 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-b/b6/b/solution

In the

S_{z}

(a, b)^{T}

b = (2 s_{x} /ℏ) a

and

a = (2 s_{x} /ℏ) b

, so a nonzero vector requires

(2 s_{x} /ℏ)^{2} = 1

. Choosing normalized phases gives

∣ + x ⟩ = \frac{∣ + z ⟩ + ∣ - z ⟩}{2} = \frac{1}{2} (11), s_{x} = \frac{ℏ}{2},

∣ - x ⟩ = \frac{∣ + z ⟩ - ∣ - z ⟩}{2} = \frac{1}{2} (1 - 1), s_{x} = - \frac{ℏ}{2} .

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