Ψ (r, t) = ψ (r) e^{- i Et /ℏ}

. Its probability density and all expectations of time-independent observables are constant. This does not require zero probability current.

i ℏ \dot{Ψ} = - (ℏ^{2} /2 m) \nabla^{2} Ψ + V Ψ

\overset{ρ}{˙} = Ψ^{*} \dot{Ψ} + \dot{Ψ}^{*} Ψ

. The potential terms cancel:

\overset{ρ}{˙} = \frac{i ℏ}{2 m} (Ψ^{*} \nabla^{2} Ψ - Ψ \nabla^{2} Ψ^{*}) = - \nabla \cdot [\frac{ℏ}{2 im} (Ψ^{*} \nablaΨ - Ψ\nabla Ψ^{*})] .

j = (ℏ/ m) Im (Ψ^{*} \nablaΨ)

\nabla \cdot j = 0

: flow may circulate without changing the density.

Here only the azimuthal phase contributes an imaginary logarithmic derivative. Since

\nabla ϕ = \hat{ϕ} / (r sin θ)

j = \frac{ℏ∣Ψ ∣ ^{2}}{m r sin θ} \hat{ϕ} = \frac{ℏ A ^{2}}{m} r e^{- r / a_{0}} sin θ \hat{ϕ} .

There is no radial or polar flow; the azimuthal current is independent of

ϕ

, so its divergence vanishes. The apparent coordinate singularity at

sin θ = 0

cancels in the final expression.

The z-component of orbital angular momentum follows from the moment of the mass current:

⟨ L_{z} ⟩ = \int m r sin θ j_{ϕ} d V = ℏ A^{2} (\int_{0}^{\infty} r^{4} e^{- r / a_{0}} d r) (\int_{0}^{π} sin^{3} θ d θ) (\int_{0}^{2 π} d ϕ) = ℏ .

Indeed

A^{2} (24 a_{0}^{5}) (4/3) (2 π) = 1

. This agrees with

L_{z} = - i ℏ \partial_{ϕ}

 Ancestors (13)

 Discussion (0)

There are no discussions about this article yet.

There are currently no matching articles.