Introduction to Quantum Mechanics · Problem 4.49

Quantum Mechanics in Three Dimensions — David J. Griffiths

Physics — Quantum Mechanics · probability_current_3d

Problem

Warning: Attempt this problem only if you are familiar with vector calculus. Define the (three-dimensional) probability current by generalization of Problem 1.14: J≡iℏ2m(Ψ ∇Ψ∗−Ψ∗ ∇Ψ)\mathbf{J} \equiv \frac{i\hbar}{2m}\left(\Psi\,\nabla\Psi^* - \Psi^*\,\nabla\Psi\right). (4.220) (a) Show that J\mathbf{J} satisfies the continuity equation ∇⋅J=−∂∂t∣Ψ∣2\nabla \cdot \mathbf{J} = -\frac{\partial}{\partial t}|\Psi|^2, (4.221) which expresses local conservation of probability. It follows (from the divergence theorem) that ∮SJ⋅da=−ddt∫V∣Ψ∣2 d3r\oint_S \mathbf{J} \cdot d\mathbf{a} = -\frac{d}{dt}\int_V |\Psi|^2\,d^3\mathbf{r}, (4.222) where V\mathcal{V} is a (fixed) volume and S\mathcal{S} is its boundary surface. In words: The flow of probability out through the surface is equal to the decrease in probability of finding the particle in the volume. (b) Find J\mathbf{J} for hydrogen in the state n=2n = 2, l=1l = 1, m=1m = 1. Answer: ℏ64πma5re−r/asin⁡θ ϕ^\frac{\hbar}{64\pi m a^5} r e^{-r/a} \sin\theta\,\hat{\phi}. (c) If we interpret mJm\mathbf{J} as the flow of mass, the angular momentum is L=m∫(r×J) d3r\mathbf{L} = m\int (\mathbf{r} \times \mathbf{J})\,d^3\mathbf{r}. Use this to calculate LzL_z for the state ψ211\psi_{211}, and comment on the result.

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