Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 / 1

A global symmetry is spontaneously broken when it preserves the action but does not preserve a chosen ground state. If $G$ is broken to the stabilizer $H$, the degenerate vacua form a vacuum manifold $G/H$. The Goldstone theorem states that a relativistic theory has one massless scalar mode for each broken continuous internal generator, so the standard counting gives $\dim G-\dim H$ Goldstone bosons.

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The continuous symmetry isStrictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are unitary matrices,Every term is therefore unchanged by cyclicity of the matrix trace:and the same conjugation argument applies to $\mathrm{Tr}(M^{†}M)$ and $\mathrm{Tr}[(M^{†}M{)}^2]$.

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Let $x_a\ge 0$ be the eigenvalues of the positive semidefinite matrix $M^{†}M$. The potential isFor $k<0$ and $\lambda >0$, each summand is minimized atso $M_0^{†}{M}_0=v^{2}1$. A symmetry transformation preserves the representative $M_0=v1$ precisely when $L=R$, givingThe number of broken generators is $2N^2-N^2=N^2$, so the Goldstone theorem predicts $N^2$ modes, each a Goldstone boson.

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Every vacuum can be written as $M=vU$ with $U(x)\in U(N)$. Freezing the massive radial modes and substituting this parameterization into the action gives, up to a constant vacuum energy,Writing $U=\exp (i{\pi }^a{T}^a/v)$ displays the $N^2$ Goldstone fields ${\pi }^a$. The omitted terms contain more derivatives or arise from integrating out radial excitations, so this is the leading nonlinear sigma model on the vacuum manifold $U(N)$.

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For Hermitian $M$, the continuous transformations preserving the field space act by conjugation,with the central $U(1)$ acting trivially; there is also the discrete symmetry $M\mapsto -M$. The vacuum equation is $M^2=v^{2}1$, so every vacuum is unitarily conjugate toBecause the integer $p$ cannot change continuously, the vacuum manifold has disconnected componentsOn the $p$th component the unbroken continuous group is $U(p)\times U(N-p)$, and the Goldstone theorem givesGoldstone bosons. The discrete sign symmetry exchanges the components $p$ and $N-p$ but produces no Goldstone mode.

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