Question

Consider the wave function ψ(x)=(x+y+3z)f(r) for a particle moving in a central potential V(r). a. Is ψ(x) an eigenstate of the squared orbital angular momentum L² ? If so, what is the eigenvalue?

Answer 1

The wave function ψ(x) is not an eigenstate of the squared orbital angular momentum operator L² because it explicitly depends on Cartesian coordinates, and thus does not have a well-defined eigenvalue for L².

Step-by-step explanation:

The wave function provided, ψ(x) = (x + y + 3z)f(r), suggests a dependency on the coordinates x, y, and z, and is also dependent on the radial distance, r, since it is multiplied by a function of r. However, for a particle in a central potential V(r), eigenstates of squared orbital angular momentum L² are typically spherical harmonics that depend only on the angular coordinates, θ and φ, and not on the radial or Cartesian coordinates explicitly. Since the given wave function explicitly depends on x, y, and z, it is not a spherical harmonic and therefore not an eigenstate of L².

As a result, ψ(x) is not an eigenstate of the orbital angular momentum operator L² and does not have a well-defined eigenvalue for L².