By Pauling L.

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1 Rigid diatomic molecule As a pilot example let us consider the quantum mechanics of a rigid diatomic molecule with nuclear masses m1 and m2 , and (equilibrium) bond length, Re (see figure). Since the molecule is rigid, its coordinates can be specified by 2 r2 its centre of mass, R = m1mr11 +m +m2 , and internal orientation, r = r2 − r1 (with |r| = Re ). 1) ˆ = −i ∇R and L ˆ =r×p ˆ denotes the angular momentum associated where P with the internal degrees of freedom. Since the internal and centre of mass degrees of freedom separate, the wavefunction can be factorized as ψ(r, R) = eiK·R Y (r), where the first factor accounts for the free particle motion of the body, and the second factor relates to the internal angular degrees of freedom.

The ground state wavefunction is therefore spherically symmetric, and the function w(ρ) = w0 is just a constant. Hence u(ρ) = ρe−ρ w0 and the actual radial wavefunction is this divided by r, and of course suitably normalized. To write the wavefunction in terms of r, we need to find κ. 2 1 With this definition, the energy levels can then be expressed as En = − 4π1 0 (Ze) 2a0 n2 . Moving on to the excited states: for n = 2, we have a choice: either the radial function w(ρ) can have one term, as before, but now the angular momentum = 1 (since n = k + + 1); or w(ρ) can have two terms (so k = 1), and = 0.

4. ATOMIC HYDROGEN 40 To further simplify the wave equation, it is convenient to introduce the dimensionless variable ρ = κr, leading to the equation ∂ρ2 u(ρ) = 1− 2ν ( + 1) + ρ ρ2 u(ρ) , where (for reasons which will become apparent shortly) we have introduced Ze2 κ . Notice that in transforming from r the dimensionless parameter ν = 4π 0 2E to the dimensionless variable ρ, the scaling factor depends on energy, so will be different for different energy bound states! Consider now the behaviour of the wavefunction near the origin.