Quantum Chemistry

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Exercise. This is the solution to exercise 7.1 in the book.

Solution. The general equation is:

2m2Ψ + U(x)Ψ(x) = EΨ(x)

(a) In the case of the harmonic oscillator

Ĥ(0) = 2 2m d2 dx2 + k 2x2

since U(x) = k 2 x2. For an oscillator governed by:

Ĥ = 2 2m d2 dx2 + k 2x2 + γ 6x3 + b 24x4

Ĥ = Ĥ(0) + Ĥ(1)

Therefore, in this case

Ĥ(0) = 2 2m2 + k 2x2

Ĥ(1) = γ 6 + b 24x4

Ψ(0) and E(0) are the wavefunction/energy pair for harmonic oscillator.

E(0) = k μ n + 1 2

Ψ(0) = 1 2n n! α π14H n(αx)eαx22

with

α = kμ 2

(b) In this case the potential is that of a particle in a box. Hence,

Ĥ(1) = 0if0 x a 2 b ifa 2 x a

Ĥ(0), Ψ(0) and E(0) are the Hamiltonian, wavefunctions, and energies of the particle in a box.

Ĥ(0) = 2 2m d2 dx2 Ψ(0) = 2 asin nπx a En(0) = n2h2 8ma2

(c) For a He atom:

Ĥ = 2 2me(12 + 22) 2e2 4πϵ0 1 r1 + 1 r2 + e2 4πϵ0 1 r12

But

2 2me12 2e2 4πϵ0 1 r1 2 2me22 2e2 4πϵ0 1 r2

are Hamiltonians for hydrogen-like systems. Therefore, we can take:

Ĥ(0) = 2 2me12 2e2 4πϵ0 1 r1 2 2me22 2e2 4πϵ0 1 r2

and

Ĥ(1) = e2 4πϵ0 1 r12

Then, the wavefunction Ψ(0) is a product of two hydrogen-like wavefunctions (since Ĥ(0) is decomposable as sum).

Ψ(0)(r 1,r2) = Ψ1(r1)Ψ2(r2)

and energy is

E(0) = E 1(0) + E 2(0)

the sum of energies of hydrogen-like atoms.

(d) In this case

Ĥ(0) = 2 2μ2 e2 eπϵ0 1 r Ĥ(1) = eζrcosθ

and Ψ(0) and E(0) are the wavefunction/energy of the hydrogen atom.

(e) In this case:

Ĥ(0) = 2 2I2 Ĥ(1) = μζcosθ

and Ψ(0) and E(0) are the wavefunction/energy of the rigid rotator.