Solution

The Hamiltonian is the sum of three commuting one-dimensional harmonic-oscillator Hamiltonians. Its product eigenstates are

Ψ_{n_{x} n_{y} n_{z}} (x, y, z) = ψ_{n_{x}} (x) ψ_{n_{y}} (y) ψ_{n_{z}} (z),

with energies

E_{n_{x} n_{y} n_{z}} = ℏ ω (n_{x} + n_{y} + n_{z} + \frac{3}{2}) .

Equivalently, the level with

N = n_{x} + n_{y} + n_{z}

has energy

E_{N} = ℏ ω (N + 3/2)

Up to normalization, the ground-state wavefunction is

Ψ_{000} = exp [- \frac{M ω}{2ℏ} (x^{2} + y^{2} + z^{2})] .

It is radial, so

\nabla Ψ_{000}

is parallel to the position vector. Since the orbital angular momentum is

L = - i ℏ r \times \nabla

, every component of

L

annihilates this state. Therefore

L_{z} Ψ_{000} = 0, L^{2} Ψ_{000} = 0,

and the ground state has

ℓ = m = 0

The first excited level has

N = 1

and is spanned by

x Ψ_{000}, y Ψ_{000}, z Ψ_{000} .

The state

Φ_{0} = z Ψ_{000}

satisfies

L_{z} Φ_{0} = 0

. Applying the ladder operators gives

L_{+} Φ_{0} = - ℏ (x + i y) Ψ_{000}, L_{-} Φ_{0} = ℏ (x - i y) Ψ_{000} .

Hence convenient

m = \pm 1

eigenstates are

Φ_{\pm 1} = (x \pm i y) Ψ_{000},

up to normalization and irrelevant overall phases.

Finally, the isotropic Hamiltonian is rotationally invariant. The rotational invariance of a central-potential Hamiltonian gives

[H, L_{z}] = [H, L^{2}] = [L_{z}, L^{2}] = 0.

These commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of

H, L^{2}, L_{z}

must exist.

Solved by gpt-5.6-sol high.

 Ancestors (11)