quantum mechanics - Why, in spin sums, we sum over final spin states and average over initial states?

I am reading Halzen's book about quarks and leptons and on page 120 he talks about spin sums.

He says that in order to calculate the amplitude between unpolarized states we have to sum over FINAL spin states and average over INITIAL states. Why is this so? why not sum over initial states and average over finals, or sum over initial and final spin states and average also over initial and final states?

Answer

For the incoming state, you don't know which spin state the particle is in, so you should average over the possible states. But you can measure the spin of the outgoing state, so to get the total cross section you should add up the cross sections for each spin.

More formally, an unpolarized incoming particle should be described as a density matrix,

$$\def\bra#1{\mathinner{\langle{#1}|}} \def\ket#1{\mathinner{|{#1}\rangle}} \rho(t=-\infty) = \frac{1}{2}( \ket{p\uparrow} \bra{p\uparrow} + \ket{p\downarrow}\bra{p\downarrow})$$ where the $\frac{1}{2}$ is for the normalization of the density matrix, $\operatorname{tr} \rho = 1$. Here $p$ is the momentum and this is of course for spin $\frac{1}{2}$ but the generalization to spin $1$ and higher will be obvious.

The density matrix will evolve to $\rho(t=\infty) = S \rho(t=-\infty) S^\dagger$ just by definition of the $S$-matrix. You want to calculate the probability to end up with momentum $q$, regardless of spin, at $t=\infty$. This is the expectation value at $t=\infty$ of the projection operator

$$P(q) = P(q,\uparrow) + P(q,\downarrow) = \ket{q\uparrow}\bra{q\uparrow} + \ket{q\downarrow}\bra{q\downarrow}.$$

The expectation value of the projection operator is

$$\langle P(q) \rangle = \operatorname{tr}(\rho P(q) ).$$ There will be four terms in the trace. One of them is $$T_{\uparrow\uparrow} = \frac{1}{2}\operatorname{tr}\big(S\ket{p\uparrow}\bra{p\uparrow}S^\dagger\ket{q\uparrow}\bra{q\uparrow}\big).$$ Note that in the middle is something that is just a number, and a familiar number too. $\bra{p\uparrow}S^\dagger\ket{q\uparrow} = (\bra{q\uparrow}S\ket{p\uparrow})^*$ is (the complex conjugate) of an S-matrix element. It is easy to realize that $$\operatorname{tr}\big(S\ket{p\uparrow}\bra{q\uparrow}\big) = \bra{q\uparrow}S\ket{p\uparrow}$$ and therefore $$T_{\uparrow\uparrow} = \frac{1}{2} |\bra{q\uparrow}S\ket{p\uparrow}|^2.$$ The other terms in the trace are similar and we end up with $$\langle P(q) \rangle = \frac{1}{2} \sum_{r,s} |S(p,r;q,s)|^2$$ where $r,s \in \{\uparrow, \downarrow\}$ are incoming and outgoing spins, respectively, and $S(p,r;q,s)$ is the $S$-matrix element. As you can see we recover the prescription average over incoming, and sum over outgoing.