Expansion of a Majorana field operator in creation and annihilation operators A Dirac spinor $\Psi=\left(\begin{array}{c}\chi_\alpha\\\psi^\dot{\alpha}\end{array} \right)$ can be expanded in the following way:$\Psi=\int \frac{d^3p}{(2\pi)^3}\sqrt{\frac{1}{2E_p}} \sum_s\left( u_s(p)a_{s p} \exp(-ipx) + v_s(p) b^{\dagger}_{s p} \exp(ipx) \right)$.
In particular we have $v(p) = u(p)^c$ where $x^c$ means the charge conjugated expression of $x$.
I was always wondering how a Majorana spinor would be expanded in a similar way. Actually, I found that this expansion is a exercise in the book of Peskin & Schroeder, exercise 3.4(e). Furthermore I even found solutions on the internet "published" by Zhong-Zhi Xianyou. He states that a 2-dim. Majorana spinor could be expanded in creation and annihilation operators in the following way ($\xi_s$ must be the spin-components of some 2-dim. spinor)
$\chi(x) =\int  \frac{d^3p}{(2\pi)^3}\sqrt{\frac{p\cdot \sigma}{2E_p}} \sum_s\left[\xi_s a_{s p}\exp(-ipx) + (-i\sigma^2)\xi^{*}_s a^{\dagger}_{s p}\exp(ipx)\right]$
However, I cannot understand it. Actually, I have the impression that $\xi_s$ and  $(-i\sigma^2)\xi^{*}_s$ don't transform in the same way under Lorentz-transformations. $\xi\equiv \xi_\alpha$ seems to transform like normal Weyl-spinor, 
whereas $(-i\sigma^2)\xi^{*}_s \equiv \xi^\dot{\alpha}$ seems to transform like a dotted Weyl-spinor but are added up both in the same expression which would mix up the transformation properties.
According to the appendix E in A.Zee's book a charge-conjugated 4-Dirac spinor has the following form:
$\Psi^c=\left(\begin{array}{c}\psi_\alpha\\\chi^\dot{\alpha}\end{array} \right)$
If I apply this representation on $u(p)$, then with $v(p)=u(p)^c$ I get the correct behaviour of $v(p)$ under Lorentz-transformations. Both upper components of $u(p)$ and $v(p)$ are undotted Weyl-spinors whereas the both lower components are both dotted Weyl-spinors. So for the expansion of the Dirac-spinor there is no contradiction (of course not!). 
From this I was tempted to write down my own expression for $\chi(x)$, but finally I realized that $u$ and $v$ are still solutions of the Dirac-equation, therefore the relation for Majorana-spinors $v(p)$=$u^c(p)$=$u(p)$ is a wrong short-cut. So at this point I don't know to proceed further. 
Can somebody help me out? Thank you.
 A: "...seems to transform like a dotted Weyl-spinor but are added up both in the same expression which would mix up the transformation properties..."
Suppose the $\left( 1/2, 0\right)$ Lorentz transformation 
$$
\psi_{a} \to \left(\text{exp}\left[ \frac{\sigma_{\mu\nu}\omega^{\mu\nu}}{2}\right]\psi\right)_{a}, \quad\text{with}\quad \sigma_{\mu\nu} = -\frac{1}{4}(\sigma_{\mu}\tilde{\sigma}_{\nu} - \sigma_{\nu}\tilde{\sigma}_{\mu}),
$$
where
$$
\sigma_\mu = (1, \sigma), \quad \tilde{\sigma}_\mu =(1,-\sigma)
$$ 
Then the spinor $(\sigma_{2}\psi^{*})_{a}$ is transformed as
$$
(\sigma_{2}\psi^{*})_{a} \to \left(\sigma_{2}\text{exp}\left[\frac{\sigma^{*}_{\mu\nu}\omega^{\mu\nu}}{2}\right]\psi^{*}\right)_{a} = \left(\text{exp}\left[\sigma_{2}\sigma_{\mu\nu}^{*}\sigma_{2}\frac{\omega^{\mu\nu}}{2}\right] \sigma_{2}\psi^{*}\right)_{a}
$$
But with the rule $\sigma_{\mu}^{*} = -\sigma_{2}\sigma_{\mu}\sigma_{2}$, $\tilde{\sigma}_{\mu}^{*} = -\sigma_{2}\tilde{\sigma}_{\mu}\sigma_{2}$ the transformation is nothing but
$$
(\sigma_{2}\psi^{*})_{a} \to \left(\text{exp}\left[ \frac{1}{2}\sigma_{\mu\nu}\omega^{\mu\nu}\right]\sigma_{2}\psi^{*}\right)_{a},
$$
which is the same as for ordinary $\psi_{a}$. 
This is quite obvious since the charge conjugation, through which we define the Majorana spinor, is just the analog of complex conjugation on the space of spinors.
...Can somebody help me out?.. 
Just start from the general expansion of the creation-destruction field for the Dirac field:
$$
\hat{\Psi}_{A}(x) = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{\sqrt{(2\pi)^{3}2E_{\mathbf p}}}\left(c_{1}u_{A}^{\sigma}(\mathbf p)\hat{a}^{\dagger}_{\sigma}(\mathbf p)e^{ipx} + c_{2}v_{A}^{\sigma}(\mathbf p)\hat{a}_{\sigma}(\mathbf p)e^{-ipx}\right) 
$$
The Majorana condition
$$
\hat{\Psi}(x) = C\hat{\bar{\Psi}}^{T}(x), \quad C = i\gamma_{0}\gamma_{2}
$$ 
which separates the subspace of the Lorentz group representations $T(\Lambda)$ for which $T(\Lambda)$ and $T^{*}(\Lambda)$ are equivalent, relates $u_{\sigma}(\mathbf p)$ to $v_{\sigma}(\mathbf p)$. After some algebra, we obtain
$$
v_{\sigma}(\mathbf p) = \bar{u}^{T}_{\sigma}(\mathbf p) = -(-1)^{1/2+\sigma}C\gamma_{5}u_{-\sigma}(\mathbf p)
$$
A: The quickest way to get to the expansion of $\chi(x)$ is to continue what you started as @Prahar also suggested. Take a 4-component Majorana field $\psi(x)$ with the Majorana condition [$a_{s}(p)=b_{s}(p)$] and apply the left-chiral projector $P_L$ or take the left-chiral component $\chi_\alpha$. Write $u_s(p)$ and $v_s(p)$ according do Peskin, eqs.(3.50) and (3.62), and extract the left-chiral component, i.e., the first two components. This leads to the expansion in your question. Beware to choose $\xi^s$ and $\eta^s$ so that the convention $u_s^c(p)=v_s(p)$ is satisfied; it is not automatic.
Now, about your impression that $i\sigma_2\xi_s^*$ is transforming in the wrong way under Lorentz, you are forgetting the factor $\sqrt{p\cdot\sigma}$ in front. This factor performs the boost in the $p$ direction and sets the Lorentz transformation behaviour: both spinors are left chiral. 
The spinors $\xi_s$ and $i\sigma_2\xi_s^*$ defines the spin direction in the rest frame and $i\sigma_2(~)^*$ just flips the spin as time reversal in quantum mechanics. For simplicity, you can use the $z$-direction and spin up and down for $\xi_s$.
