Download 4 Solved Questions on Quantum Mechanics - Homework 1 | PHYSICS 137B and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 137B, Fall 2007, Moore Problem Set 1 Solutions 1. Recall the general commutator rule [AB,C] = A[B,C] + [A,C]B . Applying this, and using [Jy, Jz] = ih̄Jx and [Jz, Jx] = ih̄Jy, we have [J2, Jz] = [J 2 x + J 2 y + J 2 z , Jz] = [J 2 x , Jz] + [J 2 y , Jz] + [J 2 z , Jz] = Jx[Jx, Jz] + [Jx, Jz]Jx + Jy[Jy, Jz] + [Jy, Jz]Jy + 0 = Jx(−ih̄Jy) + (−ih̄Jy)Jx + Jy(ih̄Jx) + (ih̄Jx)Jy = 0 . 2. (a) Consider a classical particle confined to a ring of radius a that wraps around the z-axis. Its velocity and angular frequency are related by v = ωa. Since the moment arm is fixed at a, its angular momentum (about the z- axis) is Lz = ap = µva = µa 2ω ≡ Iω. In the absence of any potential energy terms, the total energy of the particle is therefore simply E = 1 2 µv2 = 1 2 µa2ω2 = 1 2 Iω2 = L2z 2I . To obtain the Hamiltonian for the corresponding quantum system, we just interpret the dynamical variables (in this case Lz) as quantum operators. Hence the Hamiltonian should be H = L2z/2I, as claimed. 1 (b) To find the energy eigenstates and energy eigenvalues, first consider a free particle confined to a line (not a circle), i.e. a free particle in one di- mension. We know that the Hamiltonian for this system is H = p2/2µ, with energy eigenfunctions ψk(x) = e ikx and eigenvalues Ek = −h̄2k2/2µ. Now a circle of radius a is a lot like an infinite line, except that when you walk a distance 2πa you end up back where you started. So we can just use the energy eigenfunctions for the particle on the line that are consistent with this requirement, i.e. periodic with a period that is an integer multiple of 2πa. This requires choosing k = n/a with n an integer. Thus, the (normalized) energy eigenfunctions for the particle on the circle are ψn(x) = 1√ 2πa einx/a , and the eigenvalues are En = −h̄2n2 2µ . There’s an easier way to arrive at the same answer. In spherical coordinates, Lz can be represented by the differential operator Lz = −ih̄ ∂∂φ . The particle is confined to a radius r = a and a polar angle θ = π/2, so φ is the only free coordinate. The Hamiltonian is now H = − h̄ 2 2µ ∂2 ∂φ2 . This clearly has eigenfunctions ψν(φ) = e iνφ, with eigenenergies Eν = −h̄2ν2/2µ. But since φ is an angular coordinate, we demand that the wavefunctions be periodic, hence ν must be an integer. Thus we have (normalized) eigenfunc- tions ψn(φ) = 1√ 2π einφ , En = − h̄2n2 2µ for all nonzero integers n. The identification with the solutions from the previous paragraph is made by setting x = aφ. Since the Hamiltonian depends only on the operator Lz, it must commute with Lz, and we expect that the energy eigenfunctions should also be eigen- states of Lz. Indeed, we have Lzψn(φ) = −ih̄ ∂ ∂φ 1√ 2π einφ = nh̄ 1√ 2π einφ = nh̄ψn(φ) , so ψn is has angular momentum eigenvalue nh̄. 2
