Polarization Differences in the Reactions <span class="aps-inline-formula"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><mi mathvariant="normal">D</mi><mn></mn><mo>(</mo><mi>d</mi><mo>,</mo><mi>p</mi><mo>)</mo><mn></mn><mi mathvariant="normal">T</mi></math></span> and <span class="aps-inline-formula"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><mi mathvariant="normal">D</mi><mn></mn><mo>(</mo><mi>d</mi><mo>,</mo><mi>n</mi><mo>)</mo><mn></mn><mrow><mmultiscripts><mrow><mi mathvariant="normal">He</mi></mrow><mprescripts></mprescripts><mrow></mrow><mrow><mn>3</mn></mrow><mrow></mrow><mrow></mrow></mmultiscripts></mrow></math></span> and Charge Symmetry

R. A. Hardekopf; R. L. Walter; T. B. Clegg

doi:10.1103/PhysRevLett.28.760

Polarization Differences in the Reactions $D (d, p) T$ and $D (d, n) {}^{3}{He}$ and Charge Symmetry

R. A. Hardekopf, R. L. Walter, and T. B. Clegg

Phys. Rev. Lett. 28, 760 – Published 20 March 1972

Abstract

Accurate measurements have been made of the analyzing power for the reaction $T (p, d) D$ which are equivalent to proton polarizations from $D (d, p) T$ . If allowance is made for the reaction $Q$ -value difference between $D (d, p) T$ and $D (d, n) {}^{3}{He}$ , thus comparing the data at the same outgoing neutron and proton energies, the polarizations are identical. This agreement eliminates much of the concern about charge-symmetry violation in the $d + d$ reactions.