Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 57 1 b i Solution Created 2026-10-03 Updated 2026-10-07
The rotational invariance of a central-potential Hamiltonian allows a simultaneous eigenstate of energy and the two-dimensional orbital angular momentum . For a separated wavefunction , the Laplacian in polar coordinates givesThe term must be a constant; write . The angular eigenfunctions can be chosen as . Single-valuedness under imposes , hence . The radial equation in the separation of a two-dimensional central-potential eigenstate isFor a real central potential and the usual real self-adjoint radial boundary conditions, the radial equation admits a basis of real solutions: real and imaginary parts of a complex solution obey the same equation and boundary conditions. Choose a real normalized radial eigenfunction. Since the plane area element in plane polar coordinates is , the normalization isThis establishes the intended separated simultaneous eigenstate form. It is not the form of every stationary state. The radial equation depends on , so the and sectors have the same energy. For , their normalized superpositionis a single-valued stationary state with that energy but is not a single angular exponential. Quantum degeneracy is precisely why separation of variables selects a convenient eigenstate basis rather than all vectors in an energy eigenspace. The printed assertion needs this qualification. The printed polar-coordinate aid also labels a gradient component tuple as a divergence; the gradient used below is the two-dimensional vector .
Past exam of the mathematics course of the University of Cambridge 2018 ib Paper 2 17B Solution Created 2026-09-24 Updated 2026-10-03
In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential , and separation into spherical harmonics uses the orbital angular momentum eigenvalue , producing the centrifugal term .
For , substitute . Dividing the equation by and comparing powers of givesThe solution regular at the origin has , and when both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is . With and the Gamma integral, the mean radius of a circular Coulomb bound state is
At fixed principal quantum number , the allowed orbital angular momentum quantum numbers are , and each has magnetic quantum number . Hence the orbital quantum degeneracy isIncluding electron spin would double this to , but spin is absent from the stated wavefunctions.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 34B b i Solution Created 2026-09-24 Updated 2026-09-25
For occupation numbers , the Hamiltonian of the two commensurate quantum harmonic oscillators isThe five lowest distinct levels and their product-state bases areThe repeated levels exhibit quantum degeneracy.
An energy and orbital angular momentum eigenstate for a real two-dimensional central potential can be chosen with integer and real radial function . Its radial equation isQuantum degeneracy permits other stationary states, including , which are not single angular exponentials. Separation of variables supplies a basis rather than describing every superposition in a degenerate eigenspace.