 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-301/2/vii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Differentiate the Noether charge and use

\dot{ϕ}_{a} = π_{a}

and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:

\dot{Q} = \int d^{3} x [π_{2} π_{1} + ϕ_{2} (\nabla^{2} - m_{1}^{2}) ϕ_{1} - π_{1} π_{2} - ϕ_{1} (\nabla^{2} - m_{2}^{2}) ϕ_{2}] = \int d^{3} x [ϕ_{2} \nabla^{2} ϕ_{1} - ϕ_{1} \nabla^{2} ϕ_{2}] + (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

The first integral is the boundary flux of

ϕ_{2} \nabla ϕ_{1} - ϕ_{1} \nabla ϕ_{2}

, so it vanishes under the same boundary condition used for charge conservation. Hence

\frac{d Q}{d t} = (m_{2}^{2} - m_{1}^{2}) \int d^{3} x ϕ_{1} ϕ_{2} .

Conservation must hold for arbitrary configurations, rather than only a state in which this particular integral happens to vanish. Thus the internal rotation symmetry of two real scalar fields requires

m_{1}^{2} = m_{2}^{2}, and therefore m_{1} = m_{2} for nonnegative masses .

Equivalently, the local divergence is

\partial_{μ} j^{μ} = (m_{2}^{2} - m_{1}^{2}) ϕ_{1} ϕ_{2}

, exhibiting exactly how unequal masses break the symmetry.

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