Electron Scattering from Tensor-Polarized Deuterons in the VEPP-3 Electron Storage Ring

Abstract

In the plane wave impulse approximation, the differential cross section for unpolarized elastic electron-deuteron scattering may be written in the familiar Rosenbluth form:

$$ \frac{{d\sigma }}{{d\Omega }} = {\sigma _m}\left[ {A\left( {{Q^2}} \right) + B\left( {{Q^2}} \right){{\tan }^2}\left( {{\raise0.7ex\hbox{$\theta $} \!\mathord{\left/ {\vphantom {\theta 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \right)} \right] \equiv {\sigma _0} $$

(1)

where σ _m is the Mott cross section, and A and B are given in terms of the deuteron charge (G _c ), quadrupole (G _q ) and magnetic (G _m ) form factors:

$$ A\left( {{Q^2}} \right) = G_c^2 + \left( {8{\tau ^2}/9} \right)G_q^2 + \left( {2\tau /3} \right)G_m^2{\text{ }}B\left( {{Q^2}} \right) = 4\tau \left( {1 + \tau } \right)G_m^2/3 $$

(2)

with Q ² the square of the 4-momentum transfer and τ = Q ²/4m ² _d . Thus, a Rosenbluth separation may be used to extract A and B (and hence G _m ) from scattering data, but G _c and G _q are not separated. To isolate these form factors requires the use of polarization techniques. Following the Madison convention¹, the scattering of unpolarized electrons from a tensor polarized deuteron is described by the cross section:²

$$ \frac{{d\sigma }}{{d\Omega }} = {\sigma _0}\left[ {1 + {T_{20}}{t_{20}} + 2{T_{21}}\operatorname{Re} \left( {{t_{21}}} \right) + 2{T_{22}}\operatorname{Re} \left( {{t_{22}}} \right)} \right] $$

(3)

in which T _2i and t _2i are, respectively, the components of the analyzing power and polarization tensors in a spherical basis. For moderate momentum transfers and suitably chosen polarization directions, the terms involving T ₂₁ and T ₂₂ are small, and may be ignored for the moment. The tensor analyzing power, T ₂₀, is given by:

$$ {T_{20}} = - \sqrt 2 \left[ {X\left( {X + 2} \right) + {Y \mathord{\left/ {\vphantom {Y 2}} \right. \kern-\nulldelimiterspace} 2}} \right]/\left[ {2\left( {{X^2} + 1} \right) + 1} \right]{\text{ }}X = \frac{2}{3}\tau \left( {\frac{{{G_q}}}{{{G_c}}}} \right){\text{ }}Y = \frac{1}{3}\tau {\left( {\frac{{{G_m}}}{{{G_c}}}} \right)^2}\left[ {1 + 2\left( {1 + \tau } \right){{\tan }^2}\left( {{\raise0.7ex\hbox{$\theta $} \!\mathord{\left/ {\vphantom {\theta 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \right)} \right] $$

(4)

while the tensor polarization, t ₂₀, is:

$$ {t_{20}} = {p_{zz}}{P_2}\left( {\hat n \cdot \hat q} \right)/\sqrt 2 $$

(5)

in which P ₂ is the second Legendre polynomial, $\hat n$ is the polarization direction, $\hat n$ is the momentum transfer direction, and p _zz , the polarization in a cartesian basis, is 1 — 3 n ₀, with n ₀ being the fraction of deuterons with zero spin projection.