Today’s Outline - February 21, 2017csrri.iit.edu/~segre/phys406/17S/lecture_12.pdfToday’s...

Today’s Outline - February 21, 2017

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering

Covers through HW #05, equation sheet provided

Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017

C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

• Problem 6.21

• Problem 7.4

• Problem 7.14

• Problem 7.15

• Problem 7.16

• Problem 7.19

• Problem 6.39

Problem 6.21

Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.

The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels

Enj = −13.6

j + 12

in the weak Zeeman regime these energies are perturbed by the magneticfield correction

EZ = µBgJBextmj , gJ =

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 6.21

Enj = −13.6

j + 12

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 6.21

Enj = −13.6

j + 12

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 6.21

Enj = −13.6

j + 12

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 6.21

Enj = −13.6

j + 12

EZ = µBgJBextmj ,

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 6.21

Enj = −13.6

j + 12

j(j + 1)− l(l + 1) + s(s + 1)

2j(j + 1)

Problem 7.4

(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.

(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function

ψ(x) = Axe−bx2

(a) Starting with the definition ofthe arbitrary function ψ

since this function is defined to beorthogonal to the ground state ψ1

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

cn〈ψ1|ψn〉 = c1

thus the expectation value ofthe Hamiltonian is 〈H〉 =

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉

=∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

cn〈ψ1|ψn〉

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

thus the expectation value ofthe Hamiltonian is

〈H〉 =∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2

≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

Problem 7.4

ψ(x) = Axe−bx2

ψ =∞∑n=1

0 = 〈ψ1|ψ〉 =∞∑n=1

∞∑n=2

En|cn|2 ≥ Efe

∞∑n=2

|cn|2 = Efe

Problem 7.4 (cont.)

(b) start by computing the normalization constant,

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

xe−bx2(−2bx − 4bx + 4b2x3)e−bx

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx

= |A|221

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

=3~2b2m

Problem 7.4 (cont.)

1 = |A|2∫ ∞−∞

x2e−2bx2dx = |A|22

2b−→ |A|2 = 4b

the kinetic energy,

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

xe−bx2 d

(xe−bx

= − ~2

∫ ∞−∞

= − ~2

∫ ∞−∞

(−6bx2 + 4b2x4)e−2bx2dx

= −2~2bm

[−6b

2b+ 4b2

3~2b2m

Problem 7.4 (cont.)

and the potential energy

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

combining to get the expectationvalue of the Hamiltonian

taking the derivative to minimizewrt b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

=3mω2

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

−→ b =mω

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.4 (cont.)

〈V 〉 =1

∫ ∞−∞

xe−bx2x2(xe−bx

2)dx =

∫ ∞−∞

x4e−2bx2dx

2mω24b

〈H〉 =3~2b2m

+3mω2

8b∂〈H〉∂b

2m− 3mω2

8b2= 0

b2 =m2ω2

4~2−→ b =

〈H〉min =3~2

= ~ω(

Problem 7.14

If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,

V (~r) = − e2

4πε0

e−µr

where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.

First need to choose a trial function and the one which makes most senseto simplify the calculation of 〈H〉 is a hydrogen-like function

ψ =1√πb3

e−r/b

where b is a variable parameter replacing the Bohr radius a

Problem 7.14

V (~r) = − e2

4πε0

e−µr

ψ =1√πb3

e−r/b

Problem 7.14

V (~r) = − e2

4πε0

e−µr

ψ =1√πb3

e−r/b

Problem 7.14

V (~r) = − e2

4πε0

e−µr

ψ =1√πb3

e−r/b

Problem 7.14 (cont.)

With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom

computing 〈V 〉〈T 〉 =

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

computing 〈V 〉

〈T 〉 =~2

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr

= − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

〈V 〉 = − e2

4πε0

∫ ∞0

e−2r/be−µr

rr2 dr = − e2

4πε0

∫ ∞0

e−(µ+2/b)r r dr

= − e2

4πε0

(µ+ 2b )2

= − e2

4πε0

b(1 + µb2 )2

〈H〉 =~2

2mb2− e2

4πε0

b(1 + µb2 )2

∂〈H〉∂b

= − h2

4πε0

b2(1 + µb2 )2

b(1 + µb2 )3

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

= a the left side is simply the Bohr ra-dius, a

this cubic equation can be solved approximately by remembering thatµa� 1

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

the left side is simply the Bohr ra-dius, a

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

∂〈H〉∂b

= − h2

4πε0

(1 + 3µb2 )

b2(1 + µb2 )3

(4πε0e2

(1 + 3µb2 )

(1 + µb2 )3

a ≈ b

)[1− 3µb

)2]≈ b

[1− 9

4(µb)2 +

4(µb)2

]≈ b

[1− 3

4(µb)2

]−→ b ≈ a

1− 34(µb)2

4(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

]2C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

≈ ~2

[1− 2

4(µa)2

]− e2

4πε0a

[1− 3

4(µa)2

][1− 2

)2]≈ −E1

[1− 3

2(µa)2

]+ 2E1

[1− µa +

��3

4(µa)2 −

��3

4(µa)2

]≈ E1

[1− 2(µa) +

2(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

]2≈ ~2

[1− 2

4(µa)2

]− e2

4πε0a

[1− 3

4(µa)2

][1− 2

≈ −E1

[1− 3

2(µa)2

]+ 2E1

[1− µa +

��3

4(µa)2 −

��3

4(µa)2

]≈ E1

[1− 2(µa) +

2(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

]2≈ ~2

[1− 2

4(µa)2

]− e2

4πε0a

[1− 3

4(µa)2

][1− 2

)2]≈ −E1

[1− 3

2(µa)2

]+ 2E1

[1− µa +

4(µa)2 − 3

4(µa)2

≈ E1

[1− 2(µa) +

2(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

]2≈ ~2

[1− 2

4(µa)2

]− e2

4πε0a

[1− 3

4(µa)2

][1− 2

)2]≈ −E1

[1− 3

2(µa)2

]+ 2E1

[1− µa +

��3

4(µa)2 −

��

��3

4(µa)2

≈ E1

[1− 2(µa) +

2(µa)2

〈H〉min =~2

4(µa)2

]2− e2

4πε0

a[1 + 3

4(µa)2] [

1 + 12(µa)

]2≈ ~2

[1− 2

4(µa)2

]− e2

4πε0a

[1− 3

4(µa)2

][1− 2

)2]≈ −E1

[1− 3

2(µa)2

]+ 2E1

[1− µa +

��3

4(µa)2 −

��

��3

4(µa)2

]≈ E1

[1− 2(µa) +

2(µa)2

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 perturbed 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.

a. Start by solving the eigenvaluedeterminant

this is simply a quadratic equation

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(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(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣

0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(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(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(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(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb

= − h2

Eb − Ea; E

(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea;

E(2)b =

Eb − Ea

∣∣∣∣ Ea − λ hh Eb − λ

∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2

0 = λ2 − (Ea + Eb)λ+ EaEb − h2

E± =1

[(Ea + Eb)±

√(Ea + Eb)2 − 4EaEb + 4h2

[(Ea + Eb)±

√(Eb − Ea)2 + 4h2

E(2)a =

| 〈ψb|H ′|ψa〉 |2

Ea − Eb= − h2

Eb − Ea; E

(2)b =

Eb − Ea

The energies, to second order are thus

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

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

⟨ψb|H ′|ψa

⟩= 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

E− ≈ Ea −h2

Eb − Ea;

E+ ≈ Eb +h2

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

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

⟨ψa|H ′|ψb

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

= (Eb − Ea) sin 2φ+ 2h cos 2φ

−→ tan 2φ = − 2h

Eb − Ea

E− ≈ Ea −h2

Eb − Ea; E+ ≈ Eb +

Eb − Ea

⟩+ sinφ cosφ

⟨ψb|H ′|ψa

minimizing

0 =∂ 〈H〉∂φ

Eb − Ea

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 + 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 =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

− ε = tan 2φ

=sin 2φ

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

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 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ),

sin2 φ =1

2(1− cos 2φ)

defining

ε ≡ 2h

Eb − Ea

cos 2φ=

sin2 2φ = ε2(1− sin2 2φ)

sin2 2φ (1 + ε2) = ε2

sin 2φ =±ε√

1 + ε2

cos2 2φ = 1− ε2

1 + ε2=

1 + ε2

cos 2φ =∓1√1 + ε2

cos2 φ =1

2(1 + cos 2φ), sin2 φ =

2(1− cos 2φ)

cos2 φ =1

(1∓ 1√

1 + ε2

sin2 φ =1

(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

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Ea + Eb ±Eb − Ea + 2h 2h

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

sin 2φ =±ε√

1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

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

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

cos2 φ =1

(1∓ 1√

1 + ε2

), sin2 φ =

(1± 1√

1 + ε2

), sin 2φ =

±ε√1 + ε2

〈H〉min =1

(1∓ 1√

1 + ε2

(1± 1√

1 + ε2

ε√1 + ε2

[Ea + Eb ±

Eb − Ea + 2hε√1 + ε2

Eb−Ea√1 + 4h2

(Eb−Ea)2

[Ea + Eb ±

√(Eb − Ea)2 + 4h2

]C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23

Thus the upper bound on the ground state energy is

〈H〉min =1

[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

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

〈H〉min =1

[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.

E± =1

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

〈H〉min =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

E± =1

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

〈H〉min =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

E± =1

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

〈H〉min =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

E± =1

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

〈H〉min =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

E± =1

[(Ea + Eb)± (Eb − Ea)

√1 +

(Eb − Ea)2

[(Ea + Eb)± (Eb − Ea)

(Eb − Ea)2

E± ≈1

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

E− ≈1

[2Ea −

Eb − Ea

]≈ Ea −

Eb − Ea

E+ ≈1

[2Eb +

Eb − Ea

]≈ Eb +

Eb − Ea

These are simply the second order perturbation theory results.

E± ≈1

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

E− ≈1

[2Ea −

Eb − Ea

≈ Ea −h2

Eb − Ea

E+ ≈1

[2Eb +

Eb − Ea

≈ Eb +h2

Eb − Ea

E± ≈1

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

E− ≈1

[2Ea −

Eb − Ea

]≈ Ea −

Eb − Ea

E+ ≈1

[2Eb +

Eb − Ea

]≈ Eb +

Eb − Ea

E± ≈1

[(Ea + Eb)± (Eb − Ea)± 2h2

Eb − Ea

E− ≈1

[2Ea −

Eb − Ea

]≈ Ea −

Eb − Ea

E+ ≈1

[2Eb +

Eb − Ea

]≈ Eb +

Eb − Ea

Problem 7.16

Consider an electron at rest in a uniform magnetic field ~B = Bz k̂ , forwhich the Hamiltonian is

H0 =eBz

The eigenspinors, and corresponding energies are:{χ+, E+ = −(γBz~)/2

χ−, E− = +(γBz~)/2

Now we turn on a perturbation, in the form of a uniform field in the xdirection

H ′ =eBx

(a) Find the matrix elements of H ′.

(b) Using the result of problem 7.15(b) find the new ground state energy,in second order perturbation theory.

(c) Using the result of problem 7.15(c), find the variational principlebound on the ground state energy.

For an electron, γ = −e/m andthus the energies of the two eigen-spinors are

choosing the higher energy to bethe Eb as was in problem 7.15, wehave

E± = ±eBz~2m

Ea ≡ E−, and Eb ≡ E+

χb = χ+ =

)χa = χ− =

)(a) The matrix elements in the original basis set become

〈χa|H ′|χa〉 =eBx

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

〈χb|H ′|χb〉 =eBx

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

〈χa|H ′|χb〉 =eBx

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(a) The matrix elements in the original basis set become

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

=eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

=eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

=eBx~2m

(0 1)( 0

eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

=eBx~2m

E± = ±eBz~2m

χb = χ+ =

)χa = χ− =

(0 1)~

(0 11 0

eBx~2m

(0 1)( 1

(1 0)~

(0 11 0

eBx~2m

(1 0)( 0

(0 1)~

(0 11 0

eBx~2m

(0 1)( 0

eBx~2m

(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m

Egs ≈ Ea −h2

(Eb − Ea)= −eBz~

2m− (eBx~/2m)2

(eBz~/m)= − e~

)(c) Taking the variational method results (problem 7.15c), we have

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

(Eb − Ea)

= −eBz~2m

− (eBx~/2m)2

(eBz~/m)= − e~

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)

= − e~2m

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)= − e~

(c) Taking the variational method results (problem 7.15c), we have

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)= − e~

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)= − e~

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)= − e~

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Egs ≈ Ea −h2

2m− (eBx~/2m)2

(eBz~/m)= − e~

Egs =1

[Ea + Eb −

√(Eb − Ea)2 + 4h2

]= −1

√(eBz~m

(eBx~2m

= − e~2m

√B2z + B2

Problem 7.19

The fundamental problem in harnessing nuclear fusion is getting the twoparticles (say, two deutrons) close enough to each other for the attractive(but short range) nuclear force to overcome the Coulomb repulsion. The“bulldozer” method is to heat the particles up to fantastic temperatures,and allow the random collision to bring them together. A more exoticproposal is muon catalysis, in which we construct a “hydrogen moleculeion,” only with deutrons in place of protons and a muon in place of theelectron. Predict the equilibrium separation distance between the deutronsin such a structure, and explain why muons are superior to electrons forthis purpose.

The result we seek is the same as we calculated before but with theelectron mass replaced by the reduced muon mass (becuase the muonmass is a significant fraction of the proton mass)

µµ =mµmd

mµ + md=

mµ2mp

mµ + 2mp

Problem 7.19

µµ =mµmd

mµ + md=

mµ2mp

mµ + 2mp

Problem 7.19

µµ =mµmd

mµ + md

=mµ2mp

mµ + 2mp

Problem 7.19

µµ =mµmd

mµ + md=

mµ2mp

mµ + 2mp

Problem 7.19

The mass of the muon is

mµ = 207me

and the proton mass is

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule

Req =2.49a

196= 6.73× 10−13m

the muon brings the deutrons much closer and thus rasies the probabilityof fusion

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103

≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 7.19

mµ = 207me

mp ≈ 1836me

µµ =2mµmp

mµ + 2mp

µµ ≈2 · 207 · 1836m2

(207 + 2 · 1836)me

≈ 7.60104× 105me

3.879× 103≈ 196me

Req =2.49a

196= 6.73× 10−13m

Problem 6.39

In a crystal, the electric field of neighboring ions perturbs the energy levelsof an atom. As a crude model, imagine that a hydrogen atom issurrounded by three pairs of point charges, as shown in Figure 6.15.

(a) Assuming that r � d1, r � d2, and r � d3, show that

H ′ = V0 + 3(β1x2 + β2y

2 + β3z2)− (β1 + β2 + β3)r2

βi ≡ −e

4πε0

, and V0 = 2(β1d21 + β2d2

2 + β3d23)

(b) Find the lowest order correction to the ground state energy

(c) Calculate the first-order corrections to the energy of the first excitedstates (n = 2). Into how many levels does the four-fold degeneratesystem split, in the cases of (i) cubic, (ii) tetragonal, and (iii)orthorhombic symmetries?

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