Spin part of the Dirac spinors I am reading Peskin's book on QFT and he defines the spin component associated with the particle initially by $\xi^1=\begin{bmatrix}
        1 \\
        0
        \end{bmatrix}$ and $\xi^2=\begin{bmatrix}
        0 \\
        1
        \end{bmatrix}$. Then, later (on page 68 for reference), he wants to represent these in polar coordinates so he defines $\xi(\uparrow)=\begin{bmatrix}
        cos(\frac{\theta}{2}) \\
        e^{i\phi}sin(\frac{\theta}{2})
        \end{bmatrix}$ and $\xi(\downarrow) = \begin{bmatrix}
        -e^{-i\phi}sin(\frac{\theta}{2}) \\
        cos(\frac{\theta}{2}) 
        \end{bmatrix}$ and then he says $\xi^s=(\xi(\uparrow),\xi(\downarrow))$, for s=1,2. I am not sure I understand his new notation. Initially $\xi^s$ was a 2 dimensional vector, but after this redefiniton in polar coordinates, I am not sure what it is. It is like a column vector, made of 2 row vectors? Like a 2x2 matrix? And I am not sure how s=1,2 works here. What is the difference between them? Do we have just $\uparrow$ for s=1 and $\downarrow$ for s=2? I would really appreciate if someone can give me some insight into this notation. Thank you!
 A: This looks like a quantum mechanical problem that I'm used to solve: 
You work in the natural basis for the $z$ projection of the spin, that is $|\uparrow\rangle = \left(\begin{array}{c} 1 \\0 \end{array}\right), \ |\downarrow \rangle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$ with $\hat{S}_z | \sigma \rangle = | \sigma \rangle$, $\vec{S}$ is the vectorial spin operator. Now, you want to change basis for another axis along $\hat{n} = (\sin \theta \cos \phi ) \hat{x} + (\sin \theta \sin \phi)\hat{y} + \cos \theta \hat{z}$ and see how you need to write the eigenvectors of this new spin operator in the $\hat{S}_z$ basis.
What you have to do is diagonalize the $\hat{S}_{z'} = \hat{n} \cdot \vec{S}$ operator, which correspond to the projection of the spin on the general axis $\hat{n}$. The eigenvectors you find are
$$
| \uparrow' \rangle = \left( \begin{array}{c} \cos \frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2} \end{array} \right), \ |\downarrow'\rangle = \left(\begin{array}{c} - e^{-i\phi}\sin\frac{\theta}{2} \\ \cos\frac{\theta}{2} \end{array} \right),
$$
just like your spinors from Peskin and Schoeder. I guess what $\xi^s = (\xi(\uparrow), \xi(\downarrow))$ means is that in the new basis, you have $\xi^1 = \xi(\uparrow) \approx |\uparrow'\rangle$ and $\xi^2 = \xi(\downarrow)$. It means $\xi^s$ is simply the new set of spinors, projected on the general $\hat{n}$ axis, which is written in spherical coordinates.
A: I know this is too old question but I would like to write down another solution to make sure. I will use Peskin and Schoeder (P&S)'s notation.
To rotate spinor, it's enough to work in Quantum Mechanics, but we can use extended version of rotation i.e. Lorentz transformation, of course.
With P&S's notation, $\Lambda_{\frac{1}{2}}$ (generator: $S^{\mu\nu}$) represents Lorentz transf. of spinor, say,
\begin{align*}
 \psi(x) \rightarrow \psi'(x) =  \Lambda_{\frac{1}{2}} \psi(\Lambda^{-1}x).
\end{align*}
Indeed, this representation is reducible. We can form 2-dimensional representation and call two-component objects $\psi_{L, R}$ which is transformed under this as Weyl spinors.
Using explicit form of $\Lambda_{\frac{1}{2}}$, infinitesimal transformation becomes:
\begin{align*}
\psi_L \rightarrow \psi'_L = \left( 1 - i \vec{\theta} \cdot \frac{\vec{\sigma}}{2} - \vec{\beta} \cdot\frac{\vec{\sigma}}{2}\right) \psi_L \\
\psi_R \rightarrow \psi'_R = \left( 1 - i \vec{\theta} \cdot \frac{\vec{\sigma}}{2} + \vec{\beta} \cdot\frac{\vec{\sigma}}{2}\right) \psi_R 
\end{align*}
(This corresponds to (3.37) on P&S.) As we expected, we can see there is only difference in sign of boost coeffecient. Conversely, spatial rotation part is completely same. This means, we need only QM for this problem.
Let's set $\vec{\beta} = 0$. To get spin states along an axis whose polar coordinate is $(\theta, \phi)$, we should calculate these finite version of above transformations (through exponential map):
\begin{align*}
\xi_{\uparrow} = e^{-\frac{i}{2}(\phi-\pi/2) \sigma_z} e^{\frac{i}{2}\theta \sigma_x} \begin{pmatrix}1 \\ 0 \end{pmatrix}, \\
\xi_{\downarrow} = e^{-\frac{i}{2}(\phi-\pi/2) \sigma_z} e^{\frac{i}{2}\theta \sigma_x} \begin{pmatrix}0 \\ 1 \end{pmatrix} \\
\end{align*}
respectively with P&S's notation. However, this is not only way to calculate same quantities (e.g. If we rotate around $y$-axis in first, second $\phi - \pi/2$ becomes $\phi$) so we can enjoy some exercise.
