Quiz 6

Chem 5533 - Spring 1999 - Prof. Darrin York

Constants:h (Planck’s constant) = 6.6262× 10−34 J·sh = h/(2π)

Hydrogen atom:

En = − e2

8πε0a0n2and ψ1s = ψ100 =

1√πa3

0

e−r/a0

1. (20 pts) Perturbation theory provides us with an estimate of the energyand wave function for a perturbed system described by the Hamiltonian operatorH = H(0) + H(1) as a series of corrections:

En = E(0)n + E(1)

n + E(2)n · · · (1)

ψn = ψ(0)n + ψ(1)

n + ψ(2)n · · · (2)

a) What are the formulas for the first order wave function correction ψ(1)n

and second order energy correction E(2)n ?

Ψ(1)n =

∑k 6=n

H(1)kn

E(0)n − E

(0)k

Ψ(0)k

H(1)kn ≡

∫Ψ

(0)∗k H(1)Ψ(0)

n dτ = (H(1)nk )∗

E(2)n =

∫Ψ(0)∗

n H(1)Ψ(1)n dτ =

∑k 6=n

|H(1)nk |2

E(0)n − E

(0)k

b) What is the value of∫ψ

(0)∗

n ψ(1)n dτ?

zero

c) Evaluate the first order correction term for the anharmonic oscillator

H = − h2

2µd2

dx2+

12kx2 +

16γx3 (3)

taking H(0) to be the Harmonic oscillator Hamiltonian. You do not need toknow the exact equation for ψ(0)

n to solve this.

E(1)n =

∫ ∞−∞

Ψ(0)∗n H(1)Ψ(0)

n dτ

H(1) =1

6γx3

E(1)n =

1

6γ

∫ ∞−∞

Ψ(0)∗n x3Ψ(0)

n dτ = 0

(note: x3 is odd, therefore Ψ∗nx3Ψn is also odd, and its integral over all space

vanishes)

2. (20 pts) Consider the trial wave function φ = e−αr, where α is a parame-ter. State the quantum mechanical variational principle (either by a formula orin words), and use it (i.e. solve for α) to estimate the ground state energy andwave function for the hydrogen atom (show your work).

H = − h2

2µ∇2 − e2

4πε0r(4)

Some useful integrals are:∫e−αrHe−αrd3r =

e2a0

8ε0α− e2

4ε0α2(5)

where a0 is the Bohr radius, and∫e−2αrd3r =

π

α3(6)

ANSWER: The ground state energy is found by

E0 =∫ψ∗0 Hψ0 dτ∫ψ∗0ψ0 dτ

The variational principle states that if we substitute any other function (φ)for ψ0, we can calculate a new energy:

Eφ =∫φ∗Hφ dτ∫φ∗φ dτ

and that this new energy Eφ will be greater than the ground-state energyE0. We can state this more simply by the equation: Eφ ≥ E0

E[φ] =e2ao

8εoα− e2

4εoα2

πα3

=e2ao

4εoπ(

1

2α2 − α

ao)

dE

dα=

e2ao

4εoπ(α− 1

ao) = 0

α =1

ao

E[φ] = − e2

8πεoao

φ = e−r

ao