Massless limit for Dirac field I'm a little bit confused about how to take the massless limit of the Dirac field:
\begin{align}
\psi(x)=\int\frac{d^3p}{(2\pi)^2}\frac{1}{\sqrt{E_p}}\sum_{s}\left(a_p^su^s(p)e^{-i p x}+b_p^{s\dagger}v^s(p)e^{ip x}\right)\,,\label{Dirac}
\end{align}
EDITED:
What are the correct $u^s(p)$ and $v^s(p)$ if the mass is zero? For massive particles, Peskin & Schroeder gives the expression:
\begin{align}
u^s(p)=\left(\begin{array}{c}
\sqrt{p\cdot \sigma}\,\xi^s\\
\sqrt{p\cdot \overline{\sigma}}\,\xi^s
\end{array}\right)\,,\qquad
v^s(p)=\left(\begin{array}{c}
\sqrt{p\cdot \sigma}\,\eta^s\\
-\sqrt{p\cdot \overline{\sigma}}\,\eta^s
\end{array}\right)
\end{align}
If I take the massless limit for $u^s(p)$ with $p$ pointing in the $z$-direction, I find
\begin{align}
u^1(\vec{p})=\sqrt{E}\left(\begin{array}{c}
0\\ 0\\ 1\\0
\end{array}\right)\,,\quad u^2(\vec{p})=\sqrt{E}\left(\begin{array}{c}
0\\ 1\\ 0\\0
\end{array}\right)\,,
\end{align}
where I used $\xi^1=(1,0)$ and $\xi^1=(0,1)$. If I do the same for the $v$, I find
\begin{align}
v^1(\vec{p})=\sqrt{E}\left(\begin{array}{c}
0\\ 0\\ -1\\0
\end{array}\right)\,,\quad v^2(\vec{p})=\sqrt{E}\left(\begin{array}{c}
0\\ 1\\ 0\\0
\end{array}\right)\,.
\end{align}
This means that we don't have $\overline{u}^i(p)v^j(p)=0$. I thought that for every fixed momentum $p$, the spinors sets $v^i(p)$ and $u^i(p)$ span with $i=1,2$ each span a two-dimensional space, namely $\mathrm{span}(v^1(p),v^2(p))$ and $\mathrm{span}(u^1(p),u^2(p))$ that are orthogonal to each other. However, from above limits it appears as if this were not the case anymore in the massless limit (very large momentum). Did I take the limits incorrectly or did I miss a crucial point?
 A: By the definition, the coefficient $u_{s}(p)$ is the spinor in front of $e^{-ipx}$, while the coefficient $v_{s}(p)$ is the spinor in front of $e^{ipx}$. There are no continuous Lorentz transformations which change the sign of energy, and therefore your statement about mixing of $u_{s}(p)$ and $v_{s}(p)$ is not clear.
From the other side, note that for any mass, the existence of the term $v_{s}(p)e^{ipx}$  is required by the causality principle: for any two operators $A(x), B(y)$ composed from the relativistic (fermionic) fields $\psi$ the mean value $[A(x),B(y)]$ is zero for space-like intervals, which is reflected in the requirement $[\psi(x),\psi^{\dagger}(y)]_{+} = 0$ for space-like $(y-x)^{2}$'s. The latter is impossible without the $v_{s}$ term.
Note also that $u_{s}(\mathbf p)$ and $v_{s}(\mathbf p)$ are related to each other because the field $\psi(x)$ must be transformed in the definite way, as representing the particle with given spin (helicity) and mass, under the Lorentz transformations. For massive case, for example, such relation reads
$$
u_{s}(\mathbf p) = (-1)^{s+\frac{1}{2}}\gamma_{5}v_{-s}(\mathbf p)
$$
Since the Dirac equation has correct massless limit, such relation holds in the massless case.
Update
I don't understand why you are confused. Massless limit doesn't require $u, v$ to span different sub-spaces; this is because the antiparticle obviously doesn't disappear with $m\to 0$. Rather the different sub-spaces are spanned by $u_{s}, u_{-s}$ and $v_{s}, v_{-s}$ correspondingly, which is simply the statement that the Dirac equation is splitted on two Weyl equations $\sigma^{\mu}p_{\mu}\psi(p)  = 0$ and $\tilde{\sigma}^{\mu}p_{\mu}\psi(p) = 0$.
