Two projectile protons \(P_{1}\) and \(P_{2}\) both with spin up (along the \(+z\) direction) are scattered from another fixed target proton \(T\) with spin up at rest in the \(xy\) plane, as shown in the figure. They scatter one at a time. The nuclear interaction potential between both the projectiles and the target proton is \(\lambda\vec{L} \cdot \vec{S}\), where \(\vec{L}\) is the orbital angular momentum of the system with respect to the target, \(\vec{S}\) is the spin angular momentum of the system and \(\lambda\) is a negative constant in appropriate units. Which one of the following is correct?
GATE Physics - 2025,
Answer: B
Both protons are scattered upward, in the \(+y\) direction.
Both spins point along \(+z\), so \(\vec{S}\) is along \(+z\). The orbital angular momentum about the target is \(\vec{L} = \vec{r} \times \vec{p}\) with \(\vec{p}\) along \(+x\): for \(P_{1}\), passing above (\(y \gt 0\)), \(L_{z} = -yp \lt 0\); for \(P_{2}\), passing below, \(L_{z} \gt 0\).
With \(\lambda \lt 0\), the potential \(\lambda\vec{L}\cdot\vec{S}\) is positive (repulsive) for \(P_{1}\), which is pushed away from the target, that is upward. For \(P_{2}\) it is negative (attractive), so \(P_{2}\) is pulled towards the target, which is also upward.
This left-right asymmetry is how spin-orbit coupling polarises a scattered beam. The options that send the two protons in opposite directions, or both downward, get the sign of \(\lambda\vec{L}\cdot\vec{S}\) wrong for at least one of them.
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