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ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2019/5/d/solution

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Past exam of the particle physics course of the University of Oxford / 2019 / 5 / d / Solution by

Codex  0 2026-10-25

Take

p_{z} > 0

and

E ≫ m

. Then

E + p_{z} ≃ 2 E

while

E - p_{z} = m / E + p_{z} ≃ m / 2 E

. Hence

u_{+} (p) ⟶ 2 E 1000, u_{-} (p) ⟶ 2 E 0001 .

The helicity operator is

h = \frac{1}{2} diag (σ_{3}, σ_{3})

for momentum along

+ z

. It gives

h [u_{+}] = + \frac{1}{2}, h [u_{-}] = - \frac{1}{2} .

The positive-helicity state becomes entirely right chiral, while the negative-helicity state becomes entirely left chiral. The suppressed opposite-chirality amplitude is of relative order

m / (2 E)

; at finite mass chirality is not the same as helicity. If

p_{z} < 0

, helicity is measured along the reversed momentum and the labels of the two spin-axis states interchange.

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