How to understand the completeness of the Dirac spinor and why? I'm searching around to see why $$\sum u^s\bar{u}^s=(\gamma^\mu p_\mu+mc)$$ and $$\sum v^s\bar{v}^s=(\gamma^\mu p_\mu - mc)$$ is called the completeness relation.
Also wondering the same question for the polarization vector $\epsilon^s_\mu$.
 A: Usually the completeness relation for any orthonormal basis of vectors $|\psi_i \rangle$ is
$$\sum_i |\psi_i \rangle \langle \psi_i | = 1$$
...where $1$ is the identity matrix. But for vectors which aren't orthonormal this need not be $1$. Now, we don't write the $u,v$ spinors in bra/ket notation, so the whole thing may look less familiar, but each $u^s = u^{1/2}$ or  $u^{-1/2}$ is itself a vector. Thus from the transpose in the "bar", if $u^s$ was a column vector, then $\bar{u}$ is a row vector, and
$$u^s \bar{u}^s $$
is the outer product of those two vectors, just as in the sum above. The same goes for $v$ and $\epsilon$.
As for the word "completeness", this only works if you sum the entire basis together. Satisfyingly, for any orthonormal basis, the sum will "complete" to $1$. That is the justification of the word in my mind.
Edit: To answer the follow-up of why this quantity wouldn't be the identity matrix, I think the most clear example is just by considering minor modifications to the usual (1,0,0), (0,1,0), (0,0,1) basis that make it not orthonormal.
Let's say that $\{\psi_1, \psi_2, \psi_3\}$ are the usual orthonormal basis as in the previous paragraph. Here I use 3-vectors just as an example. Then
$$|\psi_1\rangle\langle \psi_1 | = \begin{bmatrix}1 & 0 &0 \\0&0&0 \\ 0&0&0\end{bmatrix} 
|\psi_2\rangle\langle \psi_2 | = \begin{bmatrix}0 & 0 &0 \\0&1&0 \\ 0&0&0\end{bmatrix} 
|\psi_3\rangle\langle \psi_3 | = \begin{bmatrix}0 & 0 &0 \\0&0&0 \\ 0&0&1\end{bmatrix}$$
This is why we have
$$|\psi_1\rangle\langle \psi_1 | + |\psi_2\rangle\langle \psi_2 | + |\psi_3\rangle\langle \psi_3 | = \begin{bmatrix}1 & 0 &0 \\0&1&0 \\ 0&0&1\end{bmatrix}$$
From this it should be clear that, for example, if the normalization of $|\psi_1\rangle$ were (2,0,0), then rather than the identity we would get
$$\sum_i |\psi_i \rangle \langle \psi_i | = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
...so it is clear that if the basis is not normalized, we can get any result. Not only that, but there is a significant freedom of choice in the result of the completeness relation if the vectors are not normalized.
But maybe your question is, why do we choose the $u^s$, $v^s$ not to be normalized in the usual way? The reason is that the factor of $2E$ in the normalization
$$\bar{u}_{s'}(p) u_s(p) = \bar{v}_{s'}(p) v_s(p) = 2E(p) \delta_{s s'}$$
is necessary in the integrand in calculations of probability to build the lorentz invariant integration measure
$$\frac{d^3p}{(2 \pi)^3 2 E(p)}$$
so the normalization is forced on us to interpret our results as probabilities. If you want to see this proof let me know.
