!18Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Assignment 11 - PHYS 5103-11/13/19-Due Wed. Nov. 20 CMwBang! Ch 4.1 thru Ch.4.4. and Lectures 21-23

Ex.1(a) Do ABCD expansion of Hamiltonian ! using p. 54 of Lect. 22 for !

(A square-root in last part of Assignment 10.) Evaluate crank vector ! and overall frequency ! =____. Evaluate beat or splitting frequency ! =____ and plot eigen-frequency levels. (See p.50-54 of Lect.23) (b) Sketch a 3D ABC-space showing the crank vector ! and the angle it makes with the A-axis. (c) Show how a rotation ! of a spin-vector S starting on the A-axis would end up.

(d) Do ABCD expansion of 2D Hooke spring matrix K. Compare resulting crank vector ! and eigenvectors of K with those of H. (e) Given initial conditions (X(0)=1,Y(0)=0,V0=0), derive and plot the resulting (Tschebycheff) path in the XY-plane .

Ex.2(a) Do ABCD expansion of Hamiltonian ! and plot eigen-frequency levels as in Ex.1.

(b) Sketch a 3D ABC-space as in Ex.1. (c) Show rotations as in Ex.1. (d) Do ABCD expansion of 2D Hooke spring matrix K as in Ex.1. (e) Given initial conditions (X(0)=1,Y(0)=0,V0=0), derive and plot a resulting (Tschebycheff) path in the XY-plane .

Ex.3 Derivation in Lect. 23 (p.47-65) of eigenvectors equates spin-vector S(α,β,.) to crank-vector ±Ω[ϕ,ϑ,.]. Try using the Projector operator method introduced in Lect. 21 p. 40-44. Are results the same? Compare ease of use.

Ex.4 The clearest example of 2-state or spin ½ resonance behavior are A, B or C type of evolution in which the spin vector moves from axis to axis as demonstrated by animations accessed on pages 71 thru 88. (a) Uncoupled oscillation of the AD type (p.71). Let Hamiltonian have A=1.5 and D=0.5 with B=0=C with initial oscillator variables x1(0)=1=x2(0) and p1(0)=0=p2(0). Describe initial spin vector (p.80 Lect.22) and its evolution. (b) Balanced coupled oscillation of the B type (p.76). Explain why phase lag is always π/2 when initial position and velocity of just one oscillator is zero. (Helpful hints on p. 94-97 or p.87 of Lect. 21.) (c) Coriolis coupled oscillation of the C type (p.87). How does this resemble motion of a Foucault pendulum?

Ex.5 For any of the apps discussed in Ex.4 you may access the control panel and reset the A, B, C, an D parameters that determine the crank vector Ω and overall frequency Ω0. See if you can set these and the spin vector S so S goes from the A-axis to the B-axis to the C-axis and back thru the A-axis. Make Ω0 at least 40 times larger than the other A, B, and C components of Ω. Show a screen clip picture of the resulting evolution.

H =Ω01+ !Ω•!S H = 1

534 −12−12 41

⎛⎝⎜

⎞⎠⎟= K

!Ω=(ΩA,ΩB ,ΩC ) Ω0

12 Ω

!Ω=(ΩA,ΩB ,ΩC )

Θ = Ω⋅t = π!Ω

H = 5 −3−3 5

⎛⎝⎜

⎞⎠⎟= K

!19Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Assignment 11. Solution to Ex.1(a) Spin expansion for driver matrices:

! Solution to Ex.1(a)-(e) views of coupled oscillation ABCD decomposition shows H and K share unit crank vector ! and e-vectors ! ,! .

!

! H-eigenvalues: !

This gives: ! and ! and ! , which relates to eigenfrequencies

found above: ! and ! and squares below.

!

K-eigenvalues:

1(e) To show the (XY)-path is a parabola let x=cosωt so y=-cos2ωt=-Re[ei2ωt]=-Re[eiωt]2=-Re[cosωt+isinωt]2 gives: y =-Re[cos2ωt-sin2ωt+i…]=1-2cos2ωt=1-2x2 {See (X1,X2) plot above left and below.} Note that the 3D sketch above has C (Y axis) pointing up while the figure below has the A (or Z) axis pointing up. (Well, C rhymes with Z.)/s

H =51 34 −12

−12 41⎛

⎝⎜⎞

⎠⎟= 15

21 00 1

⎛

⎝⎜⎞

⎠⎟+ −7

521 0

0 −21

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

−245

0 21

21 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= 215 1+ 5

−7 SA + 5−24 SB = 2

15 1+ 10−7 σ A + 10

−24 σB

= 215 1+ΩASA +ΩBSB∴

!Ω=(ΩA,ΩB ,ΩC )=( 5

−7 , 5−24,0)⎡

⎣⎤⎦

Ω0 = 215 2

1 Ω =25

H2 = K =

52 −36−36 73

⎛⎝⎜

⎞⎠⎟=

k1 + k12 −k12

−k12 k2 + k12

⎛

⎝⎜

⎞

⎠⎟

Ω̂ = (25 7 ,25

24 , 0) 3/54 /5( ) −4 /5

3/5( )H =5

1 34 −12−12 41

⎛

⎝⎜⎞

⎠⎟= 34 + 41

5·21 00 1

⎛

⎝⎜⎞

⎠⎟+ 34 − 41

5·21 00 −1

⎛

⎝⎜⎞

⎠⎟+ −125

0 11 0

⎛

⎝⎜⎞

⎠⎟+ 0 0 −i

i 0⎛

⎝⎜⎞

⎠⎟

= 755·2

1 00 1

⎛

⎝⎜⎞

⎠⎟+ −7

5·21 00 −1

⎛

⎝⎜⎞

⎠⎟+ −12

50 11 0

⎛

⎝⎜⎞

⎠⎟

ω↑ =1075 +2

5 = 10

ω↓=1075 −2

5 = 5

⎧⎨⎪

⎩⎪

Ω0=1075

21 Ω̂ =2

1 (ΩA ,ΩB ,ΩC ) = (10 -7 , 10

−24 ,0) 21 Ω =2

1 ΩA2 +ΩB

2 +ΩC2 =10

25 =25

ω↑ =Ω0 +21 Ω =10

75 +25= 10 ω↓ =Ω0 −2

1 Ω =1075 −2

5= 5

K =

52 −36−36 73

⎛⎝⎜

⎞⎠⎟= 52 + 73

21 00 1

⎛⎝⎜

⎞⎠⎟+ 52 − 73

21 00 −1

⎛⎝⎜

⎞⎠⎟− 36

0 11 0

⎛⎝⎜

⎞⎠⎟+ 0

0 −ii 0

⎛⎝⎜

⎞⎠⎟

= 125

21 00 1

⎛⎝⎜

⎞⎠⎟+ −21

21 00 −1

⎛⎝⎜

⎞⎠⎟− 36

0 11 0

⎛⎝⎜

⎞⎠⎟

K↑ = 2125 + 2

75= 100

K↓= 2125 − 2

75= 25

⎧⎨⎪

⎩⎪

!20Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

!

!21Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Solutions for driver matrices in Ex.2(a) to (e).:

! To find the (XY)-path let x=cosωt so y=-cos4ωt=-Re[ei4ωt]=-Re[eiωt]4=-Re[cosωt+isinωt]4 gives: y =-Re[cos4ωt-6cos2ωt sin2ωt+sin4ωt +i…] y =-Re[x4 –6x2 y2 +y4 +i…]=-Re[x4 –6x2 (1-x2)+ (1-x2)2 +i…] y =-1+8x2-8x4

The functions resulting from 4th harmonic is the 4th degree Tschebycheff polynomial T4. The nth harmonic is the nth degree Tschebycheff Tn.

H = 5 −3−3 5

⎛

⎝⎜⎞

⎠⎟= 5 1 0

0 1⎛

⎝⎜⎞

⎠⎟− 6

0 21

21 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= 51− 6SB = 51− 3σB = 51+ΩBSB∴

!Ω=(ΩA,ΩB ,ΩC )=(0,− 6, 0)⎡⎣ ⎤⎦

Ω0 = 5 , 21 Ω = 3, Ω0+2

1 Ω = 5+3= 8 , Ω0−21 Ω = 5-3= 2.

H2 = K =

34 −30−30 34

⎛⎝⎜

⎞⎠⎟=

k1 + k12 −k12

−k12 k2 + k12

⎛

⎝⎜

⎞

⎠⎟

!22Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Solutions for projectors in Ex.3).:

The H matrix being solved on p. 60 to 65 of Lect. 23 is

!

! !

Using eigenvalues 14 and 6 we insert them into matrix H to make projectors:

Eigenvectors match up to an overall phase and norm that would be determined by physical considerations.

H= 12 6(1− i)

6(1+ i) 8

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟= A B− iCB + iC D

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟⎟=

10+ 4cosπ3

4cosπ4sinπ3− i4 sin

π4sinπ3

4cosπ4sinπ3+ i4 sin

π4sinπ3

10− 4cosπ3

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

A= 12, B = 6, C = 6, D = 8,

eigenvalue−1

ω↑= 10+

12−82

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

2

+ 6( )2

+ 6( )2

= 10+ 4 = 14eigenvector −1

↑ =e−iπ8 cos

π6

e+iπ8 sin

π6

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

=1

eiπ4 33

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

e−iπ8 32

eigenvalue−2

ω↓= 10−

12−82

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟

2

+ 6( )2

+ 6( )2

= 10− 4 = 6 eigenvector −2

↓ =−e−iπ8 sin

π6

e+iπ8 cos

π6

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

= −eiπ4 3

31

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

e−iπ8 32

P14 =

12 − 6 6(1− i)

6(1+ i) 8 − 6

⎛

⎝⎜⎜

⎞

⎠⎟⎟

14 − 6 = 8= 18

6 6(1− i)

6(1+ i) 2

⎛

⎝⎜⎜

⎞

⎠⎟⎟P6 =

12 −14 6(1− i)

6(1+ i) 8 −14

⎛

⎝⎜⎜

⎞

⎠⎟⎟

6 −14 = −8= 18

2 − 6(1− i)

− 6(1+ i) 6

⎛

⎝⎜⎜

⎞

⎠⎟⎟

!23Unit 2-3 Assignments for Physics 5103 - Reading in Classical Mechanics with a BANG! Name________________________

Exercises 4 4(a) Uncoupled oscillation of the AD type (p.71). Let Hamiltonian have A=1.5 and D=0.5 with B=0=C with initial oscillator variables x1(0)=1=x2(0) and p1(0)=0=p2(0). Describe initial spin vector (p.80 Lect.22) and its evolution.

! , !

Initial state has α=0=γ and β=π/2 so S-vector is along B. (Below) Crank vector is along A-axis.

!

Crank Ω=(1,0,0) rotates initial B-spin vector S=(SA,SB,SC)=(0,1,0) around A-axis thru -C, -B, and +C axes.

Exercise 5 Given crank 4-vector: !

We need crank to be on the cubic (1,1,1)-diagonal ! so we set !

Solving for A and D: ! Trjectory screen shot below:

x1+ ip

1

x2+ ip

2

⎛

⎝

⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟= 1+ i01+ i0

⎛

⎝⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟=A

e−iα2 cos

β2

eiα2 sin

β2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

e−iγ2 = 2

e−i02 cos

π2

2

ei02 sin

π2

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟

ΩAΩBΩC

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

=A− D = 12B = 02C = 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ and: Ω0 = A+ D

2= 1

Asymmetry SA = 12a σ A( a) = 1

2 a1* a2

*( ) 1 00 −1

⎛⎝⎜

⎞⎠⎟

a1

a2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 12x1

2 + p12 − x2

2 − p22⎡⎣ ⎤⎦=0 = I

2cosβ

Balance SB = 12a σ B( a) = 1

2 a1* a2

*( ) 0 11 0

⎛⎝⎜

⎞⎠⎟

a1

a2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= p1 p2 + x1x2⎡⎣ ⎤⎦ =1= I2

cosα sinβ

Chirality SC = 12a σC( a) = 1

2 a1* a2

*( ) 0 −ii 0

⎛⎝⎜

⎞⎠⎟

a1

a2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= x1 p2 − x2p1⎡⎣ ⎤⎦ =0= I2

sinα sinβ

Ω0 ΩA ΩB ΩC( ) = A + D2

A − D2

B C⎛

⎝⎜

⎞

⎠⎟

ΩA =ΩB =ΩC ΩA =ΩB = B =ΩC = C = 0.1A + D = 2Ω0 or: A =Ω0 +ΩA= 4 + 0.1= 4.1A − D = 2ΩA D =Ω0 −ΩA= 4 - 0.1= 3.9