Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 301 / 3 / ii

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- Solution ii

Solution

For the displayed boost,

S [B (β)] = exp (- \frac{β}{2} γ^{0} γ^{1}) = (e^{β σ^{1} /2} 0 0 e^{- β σ^{1} /2}) .

At rest,

p_{1} \cdot σ = p_{1} \cdot \overline{σ} = m I_{2}, u (p_{1}) = m (ξ ξ) .

For

p_{2} = (m cosh β, m sinh β, 0, 0)

, the Pauli matrix identity

e^{β σ^{1}} = cosh β I_{2} + sinh β σ^{1}

gives

p_{2} \cdot σ = m e^{β σ^{1}}, p_{2} \cdot \overline{σ} = m e^{- β σ^{1}} .

Taking the positive matrix square roots,

u (p_{2}) = m (e^{β σ^{1} /2} ξ e^{- β σ^{1} /2} ξ) = S [B (β)] u (p_{1}),

as required.

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