Vol. 2 - The World of Mathematical Equations

April, 2008 PROGRESS IN PHYSICS Volume 2for i = 1; : : : ; n where we have defined:A respectively where the two independent variables z 1 ; z 3 ofW(z 1 ; z0 1 )W(z n ; z0 n )t a i t a are mixed from the two quantum Wilson lines W(zj := W(z 1 ; z0 1 )1 ; z 2 )and W(z 3 ; z 4 ) respectively and the the two independent variablesz 2 ; z 4 of are mixed from the two quantum Wilson[W(z i ; z0 i )t a ][W(z j ; z0 j )t a (47)]W(z n ; z0 n ) :lines W(z 1 ; z 2 ) and W(z 3 ; z 4 ) respectively. From this weRemark From the quantum gauge theory we derive the determine the form of A as follows.above quantum KZ equation in dual form by calculus of variation.This quantum KZ equation in dual form may be conresentan element g of SU(2) and let D(g) D(g) denoteLet D denote a representation of SU(2). Let D(g) repsideredas a quantum Euler-Lagrange equation or as a quantumYang-Mills equation since it is analogous to the classi-equation we definethe tensor product representation of SU(2). Then in the KZcal Yang-Mills equation which is derived from the classical[tYang-Mills gauge theory by calculus of variation. a t a ][D(g 1 )D(g 1 )][D(g 2 )D(g 2 )] :=:= [t a D(g 1 )D(g 1 )][t a (53)D(g 2 )D(g 2 )]5 Solving quantum KZ equation in dual formandLet us consider the following product of two quantum Wilson [D(g 1 )D(g 1 )][D(g 2 )D(g 2 )][t a t a ] :=lines::= [D(gG(z 1 ; z 2 ; z 3 ; z 4 ) := W(z 1 ; z 2 )W(z 3 ; z 4 ) ; 1 )D(g 1 )t a ][D(g 2 )D(g 2 )t a (54)] :(48)Then we let U(a) denote the universal enveloping algebrawhere a denotes an algebra which is formed by the Liewhere the quantum Wilson lines W(z 1 ; z 2 ) and W(z 3 ; z 4 )represent two pieces of curves starting at z 1 and z 3 and ending algebra su(2) and the identity matrix.at z 2 and z 4 respectively.Now let the initial operator A be of the form A 1We have that this product G(z 1 ; z 2 ; z 3 ; z 4 A 2 ) satisfies the A 3KZ equation for the variables z 1 , z 3 A 4 with A i ; i = 1; : : : ; 4 taking values in U(a). In thisand satisfies the dual case we have that in (52) the operator (z 1 z 3 ) acts on AKZ equation for the variables z 2 and z 4 . Then by solving from the left via the following formula:the two-variables-KZ equation in (44) we have that a form ofG(z 1 ; z 2 ; z 3 ; z 4 ) is given by [69–71]:t a t a A = [t a A 1 ]A 2 [t a A 3 ]A 4 : (55)e ^t log[(z 1 z 3)] Similarly the operatorC 1 ;(z 4 z 2 ) in (52) acts on A from(49)the right via the following formula:where ^t := e2 0k 0+g 0Pa t a t a and C 1 denotes a constant matrix Atwhich is independent of the variable z 1 z a 3 t a = A 1 [A 2 t a ]A 3 [A 4 t a ] : (56).We see that G(z 1 ; z 2 ; z 3 ; z 4 ) is a multi-valued analytic We may generalize the above tensor product of two quantumWilson lines as follows. Let us consider a tensor productfunction where the determination of the sign depended onthe choice of the branch.of n quantum Wilson lines: W(z 1 ; z0 1 )W(z n ; z0 n ) whereSimilarly by solving the dual two-variable-KZ equation the variables z i , z0 i are all independent. By solving the twoin (46) we have that G is of the formKZ equations we have that this tensor product is given by:C 2 e^t log[(z 4 z 2 )] W(z; 1 ; z 0 =Y(50)1)W(z n ; zn) 0 =where C(z 2 denotes a constant matrix which is independent ofi z jthe variable z 4 z 2)AY(z 0 i z 0 (57)j) ;ijij.From (49), (50) and letting:whereQij denotes a product of (z i z j ) or (z0 i z0 j )C 1 = Ae^t log[(z 4 z 2)] ; C 2 = e ^tfor i; j = 1; : : : ; n where i j. In (57) the initial operatorA is represented as a tensor product of operators A iji 0 j 0,log[(z 1 z 3)] A ; (51)i; j; i0 ; j 0 = 1; : : : ; n where each Aiji 0where A is a constant matrix we have that G(z 1 ; z 2 ; z 3 ; z 4 ) j 0 is of the form of theisinitial operator A in the above tensor product of two-Wilsonlinescase and is acted by (z i z j ) or (z0 i z0 j ) on itsgiven by:G(z 1 ; z 2 ; z 3 ; z 4 ) = e ^t log[(z 1 z 3 )] Ae^t log[(z 4 z 2 )] ; (52)two sides respectively.where at the singular case that z 1 = z 3 we define 6 Computation of quantum Wilson lineslog[(z 1 z 3 )] = 0. Similarly for z 2 = z 4 .Let us find a form of the initial operator A. We noticethat there are two operators (z 1 z 3 ) := e ^tLet us consider the following product of two quantum Wilsonlog[(z 1 z 3)] lines:and (z 4 z 2 ) = e^t log[(z 4 z 2)] acting on the two sides of G(z 1 ; z 2 ; z 3 ; z 4 ) := W(z 1 ; z 2 )W(z 3 ; z 4 ) ; (58)Sze Kui Ng. New Approach to Quantum Electrodynamics 23