1. Understand and Correct a Defective Lattice: Phasors

2. Pictures
Resonance: in blue
Distortion

3. Resonances and Distortions
A resonance is an abnormality of phase space.
A distortion is an unacceptable deviation from linearity (or mid-plane symmetry) but the map is still a rotation in disguise.
All these effects are the results of iterating a map!

4. We want to Understand MN
Globally MsN=As◦RN◦As-1 is impossible if resonances are present.
Therefore, let us raise the map “manually” to a power “N”
Phasors will emerge from that treatment!

5. Problem M = M1◦ (I + e C+…) M1 is the linear part of the map
C is the quadratic distortion of this map
“…” represents higher order terms which we will temporarily neglect
Question:
MN = M1◦ (I + e C+…) ◦ … ◦ M1◦ (I + e C+…)
Can we compute this map to first order in e?

6. Normalize what you understand already: Linear Part
A-1 ◦ MN ◦ A ={ A-1 ◦ M1◦ A ◦ A-1 ◦ (I + e C+…) ◦A }N
= {R ◦ (I + e Ĉ +…)}N
Where Ĉ = A-1C ◦ A
We will symbolically denote the effect of A on C as:
Ĉ = AC

7. Example: C is a sextupole kick
C=(0,kx2)
A is a Courant-Snyder Matrix
C ◦ A =(0,kb x2)
Ĉ =A-1 C ◦ A =(0, kb3/2 x2 )

8. MN to first order: Origin of Phasors
{R ◦ (I + e Ĉ +…)}N =?
= R◦(I + e Ĉ +…) ◦R◦(I + e Ĉ +…) ◦ … ◦ R◦(I + e Ĉ +…) ◦R◦(I + e Ĉ +…)
= R◦(I + e Ĉ +…) ◦ R-1 ◦R2 ◦(I + e Ĉ +…) ◦ R-2◦ R3 ◦(I + e Ĉ +…) R-3 ◦ … ◦ RN ◦(I + e Ĉ +…) ◦R-N ◦RN
= (I + e (R1 Ĉ+ R2 Ĉ+ R3 Ĉ+ R4 Ĉ+ … + RN Ĉ ) + Order(e2) ) ◦RN
Therefore understanding Rk Ĉ is extremely important.

9. Phasors: Eigen-operators
We will see that Ĉ is connected to a vector field and therefore in the Hamiltonian case to a simple Hamiltonian function.
For the moment, let us examine the general case of Rk Ĉ where Ĉ is a vector function. (general = non Hamiltonian)
We want to find the eigenvector field:
R Ĉ = = R-1 Ĉ ◦ R =l Ĉ

10. Resonance Basis h±=x±i p
or (x1new,x2new)= (x1old+i x2old , x1old - i x2old )
This transformation can be written in matrix form as:
The matrix R in that new basis is given by E=B-1RB which is just

11. Definition of Phasors The n-turn map is:
= (I + e (E1 G + E2 G + E3 G + E4 G + … + EN G ) + Order(e2) ) ◦ En
Ek G = E-k G ◦ Ek Eab = dab exp( (-1)a im)
In the new basis : Gi = S Gi;nmx1nx2m
Ek G = S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m

12. Sk=1,N Ek G |a = Sk=1,N S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m
Summing up
Sk=1,N Ek G |a = Sk=1,N S Gi;nm exp(ik{m-n-(-1)a} m) x1nx2m
= S Gi;nm {1-exp(iNQnm) }/{exp(-iQnm)-1} x1nx2m
Qnm = {m-n-(-1)a} m
Sextupole Order m+n=2 ; m-n-(-1)a =±3, ±1;
Let us classify the cases.

13. Phasors at 1st Order in Sextupoles
If 3m≈k2p , then G1;02 and G2;20 can get real big!
These are now quasi-secular terms and
we must treat them with care: this is best done if we reveal the Hamiltonian nature of the map explicitly.

14. Phasors at 1st Order (continue)
If 3m≈k2p , then G1;02 and G2;20 can get real big!
These are now quasi-secular terms and
we must treat them with care: this is best done if we reveal the Hamiltonian nature of the map explicitly.
Are secular terms a property of resonance? Answer: no! Tune shifts with amplitude are unavoidable in a nonlinear map! Therefore even maps which can be made into a rotation M=A◦R◦A-1 contain secular terms which we reinterpret as being part of the rotation!

15. Hamiltonian Methods
The coefficients Gi;mn in the map I+eG+… are not arbitrary. In fact, it can be shown easily (we will show that) that there are only two independent terms related to the n=k phasors in the sextupole order map.
Hamiltonian → G2;11 + 2G1;20= 0 and G1;11 + 2G2;02 = 0.
Hamiltonian → G1 = 2i ∂2 g and G2 = -2i ∂1 g .
Or G = [g,Identity] where [x,p]=1 and therefore [h+,h-]=-2i

16. From g to G g= g30 h+3+ g21 h+2 h- + g12 h+ h-2 + g03 h-3
G1=[g,h+]= 2i{ g21 h+2 +2 g12 h+ h- + 3g03 h-2 }
G2=[g,h-]= -2i{ 3g30 h+2 +2 g21 h+ h- + g12h-2 }
G1;20 = 2i g G1;11 = 4i g G1;02 = 6i g03
G2;20 = -6i g G2;11 = -4i g G2;02 = -2i g12
Therefore G2;11 + 2G1;20= 0 and G1;11 + 2G2;02 = 0.

17. Making I+eG Hamiltonian
x=I+eG+… = exp(:eg:)I
Here :g:f is defined as [g,f]
Notice that x=exp(:eg:)I is an exact solution of dx/de = [x,-g], so x is exactly Hamiltonian
Furthermore g is an invariant of x!

18. Pictures: Go back to Problem
Resonance: in blue
Distortion

19. Start with Resonance Globally Ms=As◦R◦As-1 is impossible.
Globally Ms=As◦N◦As-1 such that
N=R2p/3◦N◦R2p/3-1
is possible.

20. One-Resonance Normalization

21. In our New Phasor Language
M = exp(:g:)R where g contains the non linear effects
Remove phasors not related to k3n: New map is N = Aexp(:g:)R A-1 = exp(:r:)R
Cube the map : N3 = exp(:r3:)R3
N3 is near the identity
Therefore N3 = exp(:3H:)
H is an invariant which potentially includes islands!

22. So H is to order dodecapole
H=-(dJ+aJ2+bJ3)
+J3/2 ((A+CJ)cos(3f) + (B+DJ)sin(3f))
+J3 (Ecos(6f) + Fsin(6f))
Now for fun, we will correct a small lattice using sextupoles just to see if these tools make any sense!

23. Three corrections Standard Correction Methods
H=-(dJ+aJ2+bJ3)+J3/2 ((A+CJ)cos(3f)+(B+DJ)sin(3f))+J3 (Ecos(6f)+Fsin(6f))
Standard Correction Methods
Correct Linear Terms A=B=0
If not good enough, try A=B=C=D=0
Correct tune around island:
d+2aJ0+3bJ02 =0
A+CJ0=B+DJ0=0.
This can be done numerically or with H!

24. Next Lecture
We will look in details in the computer implementation