Product of spinors in Dirac field anticommutators

Question

I am reading a "A modern introduction to quantum field theory" by Maggiore and on page 88 it shows the anticommutators of the Dirac field:

$$ \{\psi_a(\vec{x},t),\psi_{b}^{\dagger}(\vec{y},t)\}=\delta(\vec{x}-\vec{y})\delta_{ab}. $$

So I am trying to calculate it using the anticommutators for creation and destroy operator for particle and antiparticle:

$$ \{a_{\vec{p}}^{r},{a_{\vec{q}}^{s}}^\dagger\} = \{b_{\vec{p}}^{r},{a_{\vec{q}}^{s}}^\dagger\}= (2\pi)^3\delta(\vec{p}-\vec{q})\delta^{rs} $$

and the expression for the Dirac field:

$$ \psi(x)=\int {d^3p\over (2\pi)^3\sqrt{2E_p}}\sum_{s=1,2}\left({a_{\vec{p}}^{s}} u^s(p)e^{-ipx} + {b_{\vec{p}}^{s}}^\dagger v^s(p)e^{ipx}\right) $$

where $u^s(p)$ and $v^s(p)$ are spinors and $s=1,2$ denote one of the two linear independent spinors

In calculating the filed aunticommutator I encounter products of the spinors, like $u^s(p)\bar u^r(q)$, and I am unsure about what is meant to be this product, cause if it were the usual scalar product of two vectors then I can't obtain the expected anticommtutator expression.

In the same book on page 61 it reports various formulas for $\bar u u$ and $u\bar u$, and mentions that $\bar u u$ is a complex number, while $u\bar u$ is a 4x4 matrix, but it doesn't give any details of the reason.

I desperately looked in other books and in "Quantum field theory" by Schwartz at page 191 he mentions that $u\bar u$ is a spinor outer product. I Looked for outer product in wikipedia and realized that is an easy kind of product that given two vectors it returns a matrix.

So am I correct to assume that when I have expression like $\psi\bar\psi$ the product of the spinor is meant to be an outer product?

When calculating the anticommutator of the Dirac filed I have terms like the following:

$$ \{{a_{\vec{p}}^{s}} u^s(p),{{a_{\vec{q}}^{r}}^\dagger} \bar u^r(q)\}= {a_{\vec{p}}^{s}} {{a_{\vec{q}}^{r}}^\dagger} u^s(p)\bar u^r(q) + {{a_{\vec{q}}^{r}}^\dagger} {a_{\vec{p}}^{s}}\bar u^r(q) u^s(p) $$

If $\bar u^r(q) u^s(p)$ is a complex number I can't get the correct expression for Dirac field anticommutator, I can get the correct expression if it's an 4x4 matrix as $ u^s(p) \bar u^r(q)$. That's why it seems that in expression like $\psi\bar\psi$ the product of the spinor is meant always to be an outer product.

Answer 1

You have to use the relations $$ \sum\limits_s u_a^s(p) \bar{u}_b^s(p) =(p \! \! \! /+m \, \mathbb{I})_{ab} \quad \text{and} \quad \sum\limits_s v_a^s(p) \bar{v}_b^s(p) = (p \! \! \! /-m \, \mathbb{I})_{ab},$$ where $a,b$ are Dirac indices.