Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 2

For a time-independent Hamiltonian operator and a Schrödinger picture observable with no explicit time dependence, the Heisenberg picture operator isDifferentiation gives ${\dot{O}}_H=iH{O}_H-i{O}_{H}H=i[H,O_H]$, soIf $O_S$ itself depends explicitly on time, one must add $e^{iHt}({\partial }_t{O}_S)e^{-iHt}$ to ${\dot{O}}_H$; the displayed commutator relation uses the usual convention of no explicit time dependence for the observable.

For a free real scalar field, the Lagrangian density contains its quadratic kinetic term and mass term:The field's conjugate momentum is ${\pi }_1=\partial {\mathcal{L}}_1/\partial {\dot{\varphi }}_1={\dot{\varphi }}_1$. The positive sign of the time-derivative term uses the metric convention fixed in Question 1.

Apply the Legendre transform in mechanics pointwise to the field Lagrangian density. Using ${\pi }_1={\dot{\varphi }}_1$, the Hamiltonian density isThe free Hamiltonian operator is its spatial integral:A constant vacuum-energy subtraction, if made, does not affect the commutators or the ensuing field equation.

The canonical quantization of a real scalar field promotes the classical equal-time brackets to the canonical commutation relations, in units $\hslash =1$:The Dirac delta distribution expresses locality of the conjugate variables. These are identities of operator-valued distributions, interpreted after spatial smearing where necessary.

Use the free Hamiltonian operator and the canonical commutation relations. The kinetic momentum term givesThe Heisenberg picture equation therefore yields ${\dot{\varphi }}_1={\pi }_1$.

For the momentum commutator, the mass term gives $i{m}_1^2{\varphi }_1(\mathbf{x})$. The gradient term giveswhere integration by parts differentiates the field instead of the Dirac delta distribution. ThusCombining the two first-order operator equations gives the Klein-Gordon equation:Suitable decay or spatial smearing removes the boundary term in the integration by parts.

For the two free real scalar fields, let $\mathcal{L}={\mathcal{L}}_1+{\mathcal{L}}_2$. The infinitesimal internal rotation symmetry of two real scalar fields has $\delta {\varphi }_1=\theta {\varphi }_2$ and $\delta {\varphi }_2=-\theta {\varphi }_1$, with constant $\theta$. Its kinetic terms are invariant because this is an orthogonal rotation of the two-component field.

Under the assumed invariance of the full Lagrangian density, Noether's theorem gives the current per unit transformation parameter:The Noether current and Noether charge are thereforeIndependent species commute at equal time, so these products require no extra ordering sign. With vanishing boundary flux, $Q$ is conserved. Part (vii) checks precisely what the symmetry assumption requires of the masses.

Differentiate the Noether charge and use ${\dot{\varphi }}_a={\pi }_a$ and the Klein-Gordon equation for each species. The momentum products cancel because the two species commute:The first integral is the boundary flux of ${\varphi }_2\nabla {\varphi }_1-{\varphi }_1\nabla {\varphi }_2$, so it vanishes under the same boundary condition used for charge conservation. HenceConservation 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 requiresEquivalently, the local divergence is ${\partial }_{\mu }{j}^{\mu }=(m_2^2-m_1^2){\varphi }_1{\varphi }_2$, exhibiting exactly how unequal masses break the symmetry.