Past exam of the mathematics course of the University of Cambridge 2001 ib Paper 2 9F Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2003 ib Paper 2 9A Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2018 ib Paper 3 8B Solution Created 2026-09-24 Updated 2026-10-03
A Hermitian operator satisfies on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function is Hermitian as well. Thus is Hermitian.
For a time-independent operator , the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation ,Consequently
Past exam of the mathematics course of the University of Cambridge 2022 ib Paper 4 15B a Solution Created 2026-09-24 Updated 2026-09-25
The Time-dependent Schrodinger equation is . Differentiate , use the equation and its adjoint, and include explicit time dependence of the operator. This gives Ehrenfest theorem
Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 4 4D b Solution Created 2026-09-24 Updated 2026-09-25
For a particle in the scalar potential energy ,The canonical commutation relation givesThe Ehrenfest theorem with therefore givesSince and direct action on a wavefunction gives ,Finally has no explicit time dependence and , so
The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that . These conclusions are summarized by the classical equations from Ehrenfest theorem.
Past exam of the mathematics course of the University of Cambridge 2024 ib Paper 2 15A iii Solution Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 325 4 Solution Created 2026-09-24 Updated 2026-09-25
If Bob opens the trap, Ehrenfest theorem gives branch-dependent Newtonian accelerationsso, for ,After time , the centers of Bob's two conditional wave packets differ byFor Alice's superposition, these distinguishable Bob states become entangled with and ; for Alice's mixture there was no initial coherence to entangle. Taking Bob's minimum packet width to be the Planck length , appreciable branch distinguishability begins when , atKeeping the trap closed suppresses this branch separation.
If , Bob could choose whether to destroy Alice's local coherence before a light signal from his laboratory arrived. Any procedure by which Alice distinguished the coherent superposition from the mixture in less than would then enable superluminal signalling. The largest dangerous separation is determined parametrically by , which givesCausality therefore requiresUsing for the Planck mass ,up to numerical factors.