Representation of an arbitrary beam transfer matrix
and solution of the beam matching problem
using equally spaced thin-lens quadrupoles
Sergey Orlov
XFEL Beam Dynamics Group Meeting, 6 September 2010
Acknowledgements
Being a summer student I would like to thank all members of the Machine Physics
Group for their hospitality, support, and enormous help with solving many
organizational questions.
I would like to thank the head of DESY Summer Student Program
Professor Meyer for giving me the opportunity to come here and get such
great experience not only from professional point of view.
Last but not least, I would like to thank my supervisor V.V. Balandin for
his wonderful thoughts and fantastical support in all questions that I had, for
many hours of patient explanations, corrections and clarifications, as well as
for guidance throughout the whole summer student program.
Since invention of alternating-gradient
focusing there are still several questions
which don’t have strict mathematical answers yet:
Is it possible to obtain an arbitrary block-diagonal 4×4 transfer matrix using
only drifts and quadrupoles?
If the answer to the question above is positive, could it be done
with the number of quadrupoles which does not depend
on input beam transfer matrix?
If again the answer to the question above is positive, what is the minimum
number of quadrupoles required and could we get analytical formulas for
quadrupole strengths and lengths of field free spaces (drifts)?
Surprisingly, the answers to these questions are not known not only
for quadrupoles but also for the situation when instead of quadrupoles
their thin-lens approximations are used. In this presentation we will
try to answer these questions in the case when thin-lens
approximation for quadrupole focusing matrices is used.
Mathematical formulation of the problem of
representation of an arbitrary uncoupled beam
transfer matrix by drifts and quadrupole thin lenses
Let M x and M y be arbitrary 2×2 matrices with unit determinant, and let D(l )
and Q (k ) be 2×2 matrices of drift space and thin lens quadrupole respectively,
1 l 
,
Dl   
 0 1
 1 0
.
Qk   
 k 1
The problem of representation of an arbitrary beam transfer matrix is: are
there a natural number n , real numbers k1 ,, kn , and non-negative l1 ,, ln ,
such that the following equality holds:
Q(k1 ) D(l1 ) Q(kn ) D(ln )  M x ,
Q(k ) D(l ) Q(k ) D(l )  M .
1
1
n
n
y

What is actually known about this problem
There are several papers, whose authors tried to find solution of this problem
in the form of exact analytical formulas, with most advanced probably being the
following:
1.
2.
Y-Chiu Chao and John Irwin, “Solution of a three-thin-lens system with
arbitrary transfer properties”, SLAC-PUB-5834, October 1992.
O. Napoly, “Thin lens telescopes for final focus systems”,
CERN/LEP-TH/89-69, CLIC Note 102, November 1989.
For example, authors of the paper 1 give formulas for k -values and drift
lengths in the case of three thin lenses and three variable field free spaces, i.e.,
in the case of equations:
Q(k1 ) D(l1 )Q(k2 ) D(l2 )Q(k3 ) D(l3 )  M x ,

Q(k1 ) D(l1 )Q(k2 ) D(l2 )Q(k3 ) D(l3 )  M y .
Unfortunately, the given solution contains roots and denominators, and it is
clearly visible that these denominators can be equal to zero for some input
matrices and, due to presence of roots, some solutions can become unphysical
(complex roots). And actually, as it will be shown later by example, three drifts
and three thin lenses are insufficient for representation of an arbitrary beam
transfer matrix.
What we have to solve
After multiplying all matrices on the left-hand side of the equations
Q(k1 ) D(l1 ) Q(kn ) D(ln )  M x ,

Q(k1 ) D(l1 ) Q(kn ) D(ln )  M y
we obtain the following system of eight equations (+ 2 symplecticity conditions):
 f1 (k1 , l1 ,, k n , ln )  m11x  0,


 f (k , l ,, k , l )  m y  0,
n n
22
 8 1 1
where the functions f1 ,, f8 are polynomials of variables k i and
we have to solve the system of non-linear polynomial equations.
li
, i.e.,
What is known about polynomial systems
Let us consider first the system of linear equations (system of first order
polynomial equations).
For linear systems there is a complete theory, which in algorithmic way
allows to answer the questions about the number of its solutions (unique solution,
many solutions, solution doesn’t exist at all). This algorithmic way is called
Gaussian elimination and allows to construct triangular system, which is
equivalent to the original one.
What is known about polynomial systems
Surprisingly, for general polynomial system there exists the similar algorithm,
which allows to construct the equivalent system, which is called Gröbner basis,
and it is an analogy of triangular form of linear system.
So, for general polynomial system one can also answer in algorithmic way
about absence or existence and number of its solutions.
Nowadays, the reduction of original system to its Gröbner form can be done
with the help of computers using such formula manipulators as Maple or
Mathematica.
Unfortunately, this equivalence of original system and its Gröbner form takes
place only over the field of complex numbers and appears to be not very useful
for us, as far as we are interested in physical (real) solutions.
Nevertheless, the use of computer assistance is not completely useless in our
problem. Because the absence of complex solutions means also the absence of
real solutions, computer helped us to construct examples of particular matrices,
which can not be represented with certain number of thin lenses.
Example of matrix which can not be represented
by 3 thin lenses and 3 variable drift spaces
1
1


Mx  My  

 1 0
This matrix can not be represented with less than 5 thin lenses and 4
variable drift spaces (in total, 9 parameters). This example was found with
the help of Maple and checked once more using Mathematica program.
This example clearly shows that the simple “degree-of-freedom count”
not always leads to the correct answers.
Equally Spaced Quadrupoles
As far as the hope to give the main job to the computer failed, the only way
left to get some insides into the problem is to make hand calculations. For this
purpose we consider the system constructed from equally spaced thin lenses.
Such systems, as we will see, have very pleasant symmetry, which
significantly simplifies hand calculations.
k1
k2
kn
k3
l  const
As one see, the matrix of thin lens sandwiched between two equal drift
spaces is similar to the matrix P , which depends on the single parameter only.
 1 l  1 0  1 l 


  S (l ) P(a) S 1 (l )
R  
 0 1 k 1  0 1
 a 1

P(a)  
 1 0 
a  2  2kl
S (l ) 
1  l
 l



2 1 l 1 l 
Equally Spaced Quadrupoles: Equations in new variables
The equality
R  S (l ) P(a)S 1 (l ) allows us to reformulate the original problem
R(k1 , l )  R(kn , l )  M x ,

R(k1 , l )  R(kn , l )  M y
in the more simple form
 P(a1 )  P(an )  M x ,

 P(4  a1 )  P(4  an )  M y ,
where the matrices on the right-hand side can be calculated with help of
the formulas below:
M x  S 1 (l )M x S (l ),
M y  S 1 (l )M y S (l ).
One Dimensional Case
Let us consider first the one dimensional problem, when we are interested, for
example, only in horizontal motion. This consideration is useful not only from
methodological point of view but we will use it later for the solution of complete
2D problem.
We start from consideration of three thin lenses from that physical
point of view of “degree-of-freedom count”.
 m11 m12 
.
P(a1 ) P(a2 ) P(a3 )  
 m21 m22 
Due to the simple form of the matrix P , this system can be easily solved by hand
with respect to variables a i , and we obtain the following possibilities: when m22  0 ,
we have unique solution and when m22  0 , the solution either does not exist or is
non-unique.
Geometrical interpretation: 2×2 symplectic matrix depends on 3
parameters. In the space of these 3 parameters there is a plane,
and on this plane there is a line. If matrix parameters are
outside the plane, then solution is unique. If parameters are on
the plane but outside line – solution does not exist. And if we
are on the line, we have infinite number of solutions.
In order to have solution for an arbitrary matrix M , we need to add one more
parameter. It could be the distance l between lenses or one more thin lens.
In both cases solution will be non-unique for each input matrix.
Reduction 2D  1D
Though 1D problem can be easily solved by hand, the complete 2D problem
still remains too complicated for the hand solution. It is mainly connected with
the fact that thin lens can not act in two planes independently; if it focuses beam
horizontally, then it defocuses beam vertically and vice versa. Fortunately, we
have found four-lens combination which has transfer matrix similar to the
transfer matrix of a single thin lens but its actions can be chosen independently
for both transverse planes.
Let us consider four-lens block, where two lenses in the middle are set to
the constant values a2  2  3 , a3  2  3 (or a2  2  3 , a3  2  3 )
By direct matrix multiplication one can show that

c
 wx
,
 Px ( wx )  P(a1 ) P(a2 ) P(a3 ) P(a4 )  

 1/ c 0 

 P ( w )  P(4  a ) P(4  a ) P(4  a ) P(4  a )   wy
y
y
1
2
3
4
 1/ d



d
,
0
where wx and wy are linear functions of parameters a1, a4 and c and d are
some constants. Because 2×2 matrix which connects wx and wy with a1 and
is non-degenerated, the values of parameters wx and wy can be chosen
independently.
a4
Solution of 2D problem by its reduction to the two 1D problems
For solution of 1D problem we need three thin lenses plus one additional
parameter for making m22  0. So if we take our three four-lens blocks,
we will have three independent lenses for each plane (the fact that in these
equivalent lenses we have constants c and d not equal to 1 is not essential).
x
y
In order to satisfy conditions m22
 0 and m22
 0, we need one more
parameter, which again could be chosen either as a distance between lenses or
we may add one more lens.
So we have found thin lens solution of the problem of representation of an
arbitrary transfer matrix, which requires 13 thin lenses or 12 lenses with
possibility to use distance between them as an additional parameter.
Note that in the case of 13 lenses considered solution actually uses only
7 parameters in order to represent an arbitrary transfer matrix because 6 lenses
are set to the constant strengths (in the case of 12 lenses 7 parameters depend
on input matrix and other 6 – just on distance between lenses).
Although the total number of lenses used in this solution is probably still
not the minimal needed, we have simple explicit formulas for the lens strengths
as functions of coefficients of input transfer matrix. And as concerning number
of variable parameters in our system (7 parameters in the case of 13 lenses),
it is the minimum possible value because we have an example of transfer matrix
which can not be represented by using six lenses with fixed distance between
them, namely
 0 1
.
M x  M y  
 1 0 
Beam Matching Problem
Let us assume that we have two sets of Twiss parameters given in the two
points of a beam line. The beam matching problem is the problem of finding
transfer matrix A , such that the following identity holds:
  2x  Ax 1x Ax ,
 y  y 
 x x 
.
,  y  
 x  
 2
1 




y
y 
 y  Ay  y Ay ,
x  x 

The first question is if this problem has symplectic solution at all for
arbitrary two sets of Twiss parameters. The answer is “Yes”, it is well known,
and the general symplectic solution of the matching problem can be expressed
in the following form:
1

A

T
 x
x 2 R (  x )Tx1 ,

1
A

T

y 2 R (  y )Ty1 ,
 y
 cos
R( )  
  sin 
sin  
,
cos 
 1/ 
T  
 / 
0 

 
and depends on two arbitrary parameters (phase advances).
Because our previous solution for an arbitrary transfer matrix certainly gives
also the solution of the matching problem, our current goal is to solve matching
problem with less thin lenses using freedom in choosing the phase advances.
Thin lens solution of the matching problem
We will use the same scheme as before and will find first the number of thin
lenses required for solution of 1D problem, and then we will use our four-lens
blocks for solution of 2D problem.
Matching equations for 1D problem
where
1a11  (1   2 )a12   2 a22  0,

 2 1a11  1 2 a21  1 2 a12  1 2 a22  a12  0,
aij are the coefficients of matrix A .
It is not difficult to show that this problem can be solved with 2 lenses, if and
only if 1 2  1 . So, as before, one more parameter for solution of 1D matching
problem is still needed. Again, it could be one more lens or variable distance.
Therefore, using two four-lens blocks plus one additional parameter we can
get a complete solution of 2D beam matching problem (so in our solution we
need 9 lenses with 4 again staying fixed). And also we have an example of Twiss
parameters which can not be matched with 4 thin lenses:
1 3 0 
,
    
 0 3
1
x
1
y
 1 0
.
    
 0 1
2
x
2
y
Summary
It is proven that, actually, an arbitrary beam transfer matrix
and the problem of matching of arbitrary Twiss parameters
can be solved with finite number of thin lenses.
We have found a solution for representation of transfer matrix, which uses
13 thin lenses (or 12 lenses plus variable space between them). From this 13
parameters 6 are independent from the elements of the input matrix. We also
have found an example of the matrix, which can not be represented with less
than 7 parameters.
We have found a solution of matching problem, which uses 9 thin lenses (or 8
lenses plus variable space between them). From this 9 parameters 4 are
independent from the elements of the input matrix. We also have found an
example of Twiss parameters, which can not be matched with less than 5
parameters.
Thank you very much for your
attention!