1 Problem set 9

Handout: 10/24Due date: 10/31

Problem 1

Prove that the energy to first order for the lowest-energy state of a perturbedsystem is an upper bound for the exact energy of the lowest-energy state ofthe perturbed system, that is, that E

(0)0 + E

(1)0 > E.

The ground state energy is given by

E00 =< ψ0|H0|ψ0 > (1)

and the first correction to the ground state energy is given by

E10 =< ψ0|H1|ψ0 > (2)

where the total Hamiltonian is given by H = H0 + H1. We know fromthe variational theorem that using the zeroth-order wavefunction as a trialfunction for the total Hamiltonian will give an upper bound to the energy

< ψ0|H|ψ0 >> E (3)

since ψ is not the wavefunction associated with the Hamiltonian. Therefore,we can prove that

< ψ0|H|ψ0 > > E (4)

< ψ0|H0 +H1|ψ0 > > E (5)

< ψ0|H0|ψ0 > + < ψ0|H1|ψ0 > > E (6)

E00 + E1

0 > E (7)

(8)

Problem 2

An electron moves in a harmonic potential, V = 12x2. What is the effect, to

first order, on the energies of a perturbing electric field, H1 = Fx? Explainyour reasoning. [hint: you do not need to evaluate any integrals to answerthe question.].

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Since the particle moves in a Harmonic potential the wavefunction squaredis symmetric. An electric field perturbation V 1 = Ex which is antisymmetricwill have no effect since

< ψ0|V 1|ψ0 >=

∫

dx|ψ0|2V 1 =

∫

dxsymmetric× antisymmetric = 0 (9)

Problem 3

Calculate the energy to first order of He+ in its lowest-energy state. Use thehydrogen atom in its ground state as your zeroth-order approximation. Useatomic units.

The zeroth order Hamiltonian is for the hydrogen atom

H0 = −1

2∇2 − 1

r(10)

whereas the Hamiltonian for the He+ is given by

H = −1

2∇2 − 2

r(11)

Thus, the total Hamiltonian can be split as

H = −1

2∇2 − 2

R= H0 − 1

r= H0 +H1 (12)

The energy to first order is then

E = E0 + E1 = −1

2+ < 1s| − 1

r|1s > (13)

Evaluating the first-order correction to the energy we get

E1 = − < 1s|1r|1s >= −4π

π

∫ ∞

0

exp(−2r)rdr = −41

22= −1 (14)

The energy to first-order is then E = −1/2 − 1 = −3/2.Problem 4

A hydrogen atom in its ground state is perturbed by applying a uniformelectric field, −Fz, along the z-direction.

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1. What is the first-order change in the energy?

2. For the first-order correction to the ground-state wavefunction, we canconsider a2s,1s and a2pz,1s to be the coefficients for mixing in 2s and 2pz

character. Write down the expression for these two coefficients.

3. Predict the signs of these coefficients. Explain your reasoning.

The ground state wavefunction for the hydrogen atom is 1s = 1√π

exp(−r).a) The first-order change in the energy due to the electric field is then givenby

< 1s|H1|1s >= −F < 1s|r cos θ|1s > (15)

= −F 1

π

∫ ∞

0

r3 exp(−2r)dr

∫ π

0

cos θ sin θdθ

∫ 2π

0

dφ (16)

= −F 1

π

∫ ∞

0

r3 exp(−2r)dr

∫ π

0

sym.× antisym.

∫ 2π

0

dφ = 0 (17)

b) The two cofficients are

c(1)2s,1s =

< 2s|H1|1s >E2s − E1s

= 0 (18)

c(1)2pz,1s =

< 2pz|H1|1s >E2pz

− E1s

= 0 (19)

The first coefficient are zero due to symmetry, i.e. it cannot improve theenergy due to perturbation along the Z since it is symmetric in all directions.The second coefficient is positive since the 2pz function is oriented along thedirection of the perturbation and will therefore improve the wavefunction.

Problem 5

The Hellman-Feynman theorem establishes a connection between the changein the energy and the change in the Hamiltonian of a system experiences anelectric field perturbation. The theorem is given by

dE

dF=

⟨

∂H

∂F

⟩

(20)

where E is the energy, F is the electric field strength and H is the totalhamiltonian for the system. If we apply an electric field perturbation alongthe z-direction we can write the first-order hamiltonian as

H(1) = −µzF (21)

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where µz is the electric dipole moment operator in the z-direction.1) Use the Hellman-Feynman theorem to express the change in the energydue to an electric field in the z-direction in terms of the dipole momentoperator.2) Expand the energy of a system in a Taylor expansion relative to the energy,E0, in the absence of the field.The expectation value of the dipole moment operator can also be written as

< µz >= µ0z + αzzF + · · · (22)

where µ0z is the permanent dipole moment of the system and αzz is the po-

larizability.3) Use this expansion to express the permanent dipole moment and polariz-ability in terms of derivatives of the energy with respect to the electric field.4) Use non-degenerate perturbation theory to obtain an expression for theenergy of the system in terms of the field strength to second order.5) Finally, use this expression to identify the permanent dipole moment andpolarizability of the system.[Hint: The final expression should be expression in terms of the electric dipoleoperator and the zeroth-order wavefunctions.

1.

dE

dF=

⟨

∂H

∂F

⟩

=

⟨

∂H0 − µzF

∂F

⟩

= −〈µz〉 (23)

2.

E = E0 +∂E

∂FF +

1

2

∂2E

∂F 2F 2 + · · · (24)

3.

dE

dF=∂E

∂F+∂2E

∂F 2F 2 + · · · = −µ0

z − αzzF + · · · (25)

Therefore, we can identify the dipole moment and polarizability as

µ0z = −∂E

∂F(26)

αzz = −∂2E

∂F 2(27)

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4.The energy expression from perturbations theory reads:

E = E0 + E1 + E2 + · · · (28)

= E0+ < ψ(0)0 |H1|ψ(0)

0 > +∑

m6=0

< ψ(0)m |H1|ψ(0)

0 >< ψ(0)0 |H1|ψ(0)

m >

E0 −E0m

+ · · · (29)

= E0− < ψ(0)0 |µz|ψ(0)

0 > F +∑

m6=0

< ψ(0)m |µz|ψ(0)

0 >< ψ(0)0 |µz|ψ(0)

m >

E0 − E0m

+ · · · (30)

5.We can therefore identify the terms as

µ0z = −∂E

∂F=< ψ

(0)0 |µz|ψ(0)

0 > (31)

and

αzz = −∂2E

∂F 2= −2

∑

m6=0

< ψ(0)m |µz|ψ(0)

0 >< ψ(0)0 |µz|ψ(0)

m >

E0 − E0m

(32)

Problem 6

A given unperturbed system has a doubly generate energy level for whichthe perturbation integrals have values of H1

11 = 6a, H122 = 8a, and H1

12 =2a, where a is a positive constant. We also know that the unperturbedwavefunctions are orthonormal.

1. Find the first-order correction to the energy in terms of the constant a.

2. Find the normalized correct zeroth-order wave functions.

1.The secular equation is

det(H −EI) =

∣

∣

∣

∣

H1aa − E H1

ab

H1ba H1

bb − E

∣

∣

∣

∣

=

∣

∣

∣

∣

6a−E 2a2a 8a−E

∣

∣

∣

∣

= 0 (33)

Expansion of the secular equation gives the following quadratic equation

E2 − 14Ea+ 44a2 = 0 (34)

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for which the solution is E = 7a±√

20a = 11.47a, 2.52a. The coefficients forthe lowest eigenvalue is then given by

(H1aa − E1)c1 +Habc2 = 0 (35)

(6a− 2.52a)c1 + 2ac2 = 0 (36)

c1 = −0.57c2 (37)

Using normalization we find

1 = 0.572c22 + c22 ==> c2 = 0.867 (38)

Therefore, the normalized wavefunction is

ψ1 = −0.49ψa + 0.867ψb (39)

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