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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 .
\end{displaymath} (7.336)

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

$\displaystyle \Sigma(s_i)$ $\textstyle =$ $\displaystyle \frac{1+\gamma_5\not{\;\!\!\!s}_i}{2} ,$ (7.337)
$\displaystyle \Sigma(s_i)u(p_i,s_i)$ $\textstyle =$ $\displaystyle u(p_i,s_i) ,$ (7.338)
$\displaystyle \Sigma(s_i)u(p_i,-s_i)$ $\textstyle =$ $\displaystyle 0 .$ (7.339)

We have

$\displaystyle \frac{d\sigma}{d\Omega}$ $\textstyle =$ $\displaystyle \frac{4Z^2\alpha^2m^2}{\vert\vec{q}\vert^4}
\sum_{\pm s_f,s_i} [ ...
...s_f) \gamma_0 \sum(s_i)
u(p_i,s_i)] [\overline{u}(p_i,s_i) \gamma_0 u(p_f,s_f)]$  
  $\textstyle =$ $\displaystyle \frac{4Z^2\alpha^2m^2}{\vert\vec{q}\vert^4} \textrm{Tr} \left[ \g...
...c{\not{\;\!\!\!p}_i +
m}{2m} \gamma_0 \frac{\not{\;\!\!\!p}_f + m}{2m} \right].$ (7.340)

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

    $\displaystyle \textrm{Tr} [\gamma_0 (1+\gamma_5\not{\;\!\!\!s}_i) (\not{\;\!\!\!p}_i+m) \gamma_0
  $\textstyle =$ $\displaystyle \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p...
  $\textstyle =$ $\displaystyle \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p...
..._0] +
m\textrm{Tr} [\gamma_0\gamma_5\not{\;\!\!\!s}_i\gamma_0\not{\;\!\!\!p}_f]$  
  $\textstyle =$ $\displaystyle \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p...
...not{\;\!\!\!p}_i] +
m\textrm{Tr} [\gamma_5\not{\;\!\!\!s}_i\not{\;\!\!\!p}_f^*]$  
  $\textstyle =$ $\displaystyle \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p}_f+m)] ,$ (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
\end{displaymath} (7.342)


$\displaystyle s^0 p^0 - \vec{s}\cdot\vec{p}$ $\textstyle =$ $\displaystyle 0$ (7.343)
$\displaystyle s^0 E$ $\textstyle =$ $\displaystyle \vec{s}\cdot\vec{p}$ (7.344)
$\displaystyle s^0$ $\textstyle =$ $\displaystyle \vec{s}\cdot\frac{\vec{p}}{E} =
\vec{s}\cdot\vec{\beta} .$ (7.345)


$\displaystyle -1$ $\textstyle =$ $\displaystyle (\vec{s}\cdot\vec{\beta})^2 - \vert\vec{s}\vert^2$ (7.346)
$\displaystyle \vert\vec{s}\vert^2 - (\vec{s}\cdot\vec{\beta})^2$ $\textstyle =$ $\displaystyle 1$ (7.347)
$\displaystyle \vert\vec{s}\vert^2 - \vert\vec{s}\vert^2(\hat{s}\cdot\vec{\beta})^2$ $\textstyle =$ $\displaystyle 1$ (7.348)
$\displaystyle \vert\vec{s}\vert$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{1 - (\vec{\beta}\cdot\hat{s})^2}} ,$ (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
\end{displaymath} (7.350)


\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 .
\end{displaymath} (7.351)

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

$\displaystyle \vec{\beta}\cdot\hat{s}_L$ $\textstyle =$ $\displaystyle -\beta ,$ (7.352)
$\displaystyle \vert\vec{s}_L\vert$ $\textstyle =$ $\displaystyle \frac{E}{m} ,$ (7.353)
$\displaystyle s_L^0$ $\textstyle =$ $\displaystyle -\beta\vert\vec{s}_L\vert.$ (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} ,
\end{displaymath} (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

$\displaystyle P_R$ $\textstyle =$ $\displaystyle \frac{ \vert\overline{u}(p_f,s_{fR}) \gamma_0 u(p_i,s_{iR})\vert^...
...,s_{iR})\vert^2 +
\vert\overline{u}(p_f,s_{fL}) \gamma_0 u(p_i,s_{iR})\vert^2 }$  
  $\textstyle =$ $\displaystyle \frac{
\textrm{Tr} \left[ \gamma_0 \frac{1+\gamma_5 \not{\;\!\!\!...
...\not{\;\!\!\!p}_i + m}{2m} \gamma_0
\frac{\not{\;\!\!\!p}_f + m}{2m} \right]} .$ (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]} .
\end{displaymath} (7.357)

The trace in the denominator has been evaluated before

$\displaystyle \textrm{Tr} [\gamma_0(\not{\;\!\!\!p}_i+m)\gamma_0(\not{\;\!\!\!p}_f+m)]$ $\textstyle =$ $\displaystyle 8E_iE_f -
4p_i\cdot p_f + 4m^2$  
  $\textstyle =$ $\displaystyle 4E^2 + 4\vec{p}^{\ 2}\cos\theta + 4m^2$  
  $\textstyle =$ $\displaystyle 4E^2(1+\cos\theta) + 4m^2(1-\cos\theta)$  
  $\textstyle =$ $\displaystyle 8E^2\cos^2(\theta/2) + 8m^2\sin^2(\theta/2) .$ (7.358)

The trace in the numerator becomes

    $\displaystyle \textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R(\not{\;\!\!\!p}_i+m)\gamma_0
\gamma_5 {\not{\;\!\!\!s}_f}_R(\not{\;\!\!\!p}_f+m)]$  
  $\textstyle =$ $\displaystyle \textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R\not{\;\!\!\!p}_...
...}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R\gamma_0 \gamma_5
  $\textstyle =$ $\displaystyle -\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\not{\;\!\!\!p}_i\gamma...
... +
m^2\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\gamma_0 {\not{\;\!\!\!s}_f}_R].$ (7.359)

The second term becomes

$\displaystyle m^2\textrm{Tr}[\gamma_0{\not{\;\!\!\!s}_i}_R\gamma_0{\not{\;\!\!\!s}_f}_R]$ $\textstyle =$ $\displaystyle m^2\textrm{Tr}[{\not{\;\!\!\!s}_i^*}_R{\not{\;\!\!\!s}_f}_R]$  
  $\textstyle =$ $\displaystyle 4m^2 {s_i^*}_R {s_f}_R$  
  $\textstyle =$ $\displaystyle 4m^2\frac{E_iE_f}{m^2}\left(\beta_i\beta_f + \cos\theta\right)$  
  $\textstyle =$ $\displaystyle 4(p^2 + E^2\cos\theta) ,$ (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

= 4m^2\cos\theta .
\end{displaymath} (7.361)

The total trace in the numerator of equation 7.358 now becomes

$\displaystyle \textrm{Tr}[\gamma_0\gamma_5{\not{\;\!\!\!s}_i}_R(\not{\;\!\!\!p}_i+m)\gamma_0
\gamma_5 {\not{\;\!\!\!s}_f}_R(\not{\;\!\!\!p}_f+m)]$ $\textstyle =$ $\displaystyle 4m^2\cos\theta + 4p^2 +
  $\textstyle =$ $\displaystyle 8E^2\cos(\theta/2) - 8m^2\sin(\theta/2)).$ (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] .
\end{displaymath} (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 .
\end{displaymath} (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} ,
\end{displaymath} (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} +
...;\!\!\!s}_{iR}}{2} = \frac{1+p\gamma_5\not{\;\!\!\!s}_{iR}}{2}
\end{displaymath} (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] .
\end{displaymath} (7.367)

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

next up previous contents index
Next: Higher-Order Calculations: Tree and Up: QED Processes Previous: Electron-Positron Scattering
Douglas M. Gingrich (gingrich@