Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Today’s Outline - February 19, 2014

• Problem 7.15

• WKB Approximation

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 1 / 15

Problem 7.15

Suppose you are given a quantum system whose Hamiltonian H0 admitsjust two eigenstates, ψa (with energy Ea) and ψb (with energy Eb). Theyare orthogonal, normalized and non-degenerate (assume Ea < Eb). Nowturn on a perturbation H ′, with the following matrix elements:⟨

H ′⟩

=

(0 hh 0

), h = constant

a. Find the exact eigenvalues of the perturbing Hamiltonian.

b. Estimate the energies of the perturbed system using second-orderperturbation theory.

c. Estimate the ground state energy of the pertirbed system using thevariational principle with a trial wavefuction of the form

ψ = (cosφ)ψa + (sinφ)ψb

where φ is an adjustable parameter.

d. Compare the answers to the sections above.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 2 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣

0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]

=1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]

b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements.

The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb

= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea;

E 2b =

h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

0 =

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

2

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

]=

1

2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is

E 2a =| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E 2

b =h2

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 3 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea;

E+ ≈ Eb +h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩

= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉+ sinφ cosφ

⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩

= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ

−→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

h2

Eb − Ea

c. The expectation value of the Hamiltonian using the trial wavefunction is

〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb

⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉

+ sinφ cosφ⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψa|H ′|ψb

⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)

= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h

Eb − Ea

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 4 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ

=sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ

=sin 2φ√

1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2

=1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ),

sin2 φ =1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

defining

ε ≡ 2h

Eb − Ea

rearranging to solve forsin 2φ

similarly, we can solve forcos 2φ

− ε = tan 2φ =sin 2φ

cos 2φ=

sin 2φ√1− sin2 2φ

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1

1 + ε2

cos 2φ =∓1√1 + ε2

thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

1

2(1− cos 2φ)

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 5 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

),

sin2 φ =1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

),

sin 2φ =±ε√

1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

cos2 φ =1

2

(1∓ 1√

1 + ε2

), sin2 φ =

1

2

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

substituting into the expectation value of the Hamiltonian

〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ

〈H〉min =1

2Ea

(1∓ 1√

1 + ε2

)+

1

2Eb

(1± 1√

1 + ε2

)± h

ε√1 + ε2

=1

2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

]

=1

2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

=

1

2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 6 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]

d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

Thus the upper bound on the ground state energy is

〈H〉min =1

2

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.

The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small

E± =1

2

[(Ea + Eb)± (Eb − Ea)

√1 +

4h2

(Eb − Ea)2

]

≈ 1

2

[(Ea + Eb)± (Eb − Ea)

(1 +

2h2

(Eb − Ea)2

)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 7 / 15

Problem 7.15 - solution

E± ≈1

2

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

]

E− ≈1

2

[2Ea −

2h2

Eb − Ea

]≈ Ea −

h2

Eb − Ea

E+ ≈1

2

[2Eb +

2h2

Eb − Ea

]≈ Eb +

h2

Eb − Ea

These are simply the second order perturbation theory results.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 8 / 15

Problem 7.15 - solution

E± ≈1

2

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

]

E− ≈1

2

[2Ea −

2h2

Eb − Ea

]

≈ Ea −h2

Eb − Ea

E+ ≈1

2

[2Eb +

2h2

Eb − Ea

]

≈ Eb +h2

Eb − Ea

These are simply the second order perturbation theory results.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 8 / 15

Problem 7.15 - solution

E± ≈1

2

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

]

E− ≈1

2

[2Ea −

2h2

Eb − Ea

]≈ Ea −

h2

Eb − Ea

E+ ≈1

2

[2Eb +

2h2

Eb − Ea

]≈ Eb +

h2

Eb − Ea

These are simply the second order perturbation theory results.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 8 / 15

Problem 7.15 - solution

E± ≈1

2

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

]

E− ≈1

2

[2Ea −

2h2

Eb − Ea

]≈ Ea −

h2

Eb − Ea

E+ ≈1

2

[2Eb +

2h2

Eb − Ea

]≈ Eb +

h2

Eb − Ea

These are simply the second order perturbation theory results.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 8 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

The WKB approximation

The Wentzel, Kramers, Brillouin (WKB) approximation is a valuabletechnique for obtaining approximate solutions (wavefunctions) ofone-dimensional systems where the potential V (x) varies slowly comparedto the deBroglie wavelength of the particle

Consider a particle of energy Emoving in a region of constant po-tential

this is essentially a “free” particlemoving in the positive or negativex-direction

now relax the condition of constantpotential but require that

ψ(x) = Ae±ikx

k ≡√

2m(E − V )

~

λ =2π

k� dV (x)

dx

The WKB approximation treats the wavefunction in this potential as thesame oscillatory function but with a slowly varying modulation (inamplitude and wavelength.

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 9 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential

with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential

with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential

with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

Turning points

The same argument can be madefor the case where V (x) > E ,and the wavefunction is a decayingexponential with κ and A varyingslowly.

This approximation must breakdown at the turning points, where,by definition E ≈ V and

λ,1

κ−→ ∞

ψ(x) = Ae±κx

κ ≡√

2m(V − E )

~

V(x)

x

E

turning points

classical region

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 10 / 15

The classical region

Consider a particle confined to theclassical region.

The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

The classical region

Consider a particle confined to theclassical region. The Schrodingerequation can be rewritten in termsof the classical momentum

d2ψ

dx2= −p2

~2ψ

p ≡√

2m[E − V (x)]

assume that the amplitude, A(x),and phase, φ(x), of the wavefunc-tion are real and depend on x

V(x)

x

E

turning points

classical region

ψ(x) = A(x) e iφ(x)

dψ

dx=(A′ + iAφ′

)e iφ

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2

]e iφ

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 11 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′

A2φ′ = C 2 → A =C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2

→ A =C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]

this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2

−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Amplitude and phase equations

A′′ + 2iA′φ′ + iAφ′′ − A(φ′)2 = −p2

~2A

The real equation

A′′ − A(φ′)2 = −p2

~2A

��A′′ = A

[(φ′)2 − p2

~2

]this equation cannot be solved eas-ily so we need to invoke the WKBapproximation:

The imaginary equation

0 = 2A′φ′ + Aφ′′

0 =(A2φ′

)′A2φ′ = C 2 → A =

C√φ′

Assume that A varies slowly so that A′′ � (φ′)2, p2/~2

(φ′)2 =p2

~2−→ dφ

dx= ±p

~

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 12 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]

=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Classical region solution

The WKB approximation gives asimple differential equation for φ(x)solves to an integral

combining with the solution for theamplitude

A(x) =C√φ′(x)

The probability of finding the parti-cle at a point x is inversely propor-tional to its momentum

dφ

dx= ±p

~

φ(x) = ±1

~

∫p(x) dx

ψ(x) ∼=C√p(x)

e±i~∫p(x) dx

|ψ(x)|2 ∼=|C |2

p(x)

the general solution becomes

ψ(x) ∼=1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 13 / 15

Example 8.1

Suppose we have an infinite squarewell with a bumpy bottom

V (x) =

{arb. func. 0 < x < a

∞ otherwise

assume E > V (x) inside the well

V(x)

ax

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

choosing the lower bound for the φintegral at zero φ(x) =

1

~

∫ x

0p(x ′) dx ′

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 14 / 15

Example 8.1

Suppose we have an infinite squarewell with a bumpy bottom

V (x) =

{arb. func. 0 < x < a

∞ otherwise

assume E > V (x) inside the well

V(x)

ax

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

choosing the lower bound for the φintegral at zero φ(x) =

1

~

∫ x

0p(x ′) dx ′

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 14 / 15

Example 8.1

Suppose we have an infinite squarewell with a bumpy bottom

V (x) =

{arb. func. 0 < x < a

∞ otherwise

assume E > V (x) inside the well

V(x)

ax

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

choosing the lower bound for the φintegral at zero φ(x) =

1

~

∫ x

0p(x ′) dx ′

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 14 / 15

Example 8.1

Suppose we have an infinite squarewell with a bumpy bottom

V (x) =

{arb. func. 0 < x < a

∞ otherwise

assume E > V (x) inside the well

V(x)

ax

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

choosing the lower bound for the φintegral at zero

φ(x) =1

~

∫ x

0p(x ′) dx ′

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 14 / 15

Example 8.1

Suppose we have an infinite squarewell with a bumpy bottom

V (x) =

{arb. func. 0 < x < a

∞ otherwise

assume E > V (x) inside the well

V(x)

ax

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

choosing the lower bound for the φintegral at zero φ(x) =

1

~

∫ x

0p(x ′) dx ′

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 14 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero

∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero

∫ a

0p(x) dx = nπ~

ψ(0) = 0

→ C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero

∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero

∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0

→ φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero

∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15

Example 8.1

ψ(x) ∼=1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

the boundary conditions atthe edges of the well forcethe wavefunction to zero∫ a

0p(x) dx = nπ~

ψ(0) = 0 → C2 = 0

ψ(a) = 0 → φ(a) = nπ, n = 1, 2, 3, . . .

En =n2π2~2

2ma2

C. Segre (IIT) PHYS 406 - Spring 2014 February 19, 2014 15 / 15