Past exam of the gauge field theory course of the University of Cambridge / 2016 / Paper 1 / 1 / Solution

A variational symmetry of a Lagrangian density maps admissible field configurations to configurations with the same action, up to an allowed boundary contribution. Thus (i) the action is invariant for suitable boundary conditions, while (ii) the Lagrangian density need only change by a total divergence, rather than be pointwise invariant. For an internal symmetry of a classical field theory, the coordinates are fixed. A spacetime transformation also changes the integration measure and must be treated accordingly.

A conserved current satisfies ${\partial }_{\mu }{j}^{\mu }=0$ on shell. The Noether charge $Q=\int j^0{d}^{3}x$ then obeys $\dot{Q}=-{\int }_{\partial V}\mathbf{j}\cdot d\mathbf{S}$, so conservation of the integrated charge additionally requires vanishing net boundary flux.

For several real field components $q_A$ and a constant infinitesimal parameter, write $\delta q_A=\epsilon X_A$ and $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$. The product rule givesSubtract the specified divergence. The Euler-Lagrange field equations make the first term vanish, proving the Noether theorem withFor a complex scalar field, treat the field and its complex conjugate as independent variables in this calculation; this is equivalent to using its two real components. This is the Noether current for a first-derivative scalar field, summed over components.

Collect the two complex fields into $\Phi =({\varphi }_1,{\varphi }_2{)}^T$. Their Lagrangian density has the radial formFor a constant Hermitian matrix $T$, the unitary transformation $\delta \Phi =i\epsilon T\Phi$ preserves both the kinetic contraction and ${\Phi }^{†}\Phi$. In particular the two specified transformations have generators $T_1={\sigma }_1$ and $T_2={\sigma }_2$, the first two Pauli matrices. Their Noether currents are instances of the unitary scalar multiplet currentExplicitly,An overall change of current sign corresponds to reversing the associated symmetry parameter; the signs here follow the two transformations as written.

The matrix commutator of their infinitesimal matrices is $[i{\sigma }_1,i{\sigma }_2]=-2i{\sigma }_3$. The antisymmetric difference of successive transformations therefore supplies a third symmetry, after a nonzero rescaling of its parameter:Its current isThe three traceless matrices ${\sigma }_a/2$ obey $[{\sigma }_a/2,{\sigma }_b/2]=i{ϵ}_{abc}{\sigma }_c/2$. Their exponentials act as SU(2) on the complex doublet. This identifies both the Lie algebra and the group acting on these fields: the central element $-I$ changes $\Phi$ and is not the identity transformation, so the doublet action is not an action of $SO(3)$ obtained by identifying $I$ with $-I$.

There is also a common global phase symmetry of a complex scalar field, $\Phi \mapsto e^{i\alpha }\Phi$, with current $j_I^{\mu }$. Together with SU(2) this gives $U(2)\simeq (SU(2)\times U(1))/{\mathbb{Z}}_2$ among complex-linear transformations. Independent constant phases of ${\varphi }_1$ and ${\varphi }_2$, and field interchange, are already contained in this group.

In fact this particular scalar potential has a larger real internal symmetry. Write ${\varphi }_1=(q_1+i{q}_2)/\sqrt{2}$ and ${\varphi }_2=(q_3+i{q}_4)/\sqrt{2}$; thenThe O(4) symmetry of a complex scalar doublet includes transformations mixing $\Phi$ with ${\Phi }^{\ast }$, which need not be complex-linear. Its connected part has six rotation generators, so listing $U(2)$ alone would not exhaust the internal symmetries of the displayed isolated theory. Individual field conjugations are examples of its discrete transformations. Finally the spacetime-independent scalar coefficients and Lorentz contractions give Poincaré symmetry, and the real scalar action also permits spatial reflection and time reversal. These statements concern the action; a chosen vacuum or external boundary conditions can preserve fewer symmetries. Special parameter choices can introduce additional symmetries, but are not required here.