Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 14D (Quantum Mechanics)

For a normalized state, writeThen ${\Delta }_{\psi }A=\Vert A^{\prime }\psi \Vert$ and ${\Delta }_{\psi }B=\Vert B^{\prime }\psi \Vert$. The Schwarz inequality givesSince $A^{\prime }$ and $B^{\prime }$ are Hermitian and their commutator equals $[\hat{A},\hat{B}]$,Thereforewhich is the Robertson uncertainty principle.

Because $\hat{X}-i\lambda \hat{Y}$ is the adjoint of $\hat{X}+i\lambda \hat{Y}$,Expanding gives the real quadraticIts discriminant must be nonpositive. Since the expectation of a commutator of Hermitian operators is purely imaginary,Taking $\hat{X}=\hat{A}-⟨\hat{A}⟩$ and $\hat{Y}=\hat{B}-⟨\hat{B}⟩$ and then taking square roots gives the stated uncertainty relation. This is the quadratic-norm proof of the Heisenberg uncertainty relation.

For a differentiable wave function,Thus $[\hat{x},\hat{p}]=i\hslash I$. Substitution into part (a) yields

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrodinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.