Polarization in Electron Scattering

Consider the calculation for an incident beam of polarized electrons. The Coulomb scattering of an electron incident with momentum $p_i$ and spin $s_i$, where $s_i\cdot p_i=0$, and summed over final spin states $\pm s_f$, is given by

$$\frac{d\sigma}{d\Omega} = \frac{4Z^2\alpha^2m^2}{\vert\vec{q... ... s_f} \vert\overline{u}(p_f,s_f) \gamma^0 u(p_i,s_i)\vert^2 .$$ (7.336)

In order to take advantage of the trace technique we introduce the spin projection operator

$$\Sigma(s_i) = \frac{1+\gamma_5\not{\;\!\!\!s}_i}{2} ,$$ (7.337)

$$\Sigma(s_i)u(p_i,s_i) = u(p_i,s_i) ,$$ (7.338)

$$\Sigma(s_i)u(p_i,-s_i) = 0 .$$ (7.339)

We have

$$\frac{d\sigma}{d\Omega} = \frac{4Z^2\alpha^2m^2}{\vert\vec{q}\vert^4} \sum_{\pm s_f,s_i} [ ... ...s_f) \gamma_0 \sum(s_i) u(p_i,s_i)] [\overline{u}(p_i,s_i) \gamma_0 u(p_f,s_f)]$$

$$= \frac{4Z^2\alpha^2m^2}{\vert\vec{q}\vert^4} \textrm{Tr} \left[ \g... ...c{\not{\;\!\!\!p}_i + m}{2m} \gamma_0 \frac{\not{\;\!\!\!p}_f + m}{2m} \right].$$ (7.340)

We could also introduce $\Sigma(s)$ twice, both into matrix element and into adjoint, but this is unnecessary. The trace becomes

$$\textrm{Tr} [\gamma_0 (1+\gamma_5\not{\;\!\!\!s}_i) (\not{\;\!\!\!p}_i+m) \gamma_0 (\not{\;\!\!\!p}_f+m)]$$

$$= \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p... ...a_0\gamma_5\not{\;\!\!\!s}_i(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p}_f+m)]$$

$$= \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p... ..._0] + m\textrm{Tr} [\gamma_0\gamma_5\not{\;\!\!\!s}_i\gamma_0\not{\;\!\!\!p}_f]$$

$$= \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p... ...not{\;\!\!\!p}_i] + m\textrm{Tr} [\gamma_5\not{\;\!\!\!s}_i\not{\;\!\!\!p}_f^*]$$

$$= \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p}_f+m)] ,$$ (7.341)

where $p_f^*=(p_f^0,-\vec{p}_f)$ and we have used $\textrm{Tr}[\gamma_5\not{a}\not{b}]=0$ in the last two terms. The additional trace involving the spin vector vanishes and we once again return to the Mott formula. Our result, that the differential cross-section is the same for a polarized incident beam as for an unpolarized incident beam is a special consequence of the use of lowest order perturbation theory only and is not true generally. In other words, Coulomb scattering of electrons does not lead to a polarization of the beam to lowest order.

We now develop some useful relationships for the polarization vector of an electron. The polarization vector $s$ satisfies

$$s^2 = -1 = (s^0)^2 - \vec{s}\cdot\vec{s} \quad\textrm{and}\quad s\cdot p = 0$$ (7.342)

or

$$s^0 p^0 - \vec{s}\cdot\vec{p} = 0$$ (7.343)

$$s^0 E = \vec{s}\cdot\vec{p}$$ (7.344)

$$s^0 = \vec{s}\cdot\frac{\vec{p}}{E} = \vec{s}\cdot\vec{\beta} .$$ (7.345)

Therefore

$$-1 = (\vec{s}\cdot\vec{\beta})^2 - \vert\vec{s}\vert^2$$ (7.346)

$$\vert\vec{s}\vert^2 - (\vec{s}\cdot\vec{\beta})^2 = 1$$ (7.347)

$$\vert\vec{s}\vert^2 - \vert\vec{s}\vert^2(\hat{s}\cdot\vec{\beta})^2 = 1$$ (7.348)

$$\vert\vec{s}\vert = \frac{1}{\sqrt{1 - (\vec{\beta}\cdot\hat{s})^2}} ,$$ (7.349)

where $\hat{s}$ is a unit vector along $\vec{s}$.

For the electron spin polarization lined up along $\vec{\beta}$, denoting a right-handed electron with polarization $s_R$, we have

$$\vec{\beta}\cdot\hat{s}_R = \beta$$ (7.350)

and

$$\vert\vec{s}_R\vert = \frac{1}{\sqrt{1-\beta^2}} = \frac{E}{m} \quad\textrm{and}\quad s_R^0 = \beta \vert\vec{s}_R\vert .$$ (7.351)

Similarly, for spin polarizaton antiparallel to $\vec{\beta}$, denoted a left-handed electron with $s_L = -s_R$, we have

$$\vec{\beta}\cdot\hat{s}_L = -\beta ,$$ (7.352)

$$\vert\vec{s}_L\vert = \frac{E}{m} ,$$ (7.353)

$$s_L^0 = -\beta\vert\vec{s}_L\vert.$$ (7.354)

Similar formulas apply to ingoing and outgoing scattered electrons. The eigenstates of $\Sigma(s)$ with $s=\pm s_R=\mp s_L$ are known as positive- and negative-helicity eigenstates.

We now consider an incident electron with spin lined up along its direction of motion and compute the polarization of the scattered electron as a function of the scattering angle. The polarizaton of scattered electrons is measured by

$$P = \frac{N_R-N_L}{N_R+N_L} ,$$ (7.355)

where $N_R$ denotes the number emerging with positive helicity (or polarized right-handed) and $N_L$ the number with negative helicity. $N_R,N_L,P$ are generally functions of the scattering energy and angle.

The polarization for Coulomb scattering of a right-handed electron, ${s_i}_R$, is given by

$$P_R = \frac{ \vert\overline{u}(p_f,s_{fR}) \gamma_0 u(p_i,s_{iR})\vert^... ...,s_{iR})\vert^2 + \vert\overline{u}(p_f,s_{fL}) \gamma_0 u(p_i,s_{iR})\vert^2 }$$

$$= \frac{ \textrm{Tr} \left[ \gamma_0 \frac{1+\gamma_5 \not{\;\!\!\!... ...\not{\;\!\!\!p}_i + m}{2m} \gamma_0 \frac{\not{\;\!\!\!p}_f + m}{2m} \right]} .$$ (7.356)

Since we have just shown that the trace involving a single $\gamma_5\not{\;\!\!\!s}$ vanishes, we have

$$P_R = \frac{\textrm{Tr} \left[ \gamma_0 \gamma_5 \not{\;\!\!... ... + m}{2m} \gamma_0 \frac{\not{\;\!\!\!p}_f + m}{2m} \right]} .$$ (7.357)

The trace in the denominator has been evaluated before

$$\textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p}_f+m)] = 8E_iE_f - 4p_i\cdot p_f + 4m^2$$

$$= 4E^2 + 4\vec{p}^{\ 2}\cos\theta + 4m^2$$

$$= 4E^2(1+\cos\theta) + 4m^2(1-\cos\theta)$$

$$= 8E^2\cos^2(\theta/2) + 8m^2\sin^2(\theta/2) .$$ (7.358)

The trace in the numerator becomes

$$\textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R(\not{\;\!\!\!p}_i+m)\gamma_0 \gamma_5 {\not{\;\!\!\!s}_f}_R(\not{\;\!\!\!p}_f+m)]$$

$$= \textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R\not{\;\!\!\!p}_... ...}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R\gamma_0 \gamma_5 {\not{\;\!\!\!s}_f}_R]$$

$$= -\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\not{\;\!\!\!p}_i\gamma... ... + m^2\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\gamma_0 {\not{\;\!\!\!s}_f}_R].$$ (7.359)

The second term becomes

$$m^2\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\gamma_0{\not{\;\!\!\!s}_f}_R] = m^2\textrm{Tr}[{\not{\;\!\!\!s}_i^*}_R{\not{\;\!\!\!s}_f}_R]$$

$$= 4m^2 {s_i^*}_R {s_f}_R$$

$$= 4m^2\frac{E_iE_f}{m^2}\left(\beta_i\beta_f + \cos\theta\right)$$

$$= 4(p^2 + E^2\cos\theta) ,$$ (7.360)

where ${s_i^*}_R=({s_i^0}_R,-{\vec{s_i}}_R)$. The first term in equation 7.360 is a rather involved calculation and we have used FORM to evaluate it. The result is

$$-\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\not{\;\!\!\!p}_i\... ...a_0{\not{\;\!\!\!s}_f}_R\not{\;\!\!\!p}_f] = 4m^2\cos\theta .$$ (7.361)

The total trace in the numerator of equation 7.358 now becomes

$$\textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R(\not{\;\!\!\!p}_i+m)\gamma_0 \gamma_5 {\not{\;\!\!\!s}_f}_R(\not{\;\!\!\!p}_f+m)] = 4m^2\cos\theta + 4p^2 + 4E^2\cos\theta$$

$$= 8E^2\cos(\theta/2) - 8m^2\sin(\theta/2)).$$ (7.362)

The polarization thus becomes

$$P_R = 1 - \left[ \frac{2m^2\sin^2(\theta/2)}{E^2\cos^2(\theta/2) + m^2\sin^2(\theta/2)} \right] .$$ (7.363)

In the relativistic, limit $m/E\rightarrow 0$ or $\beta\rightarrow 1$ and we find $P_R\rightarrow 1$. Thus no de-polarization of incident electrons occurs in the high energy limit of Coulomb scattering.

In the nonrelativistic limit, $E\rightarrow m$ and this reduces to

$$P\approx 1 - 2\sin^2\frac{\theta}{2} = \cos\theta .$$ (7.364)

This is just the geometric overlap between the initial and final quantization axis

$$\cos\theta = \frac{\vec{p}_i\cdot\vec{p}_f}{\vert\vec{p}\vert^2} ,$$ (7.365)

and indicates, that the spin is not influenced at all by the collision, when viewed from a fixed system.

For an incident electron beam that is not completely polarized, but only partially polarized along its direction of motion, we define $p_R$ as the fraction with positive helicity and $p_L=1-p_R$ the fraction with negative helicity. Using $p\equiv p_R - p_L$ to denote the polarization of the incident electrons, we can show that $P = pP_R$. Using

$$p_R\frac{1+\gamma_5\not{\;\!\!\!s}_{iR}}{2} + p_L\frac{1-\ga... ...;\!\!\!s}_{iR}}{2} = \frac{1+p\gamma_5\not{\;\!\!\!s}_{iR}}{2}$$ (7.366)

and inserting this identity into the expression for $P_R$ gives

$$P = p\left[ 1 - \frac{2m^2\sin^2(\theta/2)}{E^2\cos^2(\theta/2) + m^2\sin^2(\theta/2)} \right] .$$ (7.367)

For the special case of $p=0$ we see that any initially unpolarized beam of electrons remains unpolarized in Coulomb scattering.