

Then, they summarize: The general solution of the Dirac equation can be written as a linear combination of plane waves. The positive frequency waves are of the form $$\psi(x) = u(p)e^{-ip\cdot x}, \ \ \ p^2 = m^2, \ \ \ p^0 >0.$$ And there are two linearly independent solutions for $u(p),$ $$u^s (p) = \begin{pmatrix} \sqrt{p \cdot \sigma}\ \xi^s \\ \sqrt{p\cdot \bar{\sigma}}\ \xi^s \end{pmatrix}, \ \ \ s=1, 2, $$ which are normalized: $$\bar{u}^r (p) u^s (p) = 2m \delta^{rs}.$$

Next, we can consider the unpolarized cross section for $e^+e^{-} \to \mu^+ \mu^-$ to lowest order. The amplitude is given by: $$\bar{v}^{s'} \left(p'\right) \left(-ie\gamma^{\mu}\right)u^s\left(p\right)\left(\frac{-ig_{\mu \nu}}{q^2}\right)\bar{u}^{r} \left(k\right) \left(-ie\gamma^{\nu}\right)v^{r'}\left(k'\right)$$

Then, I quote

In most experiments the electron and positron beams are unpolarized, so the measured cross section is an average over the electron and positron spins $s$ and $s'$. Muon detectors are normally blind to polarization, so the measured cross section is a sum over the muon spins $r$ and $r'.$ ... We want to compute $$\frac{1}{2}\sum_s \frac{1}{2} \sum_{s'} \sum_r \sum_{r'}|M(s, s' \to r, r')|^2.$$

Why, in order to take the average and the sum, do we only need to sum, rather than integrate, over the spin phase space? Doesn't each incoming particle have an infinite number of spinors? Assuming it is unpolarized, the probability distribution will be uniform, but in principle it still seems like we should integrate over some $\theta$ for spinors $\xi = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$. Instead, what has been done is to assume each incoming particle is in a definite state of "spin-up" or "spin-down" and assign a $50/50$ probability to each.

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The best analogy I can give is that the incoming and outgoing spins are in a mixed state. In a mixed state that is 50% $+z$ and 50% $-z$ for example, expectation values are equal to \begin{equation} \left < A \right > = \mathrm{Tr} [\rho A] = \frac{1}{2} \left < + \right | A \left | + \right > + \frac{1}{2} \left < - \right | A \left | - \right > \end{equation} and it is not correct to integrate over the possible relative coefficients between $\left | + \right >$ and $\left | - \right >$. Note that choosing this basis was merely a matter of convenience. Something which is 50% $+x$ and 50% $-x$ is the exact same mixed state.

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Just to be clear, we can take \begin{equation} \rho_1 = \frac{1}{2} \left | +x \right > \left < +x \right | + \frac{1}{2} \left | -x \right > \left < -x \right | \end{equation} and plug in $\left | \pm x \right > = \frac{1}{\sqrt{2}} \left ( \left | +z \right > \pm \left | -z \right > \right )$ to get \begin{equation} \rho_2 = \frac{1}{2} \left | +z \right > \left < +z \right | + \frac{1}{2} \left | -z \right > \left < -z \right |. \end{equation} Therefore, when we compute expectations with $\rho_2$, we are not missing the contributions of $\rho_1$ but rather doing something equivalent to if we had used $\rho_1$.

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We can't solve it analytically, so the average is taken as a good approximation.We also make the assumption that each particle has finite spinors.

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Because The spine can only have two possible (discrete) values, (+1/2 and -1/2), the integration is appropriate for continuous spectrum.

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