A Modern Approach to Quantum Mechanics · Problem 9.12

Bound States of Central Potentials — John S. Townsend

Physics — Quantum Mechanics · angular-momentum-measurement-wave-function

Problem

The wave function for a particle is of the form ψ(r)=(x+y+z)f(r)\psi(\mathbf{r}) = (x + y + z)f(r). What are the values that a measurement of L2\mathbf{L}^2 can yield? What values can be obtained by measuring LzL_z? What are the probabilities of obtaining these results? Suggestion: Express the wave function in spherical coordinates and then in terms of the Yl,mY_{l,m}'s.
Intuitive explanation

The key is recognizing that ψ\psi must be decomposed into eigenstates of the angular momentum operators L2\mathbf{L}^2 and LzL_z—these eigenstates are the spherical harmonics Yl,mY_{l,m}. When you convert x+y+zx + y + z to spherical coordinates, you'll find it's a linear combination of specific harmonics, each with definite ll and mm quantum numbers. Measuring L2\mathbf{L}^2 gives ℏ2l(l+1)\hbar^2 l(l+1) and LzL_z gives ℏm\hbar m, but only for the harmonics present in your expansion; the probability of each outcome is determined by the squared coefficient of that harmonic in the decomposition.

Intuitive explanation

The key is recognizing that ψ\psi must be decomposed into eigenstates of the angular momentum operators L2\mathbf{L}^2 and LzL_z—these eigenstates are the spherical harmonics Yl,mY_{l,m}. When you convert x+y+zx + y + z to spherical coordinates, you'll find it's a linear combination of specific harmonics, each with definite ll and mm quantum numbers. Measuring L2\mathbf{L}^2 gives ℏ2l(l+1)\hbar^2 l(l+1) and LzL_z gives ℏm\hbar m, but only for the harmonics present in your expansion; the probability of each outcome is determined by the squared coefficient of that harmonic in the decomposition.

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