Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 34A (Principles of Quantum Mechanics)

${\Delta }_{\psi }Q=\sqrt{⟨Q^2⟩-⟨Q{⟩}^2}$. The Robertson uncertainty principle is

The displacement operator is unitary, so $\Vert |\alpha ⟩\Vert =1$. By coherent state, $A|\alpha ⟩=\alpha |\alpha ⟩$, hence $\Vert A|\alpha ⟩\Vert =|\alpha |$.

Using $X=(A+A^{†})/\sqrt{2\omega }$ and $H=\omega (A^{†}A+1/2)$,

Since $[X,H]=iP$ and $⟨P⟩=\sqrt{2\omega }\Im \alpha$, the bound is
$\Delta X\Delta H\ge \sqrt{\omega /2}|\Im \alpha |$. The left side is $\sqrt{\omega /2}|\alpha |$, so equality holds exactly when $\Re \alpha =0$.

Coherent state evolution givesThe condition from part (d) becomes