 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2017/1/b/solution

Let

χ = (5 /3, 2/3)^{T}

, so

χ^{†} χ = 1

(σ \cdot p)^{2} = ∣ p ∣^{2} I

gives

ψ^{†} ψ = ∣ N ∣^{2} (1 + \frac{∣ p ∣ ^{2}}{( E + m ) ^{2}}) = ∣ N ∣^{2} \frac{2 E}{E + m},

using

E^{2} = ∣ p ∣^{2} + m^{2}

. Thus unit spinor probability density is obtained with

N = \frac{E + m}{2 E} .

An arbitrary constant phase is immaterial. For the alternative Lorentz normalization

u^{†} u = 2 E

, choose

N = E + m

. Neither convention makes an infinite plane wave square-integrable; a box-normalized state includes an additional

V^{- 1/2}

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