Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 2

Under spatial reflection, ${\varphi }^{\prime }(t,\mathbf{x})={\gamma }_p\varphi (t,-\mathbf{x})$, soBecause ${\gamma }_p^2=1$, the Lorentz scalar ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi$ and the mass term $m^2{\varphi }^2$ are unchanged after evaluating at the reflected point. The spatial change of variables $\mathbf{x}\mapsto -\mathbf{x}$ has unit absolute Jacobian, so the action of the free real scalar field is invariant.

Solved by gpt-5.6-sol high.

Insert the mode expansion of a free field into the parity law and change integration variable $\mathbf{k}\mapsto -\mathbf{k}$. Independence of the plane waves givesSince $\mathcal{P}|0⟩=|0⟩$, an $n$-particle Fock state therefore transforms as

Solved by gpt-5.6-sol high.

LetThe canonical commutation relations give $[N,a_{\mathbf{k}}]=-a_{\mathbf{k}}$ and $[K,a_{\mathbf{k}}]=-a_{-\mathbf{k}}$. Hence exponentiating the adjoint action givesWriting $R{a}_{\mathbf{k}}=a_{-\mathbf{k}}$, one has $R^2=1$ and ${\mathrm{ad}}_K=-R$ on annihilation operators. Therefore

Solved by gpt-5.6-sol high.

The number operator $N$ is self-adjoint. Alsoafter $\mathbf{q}\mapsto -\mathbf{q}$. Thus both exponents are anti-Hermitian and ${\mathcal{P}}_1,{\mathcal{P}}_2$ are unitary. Their product obeysand likewise for creation operators. Both generators annihilate the vacuum, so the product leaves it invariant. Substitution in the free-field mode expansion yieldsHence $\mathcal{P}={\mathcal{P}}_1{\mathcal{P}}_2$ implements parity with intrinsic parity ${\gamma }_p$.

Solved by gpt-5.6-sol high.