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Let's say that I see a particle (i.e. an electron) is moving along the y-axis and has a spin of -1/2 along the z-direction.

Now, if another person who is moving with certain velocity v along the y-direction, how will that person observe the spin of the particle? Will it become a combination of the projections in (x,y,z)? Or will it stay the same?

If we want to be more explicit, let's assume the initial state of the particle, as observed from rest is $|p=p_2,s_z=-1/2 \rangle$ and $\Lambda$ is the Lorentz transform.

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  • $\begingroup$ Read Peskin’s intro on the unitary realization of the Lorentz group on Fock space in his QFT book $\endgroup$ – InertialObserver Sep 7 '18 at 5:55
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Lorentz transformation of spin particles \begin{align*} & \text{The general Lorentz Transformation matrix with rotation is: }\\\\ &\begin{pmatrix} t' \\ \vec{x}' \\ \end{pmatrix} =\begin{bmatrix} \gamma & \gamma \,\vec{v}^T \\ \gamma \,\vec{v}^T & I+\frac{\gamma-1}{\vec{v}^2}\,\vec{v}\,\vec{v}^T \\ \end{bmatrix} \begin{pmatrix} t \\ R\,\vec{x} \\ \end{pmatrix} &(1) \\\\ &\text{with:}\\ \gamma&=\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}}\\\\ \vec{v}&=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\,,\quad\text{velocity boost vector}\\ I&=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix}\,,\quad\text{$3\times3$ identity matrix}\\ R&=3\times 3 \quad \text{ Rotation Matrix} \quad (R^TR=I)\\\\\\ &\text{For your case (I am not sure?):}\\\\ \vec{v}&=\begin{bmatrix} 0 \\ v_y \\ 0 \\ \end{bmatrix}\quad \vec{x}=\begin{bmatrix} 0 \\ y \\ 0 \\ \end{bmatrix}\\\\ R&=\begin{pmatrix} \cos(\psi) & -\sin(\psi) & 0 \\ \sin(\psi) & \cos(\psi) & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\quad \,,\psi=\int \omega_z\,dt=s_z\,t\\\\ &\text{Results } \\\\ & \begin{pmatrix} t' \\ x' \\ y' \\ z' \\ \end{pmatrix}= \begin{pmatrix} \gamma(v_y)\left[t+v_y\,\cos(s_z\,t)\,y\right] \\ -\sin(s_z\,t)\,y \\ \gamma(v_y)\left[v_y\,t+\cos(s_z\,t)\,y\right] \\ 0 \\ \end{pmatrix} \end{align*}

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  • $\begingroup$ The op was talking about unitary representations of the Lorentz group on one-particle quantum states. Your answer is off topic. $\endgroup$ – Giorgio Comitini Sep 7 '18 at 15:13

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