The condensate transforms as $\langle\bar q_Lq_R\rangle\mapsto L\langle\bar q_Lq_R\rangle R^\dagger$, so its value proportional to the identity produces <chiral symmetry breaking>
$$
SU(2)_L\times SU(2)_R\longrightarrow SU(2)_V.
$$
The vacuum manifold is the coset $(SU(2)_L\times SU(2)_R)/SU(2)_V\cong SU(2)$ and has three Goldstone modes. A field $U(x)\in SU(2)$ transforms as $U\mapsto LUR^\dagger$. Lorentz invariance and chiral symmetry leave, at two-derivative order, one invariant metric on this coset, so
$$
\mathcal L_\pi=\frac{f_\pi^2}{4}\operatorname{tr}(\partial_\mu U^\dagger\partial^\mu U).
$$
Since the Lagrangian density has dimension four and $U$ is dimensionless, the <pion decay constant> $f_\pi$ has mass dimension one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 305 / 2 / ii / Solution

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{wiki}

The pion decay constant normalizes the axial current matrix element and the leading pion nonlinear-sigma-model action. It has mass dimension one in four spacetime dimensions.


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