Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 9D (Quantum Mechanics)

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- Solution 9D

Solution

Use the position-space momentum operator

P_{j} = - i ℏ \partial_{j}

and set

F = (x + i y) z

. Direct differentiation gives

L_{z} F = - i ℏ (x \partial_{y} - y \partial_{x}) F = - i ℏ (i x - y) z = ℏ F .

To compute the squared orbital angular momentum, write

D_{x} = y \partial_{z} - z \partial_{y}

,

D_{y} = z \partial_{x} - x \partial_{z}

,

D_{z} = x \partial_{y} - y \partial_{x}

, so

L_{j} = - i ℏ D_{j}

. Expanding the products, including derivatives of their variable coefficients, gives

D_{x}^{2} + D_{y}^{2} + D_{z}^{2} = r^{2} Δ - E^{2} - E, E = x \partial_{x} + y \partial_{y} + z \partial_{z} .

For example

D_{x}^{2} = y^{2} \partial_{z}^{2} + z^{2} \partial_{y}^{2} - 2 yz \partial_{y} \partial_{z} - y \partial_{y} - z \partial_{z}

; summing the three cyclic expressions proves the identity.

Here

Δ F = 0

and

EF = 2 F

, so

E^{2} F = 4 F

. Therefore

L^{2} F = 6 ℏ^{2} F, L_{z} F = ℏ F .

This is the angular momentum of a homogeneous harmonic polynomial, with angular quantum numbers

ℓ = 2, m = 1

. The polynomial is a nonzero angular eigenfunction; by itself it is not a normalizable full-space wavefunction.

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