Consider the process in which electrons of very high energy scatter from a target. In leading order in $\alpha$, the electron is connected to the target by one photon propagator:
where the blob is, in general, complicated coupling of the virtual photon to the target.
If the initial and final energies of the electron are $E$ and $E'$. Under the high-energy limit, \[p = (E, 0,0,E), \quad p'=(E',E'\sin\theta,0,E'\cos\theta),\tag{1.2}\] which implies \[q^2 = (p'-p)^2 = -2EE'(1-\cos\theta).\tag{1.3}\] In the limit of forward scattering, whatever the energy loss, the photon momentum approaches $q^2 = 0$; thus the reaction is highly peaked in the forward direction. It is tempting to guess that, in this limit, the virtual photon becomes a real photon. Let us investigate in what sense that is true.
(a) The matrix element of the above process is given by \[\mathcal{M} = (-ie)\bar{u}(p')\gamma^\mu u(p)\frac{-ig_{\mu\nu}}{q^2}\hat{\mathcal{M}}^\nu(q),\tag{1.4}\] where $\hat{\mathcal{M}}^\nu(q)$ represents the (in general, complicated) coupling of the virtual photon to the target. Let us analyse the structure of the piece $\bar{u}(p')\gamma^\mu u(p)$. Let $q = (q^0,\vec{q})$ and $\Tilde{q}=(q^0,-\vec{q})$. We can expand the spinor products as \[\bar{u}(p')\gamma^\mu u(p) = Aq^\mu + B\Tilde{q}^\mu + C \epsilon_1^\mu + D\epsilon_2^\mu,\tag{1.5}\] where $A, B, C , D$ are functions of the scattering angle and energy loss and $\epsilon_1, \epsilon_2$ are two unit vectors transverse to $\vec{q}$. By dotting the expression with $q^\mu$, we find \[q_\mu \cdot [\bar{u}(p')\gamma^\mu u(p)] = Aq^2 + Bq\cdot \Tilde{q}.\tag{1.6}\] By Ward identity, we have $q_\mu \bar{u}(p')\gamma^\mu u(p)]=0$. Thus, we have \[B = -A\frac{q^2}{q\cdot \Tilde{q}} \approx A \frac{2EE'(1-\cos\theta)}{(q^0)^2 + \vec{q}^2} \sim \theta^2.\tag{1.7}\] Thus, $B$ can be ignored in the case of forward scattering.
Again, by Ward identity, we have $q\cdot \hat{\mathcal{M}}(q) = 0$. Thus, $A$ can be ignored as well.
(b) In the frame where $p=(E,0,0,E)$, the spinor $u_s(p)$ is given by \[u_s(p) = \begin{pmatrix} \sqrt{p\cdot \sigma}\xi_s\\ \sqrt{p\cdot \bar{\sigma}}\xi_s, \end{pmatrix}\tag{1.8}\] where \[\sigma^3\xi_\pm = \pm\xi_\pm \implies \xi_+ = \begin{pmatrix} 1\\0 \end{pmatrix}, \quad \xi_-=\begin{pmatrix} 0\\1 \end{pmatrix},\tag{1.9}\] and \[\sqrt{p\cdot \sigma} = \sqrt{E(1 - \sigma^3)}, \quad \sqrt{p\cdot \sigma} = \sqrt{E(1 + \sigma^3)}.\tag{1.10}\] Thus, \[\boxed{ u_+(p) = \sqrt{2E}\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix},\quad u_-(p)=\begin{pmatrix} 0\\1\\0\\0 \end{pmatrix}.}\tag{1.11}\]
Next, let us evaluate $u(p')$. Using $p' = (E', E'\sin\theta,0,E'\cos\theta)$, we find \[\hat{p}'\cdot \vec{\sigma} \xi_\pm = \pm \xi_\pm \implies \xi_+ = \begin{pmatrix} \cos\frac{\theta}{2}\\\sin\frac{\theta}{2} \end{pmatrix},\quad \xi_- = \sqrt{2E}\begin{pmatrix} -\sin\frac{\theta}{2}\\\cos\frac{\theta}{2} \end{pmatrix}\tag{1.12}\] and \[\sqrt{p\cdot \sigma} = \sqrt{E'(1-\hat{p}\cdot \vec{\sigma})}, \quad \sqrt{p\cdot \bar{\sigma}} = \sqrt{E'(1+\hat{p}\cdot \vec{\sigma}).}\tag{1.13}\] Thus, we find \[\boxed{ u_+(p') = \sqrt{2E'}\begin{pmatrix} 0\\0\\ \cos\frac{\theta}{2}\\\sin\frac{\theta}{2} \end{pmatrix},\quad u_-(p') = \sqrt{2E'}\begin{pmatrix} -\sin\frac{\theta}{2}\\\cos\frac{\theta}{2}\\0\\0 \end{pmatrix} .}\tag{1.14}\]
The polarisation vectors are defined to be parallel and horizontal to the scattering plane ($x-z$ plane). Thus, \[\boxed{ \epsilon_1^\mu = \frac{1}{\sqrt{E^2+E'^2-2EE'\cos\theta}}\begin{pmatrix} 0, E'\cos\theta-E, 0, -E'\sin\theta \end{pmatrix},\quad \epsilon_2^\mu = (0,0,1,0). }\tag{1.15}\]
Now, let us evaluate $\bar{u}(p')\vec{\gamma}\cdot\vec{\epsilon_i}u(p)$. With some algebra, one should find \[\boxed{ \begin{aligned} &\bar{u}_\pm(p')\vec{\gamma}\cdot\vec{\epsilon_1}u_\pm(p) = -\frac{E'+E}{\abs{E'-E}}\theta\sqrt{EE'} = C,\\ &\bar{u}_\pm(p')\vec{\gamma}\cdot\vec{\epsilon}_2u_\pm(p) = \pm\theta\sqrt{EE'} = D_\pm,\\ &\bar{u}_\pm(p')\vec{\gamma}\cdot\vec{\epsilon}_1u_\mp(p) = \bar{u}_\pm(p')\vec{\gamma}\cdot\vec{\epsilon}_2u_\mp(p) = 0. \end{aligned}}\tag{1.16}\]