I'm trying to understand a result in Peskin and Schroeders chapter 3 on the dirac equation, during the free particle calculation for the case $p^0=E>0$. Peskin and schroeder derived this expression: $$u(p)=\left[\begin{matrix}\left((\sqrt{(E+p^3)}\times \frac{(I_{2\times 2}-\sigma^3)}{2}+\sqrt{(E-p^3)}\times \frac{(I_{2\times 2}+\sigma^3)}{2}\right) \xi \\ \left((\sqrt{(E+p^3)}\times \frac{(I_{2\times 2}+\sigma^3)}{2}+\sqrt{(E-p^3)}\times \frac{(I_{2\times 2}-\sigma^3)}{2}\right) \xi \end{matrix}\right].\tag{3.49} $$
And they claim it can be simplified to: $$u(p)=\left[\begin{matrix}\sqrt{p\cdot \sigma}\ \xi \\ \sqrt{p \cdot \bar{\sigma}}\ \xi \end{matrix}\right].\tag{3.50}$$ Here, $$\sigma^{\mu}= (I_{2\times 2}, \sigma^i)\quad\text{and}\quad\bar{\sigma}^{\mu}=(I_{2\times 2},-\sigma^i).\tag{3.41}$$
I tried this: Start by simplifying the projection operators $$(I_{2\times 2}-\sigma^3)/2=\frac{1}{2}\times\left(\left[\begin{matrix}1 &0 \\ 0 & 1\end{matrix}\right] - \left[\begin{matrix}1 &0 \\ 0 &-1\end{matrix}\right]\right)=\left[\begin{matrix}0 & 0 \\0 &1\end{matrix}\right]$$ Similarly, $$(I_{2\times 2}+\sigma^3)/2= \left[\begin{matrix}1 & 0 \\0 &0\end{matrix}\right]$$
Thus, my original expression simplifies down to this:
$$u(p)=\left[\begin{matrix}\left(\left[\begin{matrix}0 &0 \\ 0 &\sqrt{E+p^3} \end{matrix}\right]+ \left[\begin{matrix}\sqrt{E-p^3}& 0\\ 0& 0\end{matrix}\right]\right)\xi \\ \left(\left[\begin{matrix}\sqrt{E+p^3}& 0\\ 0& 0\end{matrix}\right]+ \left[\begin{matrix}0 &0 \\ 0 &\sqrt{E-p^3} \end{matrix}\right]\right)\xi\end{matrix}\right] $$
I can simplify that expression: $$u(p)=\left[\begin{matrix}\left[\begin{matrix}\sqrt{E-p^3} & 0 \\ 0 &\sqrt{E+p^3}\end{matrix}\right] \xi\\ \left[\begin{matrix}\sqrt{E+p^3}& 0 \\ 0 &\sqrt{E-p^3}\end{matrix}\right]\xi\end{matrix}\right] $$
Now, I will try modify the expected result to match this. I know that $p\cdot \sigma =(E,p)\cdot(I_{2\times 2},\sigma^i)$, and in this case, a boost in the 3-direction, I have:
$$p\cdot \sigma=E\cdot I_{2\times 2}+p^3\cdot \sigma ^3=\left[\begin{matrix}E& 0\\ 0& E\end{matrix}\right]+\left[\begin{matrix}p^3 &0 \\ 0& -p^3\end{matrix}\right]=\left[\begin{matrix}E+p^3& 0 \\ 0& E-p^3\end{matrix}\right]$$
I know that the square root of a matrix is the square root of its eigenvalues, so $$\sqrt{p\cdot \sigma}= \left[\begin{matrix}\sqrt{E+p^3}& 0 \\ 0 &\sqrt{E-p^3}\end{matrix}\right]$$ Similarly, $$p\cdot \bar{\sigma}=\left[\begin{matrix}\sqrt{E-p^3}& 0 \\ 0& \sqrt{E+p^3}\end{matrix}\right]$$
So, I have that
$$u(p)=\left[\begin{matrix}\left[\begin{matrix}\sqrt{E+p^3} & 0 \\ 0 &\sqrt{E-p^3}\end{matrix}\right] \xi\\ \left[\begin{matrix}\sqrt{E-p^3}& 0 \\ 0 &\sqrt{E+p^3}\end{matrix}\right]\xi\end{matrix}\right] $$
This is very similar to what I got to, but the signs are switched, and I cannot figure out why. I have checked my algebra and I think it is correct, is there a step missing or are these things equivalent in any way? I'm genuinely lost here, and would love some guidance
