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A configuration space approach to the nuclear four-body problem
ANNALS
OF PHYSICS
111,
A Configuration
162-200 (1978)
Space Approach
to the
Nuclear
Four-Body
Problem
W. ZICKENDRAHT Fachbereich
Physik,
Phi&p+Universitat,
Marburg,
3550
Marburg,
Germany
Received November 6, 1976
The derivation of a complete orthogonal system for the nuclear four-body problem is discussed, which will be especially useful in the case of identical particles. A method for dealing with four-body bound states is proposed. Part of the orthogonal system is given explicitly.
1. INTRODUCTION The method of the hyperspherical harmonics has been treated extensively during the last decade. The earliest publications dealt with the three-particle problem [I, 21, for which it was possible to derive orthogonal functions with extremely simple properties with respect to exchange of identical particles. A paper on the four-particle problem which started from the samepoint of view has been published by the author [3]. The work is continued with the present paper. A number of difficulties which will be discussedin this introduction delayed the continuation of the work at first. The emphasis of this paper is on constructing orthogonal functions with simple properties with respect to exchanges of identical particles and to distinguish between collectiye and single-particle degreesof freedom as in the general A-nucleon problem. The price one has to pay for this gain in simplicity is a tedious method for finding the orthogonal functions. Other authors put the emphasis on giving the explicit forms of a complete orthogonal system for the four-particle problem [46]. These authors do not distinguish between collective and single-particle degrees of freedom and their functions are difficult to handle for identical particles. Surkov [4] started with Jacobi coordinates. The polar anglesof the three vectors are six of the coordinates used. The other three coordinates are derived from the three lengths of the vectors as is usually done with the method of hyperspherical harmonics. Only part of the orthogonal system is given explicitly in the paper of Surkov. Galbraith [5] and Chacon and Amaya [6] start with three coordinate vectors defined at first by Levy-Leblond [7]. Except for this difference the hyperspherical coordinates are defined in the same way as in the paper by Surkov. In the papers [5,6] complete orthogonal systems are derived, complete sets of operators are given, and the properties with respect to exchanges of identical particles are examined. Eight of the coordinates used in the papers [4-61 are not symmetric with respect to arbitrary exchanges of identical particles, they change under at least one of the possible exchanges. Practical calcu162 0003-4916/78/1111-0162$05.00/0 Copyright All rights
0 1978 by Academic Press, Inc. of reproduction in any form reserved.
FOUR-BODY
PROBLEM
163
lations with such orthogonal functions might be difficult. There are a number of problems in which the space part of the four-particle wavefunction will have a high degree of symmetry with respect to particle exchanges. The ground state and the first excited 0+-state of the N particle, for example, are described in a good approximation as completely antisymmetric in the spin-isospin coordinates and completely symmetric in the space coordinates. The four-a model of I60 (with OLparticles in the ground state or in the first excited state) is another problem which is simplified by the use of symmetric coordinates and the orthogonal functions derived in this paper. This problelm is presently considered by the author as a continuation of earlier work on three (y.particles-that is the three-a-model of 12C-which could be treated in an extremely simple way [B]. Further four-particle problems of interest in this connection are other four-cluster problems in nuclei with two or more identical clusters, also the problem of three nucleons outside a closed shell, which together with the core make a four-particle problem. There is also the possibility of applications in molecular physics for which one may be encouraged by simple three-particle calculations with the corresponding three-particle coordinates and orthogonal functions [9, lo]. Generalizations of the transformations given in [3] to the A-particle problem have been given by several authors [l l-131. The symmetric four-particle coordinates are special cases of the A-particle coordinates given in these papers. One purpose, among others, of these papers is to derive the collective degrees of freedom from the single-particle degrees of freedom, also the coupling of the collective degrees to the remaining single-particle degrees. Thus it was possible in a simple way to derive the Bohr equations for the collective motion from the A-particle Schrodinger equation [15]. The success of this generalization to the A-particle problem is another reason for continuing the work started in [3]: For four particles there are six collective coordinates and three coordinates which must be considered as the remaining single-particle coordinates. So there is the clear distinction between these different degrees of freedom for four particles as for A particles. Recently, Vanagas [16] has given a method for the irreducible expansion of an arbitrary microscopic Hamiltonian in a series, the first term of which gives the collective part. Vanagas was able to prove that there can be no more than six collective coordinates for a system of identical particles (like nucleons) if one requires these coordinates to be invariant under the exchange of the particles. This proof brings the distinction between six collective coordinates and 3A-9 single-particle coordinates to a firm basis. The list of references given at the end of this paper does by no means contain all the papers on hyperspherical harmonics. Most of the papers on this problem which appeared before 1972 are contained in a review article by Louck and Galbraith [17]. There were two problems which had to be solved after the publication of [3]: One is concerned with the range of two angular variables, the other with the derivation of that part of the complete orthogonal system which depends on these two angular variables. It is shown in this article how these problems are solved. The main results of [3] which are necessary for understanding will be given in Section 2.
164
W. ZICKENDRAHT
2. THE SCHR~DINGER
EQUATION
IN THE SPECIAL COORDINATES
The space vectors of the four particles are q (i = I, 2, 3,4). In the center of mass system the vectors x1 , x, , xB are used [7]: x1 = Q(rI 4 r2 - r3 - r4), x2 = H--r1 + r2 - fs + q), x3 = &(-rl
In the first step of the transformation are introduced [4]:
(1)
+ r2 + r, - r4).
complex variables ljk (spherical components)
t,l = (x11 + ix21 + ix,, - x*,)/2, to1 =
-(x1,
+
ix23)/2"",
tl, = (-x11 - ix,1 + ix,, - x,,)/2 t,O = -(x31 + ix,,)/21iz,
too= x33 tO, =
(2)
3
(XQ1
iX32)/Z!‘12,
-
t;l = (-x11 + ix,, - ix,,
-
x22)/2,
to1 = (XI3 - ix‘J21/2, t1: = (XI1 - ixzl - ix,, - x2,)/2.
The second step transforms the tik to six angular variables LY,/3, y, #, 8, y and to three variables yr , y2 , y, , which will be replaced later on by two angular variables and one length: (3) SI1= s1: = $(y1 - y,);
ST, = s;l = -iK.h
+ id;
(4)
so0= Y3; so
1
=
s1 o =
s-l0
=
s",
zzz 0.
The angles #, 8, q define the orientation of a body-fixed coordinate system with respect to the center of mass system [18]. This body-fixed system consists of the three axes of the principal moments of inertia. This can be understood easily: The
FOUR-BODY
165
PROBLEM
moment of inertia with respect to an arbitrary axis (the direction of which is given by the unit vector e) in the center of mass system is (rs is the space vector of the center of mass; equal masses m of the particles are assumed): J = m
i
i i=l
(rt - rs)” -
i
[(ri - rJ
i=l
4” I
(5) = m i
[xi2 - (x,e)“].
i-1
Applying
tlhe transformations
(2) and (3) one obtains:
J = m{Y12+ ~22+ ~32- (yle>” - (Y2e)2- (y,e)“>.
(6)
Here yr , ya , ys are the three orthogonal vectors the directians of which with respect to the center of mass coordinate system are given by the three Euler angles ~4, 9, CJJ. The lengths of the three vectors are y1 , y2, y3 . The distinction between collective and single-particle coordinates can be made here: While y1 , y2 , y, , CJJ,9, $ are of a collective nature, 01,/3, y are the remaining single-particle coordinates [ll-13, 161. One obtains for the principal moments of inertia:
4 = m(y22+ ~~~1; J2 = m(y32+ y12); J3 = m(y12 + ~2).
(7)
The sum of the d operators in the Schrodinger equation has the following form in the new coordimates:
(8)
166
W. ZICKENDRAHT
Lo1, L,, , L,, are the components of the total orbital angular momentum with respect to the body-fixed coordinate system [18]. Li,, Liz, Li, are of the same form as L,, , L,, , Le3 with $, 9, q~replaced by 01,/3, y.
3. INVARIANCE PROPERTIES OF THE EIGENFUNCTIONS UNDER COORDINATE TRANSFORMATIONS It is the purpose of this paper to construct eigenfunctions
of the operators
Li5 = --%(a/+). The last of the five with y2 = y12 + y22 + ys2, L,, = -E(a/+), operators (9) is the angular part of the sum (A, + A, + A3), that is the Casimir operator N(9). It contains eight angular variables. These are $, 8, y, LY,/3, y and two more variables which depend on the ratios yl/y and y2/y. The latter two variables would be one possible choice for the remaining two angular variables. The operators (9) all commute with each other. These five operators do not form a complete set of commutable operators in the eight angle variables. Three more operators would be needed. The experience with the three-particle problem shows that it is best to do without these missing operators. In the three-particle case one has four simple commuting operators in a five-dimensional angular space. The fifth operator was derived (unpublished) and turned out to be complicated containing third derivatives in the angles. So it must be expected that the three missing operators for the four-particle problem are complicated too. Doing without these missing operators will mean in general that one has degenerate eigenfunctions for fixed eigenvalues of the operators (9). The tables at the end of this paper show that for X < 3 there are no degenerate functions. For those cases with X 2 4 in which one has degenerate functions one has to orthogonalize them. L,, is the z component of the total orbital angular momentum. For all four-particle problems with central interactions the wavefunction will be an eigenfunction of L,, . The corresponding quantity for the internal orbital angular momentum is Lit . The interactions of the four particles will always depend on 01,/3, y whether central or noncentral. Hence the wavefunction will not be an eigenfunction of Lit . The hyperspherical harmonics which will be defined in detail below will be designated as En with h(X + 7) as the eigenvalue of the Casimir operator. FA shall be an eigenfunction of all the operators (9). Strictly taken the quantum numbers of all these operators should appear as indices of FA . They were left out because the formulas would get too lengthy with all these quantum numbers as indices. The eigenfunctions of Le2 and L,, are the elements of the rotation matrices D&&4 $7 v) L e‘DLe KS4 = fi2Le(L, + 1) D;:Me ,
(10)
FOUR-BODY
Similarly
167
PROBLEM
one has for Liz and Lit the eigenfunctions D&,&: L,‘D&,i
= A2L,(Li + 1) D&
,
(11) L,,D;f,, I 1 = H4iD;~,, I I . D&,,, is also an eigenfunction for L,, = -ifi a/h+h, D&,+ for Liz = -it? a/&.. L,, and Lizi do not commute with the Casimir operator. Hence FA will contain sums over K, and Ki . yAFA is an eigenfunction of A, + A, + A, with eigenvalue zero.
This follows immediately
from writing
A, + A, + A, = &
+ $ $ - -$ . (Casimir operator).
(12)
As F,, is the eigenfunction of the Casimir operator with eigenvalue X(X + 7) and depends on the eight angular variables but not on y, one has
(A, + FA is a hornogeneous polynomial
+
A,
A,)(Y~ * FJ = 0.
(13)
in y1 , yZ , y3 of degree h. The Ansatz for yAFA is
now: (14)
With this Ansatz yAFA is an eigenfunction of La2, L,, , Li2, LiE with eigenvalues + l), M4,, fi2Li(Li + l), ?iM, . The functions G$ are homogeneous polynomials in ,yl , y2 , y3 of degree X. Strictly taken these functions should have indices Li , L, , h; they were left out for the same reasons as above. With Eq. (13) one obtains a system of coupled equations for the functions G2 :
ii2L,(L,
a2
I 8Y12
,
1 i 2y1 . i Y12 -
+ -
Y22
Y22
l ) -& - Y12
+
l ) -& Y12 - Y32
+ &
+
2Y3
- (
+ +-
y32
1
2
y12
+ +
3%
y32
-(
y22
: y3
A
y22
) ’&
- if [L,(L, + 1) + Li(Lc + 1) - Ke2- Ki2] * [ (;:I _’ $~2 +
y32+ ‘12 ] (Y32
y12 + y22 (Ke2+ K;) (Y12 - Y22)2 -1 _
= 4[
Y22
(Y22
+ -
Y32 Y32)2
-
-
( yl:y~y;22~ K&i]
Y32 + Y12 (Y32 - Y1212
- Y1212
GE
+g?~e+&a+,G& 1*[&+,g~+,G~+,
168
W.
ZICKENDRAHT
gKL = [(L + K)(L - K + l)]““.
(15)
In principal Eqs. (15) would be sufficient to determine the functions G? . This task is simplified by making use of invariance properties: Transformation (3) defines a system of body-fixed axes y1 , y2 , y3 . The directions of these three axes, that is the directions of the vectors y1 , y2 , y3 , are orthogonal. The position of this body-fixed Cartesian coordinate system with respect to the center of mass system is described by the three Euler angles. The designations of the axes as 1, 2, 3 axes are arbitrary; the same is true for the designation of positive or negative directions; but only righthand coordinate systems are considered. This arbitrariness in the designation of the axes was already treated in the paper by Bohr [ 141.It has as a consequence that there are a number of rotations in q, 8, # space and exchanges of yl, y, , y3 which leave the Lagrangian invariant. The same arbitrariness as with respect to y, 9, zj exists with respect to the internal Euler angles 01,/3, y. If we call another possible designation of the new coordinates a’, p’, y’, etc., then one has tj” =
f
f
p=-1 *z--1
D;&x’, /3’, y’) D&h’,
Y, y’)(S,“)‘.
(16)
Equating (3) and (16) one has:
The transformation three parts:
from the unprimed
to the primed coordinates is composed of
(1) A rotation in 01,p, y space with the Euler angles Z, p, 7: D1,k(a,
(2)
fig 7)
=
;
D1,,‘&
ft
7)
D;‘k((y’/%‘)a
(18.1)
A rotation in Ifi, 9, y space with the Euler angles I,& 8, I$ &(#,
8, yJ) = c ~W, 9’
$2 q) @j($-c 8’7 $1.
(3) With (18.1) and (18.2) one obtains from (17) the transformation to (Sq3’:
(18.2) from SqP (18.3)
FOUR-BODY
169
PROBLEM
The angles I%,p, 7, 4, 8, q are determined from the requirement that the primed axes yl’, y2’, y3’ are just another possible designation of the body-fixed axes. That means that there are only three independent quantities (S,*)’ just as in (4): (Soy’ = (S(p)’ = (SZ)’ = (2,)
= 0;
(19.1)
(s;)’ = (SI:)‘;
(19.2)
(Sl,)’ = (s;l)‘.
(19.3)
Replacing the quantities in (19.1) to (19.3) by the right-hand sides of (18.3) and SQP by yr , yz , y3 as in (4) one obtains a number of equations for the six angles. The solutions tr, p, 7, $, 8, q yield all possible transformations in the space of the six angles which leave the Laplacian invariant. The transformation between y1 , y3 , y3 and yl’, yz’, y3’ for a special set 5, etc., is given by (18.3). To illustrate the procedure for determining &, etc., somewhat more, the case (SIo)’ = 0 will be considered as an example: There are three independent terms on the right-hand side of Eq. (18.3), namely, the three terms containing y, - yz , y, + yz , y3. The factors with each of these terms must vanish: D;,(&7)
. D&&)
D:,(&)
. D’&,J9g
= 0;
+ I?,,(&)
(20.1)
- D:,<$J8q>= 0.
(20.3)
Equation (20.1) has two possible solutions: (1) cos p = 0;
p = 71.12;
(2)
iJ=Oorfi=rr.
sin B = 0;
(21)
Equations (20.2) and (20.3) yield further restrictions on ol, etc. We are now able to derive the properties of the functions GE; which follow from the requirement that the solutions (14) are invariant when the unprimed coordinates are replaced buythe primed ones. This requirement is justified because the five operators of which JJ~F,, is an eigenfunction do not change when the body-fixed designation of axes is changed. Hence one obtains from equating (14) for the primed and the unprimed coordinates:
Equation (22) is evident when there are no degenerate eigenfunctions for fixed eigenvalues A, 1\, , M, , Li , Mi . When there are two or more degenerate functions one will have two or more relations like (22) where the G2( y,‘y,‘y,‘) on the right-hand side will be linear combinations containing contributions from the several degenerate
170
W.
ZICKENDRAHT
functions. But by linear combination of the two or more relations one can again derive relations like (22). Evaluating (22) for all possible transformations, i.e., all the ol etc., one obtains the following relations: K, + Ki must be an even number;
(-l(Ke+K3’2 GI;;(y,
c c &,K.
, yl , Y,);
~2 , ~3) = (-)Le+Li * G:$Y, , u2 , Y,>;
(+’
K,’
* G~(Y, , y2 3Y,) = GUY,
G;;(yl
(F) * &i
(23. I) (23.2) (23.3)
3yz 3y3) = ‘$;C-Y~ >-Y, , Y,>;
(23.4)
(5) * G$(Y, 3-‘Jo>Y,> = G;;C4; 71’2 , I;>.
(23.5)
KS’
A very useful relation can be derived from the above: C-1 “+G - G;;f(d(yl >y2 , y3) = GF(-Y,
, y2 , -Y,).
(24)
The fact that the functions G2 are polynomials in the variables yr , yz , y3 and the relations (23) for L, = L, = 0 will yield us two angular variables u and u which will be useful in the evaluation of matrix elements later on. We will then replace the variables y1 , y2 , y3 by Y = ( y12 + y22 + y3 2>112,u and v. For Li = L, = 0 there is only one function G,,O(y1 y2 y3). For this, one has G,'(Y,
, ~2 3 ~3)
=
Go"(u2
, ~1,
G,'(Y,
3 ~2 2 ~3)
=
Go'(Y,
3 ~2 > VA
G,'(Y,
7 YZ 9 ~3)
=
Go"(-YI
(25.1)
~31,
(25.2) 3 Y&,
(25.3)
Go”(Y, , YZ3~3) = Go’<-Y, , yz 3 -~a).
(25.4)
, -YZ
Instead of yl, y2, y3 we introduce: Y2
=
Y12
+Y22
+
Y32;
24 =
Yl * Yz * Y3/Y3i
v =
(Y12 - Y22 + Y22 ' Y32 +
(26) Y32 - Y12YY4.
As these variables are invariant under the transformations (25), it is possible to expand Go0 in a series Go0 = y” C,,, a,, * zP * P. For low values of A there are a
171
FOUR-BODY PROBLEM
few terms in this series only. The Schrodinger equation (15) for Li = L, = 0 has the following form after separation of the y dependence !
cv __ 9g) . -$
+ (4~ + 12u2 - 1621~). $
_ 30~ . ; + (8 - 4411). $Goo
+ 8~41- 3~) * -
a2
auav
. y-“) = --X0 + 7) Go0 * Y-“a
(27)
One can find the polynomial solutions for Go0 from (27) very quickly. Unfortunately the coordinates u and ZIare of no help in deriving the basis functions G% for Li # 0, L, # 0. But they are of big help in evaluating matrix elements. The range of these variables is needed for that problem.
4. THE RANGE OF THE VARIABLES
u AND u AND INTEGRATIONS
IN Z.&Y SPACE
The volume element has the following form in y1 , etc.: dT = /Y,~ - ~2~ I - 1~2~ - ys2 I * lys2 -YIP
I
- sin 6 - sin /? . dy, * dy, . dy, . dcp. d8 - d$ . da * d/3 - dy.
(28)
A serious difficulty in evaluating integrals is caused by the factors 1yi2 - yk2 I. One could be tempted to use other variables for the integrations, for example, y, c1 , E,: y, = y . sin Ed. cos l 2 ; ye = y - sin Ed * sin Ed;
(29)
y3 = y * cos El . The experience has shown that integrations with such variables are extremely complicated as a consequence of the factors 1yr2 - y22 1, etc. It turns out that the variables u and v are of use in integrations. One obtains for the y1 y2ya-dependent part of the volume element: 1y12 - y22/ . 1y22 - y32 1 . 1y22 - y121. dyl * dy3 * dy, cc y8 * dy - du . dv.
The variables yi range from -co to + co. The derivation of the domains for the variables u, u is rather lengthy and is given in Appendix I. The final results will be given here: The expansion of a four-particle wavefunction in an orthogonal system as proposed in this paper will lead to the evaluation of matrix elements M, that is, integrations over eight angular variables. The integrand after integrations over CL,/3, y, #, 8, g,
172
W. ZICKENDRAHT
will be calledf(u, form :
t”) with t = (1 - 30)1/z. Then the matrix element has the following M = j’
dt . t j;;::, du .f(u,
t2)
(30)
* 319.
(31)
-l/2
with u(t) = (1 - t)(l + 2tyq3
A special method to deal with the four-particle problem is to expand the interaction of the particles in the system of functions of this paper. For four identical spinless particles, for example, one would have with central interactions only V = C V,(l rk - ri I) = V,(y) + V,(y) . (1 -
i
llz~j2) + ....
In this expansion all the functions which are completely symmetric with respect to particle exchanges will occur. The factors V,,(y), V,(y), etc., are to be taken as expansion coefficients and are determined from V, . A discussion on the completely symmetric functions will follow in Section 7. Two of these functions are contained in (32), namely, 1 and 1 - llv/2. They are orthogonal to each other, which can be checked by using formula (34). With an expansion like (32) the functionfin (30) wil1 be a polynomial in the variables u and t 2 or u and v. Withf = z&P one finds from (23) (abbreviation for M : M,,): Iodd :M,,=O;
(33)
I even: MI, = [2/(1 + I)] 2 [n !/i!(n - i) !I(-)“-“”
2i(2n - i - 1)
i=O
x 3"++l(2n
+ I -
i + l)! (I + 1)!!/(31+
4n - 2i + 7)!!.
(34)
5. DERIVATION OF THE ORTHOGONALSYSTEMFROM THE SCHR~DINGEREQUATION AND THE INVARIANCE RELATIONS
The functions Gg; are derived with the help of the relations (23) from the coupled system of Eqs. (15). The singularities of the differential equations (15) at y, = yz and y1 = -y2 lead to the general Ansatz:
One obtains another system of coupled differential equations for the functions G$ . It is necessary only to derive the functions with KS 3 0 and K, > 0. Functions for which one or both of Ki and K, are
FOUR-BODY
173
PROBLEM
(23.2) and (23.4) (a form is chosen which is convenient Appendix; p is a constant factor independent of X):
for tabulation,
see the
(1) If Ki and K, are even and X is even, or if Ki and K, are odd and X is odd: GK” Z.Yp(y, - y2)(Ki+Kc)‘2 (y, + y2)‘Ki-Ke”2f~(y1y2y3); Ke I f3JkY2Yd
=
%(Y,2n
+
+
J’i?
+
b,Y,2(Y:n-2
WIJ’~~‘?+~
+
YF”>
+
+
Vi”-“>
+
“’
bJ22Y,y2(y~n-4
(36) +
GzJ’I~Y~~
+ YF4)
+ ... + bn-ly:y;-ly;-* + c,y,4(Yf”-4+ YF”> + *‘a + d,y&Jy
+ YF>
(37)
with 2n = h - (Ki + Ke)/2 -
[ Ki - Ke I/2*
(38)
(2) If Ki and K, are even and h is odd, or if Ki and K& are odd and X is even: (3;; = P * y3(y1 - y2)(Ki+Ke)‘2 (y, + ~~)‘~“-~~“~f~(y,
v, v,),
(39)
f2: as in (3’7), but 2n = X -
1 - (Ki + K,)/2 -
j Ki - KA l/2.
(40)
As fit is a polynomial in y1 , y, , y, it is obvious that ala2 ,..., b, se- b,-, , etc., are 0 in many cases, namely, in all cases in which in (37) an exponent would be negative. For example, if n = 0 only a,, # 0, all other coefficients are zero. If either .Ki or K, or both are zero either the coefficients with even indices or those with odd indices are zero. This follows from relation (24). Tn detail: X odd and L, even or X even and L, odd:
fti = Y~Y~{~,(Y,~~-~ + $-3 + a3.-.I; h odd and L, odd or X even and L, even: f?
= a,(.$
+ $7
+ a, *-a;
(41)
X odd and Li even or h even and Li odd: fje = yly2{al(yF2
+ A”-“>
h odd ,and Li odd or X even and Li even: fle = ady:”
+
~3
+ a2 -.a.
+ a3 ...I;
174
W. ZICKENDRAHT
The numbers of coefficients have been reduced by these relations. Only those coefficients which may be different from zero due to the relations (37) for Ki > 0 and K, > 0 and due to (41) for the remaining cases are tabulated in Appendix IT up to X = 6. Since the set of operators used is not complete one has in general degenerate functions for fixed values of h, Li , Mi, L, , M, . One has to make sure that no functions are missing. This is simply done by considering the four particles moving in oscillator potentials. The potential energy is then proportional to y2. This four-particle problem is then solved with two different coordinate systems: Jacobi coordinates and the coordinates used in this paper. By comparing the numbers of solutions in both coordinate systems one makes sure that the system derived in this paper is complete, that is that there are no functions missing for fixed values of X, Li , Mi , L, , M, .
6. SYMMETRY RELATIONS UNDER EXCHANGE OF IDENTICAL PARTICLES The exchange of identical particles means reflection of the coordinates yi and a rotation in CY,/3, y space [3]. The angles of rotation will be called (or , j?r , yr . The results of [3] need a revision due to a mistake. The final results for the exchanges are : Exchange l-2 1-3 I-4 2-3 204 3-4
%
742
0 0 5-P
Pl
42 42
7? 7r 75-P
42
Yl
742
77 42
0 0 3~12
7. FOUR IDENTICAL SPINLESS PARTICLES AS AN EXAMPLE The case of four identical spinless particles is considered in this section because the simplicity of the method is best demonstrated with it. The wavefunction has to be symmetric with respect to all possible exchanges. This leads to the following functions which are completely symmetric:
where the coefficients a,. have to be determined from the requirement that Y/ is invariant under all rotations in 01,p, y space given in Section 6. The results for the coefficients aMi are: aMi f 0
for even Mi only;
(43.1)
FOUR-BODY
aMi = (-)^
175
PROBLEM
C amz(-)(Mi+m@ mi
. dzMi (+)
;
(43.2)
The functions Y of Eq. (42) are symmetric with respect to all possible exchanges if the coefficients aMi are solutions of Eqs. (43). (For the rest of this section these functions will simply be called symmetric.) For Li < 6 one obtains the following results. (Up to Li = 3 some details of the evaluation of Eqs. (43) will be given for illustration. This will not be done for L 3 4.) Li = 0: All solutions with even X are symmetric. Li = 1: One obtains from (43.2) a, = 0, as d&(n/2) solution for Li = 1.
= 0; so there is no symmetric
Li = 2: a, = a-2 = a, = 0; no symmetric solution. Li = 3: (43.4) yields a2 = (-)^ u-~ , a, = (-)n+1 a,, ; that is a,, = 0 for even A. For even X Eq. (43.2) yields a2 = a,{1 - (-)A}/2; so there is no symmetric
solution for even h. For odd h one obtains from (43.2) and (43.3) resp.: a,, = a,{d$(n/2)
- d!,,(r/2))
a,, = -az{d;&r/2)
- d!,,(r/2))
= 2a,d&/2), = -2a2d&,(r/2);
so again a, = a-, = 0. So there is no symmetric solution for Li = 3. Li = 4: Symmetric
solutions are found for even X only. They are of the form:
Li = 5: N’o symmetric solution. Li = 6: Symmetric
solutions are found for even and odd A:
A odd : c c GzD&, Ki 4
{D&z + D&m, - (551’2/ll)[D’&
+ D6K,-JI.
(46)
In practical calculations one could expand the two-particle interactions in a series containing the symmetric functions with Li = 0, 4, 6, etc., and L, = 0. The matrix elements are evaluated by integrating over #, 8, y, ~1, /3, y at first. The remaining integrations in u, a space are extremely simple and are performed as discussed in 595/1x1/1-12
176
W. ZICKENDRAHT
Section 4. Calculations are simple if one knows the functions G2 . The main problem consists in deriving these functions, as pointed out already. One could be tempted to apply the method to the 401model of 160. But one would need many functions with X > 6 for that case. All the symmetric functions given above with X < 6 are of even parity (reflection of the coordinate system means reflection of yi . As the GE: are polynomials of order h in yi , the parity is equal to (-)“). For A = 7 one has functions of odd parity with Li = 6. This means that the odd parity states of I60 would contain functions with X > 7 only. The relative motion of the 01particles would thus correspond to a rather high excitation. This is simply a consequence of the antisymmetrization with respect to all nucleons [19].
APPENDIX
I: THE VARIABLES u AND u
In Section 4, the final results for the domains of u and 2,have been given. How this result is obtained will be shown in this Appendix. First y, , yZ , y3 are replaced by y, e1 , Edof Eq. (29) in the expressions (26) for u and Y: u2 = (sin* Ed * co? e1 * sin2 26,)/4,
(47.1)
0 = (sin* c1 - sin2 2~,)/4 + sin2 l 1 * cos2 l 1 .
(47.2)
For v one obtains from (47): O
(40
The range of u will depend on U. The rest of this section will be concerned with finding the range of U. At first (47.2) is written as: sin2 2~~ = (40 - sin2 e1 * cos2 E&in*
l 1.
(4%
From (49) the range of E~for fixed u is derived: As sin2 2r, < 1 it follows: i
22 sin2 s1 - - -;++o. 31
(50)
In (50) one has to consider the two cases of sin2 Ed < $ and of sin2 Ed > 8: (1) sin2 Ed 3 # yields: 2 4(1 - 321) l/2* j < sin2 l 1 < f + ( ) ’ 9
(51)
(2) sin2 c1 < Q yields: --2 3
( 4(1 - 3v) >‘Ia ~ sin2 B < 2 9 11 3’
(52)
FOUR-BODY
177
PROBLEM
Equations (51) and (52) can be written as one necessary inequality: 2(1 - t) < sin2 c1 < 2(1 3’ t, 3
(53)
with t = (1 - 3a)l12 and 0 < t < 1. Equation (53) does not give the range of sin2 Ed yet, as we have used the condition sin2 2~~ < 1 alone. The lower limit sin2 2~~ 3 0 will yield further restrictions on sin2 Q: v - sin2 Ed * cos2 El >, 0
(54)
(sin2 l 1 - 4))” - 4 + v > 0.
(55)
or
One has to ‘consider two cases again: (1)
For sin2 l 1 > s Eq. (55) yields sin2 Ed 3 4 + (Q - v)li2.
(2)
(56)
For sin2 l 1 < + one obtains sin2 c1 < 4 - (t - v)l12.
(57)
These restrictions hold for 0 < v < 4 only, as from (54) it follows immediately: sin2 2~~ < 4v.
(58)
This means that for v > $ the variable 6Xcan have any value between 0 and 7r. The results obta.ined so far in the range 0 < v < & are combined in (59): For0
<$: 2(1 - t) < sin2 e1 < 2(1 3’ t, 3
with either sin2 E > 1 + (1 - 4W2 1, 2
(5%
or sin2 E ( 1 - (1 - 4v)1/2 1-L
2
*
As in this range for v one has always 2[1 - (1 - 30)1/2-j ~ 1 - (1 - 4v)1/2 3 2
178
W.
ZICKENDRAHT
and 2(1 + t) > 1
3
”
one obtains finally two separate ranges for sin2 Q:
o
,
2[1 - (1 - 3v)‘/2] ~ Sin2 E1 < 1 - (1 - 4791/z 3 2 (60.1)
and
l + (l -
4u)1’2
2 For the remaining
<
sin2
E l---
<
1
a
range of ZJthe relation (53) holds with no further restrictions: 2[1 - (1 - 3V)1/2] < sin2 E < 3 -.. I\
2[1 +
(1 -
3
3’)1’21.
The range for u2 is derived from (60). As a first step sin2 2~ is eliminated using (47.2):
(602)
from (47.1)
u2 = v . cos2 cl - sin2 l 1 . cos 461.
(61)
The derivative of u2 with respect to e1 is taken now to find upper and lower limits: au2 zg=
o
yields
co9 El = y
)
, sin2 cl = EjJJ
,
(62)
where for convenience the variable t has been used again instead of v. We consider the range & 2 t > 0 at first which corresponds to the range (60.2) for v. We have to find out whether the values (62) for cos2 l 1 are in the allowed range given in (60.2) or not. From (62) one obtains
sin2e1= i 7 i =
2(1? t, * 5
(63)
which is within the range (60.2). Thus we obtain from (61) and (63)
21 >t>o:
(1 - t)” (1 + 2t) 21 ” $ > (1 + 0”27(1 - 20 ’ >
(64)
The range 0 < v < $ corresponds to 4 < t < 1. There are two different regions for sin2 Edin this range; these are given by (60.1). One finds that sin2 l 1 = (2 - t)/3 of Eq. (62) is not within the allowed regions given
FOUR-BODY
179
PROBLEM
by (60.1), while sin2 Em= (2 + t)/3 is within the second one of these regions yielding an upper limit for z? (1 - q2 (1 + 20 >, u2.
(65)
27
The lower limit in this case is not found by a solution of the equation au2/&, = 0. It is found as u = 0 which one finds from sin2 cl =
1 * (1 - 4uy 2
.
So the result is: 1 > t > 1 (1 - 0” (1 + 20 > g > 0 ’
‘2
27
“*
(66)
Equations (64) and (66) show the limits of the variables which are to be observed in integrations in U, t space. The rest of this section will be used to show how integrations in U, t space can be performed very simply. The expansion of a four-particle wavefunction in an orthogonal system as proposed in this paper will lead to the evaluation of matrix elements, that is, integrations over eight angular variables. After the integration over 01,/$ y, #, I?, 9 one will have integrals of the form (where we do not write down the limits yet) s
dt - t duf(u, t2). s
(67)
Here v was replaced by the variable t. f is the integrand of the matrix element after integration over 01,p, y, #, 8, v. It must have the invariance properties as Go0 for Li = L, = 0 in Eqs. (25). The integration limits in (67) have to be determined from (64) and (66). From (64) one obtains two regions for U. With the abbreviation a(t) = (1 - t) . (1 + 2w2
658)
3 . 3112
these are:
&>ttO:
(1) (2)
40 -a(-t)
> u2
a(-t);
3 24>, -a(t).
(69)
And from (66) one obtains: l>t>$:
a(t) > 243 -a(t).
(70)
180
W. ZICKENDRAHT
The integral (67) with its limits is then: (71)
After combining
these integrals differently, one obtains finally M = jTl,2 dt t j”‘“’ duf(u, t2). -a(t)
APPENDIX
(72)
II: THE COMPLETE SYSTEMOF FUNCTIONS WITHX < 6
In the following tables the factors p and the functions fz of Eq. (36) are given. Only Ki 3 0 and K, 2 0 appear as functions with Ki < 0 or K, < 0 can be derived with the relations of Section 5. In the cases where Ki > 0 and K, > 0 the sets of coefficients a, , a, , etc., are given. In the cases where either Ki or K, or both are zero only a,, a2, etc., or a,, a,, etc., are given; that is, those coefficients which vanish due to the relations (41) do not appear in the tables. Above the tables the values of Li and L, appear. Li was always chosen as Li > L, . The coefficients for the case Li < L, are found by exchanging Ki and K, . The third column in the tables contains the factors p of Eq. (36). The columns below the h values contain the factors a,, etc. As an example for the use of the tables, the case Li = 5, L, = 1 with h = 5 is considered. For Ki = 4, K, = 0 we have case 2 of Section 5, hence Go4
= Y,(YI - YZ)~(Y~+ ~2)~ *fo4,
2n = 0.
(73) (74)
In this case only a, # 0, all other coefficients vanish and do not appear in the table. a, = 1 is taken from the table. For Ki = 2, K, = 0 we have case 2 again and (p is taken from the table) G ? = 2 . 3'12 - 3 J&l2 0”
-
Y22)h2,
2n = 2.
(75) (76)
Hence the coefficients a,, b, are different from zero and appear in the table. We have fo2 = 4Y12 +
Y22)
+
2h2.
(77)
FOUR-BODY
181
PROBLEM
For Ki = K, = 0:
G,O= yy3
.fo",
(78)
2n = 4.
(7%
The coefficients ao, a2 , b, , co appear in the table, thus: ho = 3(y14 + ~2~) +
For Ki
-
2~1~~2~
+
Q,2h2
~2~)
8~:.
(80)
5, K, = 1 we have case 1 of Section 5:
G5 = T(Y~
+~,)~f,5,
-Y~)~(YI
2n = 0.
(81) (82)
Hence
fi" = 1. For Ki
(83)
.3, K, = 1: G3 = B(YI -
~21~01
+
(84)
v2)fi3;
2n = 2.
035)
a, , q , b, appear in the tables as the only nonvanishing .A” = -5(y12
+
~2~)
-
2~1~2
+
coefficients, hence 8~3~.
(86)
For Ki = K, = 1 finally:
G1 = 'g
(~1
- YAP;
(87)
2n = 4.
(88)
a, , a, , a2 , b. , bl , co appear in the table:
fll = 5(y14 + y24) + ~YIY~(YI~ + y22) + 6~~~~23 - ~~Yz~(YI~+
~2~)
-
8~3~~1~2
+ 8~:.
In general. there will be several eigenfunctions for given values of Li, L, , X as our set of commutable operators is not complete [3]. This is the case for Li = L, = 1 for h = 5; and 6, for example. In both cases there are two eigenfunctions. These eigenfunctions have not been orthogonalized. This is a somewhat tedious procedure and will take a lot of time if one wants to do it for all the cases below. As in general
182
W.
ZICKENDRAHT
it is hoped that one will need only a few functions from the table, the orthogonalization procedure is best carried out for each case, as required. Functions with X < 6 (for L, = Li = 0 they are given explicitly on account of their simplicity): Li = L, = 0: h = 0,
Go0 = 1;
h = 3,
Go0 = u . y3; 1 - llv
h = 4,
Go0 = y4 2
X = 6,
Go’+1
;
39v
+F-).
Lc=L,=l: Ki
1
K,
p
h=l
11
h=2
1
001
X=3
1
2
-2
X=4
4 11 -7
-14 8
h=5
1 13 1
X=5
12 -13 -67 -54 0 25
-2 24
A=6
0 1 1 0 -12 --I
50 -108 -108 24
-2 24 0 0
X=6
0 4 9 -1 -6 0 0 2 -10 8
Li = 1, L, = 0: no solutions for all values of A; Li = 2, L, = 0: Ki
K,
P
x=2
2
0
1
1
A=4
x=5 2
A=6 1
-11
2 -13 -9 6
0
0
6rJ2J3
--I 2
-2 22 -11 4
--I
-2 2
3 21 0 -24 4
1 -13 -31 0 18 1 -2 0 36 -28
FOUR-BODY
183
PROBLEM
Li=2,Le=1: Ki
Ke
P
2
0
1
x=3
x=4 1
A=5 1
X=6 1
1
-2 1
1
21J2/2
0
-1 -1
-1 -1
-16 0
-1 -1 15 0 16 1
-1 -3
1
2 2 -1
Li = L., = 2, G2 = G% except for the last solution where we have G$ = -G$ Ki Ke 22
2
0
11
P
X=2
X=3
X=4
A=4
X=5
1
1
1
0 1 4
1 3 0
2 5
0 61i2/3
0
2/3
1
-1
0
1 4
-6
2
0 6
0 18 0 0
0 -1 1
-1 -9 -1
-2
-1
3
A=5
X=6 1
-22 -6
36
1 4
9 0
-108
-8 4
-1 -1 -1
0 -2 -4 -2 4 2
0
1
X=6
0 4 9
0 0 1 0 10 2
0 0 0 0 0 0
-1 -8 0
0
-4 23 15 -12
0 1
04000 -39 -3 438 -84 16
-1 -7 -6 -52
X=6
4 0 -19 -27 -100
4
X=6
-1 -2
0 2 1 -1
-3 3 42 0 0
8-4 8 -4 54 2 -38 10 -84 4 8 -4
0 -1 1 -1
3 6 -60 0 0 0 -4 -14 2 12 0
0 0 0 0 0 0 0 0 0 0 0
184
W.
ZICKENDRAHT
Li = 3, L, = 0, X = 6: Go2 = (y12 - y22)(y,2 - y32)(y32 - ylz); L;=3,L,=
1:
Ki
Ke
P
2
0
3112
0
0
10915
1
3
1
1
A=3
x=4
1
1
-9 18
-9 18
9
116
151/2/30
A=5 1 0
0
-2 -2
0 6 -15 24
-9 39
11
0 12 3
39
2 2
-6 -13 -26 12
0 1
-9 78 -78 96
13 -4
4 7
-6 -12
X=6
1 0
-22
-4 -13 13 3
-6 12
X=6
0 1
-9 -26 26 -8
-9
-9
A=5
0
3 -45 -84
-12
1 -10 0
-12 -12
39
3 18
78 -12
0
Li = 3yL, = 2: Ki
Ke
2
2
P
A=4
696
A=5 0
-2 -2
2
0
0
2
l/3
5y5
x=5
-3 -5
0 1 I
0 6
3
2
3 -5
X=6
1
3
A=6
0 -5 -16 22 22 -16 0 38 10 -28
0
5
1
0
0 0 1 -7 -7 6 0 -13 -5 8 0 -5
FOUR-BODY
Ki
Ke
P
3
1
l/2
1
1
X=5
A=4
X=5
2
15930
0 3 2
-2 -12 -8
185
PROBLEM
A=6 0 0
5
0 6 1
-2 -34 -56 22 44 -8
0 30 40 -5 -30 0
-11 -11
-1
0
0 0
-3 2
-4 -1 -6
-10 0 12
0
A=6
Li = L, = 39 G2 = G2 z Ki 2
K, 2:
P 1
A=;
X=4
-5
A=5 0
0
0
301q5 l/5
3
3
l/2
3
I
15y10
1
1
l/l0
5 -30 -40
-5
10 0
2 8
0 48 -32 16
-60 1560 640 -320
-12 -144 96 -264
0 12 0 28
2 6 2
-40 -90 110
6 0 0
0 2 0
0 0 2
-2 56 48
0
6
40 30 -30
0 4 6
6 6 24 -42 -60 0
-120 -110 380 490 -1220 -160
2 28 132 -96 -320 32
0
0 8 48 -2 -12 0
6
5
-15 -10 -80
-2
2 -84 32
-2 -2
0
X=6
0 4
2 -12 -96
X=6
-10 -100 40
4
0
X=6
0 0 0
-2
2
X=5
0 0 1 1 6 1
-2 -14 -10 -52 -10
0 0 -1 -1 -6 -1
0 -2
0 -2
-6 -4 -2 -12 8 48 0
0 -28 0 -12
186
W.
ZICKENDRAHT
Li = 4, L, = 0: Ki
Ke
P
4
0
l/2
A=4
A=6 2
2 -15
2
0
w/7
-2 4
0
0
701f2/70
6 4 -16 16
Li=4,L,=
-2 13 2 -13 6 -58 7 210 -68 16
1: Ki
K,
P
4
0
1
2
0
71/s/7
x=5
X=6 1
1 -1
-1 2
3
1
0
112
1
71q14
-1 -1
-1
1
2
1
1
0 3 2 1
3 3 4 -7 -6 4
-6 -4
FOUR-BODY
187
PROBLEM
Li = 4, L, = 2: Ki
Kc
P
4
2
1
4
0
2
2
x=4
x=5
-1
-1
6112/3
1
7y7
0
0
2
2 +42112/21
7oy3
0
5
1 0 -2
-1 -1
3
-3
2 4
0
0
2 . lo51ylo5
3 2 4
-12 24
-16
3
1
1
I.
2y2
-2
14914
6 4 -8
X=6 1 2 0
2 2 -2
2
A=6
0 1
-1 18 -2 -4 12 4 40 16 1 10 -4 -4
A=6 0 1 0
0 0 1
0
0
-6
3
0
-1 -8 2 -10 -8 0 -3 -3 6
3 -30 -8 40
0 14 -4 -8
-4 -1
-3 -37 -2 180 -56 16
0 12 18 -60 0 0
0 -12 -3 30 0 0
0
-1
-16 -16
0 -3 -2 1 2 -4
0 16 32 -16 -32 0
0
8
-1
0 7 7
3 -5 -12 7 14 -4
-4 -2 0 4 0 -2 -12 0
188
W.
Lj = 4,L,
ZICKENDRAHT
= 3: Ki
Ke
P
4
2
1
A=5
X=6 I
A=6 0
0 0 7
-5 -5 4
0
3095
-1
5
0
2
2
71y7
-1 -6 -4
0 5 0 20 20 -10
0 0 14 0 -14
2
0
2101J2/35
0
3
2
3
70935
1 8
-5 10
-70
10
0 10 -35 60
-28 -7 -28
5 10 0
-7 -7
0
3112/2 --I -1
0
0
0
3
1
51910
0 1 11
-5 -30 -20
1
3
211/2/14
0 3 1
-15 -10 20
-7
0
15 40 -30 40 -80 -80
0 35 210 35 210 0
1
1
35970 -3 -2 -11 -34 -16
0 -35 35 0 7
Ki
Ka
P
4
4
1
FOUR-BODY
PROBLEM
x=4
x=5
-1
189
A=6
-2
-1 -2
2 0 4
2
71/z/7
X=6 0 4
2
2
2 2 -2
4
0
7oy35
-3
0
2
l/7
12
0
1oy35
10
0
l/35
3
3
l/2
-4 -4 -24
6 48
-2 28 -24
-60
-18 -12 -192 -128
240 640
-4
0 2 -6
3
1
71q14
12
-3
1
1
l/14
-36 -24 -64
-4 -6
0
0 10
-80 12
0 -1 -26 16 132 -16
6 -24 -6 24
0 50 -30 -80
-18 114 24 1680 224 - 128
0 -200 100 -2600 0 0
8 4
-2 -12 8
6
6 12 0
0 18 12 -6 156 -32
-18 -24 180 56 -176 -32
-6
0 1 -2
0
2
-1 -6
1 11 -11 -3 -11 -23 9 17 -316 61 366 16
190
W.
ZICKENDRAHT
Li = 5, L, = 0: no solution for h < 6; L,=5,L,=
1: Ki
KC.
4
0
2
0
0
x=5
P
A=6 1
2 . 3y3
0
-1
701J2/21
-1 2
2
3 2
3 2
-8
-8 8
5
1
5112/2
3
1
116
1
1
8
1
-1
8
-2 -8
-5 -2
421J2/42
3
5 4 6
-1 4 2 4
-12 -8
-8 -8
8
Li = 5, L, = 2:
Ki
K?
P
4
2
- 101y5
0 2
2
A=6 0 1 1
0
2 * 15y5 3011y 5
0 2 2 1 -2 -2
FOUR-BODY Ki
K?
2
0
191
PROBLEM
X=6
P 2 * w2/5
0 0 -1 2
0
2
2 * w2/7
-1 2
5
1
1
1
3
1
- 51/2/5
1 2 0
1
1
2101/2/105
1 8 6 0 -16 -8
Li = .5, L, = 3: Ki
K
P
A=5
4
2
1
-2
A=6 0 2 -4
4
0
2 * 301y5
1
2
2
31/z/3
4 4 -4
-1 0 -4 -4 4 4 -8
2
0
0
2
4 . 101y15
2 - 701J2/21
-3 1
3 9
4
3 0
-3
192
W. Ki
A=5
P
K,
0
ZICKENDRAHT
0
4 * 211J2/63
A=6 9 6
-4 -16
-9 -6 -36 96
-1
-3
5
3
15y3
5
1
1
1
1
3
3
3y9
5 6
9 6
-4
-12
3
1
51/2/l 5
-5 -2 -12
-3 42 -12
1
3
141J2/21
-5 -2
-3 6 6
6 1
1
2101/2/105
5 4 6 18 12 -32
1 -64 -42 6 68 -32
Li = 59L, = 4: P
X=6
JG
K,
4
4
-2
4
2
2 - 35y35
0 1 8
4
0
-6
0 1
’ 51q5
. 14112/7
0 1 1
FOUR-BODY
2
4
2
2
193
PROBLEM
2 * 15q5
0 2 1
4 * 1051/2/105
0 -1 -1 -4 -13 -6
2
0
2 * 421j2/21
0 0 3 8
0
4
-2
* 141y7
1
0
2
2 * 21j2/7
1 12
5
1
3
3
5y5
3
1
35y35
1
3
2101/2/105
1
1
301y105
-3
* 7q7
1 -1 -6 -4 3 6 28 1 22 6 -3 -24 -18 -42 -92 -32
194
W.
ZICKENDRAHT
Li=L,=5,G2=G$: Ki
Ke
P
4
4
215
A=5
A=6 5
0 -1 4
4
2
4 * 31/z/15
-5
0 1 -2
4
0
2
2
--I
2 * 7oy35 s/15
5 5 10
0 -1
-1 1 -8 4
2
0
4 . 2101f2/315
0
0
4163
5
5
1
5
3
51/2/l 5
5
1
21O1/2/1O5
3
3
l/45
3
I
42112/315
1
I
21105
-15 -40
3 36
45 30 160 64
-9 -6 -144 -192
1 -5
1 -3
5
1
25 30 160
9 -66 96
-25 -10 -240 25 20 30 360 240 320
-3 114 -48 1 -172 -114 24 -496 64
FOUR-BODY
195
PROBLEM
Li = 69L, = 0: &
K?
P
4
0
661/2/l 1
X=6
-1 2
2
0
5sy55
5 6 -16 16
0
0
2 . 231112/231
-5 -3 18 12 -24 16
Li = 6, L, = 1: no solution for X < 6; Li = Ci,L, = 2: Ki
Kt?
P
6
2
1
6
0
6112/3
-1
4
2
661j2/33
-3 -2
X=6 1
4 4
0
2 * 11’/2/11
1 0
2
2
551j2/165
15 20 26 -32 -32 16
196
W.
2
0
ZICKENDRAHT
330112/165
-5 -6 0 16
0
2
2 * 23111a/231
-5 -6 12 -8
0
0
2 - 15W2/23 1
5 3 0 0 -24 32
3
1
2 * 551J2/55
-5 -2 8
1
1
2 - 221Ja/33
5 4 6 -12 -8 8
Li = 6, L, = 3: no solution for A < 6; Li = 6, L, = 4: Ki
Ke
P
6
4
1
6
2
2 * 71/2/7
6
0
3 * 7oy35
A=6 1 -1 1
FOUR-BODY
197
PROBLEM
Ki
Ke
P
4
4
66112/33
X=6 -3 -4 2
4
2
2 - 462112/231
4
0
2 * 1 155112/385
2
4
55112/165
3 2 10 -3 -14 15 10
-16 2
2
2 . 3851/2/1155
-15 -20 -26 -80 -80 96
2
0
1541y1155
45 54 336 -256
0
4
2 . 2311/2/231
-5 6
0
2
4 . 33112/231
5 6 30 -48
0
-0
2 * 3301q1155
-15 -9 -126 -84 128 128
5
3
2 +6112/3
1
198
W.
Ki
K,
ZICKENDRAHT
P
X=6
5
1
2 . 421J2/7
--I
3
3
2 * 1101y55
-5 -6 4
3
1
2 * 770112J385
15 6 4
1
3
4 * 11112/33
5 2 -6
1
1
4 . 771J2/231
-15 -12 -18 -6 -4 32
L, =:6, La = 5: no solution for h < 6; L,=Li = 6,G2 = G$: Ki
Ke
6
6
6
4
P
66112/l 1
X=6
-1
551/2/l 1 6
0
10 . 231112/231
4
4
2/11
4
2
301y33
-1 3 4 20 -3 -2 -32
FOUR-BODY
199
PROBLEM
Ki
Ke
4
0
2
2
l/33
2
0
1051J2/231
0
0
l/231
5
5
2
5
3
2 - 16W2/1 I
5
1
10 . 661J2/33
1
3
3
l/11
30 36 64
3
1
101q11
1
1
4133
P 10 . 14y77
X=6 1 12 15 20 26 256 256 256 -10 -12 -192 -256 100 60 2160 1440 3840 1024 1 -1
-10 -4 -32 25 20 30 120 80 64
200
W.
ZICKENDRAHT REFERENCES
Ann. Phys. (N.Y.) 35 (1965), 18. Phys. 3 (1966), 461. 3. W. ZICKENDRAHT, J. Math. Phys. 10 (1968), 30. 4. E. L. SURKOV, Sov. J. Nucl. Phys. 5 (1967), 644. 5. H. W. GALBRAITH, J. Math. Phys. 12 (1971), 782. 6. E. -CON AND A. AMAYA, Ann. Phys. (N.Y.) 97 (1976), 266. 7. J. M. L&Y-LEBLOND, J. Math. Phys. 7 (1966), 2217. 8. W. ZICKENDRAHT, Z. Phys. 251 (1972), 365. 9. W. ZICKENDRAHT AND H. STENSCHKE, Phys. Lett. 17 (1965), 243. 10. L. W. BRUCH AND H. STENSCHKE, J. Chem. Phys. 57 (1972), 1019. 11. W. ZICKENDRAHT, J. Math. Phys. 12 (1971), 1663. 1. W.
2. Yu.
ZICKENDRAHT,
A. SIMONOV,
Sov. J. Nucl.
12. A. Y. DZYUBLIK, V. I. OVCHARENKO, A. I. STESHENKO, AND G. F. FILIPPOV, Sov. J. Nucl. Phys. 15 (1972), 487. 13. P. GUI~HANI AND D. J. ROWE, Canad. J. Phys. 54 (1976), 970. 14. A. BOHR, Kgl. Dan. Vidensk. Selsk., Mat. Fys. Medd. 26 (1952), 14. 15. W. ZICKENDRAHT, Z. Phys. 253 (1972), 356. 16. V. V. VANAGAS, Sov. J. Nucl. Phys. 23 (1976), 505. 17. J. D. LOUCK AND H. W. GALBRAITH, Rev. Mod. Phys. 44 (1972), 540. 18. M. E. ROSE, “Elementary Theory of Angular Momentum,” Wiley, New York, 1957. 19. K. WILDERMUTH AND W. MCCLURE, “Cluster Representations of Nuclei,” p. 1, Springer Tracts in Modern Physics 41, Springer, New York, 1966.
OF PHYSICS
111,
A Configuration
162-200 (1978)
Space Approach
to the
Nuclear
Four-Body
Problem
W. ZICKENDRAHT Fachbereich
Physik,
Phi&p+Universitat,
Marburg,
3550
Marburg,
Germany
Received November 6, 1976
The derivation of a complete orthogonal system for the nuclear four-body problem is discussed, which will be especially useful in the case of identical particles. A method for dealing with four-body bound states is proposed. Part of the orthogonal system is given explicitly.
1. INTRODUCTION The method of the hyperspherical harmonics has been treated extensively during the last decade. The earliest publications dealt with the three-particle problem [I, 21, for which it was possible to derive orthogonal functions with extremely simple properties with respect to exchange of identical particles. A paper on the four-particle problem which started from the samepoint of view has been published by the author [3]. The work is continued with the present paper. A number of difficulties which will be discussedin this introduction delayed the continuation of the work at first. The emphasis of this paper is on constructing orthogonal functions with simple properties with respect to exchanges of identical particles and to distinguish between collectiye and single-particle degreesof freedom as in the general A-nucleon problem. The price one has to pay for this gain in simplicity is a tedious method for finding the orthogonal functions. Other authors put the emphasis on giving the explicit forms of a complete orthogonal system for the four-particle problem [46]. These authors do not distinguish between collective and single-particle degrees of freedom and their functions are difficult to handle for identical particles. Surkov [4] started with Jacobi coordinates. The polar anglesof the three vectors are six of the coordinates used. The other three coordinates are derived from the three lengths of the vectors as is usually done with the method of hyperspherical harmonics. Only part of the orthogonal system is given explicitly in the paper of Surkov. Galbraith [5] and Chacon and Amaya [6] start with three coordinate vectors defined at first by Levy-Leblond [7]. Except for this difference the hyperspherical coordinates are defined in the same way as in the paper by Surkov. In the papers [5,6] complete orthogonal systems are derived, complete sets of operators are given, and the properties with respect to exchanges of identical particles are examined. Eight of the coordinates used in the papers [4-61 are not symmetric with respect to arbitrary exchanges of identical particles, they change under at least one of the possible exchanges. Practical calcu162 0003-4916/78/1111-0162$05.00/0 Copyright All rights
0 1978 by Academic Press, Inc. of reproduction in any form reserved.
FOUR-BODY
PROBLEM
163
lations with such orthogonal functions might be difficult. There are a number of problems in which the space part of the four-particle wavefunction will have a high degree of symmetry with respect to particle exchanges. The ground state and the first excited 0+-state of the N particle, for example, are described in a good approximation as completely antisymmetric in the spin-isospin coordinates and completely symmetric in the space coordinates. The four-a model of I60 (with OLparticles in the ground state or in the first excited state) is another problem which is simplified by the use of symmetric coordinates and the orthogonal functions derived in this paper. This problelm is presently considered by the author as a continuation of earlier work on three (y.particles-that is the three-a-model of 12C-which could be treated in an extremely simple way [B]. Further four-particle problems of interest in this connection are other four-cluster problems in nuclei with two or more identical clusters, also the problem of three nucleons outside a closed shell, which together with the core make a four-particle problem. There is also the possibility of applications in molecular physics for which one may be encouraged by simple three-particle calculations with the corresponding three-particle coordinates and orthogonal functions [9, lo]. Generalizations of the transformations given in [3] to the A-particle problem have been given by several authors [l l-131. The symmetric four-particle coordinates are special cases of the A-particle coordinates given in these papers. One purpose, among others, of these papers is to derive the collective degrees of freedom from the single-particle degrees of freedom, also the coupling of the collective degrees to the remaining single-particle degrees. Thus it was possible in a simple way to derive the Bohr equations for the collective motion from the A-particle Schrodinger equation [15]. The success of this generalization to the A-particle problem is another reason for continuing the work started in [3]: For four particles there are six collective coordinates and three coordinates which must be considered as the remaining single-particle coordinates. So there is the clear distinction between these different degrees of freedom for four particles as for A particles. Recently, Vanagas [16] has given a method for the irreducible expansion of an arbitrary microscopic Hamiltonian in a series, the first term of which gives the collective part. Vanagas was able to prove that there can be no more than six collective coordinates for a system of identical particles (like nucleons) if one requires these coordinates to be invariant under the exchange of the particles. This proof brings the distinction between six collective coordinates and 3A-9 single-particle coordinates to a firm basis. The list of references given at the end of this paper does by no means contain all the papers on hyperspherical harmonics. Most of the papers on this problem which appeared before 1972 are contained in a review article by Louck and Galbraith [17]. There were two problems which had to be solved after the publication of [3]: One is concerned with the range of two angular variables, the other with the derivation of that part of the complete orthogonal system which depends on these two angular variables. It is shown in this article how these problems are solved. The main results of [3] which are necessary for understanding will be given in Section 2.
164
W. ZICKENDRAHT
2. THE SCHR~DINGER
EQUATION
IN THE SPECIAL COORDINATES
The space vectors of the four particles are q (i = I, 2, 3,4). In the center of mass system the vectors x1 , x, , xB are used [7]: x1 = Q(rI 4 r2 - r3 - r4), x2 = H--r1 + r2 - fs + q), x3 = &(-rl
In the first step of the transformation are introduced [4]:
(1)
+ r2 + r, - r4).
complex variables ljk (spherical components)
t,l = (x11 + ix21 + ix,, - x*,)/2, to1 =
-(x1,
+
ix23)/2"",
tl, = (-x11 - ix,1 + ix,, - x,,)/2 t,O = -(x31 + ix,,)/21iz,
too= x33 tO, =
(2)
3
(XQ1
iX32)/Z!‘12,
-
t;l = (-x11 + ix,, - ix,,
-
x22)/2,
to1 = (XI3 - ix‘J21/2, t1: = (XI1 - ixzl - ix,, - x2,)/2.
The second step transforms the tik to six angular variables LY,/3, y, #, 8, y and to three variables yr , y2 , y, , which will be replaced later on by two angular variables and one length: (3) SI1= s1: = $(y1 - y,);
ST, = s;l = -iK.h
+ id;
(4)
so0= Y3; so
1
=
s1 o =
s-l0
=
s",
zzz 0.
The angles #, 8, q define the orientation of a body-fixed coordinate system with respect to the center of mass system [18]. This body-fixed system consists of the three axes of the principal moments of inertia. This can be understood easily: The
FOUR-BODY
165
PROBLEM
moment of inertia with respect to an arbitrary axis (the direction of which is given by the unit vector e) in the center of mass system is (rs is the space vector of the center of mass; equal masses m of the particles are assumed): J = m
i
i i=l
(rt - rs)” -
i
[(ri - rJ
i=l
4” I
(5) = m i
[xi2 - (x,e)“].
i-1
Applying
tlhe transformations
(2) and (3) one obtains:
J = m{Y12+ ~22+ ~32- (yle>” - (Y2e)2- (y,e)“>.
(6)
Here yr , ya , ys are the three orthogonal vectors the directians of which with respect to the center of mass coordinate system are given by the three Euler angles ~4, 9, CJJ. The lengths of the three vectors are y1 , y2, y3 . The distinction between collective and single-particle coordinates can be made here: While y1 , y2 , y, , CJJ,9, $ are of a collective nature, 01,/3, y are the remaining single-particle coordinates [ll-13, 161. One obtains for the principal moments of inertia:
4 = m(y22+ ~~~1; J2 = m(y32+ y12); J3 = m(y12 + ~2).
(7)
The sum of the d operators in the Schrodinger equation has the following form in the new coordimates:
(8)
166
W. ZICKENDRAHT
Lo1, L,, , L,, are the components of the total orbital angular momentum with respect to the body-fixed coordinate system [18]. Li,, Liz, Li, are of the same form as L,, , L,, , Le3 with $, 9, q~replaced by 01,/3, y.
3. INVARIANCE PROPERTIES OF THE EIGENFUNCTIONS UNDER COORDINATE TRANSFORMATIONS It is the purpose of this paper to construct eigenfunctions
of the operators
Li5 = --%(a/+). The last of the five with y2 = y12 + y22 + ys2, L,, = -E(a/+), operators (9) is the angular part of the sum (A, + A, + A3), that is the Casimir operator N(9). It contains eight angular variables. These are $, 8, y, LY,/3, y and two more variables which depend on the ratios yl/y and y2/y. The latter two variables would be one possible choice for the remaining two angular variables. The operators (9) all commute with each other. These five operators do not form a complete set of commutable operators in the eight angle variables. Three more operators would be needed. The experience with the three-particle problem shows that it is best to do without these missing operators. In the three-particle case one has four simple commuting operators in a five-dimensional angular space. The fifth operator was derived (unpublished) and turned out to be complicated containing third derivatives in the angles. So it must be expected that the three missing operators for the four-particle problem are complicated too. Doing without these missing operators will mean in general that one has degenerate eigenfunctions for fixed eigenvalues of the operators (9). The tables at the end of this paper show that for X < 3 there are no degenerate functions. For those cases with X 2 4 in which one has degenerate functions one has to orthogonalize them. L,, is the z component of the total orbital angular momentum. For all four-particle problems with central interactions the wavefunction will be an eigenfunction of L,, . The corresponding quantity for the internal orbital angular momentum is Lit . The interactions of the four particles will always depend on 01,/3, y whether central or noncentral. Hence the wavefunction will not be an eigenfunction of Lit . The hyperspherical harmonics which will be defined in detail below will be designated as En with h(X + 7) as the eigenvalue of the Casimir operator. FA shall be an eigenfunction of all the operators (9). Strictly taken the quantum numbers of all these operators should appear as indices of FA . They were left out because the formulas would get too lengthy with all these quantum numbers as indices. The eigenfunctions of Le2 and L,, are the elements of the rotation matrices D&&4 $7 v) L e‘DLe KS4 = fi2Le(L, + 1) D;:Me ,
(10)
FOUR-BODY
Similarly
167
PROBLEM
one has for Liz and Lit the eigenfunctions D&,&: L,‘D&,i
= A2L,(Li + 1) D&
,
(11) L,,D;f,, I 1 = H4iD;~,, I I . D&,,, is also an eigenfunction for L,, = -ifi a/h+h, D&,+ for Liz = -it? a/&.. L,, and Lizi do not commute with the Casimir operator. Hence FA will contain sums over K, and Ki . yAFA is an eigenfunction of A, + A, + A, with eigenvalue zero.
This follows immediately
from writing
A, + A, + A, = &
+ $ $ - -$ . (Casimir operator).
(12)
As F,, is the eigenfunction of the Casimir operator with eigenvalue X(X + 7) and depends on the eight angular variables but not on y, one has
(A, + FA is a hornogeneous polynomial
+
A,
A,)(Y~ * FJ = 0.
(13)
in y1 , yZ , y3 of degree h. The Ansatz for yAFA is
now: (14)
With this Ansatz yAFA is an eigenfunction of La2, L,, , Li2, LiE with eigenvalues + l), M4,, fi2Li(Li + l), ?iM, . The functions G$ are homogeneous polynomials in ,yl , y2 , y3 of degree X. Strictly taken these functions should have indices Li , L, , h; they were left out for the same reasons as above. With Eq. (13) one obtains a system of coupled equations for the functions G2 :
ii2L,(L,
a2
I 8Y12
,
1 i 2y1 . i Y12 -
+ -
Y22
Y22
l ) -& - Y12
+
l ) -& Y12 - Y32
+ &
+
2Y3
- (
+ +-
y32
1
2
y12
+ +
3%
y32
-(
y22
: y3
A
y22
) ’&
- if [L,(L, + 1) + Li(Lc + 1) - Ke2- Ki2] * [ (;:I _’ $~2 +
y32+ ‘12 ] (Y32
y12 + y22 (Ke2+ K;) (Y12 - Y22)2 -1 _
= 4[
Y22
(Y22
+ -
Y32 Y32)2
-
-
( yl:y~y;22~ K&i]
Y32 + Y12 (Y32 - Y1212
- Y1212
GE
+g?~e+&a+,G& 1*[&+,g~+,G~+,
168
W.
ZICKENDRAHT
gKL = [(L + K)(L - K + l)]““.
(15)
In principal Eqs. (15) would be sufficient to determine the functions G? . This task is simplified by making use of invariance properties: Transformation (3) defines a system of body-fixed axes y1 , y2 , y3 . The directions of these three axes, that is the directions of the vectors y1 , y2 , y3 , are orthogonal. The position of this body-fixed Cartesian coordinate system with respect to the center of mass system is described by the three Euler angles. The designations of the axes as 1, 2, 3 axes are arbitrary; the same is true for the designation of positive or negative directions; but only righthand coordinate systems are considered. This arbitrariness in the designation of the axes was already treated in the paper by Bohr [ 141.It has as a consequence that there are a number of rotations in q, 8, # space and exchanges of yl, y, , y3 which leave the Lagrangian invariant. The same arbitrariness as with respect to y, 9, zj exists with respect to the internal Euler angles 01,/3, y. If we call another possible designation of the new coordinates a’, p’, y’, etc., then one has tj” =
f
f
p=-1 *z--1
D;&x’, /3’, y’) D&h’,
Y, y’)(S,“)‘.
(16)
Equating (3) and (16) one has:
The transformation three parts:
from the unprimed
to the primed coordinates is composed of
(1) A rotation in 01,p, y space with the Euler angles Z, p, 7: D1,k(a,
(2)
fig 7)
=
;
D1,,‘&
ft
7)
D;‘k((y’/%‘)a
(18.1)
A rotation in Ifi, 9, y space with the Euler angles I,& 8, I$ &(#,
8, yJ) = c ~W, 9’
$2 q) @j($-c 8’7 $1.
(3) With (18.1) and (18.2) one obtains from (17) the transformation to (Sq3’:
(18.2) from SqP (18.3)
FOUR-BODY
169
PROBLEM
The angles I%,p, 7, 4, 8, q are determined from the requirement that the primed axes yl’, y2’, y3’ are just another possible designation of the body-fixed axes. That means that there are only three independent quantities (S,*)’ just as in (4): (Soy’ = (S(p)’ = (SZ)’ = (2,)
= 0;
(19.1)
(s;)’ = (SI:)‘;
(19.2)
(Sl,)’ = (s;l)‘.
(19.3)
Replacing the quantities in (19.1) to (19.3) by the right-hand sides of (18.3) and SQP by yr , yz , y3 as in (4) one obtains a number of equations for the six angles. The solutions tr, p, 7, $, 8, q yield all possible transformations in the space of the six angles which leave the Laplacian invariant. The transformation between y1 , y3 , y3 and yl’, yz’, y3’ for a special set 5, etc., is given by (18.3). To illustrate the procedure for determining &, etc., somewhat more, the case (SIo)’ = 0 will be considered as an example: There are three independent terms on the right-hand side of Eq. (18.3), namely, the three terms containing y, - yz , y, + yz , y3. The factors with each of these terms must vanish: D;,(&7)
. D&&)
D:,(&)
. D’&,J9g
= 0;
+ I?,,(&)
(20.1)
- D:,<$J8q>= 0.
(20.3)
Equation (20.1) has two possible solutions: (1) cos p = 0;
p = 71.12;
(2)
iJ=Oorfi=rr.
sin B = 0;
(21)
Equations (20.2) and (20.3) yield further restrictions on ol, etc. We are now able to derive the properties of the functions GE; which follow from the requirement that the solutions (14) are invariant when the unprimed coordinates are replaced buythe primed ones. This requirement is justified because the five operators of which JJ~F,, is an eigenfunction do not change when the body-fixed designation of axes is changed. Hence one obtains from equating (14) for the primed and the unprimed coordinates:
Equation (22) is evident when there are no degenerate eigenfunctions for fixed eigenvalues A, 1\, , M, , Li , Mi . When there are two or more degenerate functions one will have two or more relations like (22) where the G2( y,‘y,‘y,‘) on the right-hand side will be linear combinations containing contributions from the several degenerate
170
W.
ZICKENDRAHT
functions. But by linear combination of the two or more relations one can again derive relations like (22). Evaluating (22) for all possible transformations, i.e., all the ol etc., one obtains the following relations: K, + Ki must be an even number;
(-l(Ke+K3’2 GI;;(y,
c c &,K.
, yl , Y,);
~2 , ~3) = (-)Le+Li * G:$Y, , u2 , Y,>;
(+’
K,’
* G~(Y, , y2 3Y,) = GUY,
G;;(yl
(F) * &i
(23. I) (23.2) (23.3)
3yz 3y3) = ‘$;C-Y~ >-Y, , Y,>;
(23.4)
(5) * G$(Y, 3-‘Jo>Y,> = G;;C4; 71’2 , I;>.
(23.5)
KS’
A very useful relation can be derived from the above: C-1 “+G - G;;f(d(yl >y2 , y3) = GF(-Y,
, y2 , -Y,).
(24)
The fact that the functions G2 are polynomials in the variables yr , yz , y3 and the relations (23) for L, = L, = 0 will yield us two angular variables u and u which will be useful in the evaluation of matrix elements later on. We will then replace the variables y1 , y2 , y3 by Y = ( y12 + y22 + y3 2>112,u and v. For Li = L, = 0 there is only one function G,,O(y1 y2 y3). For this, one has G,'(Y,
, ~2 3 ~3)
=
Go"(u2
, ~1,
G,'(Y,
3 ~2 2 ~3)
=
Go'(Y,
3 ~2 > VA
G,'(Y,
7 YZ 9 ~3)
=
Go"(-YI
(25.1)
~31,
(25.2) 3 Y&,
(25.3)
Go”(Y, , YZ3~3) = Go’<-Y, , yz 3 -~a).
(25.4)
, -YZ
Instead of yl, y2, y3 we introduce: Y2
=
Y12
+Y22
+
Y32;
24 =
Yl * Yz * Y3/Y3i
v =
(Y12 - Y22 + Y22 ' Y32 +
(26) Y32 - Y12YY4.
As these variables are invariant under the transformations (25), it is possible to expand Go0 in a series Go0 = y” C,,, a,, * zP * P. For low values of A there are a
171
FOUR-BODY PROBLEM
few terms in this series only. The Schrodinger equation (15) for Li = L, = 0 has the following form after separation of the y dependence !
cv __ 9g) . -$
+ (4~ + 12u2 - 1621~). $
_ 30~ . ; + (8 - 4411). $Goo
+ 8~41- 3~) * -
a2
auav
. y-“) = --X0 + 7) Go0 * Y-“a
(27)
One can find the polynomial solutions for Go0 from (27) very quickly. Unfortunately the coordinates u and ZIare of no help in deriving the basis functions G% for Li # 0, L, # 0. But they are of big help in evaluating matrix elements. The range of these variables is needed for that problem.
4. THE RANGE OF THE VARIABLES
u AND u AND INTEGRATIONS
IN Z.&Y SPACE
The volume element has the following form in y1 , etc.: dT = /Y,~ - ~2~ I - 1~2~ - ys2 I * lys2 -YIP
I
- sin 6 - sin /? . dy, * dy, . dy, . dcp. d8 - d$ . da * d/3 - dy.
(28)
A serious difficulty in evaluating integrals is caused by the factors 1yi2 - yk2 I. One could be tempted to use other variables for the integrations, for example, y, c1 , E,: y, = y . sin Ed. cos l 2 ; ye = y - sin Ed * sin Ed;
(29)
y3 = y * cos El . The experience has shown that integrations with such variables are extremely complicated as a consequence of the factors 1yr2 - y22 1, etc. It turns out that the variables u and v are of use in integrations. One obtains for the y1 y2ya-dependent part of the volume element: 1y12 - y22/ . 1y22 - y32 1 . 1y22 - y121. dyl * dy3 * dy, cc y8 * dy - du . dv.
The variables yi range from -co to + co. The derivation of the domains for the variables u, u is rather lengthy and is given in Appendix I. The final results will be given here: The expansion of a four-particle wavefunction in an orthogonal system as proposed in this paper will lead to the evaluation of matrix elements M, that is, integrations over eight angular variables. The integrand after integrations over CL,/3, y, #, 8, g,
172
W. ZICKENDRAHT
will be calledf(u, form :
t”) with t = (1 - 30)1/z. Then the matrix element has the following M = j’
dt . t j;;::, du .f(u,
t2)
(30)
* 319.
(31)
-l/2
with u(t) = (1 - t)(l + 2tyq3
A special method to deal with the four-particle problem is to expand the interaction of the particles in the system of functions of this paper. For four identical spinless particles, for example, one would have with central interactions only V = C V,(l rk - ri I) = V,(y) + V,(y) . (1 -
i
llz~j2) + ....
In this expansion all the functions which are completely symmetric with respect to particle exchanges will occur. The factors V,,(y), V,(y), etc., are to be taken as expansion coefficients and are determined from V, . A discussion on the completely symmetric functions will follow in Section 7. Two of these functions are contained in (32), namely, 1 and 1 - llv/2. They are orthogonal to each other, which can be checked by using formula (34). With an expansion like (32) the functionfin (30) wil1 be a polynomial in the variables u and t 2 or u and v. Withf = z&P one finds from (23) (abbreviation for M : M,,): Iodd :M,,=O;
(33)
I even: MI, = [2/(1 + I)] 2 [n !/i!(n - i) !I(-)“-“”
2i(2n - i - 1)
i=O
x 3"++l(2n
+ I -
i + l)! (I + 1)!!/(31+
4n - 2i + 7)!!.
(34)
5. DERIVATION OF THE ORTHOGONALSYSTEMFROM THE SCHR~DINGEREQUATION AND THE INVARIANCE RELATIONS
The functions Gg; are derived with the help of the relations (23) from the coupled system of Eqs. (15). The singularities of the differential equations (15) at y, = yz and y1 = -y2 lead to the general Ansatz:
One obtains another system of coupled differential equations for the functions G$ . It is necessary only to derive the functions with KS 3 0 and K, > 0. Functions for which one or both of Ki and K, are
FOUR-BODY
173
PROBLEM
(23.2) and (23.4) (a form is chosen which is convenient Appendix; p is a constant factor independent of X):
for tabulation,
see the
(1) If Ki and K, are even and X is even, or if Ki and K, are odd and X is odd: GK” Z.Yp(y, - y2)(Ki+Kc)‘2 (y, + y2)‘Ki-Ke”2f~(y1y2y3); Ke I f3JkY2Yd
=
%(Y,2n
+
+
J’i?
+
b,Y,2(Y:n-2
WIJ’~~‘?+~
+
YF”>
+
+
Vi”-“>
+
“’
bJ22Y,y2(y~n-4
(36) +
GzJ’I~Y~~
+ YF4)
+ ... + bn-ly:y;-ly;-* + c,y,4(Yf”-4+ YF”> + *‘a + d,y&Jy
+ YF>
(37)
with 2n = h - (Ki + Ke)/2 -
[ Ki - Ke I/2*
(38)
(2) If Ki and K, are even and h is odd, or if Ki and K& are odd and X is even: (3;; = P * y3(y1 - y2)(Ki+Ke)‘2 (y, + ~~)‘~“-~~“~f~(y,
v, v,),
(39)
f2: as in (3’7), but 2n = X -
1 - (Ki + K,)/2 -
j Ki - KA l/2.
(40)
As fit is a polynomial in y1 , y, , y, it is obvious that ala2 ,..., b, se- b,-, , etc., are 0 in many cases, namely, in all cases in which in (37) an exponent would be negative. For example, if n = 0 only a,, # 0, all other coefficients are zero. If either .Ki or K, or both are zero either the coefficients with even indices or those with odd indices are zero. This follows from relation (24). Tn detail: X odd and L, even or X even and L, odd:
fti = Y~Y~{~,(Y,~~-~ + $-3 + a3.-.I; h odd and L, odd or X even and L, even: f?
= a,(.$
+ $7
+ a, *-a;
(41)
X odd and Li even or h even and Li odd: fje = yly2{al(yF2
+ A”-“>
h odd ,and Li odd or X even and Li even: fle = ady:”
+
~3
+ a2 -.a.
+ a3 ...I;
174
W. ZICKENDRAHT
The numbers of coefficients have been reduced by these relations. Only those coefficients which may be different from zero due to the relations (37) for Ki > 0 and K, > 0 and due to (41) for the remaining cases are tabulated in Appendix IT up to X = 6. Since the set of operators used is not complete one has in general degenerate functions for fixed values of h, Li , Mi, L, , M, . One has to make sure that no functions are missing. This is simply done by considering the four particles moving in oscillator potentials. The potential energy is then proportional to y2. This four-particle problem is then solved with two different coordinate systems: Jacobi coordinates and the coordinates used in this paper. By comparing the numbers of solutions in both coordinate systems one makes sure that the system derived in this paper is complete, that is that there are no functions missing for fixed values of X, Li , Mi , L, , M, .
6. SYMMETRY RELATIONS UNDER EXCHANGE OF IDENTICAL PARTICLES The exchange of identical particles means reflection of the coordinates yi and a rotation in CY,/3, y space [3]. The angles of rotation will be called (or , j?r , yr . The results of [3] need a revision due to a mistake. The final results for the exchanges are : Exchange l-2 1-3 I-4 2-3 204 3-4
%
742
0 0 5-P
Pl
42 42
7? 7r 75-P
42
Yl
742
77 42
0 0 3~12
7. FOUR IDENTICAL SPINLESS PARTICLES AS AN EXAMPLE The case of four identical spinless particles is considered in this section because the simplicity of the method is best demonstrated with it. The wavefunction has to be symmetric with respect to all possible exchanges. This leads to the following functions which are completely symmetric:
where the coefficients a,. have to be determined from the requirement that Y/ is invariant under all rotations in 01,p, y space given in Section 6. The results for the coefficients aMi are: aMi f 0
for even Mi only;
(43.1)
FOUR-BODY
aMi = (-)^
175
PROBLEM
C amz(-)(Mi+m@ mi
. dzMi (+)
;
(43.2)
The functions Y of Eq. (42) are symmetric with respect to all possible exchanges if the coefficients aMi are solutions of Eqs. (43). (For the rest of this section these functions will simply be called symmetric.) For Li < 6 one obtains the following results. (Up to Li = 3 some details of the evaluation of Eqs. (43) will be given for illustration. This will not be done for L 3 4.) Li = 0: All solutions with even X are symmetric. Li = 1: One obtains from (43.2) a, = 0, as d&(n/2) solution for Li = 1.
= 0; so there is no symmetric
Li = 2: a, = a-2 = a, = 0; no symmetric solution. Li = 3: (43.4) yields a2 = (-)^ u-~ , a, = (-)n+1 a,, ; that is a,, = 0 for even A. For even X Eq. (43.2) yields a2 = a,{1 - (-)A}/2; so there is no symmetric
solution for even h. For odd h one obtains from (43.2) and (43.3) resp.: a,, = a,{d$(n/2)
- d!,,(r/2))
a,, = -az{d;&r/2)
- d!,,(r/2))
= 2a,d&/2), = -2a2d&,(r/2);
so again a, = a-, = 0. So there is no symmetric solution for Li = 3. Li = 4: Symmetric
solutions are found for even X only. They are of the form:
Li = 5: N’o symmetric solution. Li = 6: Symmetric
solutions are found for even and odd A:
A odd : c c GzD&, Ki 4
{D&z + D&m, - (551’2/ll)[D’&
+ D6K,-JI.
(46)
In practical calculations one could expand the two-particle interactions in a series containing the symmetric functions with Li = 0, 4, 6, etc., and L, = 0. The matrix elements are evaluated by integrating over #, 8, y, ~1, /3, y at first. The remaining integrations in u, a space are extremely simple and are performed as discussed in 595/1x1/1-12
176
W. ZICKENDRAHT
Section 4. Calculations are simple if one knows the functions G2 . The main problem consists in deriving these functions, as pointed out already. One could be tempted to apply the method to the 401model of 160. But one would need many functions with X > 6 for that case. All the symmetric functions given above with X < 6 are of even parity (reflection of the coordinate system means reflection of yi . As the GE: are polynomials of order h in yi , the parity is equal to (-)“). For A = 7 one has functions of odd parity with Li = 6. This means that the odd parity states of I60 would contain functions with X > 7 only. The relative motion of the 01particles would thus correspond to a rather high excitation. This is simply a consequence of the antisymmetrization with respect to all nucleons [19].
APPENDIX
I: THE VARIABLES u AND u
In Section 4, the final results for the domains of u and 2,have been given. How this result is obtained will be shown in this Appendix. First y, , yZ , y3 are replaced by y, e1 , Edof Eq. (29) in the expressions (26) for u and Y: u2 = (sin* Ed * co? e1 * sin2 26,)/4,
(47.1)
0 = (sin* c1 - sin2 2~,)/4 + sin2 l 1 * cos2 l 1 .
(47.2)
For v one obtains from (47): O
(40
The range of u will depend on U. The rest of this section will be concerned with finding the range of U. At first (47.2) is written as: sin2 2~~ = (40 - sin2 e1 * cos2 E&in*
l 1.
(4%
From (49) the range of E~for fixed u is derived: As sin2 2r, < 1 it follows: i
22 sin2 s1 - - -;++o. 31
(50)
In (50) one has to consider the two cases of sin2 Ed < $ and of sin2 Ed > 8: (1) sin2 Ed 3 # yields: 2 4(1 - 321) l/2* j < sin2 l 1 < f + ( ) ’ 9
(51)
(2) sin2 c1 < Q yields: --2 3
( 4(1 - 3v) >‘Ia ~ sin2 B < 2 9 11 3’
(52)
FOUR-BODY
177
PROBLEM
Equations (51) and (52) can be written as one necessary inequality: 2(1 - t) < sin2 c1 < 2(1 3’ t, 3
(53)
with t = (1 - 3a)l12 and 0 < t < 1. Equation (53) does not give the range of sin2 Ed yet, as we have used the condition sin2 2~~ < 1 alone. The lower limit sin2 2~~ 3 0 will yield further restrictions on sin2 Q: v - sin2 Ed * cos2 El >, 0
(54)
(sin2 l 1 - 4))” - 4 + v > 0.
(55)
or
One has to ‘consider two cases again: (1)
For sin2 l 1 > s Eq. (55) yields sin2 Ed 3 4 + (Q - v)li2.
(2)
(56)
For sin2 l 1 < + one obtains sin2 c1 < 4 - (t - v)l12.
(57)
These restrictions hold for 0 < v < 4 only, as from (54) it follows immediately: sin2 2~~ < 4v.
(58)
This means that for v > $ the variable 6Xcan have any value between 0 and 7r. The results obta.ined so far in the range 0 < v < & are combined in (59): For0
<$: 2(1 - t) < sin2 e1 < 2(1 3’ t, 3
with either sin2 E > 1 + (1 - 4W2 1, 2
(5%
or sin2 E ( 1 - (1 - 4v)1/2 1-L
2
*
As in this range for v one has always 2[1 - (1 - 30)1/2-j ~ 1 - (1 - 4v)1/2 3 2
178
W.
ZICKENDRAHT
and 2(1 + t) > 1
3
”
one obtains finally two separate ranges for sin2 Q:
o
,
2[1 - (1 - 3v)‘/2] ~ Sin2 E1 < 1 - (1 - 4791/z 3 2 (60.1)
and
l + (l -
4u)1’2
2 For the remaining
<
sin2
E l---
<
1
a
range of ZJthe relation (53) holds with no further restrictions: 2[1 - (1 - 3V)1/2] < sin2 E < 3 -.. I\
2[1 +
(1 -
3
3’)1’21.
The range for u2 is derived from (60). As a first step sin2 2~ is eliminated using (47.2):
(602)
from (47.1)
u2 = v . cos2 cl - sin2 l 1 . cos 461.
(61)
The derivative of u2 with respect to e1 is taken now to find upper and lower limits: au2 zg=
o
yields
co9 El = y
)
, sin2 cl = EjJJ
,
(62)
where for convenience the variable t has been used again instead of v. We consider the range & 2 t > 0 at first which corresponds to the range (60.2) for v. We have to find out whether the values (62) for cos2 l 1 are in the allowed range given in (60.2) or not. From (62) one obtains
sin2e1= i 7 i =
2(1? t, * 5
(63)
which is within the range (60.2). Thus we obtain from (61) and (63)
21 >t>o:
(1 - t)” (1 + 2t) 21 ” $ > (1 + 0”27(1 - 20 ’ >
(64)
The range 0 < v < $ corresponds to 4 < t < 1. There are two different regions for sin2 Edin this range; these are given by (60.1). One finds that sin2 l 1 = (2 - t)/3 of Eq. (62) is not within the allowed regions given
FOUR-BODY
179
PROBLEM
by (60.1), while sin2 Em= (2 + t)/3 is within the second one of these regions yielding an upper limit for z? (1 - q2 (1 + 20 >, u2.
(65)
27
The lower limit in this case is not found by a solution of the equation au2/&, = 0. It is found as u = 0 which one finds from sin2 cl =
1 * (1 - 4uy 2
.
So the result is: 1 > t > 1 (1 - 0” (1 + 20 > g > 0 ’
‘2
27
“*
(66)
Equations (64) and (66) show the limits of the variables which are to be observed in integrations in U, t space. The rest of this section will be used to show how integrations in U, t space can be performed very simply. The expansion of a four-particle wavefunction in an orthogonal system as proposed in this paper will lead to the evaluation of matrix elements, that is, integrations over eight angular variables. After the integration over 01,/$ y, #, I?, 9 one will have integrals of the form (where we do not write down the limits yet) s
dt - t duf(u, t2). s
(67)
Here v was replaced by the variable t. f is the integrand of the matrix element after integration over 01,p, y, #, 8, v. It must have the invariance properties as Go0 for Li = L, = 0 in Eqs. (25). The integration limits in (67) have to be determined from (64) and (66). From (64) one obtains two regions for U. With the abbreviation a(t) = (1 - t) . (1 + 2w2
658)
3 . 3112
these are:
&>ttO:
(1) (2)
40 -a(-t)
> u2
a(-t);
3 24>, -a(t).
(69)
And from (66) one obtains: l>t>$:
a(t) > 243 -a(t).
(70)
180
W. ZICKENDRAHT
The integral (67) with its limits is then: (71)
After combining
these integrals differently, one obtains finally M = jTl,2 dt t j”‘“’ duf(u, t2). -a(t)
APPENDIX
(72)
II: THE COMPLETE SYSTEMOF FUNCTIONS WITHX < 6
In the following tables the factors p and the functions fz of Eq. (36) are given. Only Ki 3 0 and K, 2 0 appear as functions with Ki < 0 or K, < 0 can be derived with the relations of Section 5. In the cases where Ki > 0 and K, > 0 the sets of coefficients a, , a, , etc., are given. In the cases where either Ki or K, or both are zero only a,, a2, etc., or a,, a,, etc., are given; that is, those coefficients which vanish due to the relations (41) do not appear in the tables. Above the tables the values of Li and L, appear. Li was always chosen as Li > L, . The coefficients for the case Li < L, are found by exchanging Ki and K, . The third column in the tables contains the factors p of Eq. (36). The columns below the h values contain the factors a,, etc. As an example for the use of the tables, the case Li = 5, L, = 1 with h = 5 is considered. For Ki = 4, K, = 0 we have case 2 of Section 5, hence Go4
= Y,(YI - YZ)~(Y~+ ~2)~ *fo4,
2n = 0.
(73) (74)
In this case only a, # 0, all other coefficients vanish and do not appear in the table. a, = 1 is taken from the table. For Ki = 2, K, = 0 we have case 2 again and (p is taken from the table) G ? = 2 . 3'12 - 3 J&l2 0”
-
Y22)h2,
2n = 2.
(75) (76)
Hence the coefficients a,, b, are different from zero and appear in the table. We have fo2 = 4Y12 +
Y22)
+
2h2.
(77)
FOUR-BODY
181
PROBLEM
For Ki = K, = 0:
G,O= yy3
.fo",
(78)
2n = 4.
(7%
The coefficients ao, a2 , b, , co appear in the table, thus: ho = 3(y14 + ~2~) +
For Ki
-
2~1~~2~
+
Q,2h2
~2~)
8~:.
(80)
5, K, = 1 we have case 1 of Section 5:
G5 = T(Y~
+~,)~f,5,
-Y~)~(YI
2n = 0.
(81) (82)
Hence
fi" = 1. For Ki
(83)
.3, K, = 1: G3 = B(YI -
~21~01
+
(84)
v2)fi3;
2n = 2.
035)
a, , q , b, appear in the tables as the only nonvanishing .A” = -5(y12
+
~2~)
-
2~1~2
+
coefficients, hence 8~3~.
(86)
For Ki = K, = 1 finally:
G1 = 'g
(~1
- YAP;
(87)
2n = 4.
(88)
a, , a, , a2 , b. , bl , co appear in the table:
fll = 5(y14 + y24) + ~YIY~(YI~ + y22) + 6~~~~23 - ~~Yz~(YI~+
~2~)
-
8~3~~1~2
+ 8~:.
In general. there will be several eigenfunctions for given values of Li, L, , X as our set of commutable operators is not complete [3]. This is the case for Li = L, = 1 for h = 5; and 6, for example. In both cases there are two eigenfunctions. These eigenfunctions have not been orthogonalized. This is a somewhat tedious procedure and will take a lot of time if one wants to do it for all the cases below. As in general
182
W.
ZICKENDRAHT
it is hoped that one will need only a few functions from the table, the orthogonalization procedure is best carried out for each case, as required. Functions with X < 6 (for L, = Li = 0 they are given explicitly on account of their simplicity): Li = L, = 0: h = 0,
Go0 = 1;
h = 3,
Go0 = u . y3; 1 - llv
h = 4,
Go0 = y4 2
X = 6,
Go’+1
;
39v
+F-).
Lc=L,=l: Ki
1
K,
p
h=l
11
h=2
1
001
X=3
1
2
-2
X=4
4 11 -7
-14 8
h=5
1 13 1
X=5
12 -13 -67 -54 0 25
-2 24
A=6
0 1 1 0 -12 --I
50 -108 -108 24
-2 24 0 0
X=6
0 4 9 -1 -6 0 0 2 -10 8
Li = 1, L, = 0: no solutions for all values of A; Li = 2, L, = 0: Ki
K,
P
x=2
2
0
1
1
A=4
x=5 2
A=6 1
-11
2 -13 -9 6
0
0
6rJ2J3
--I 2
-2 22 -11 4
--I
-2 2
3 21 0 -24 4
1 -13 -31 0 18 1 -2 0 36 -28
FOUR-BODY
183
PROBLEM
Li=2,Le=1: Ki
Ke
P
2
0
1
x=3
x=4 1
A=5 1
X=6 1
1
-2 1
1
21J2/2
0
-1 -1
-1 -1
-16 0
-1 -1 15 0 16 1
-1 -3
1
2 2 -1
Li = L., = 2, G2 = G% except for the last solution where we have G$ = -G$ Ki Ke 22
2
0
11
P
X=2
X=3
X=4
A=4
X=5
1
1
1
0 1 4
1 3 0
2 5
0 61i2/3
0
2/3
1
-1
0
1 4
-6
2
0 6
0 18 0 0
0 -1 1
-1 -9 -1
-2
-1
3
A=5
X=6 1
-22 -6
36
1 4
9 0
-108
-8 4
-1 -1 -1
0 -2 -4 -2 4 2
0
1
X=6
0 4 9
0 0 1 0 10 2
0 0 0 0 0 0
-1 -8 0
0
-4 23 15 -12
0 1
04000 -39 -3 438 -84 16
-1 -7 -6 -52
X=6
4 0 -19 -27 -100
4
X=6
-1 -2
0 2 1 -1
-3 3 42 0 0
8-4 8 -4 54 2 -38 10 -84 4 8 -4
0 -1 1 -1
3 6 -60 0 0 0 -4 -14 2 12 0
0 0 0 0 0 0 0 0 0 0 0
184
W.
ZICKENDRAHT
Li = 3, L, = 0, X = 6: Go2 = (y12 - y22)(y,2 - y32)(y32 - ylz); L;=3,L,=
1:
Ki
Ke
P
2
0
3112
0
0
10915
1
3
1
1
A=3
x=4
1
1
-9 18
-9 18
9
116
151/2/30
A=5 1 0
0
-2 -2
0 6 -15 24
-9 39
11
0 12 3
39
2 2
-6 -13 -26 12
0 1
-9 78 -78 96
13 -4
4 7
-6 -12
X=6
1 0
-22
-4 -13 13 3
-6 12
X=6
0 1
-9 -26 26 -8
-9
-9
A=5
0
3 -45 -84
-12
1 -10 0
-12 -12
39
3 18
78 -12
0
Li = 3yL, = 2: Ki
Ke
2
2
P
A=4
696
A=5 0
-2 -2
2
0
0
2
l/3
5y5
x=5
-3 -5
0 1 I
0 6
3
2
3 -5
X=6
1
3
A=6
0 -5 -16 22 22 -16 0 38 10 -28
0
5
1
0
0 0 1 -7 -7 6 0 -13 -5 8 0 -5
FOUR-BODY
Ki
Ke
P
3
1
l/2
1
1
X=5
A=4
X=5
2
15930
0 3 2
-2 -12 -8
185
PROBLEM
A=6 0 0
5
0 6 1
-2 -34 -56 22 44 -8
0 30 40 -5 -30 0
-11 -11
-1
0
0 0
-3 2
-4 -1 -6
-10 0 12
0
A=6
Li = L, = 39 G2 = G2 z Ki 2
K, 2:
P 1
A=;
X=4
-5
A=5 0
0
0
301q5 l/5
3
3
l/2
3
I
15y10
1
1
l/l0
5 -30 -40
-5
10 0
2 8
0 48 -32 16
-60 1560 640 -320
-12 -144 96 -264
0 12 0 28
2 6 2
-40 -90 110
6 0 0
0 2 0
0 0 2
-2 56 48
0
6
40 30 -30
0 4 6
6 6 24 -42 -60 0
-120 -110 380 490 -1220 -160
2 28 132 -96 -320 32
0
0 8 48 -2 -12 0
6
5
-15 -10 -80
-2
2 -84 32
-2 -2
0
X=6
0 4
2 -12 -96
X=6
-10 -100 40
4
0
X=6
0 0 0
-2
2
X=5
0 0 1 1 6 1
-2 -14 -10 -52 -10
0 0 -1 -1 -6 -1
0 -2
0 -2
-6 -4 -2 -12 8 48 0
0 -28 0 -12
186
W.
ZICKENDRAHT
Li = 4, L, = 0: Ki
Ke
P
4
0
l/2
A=4
A=6 2
2 -15
2
0
w/7
-2 4
0
0
701f2/70
6 4 -16 16
Li=4,L,=
-2 13 2 -13 6 -58 7 210 -68 16
1: Ki
K,
P
4
0
1
2
0
71/s/7
x=5
X=6 1
1 -1
-1 2
3
1
0
112
1
71q14
-1 -1
-1
1
2
1
1
0 3 2 1
3 3 4 -7 -6 4
-6 -4
FOUR-BODY
187
PROBLEM
Li = 4, L, = 2: Ki
Kc
P
4
2
1
4
0
2
2
x=4
x=5
-1
-1
6112/3
1
7y7
0
0
2
2 +42112/21
7oy3
0
5
1 0 -2
-1 -1
3
-3
2 4
0
0
2 . lo51ylo5
3 2 4
-12 24
-16
3
1
1
I.
2y2
-2
14914
6 4 -8
X=6 1 2 0
2 2 -2
2
A=6
0 1
-1 18 -2 -4 12 4 40 16 1 10 -4 -4
A=6 0 1 0
0 0 1
0
0
-6
3
0
-1 -8 2 -10 -8 0 -3 -3 6
3 -30 -8 40
0 14 -4 -8
-4 -1
-3 -37 -2 180 -56 16
0 12 18 -60 0 0
0 -12 -3 30 0 0
0
-1
-16 -16
0 -3 -2 1 2 -4
0 16 32 -16 -32 0
0
8
-1
0 7 7
3 -5 -12 7 14 -4
-4 -2 0 4 0 -2 -12 0
188
W.
Lj = 4,L,
ZICKENDRAHT
= 3: Ki
Ke
P
4
2
1
A=5
X=6 I
A=6 0
0 0 7
-5 -5 4
0
3095
-1
5
0
2
2
71y7
-1 -6 -4
0 5 0 20 20 -10
0 0 14 0 -14
2
0
2101J2/35
0
3
2
3
70935
1 8
-5 10
-70
10
0 10 -35 60
-28 -7 -28
5 10 0
-7 -7
0
3112/2 --I -1
0
0
0
3
1
51910
0 1 11
-5 -30 -20
1
3
211/2/14
0 3 1
-15 -10 20
-7
0
15 40 -30 40 -80 -80
0 35 210 35 210 0
1
1
35970 -3 -2 -11 -34 -16
0 -35 35 0 7
Ki
Ka
P
4
4
1
FOUR-BODY
PROBLEM
x=4
x=5
-1
189
A=6
-2
-1 -2
2 0 4
2
71/z/7
X=6 0 4
2
2
2 2 -2
4
0
7oy35
-3
0
2
l/7
12
0
1oy35
10
0
l/35
3
3
l/2
-4 -4 -24
6 48
-2 28 -24
-60
-18 -12 -192 -128
240 640
-4
0 2 -6
3
1
71q14
12
-3
1
1
l/14
-36 -24 -64
-4 -6
0
0 10
-80 12
0 -1 -26 16 132 -16
6 -24 -6 24
0 50 -30 -80
-18 114 24 1680 224 - 128
0 -200 100 -2600 0 0
8 4
-2 -12 8
6
6 12 0
0 18 12 -6 156 -32
-18 -24 180 56 -176 -32
-6
0 1 -2
0
2
-1 -6
1 11 -11 -3 -11 -23 9 17 -316 61 366 16
190
W.
ZICKENDRAHT
Li = 5, L, = 0: no solution for h < 6; L,=5,L,=
1: Ki
KC.
4
0
2
0
0
x=5
P
A=6 1
2 . 3y3
0
-1
701J2/21
-1 2
2
3 2
3 2
-8
-8 8
5
1
5112/2
3
1
116
1
1
8
1
-1
8
-2 -8
-5 -2
421J2/42
3
5 4 6
-1 4 2 4
-12 -8
-8 -8
8
Li = 5, L, = 2:
Ki
K?
P
4
2
- 101y5
0 2
2
A=6 0 1 1
0
2 * 15y5 3011y 5
0 2 2 1 -2 -2
FOUR-BODY Ki
K?
2
0
191
PROBLEM
X=6
P 2 * w2/5
0 0 -1 2
0
2
2 * w2/7
-1 2
5
1
1
1
3
1
- 51/2/5
1 2 0
1
1
2101/2/105
1 8 6 0 -16 -8
Li = .5, L, = 3: Ki
K
P
A=5
4
2
1
-2
A=6 0 2 -4
4
0
2 * 301y5
1
2
2
31/z/3
4 4 -4
-1 0 -4 -4 4 4 -8
2
0
0
2
4 . 101y15
2 - 701J2/21
-3 1
3 9
4
3 0
-3
192
W. Ki
A=5
P
K,
0
ZICKENDRAHT
0
4 * 211J2/63
A=6 9 6
-4 -16
-9 -6 -36 96
-1
-3
5
3
15y3
5
1
1
1
1
3
3
3y9
5 6
9 6
-4
-12
3
1
51/2/l 5
-5 -2 -12
-3 42 -12
1
3
141J2/21
-5 -2
-3 6 6
6 1
1
2101/2/105
5 4 6 18 12 -32
1 -64 -42 6 68 -32
Li = 59L, = 4: P
X=6
JG
K,
4
4
-2
4
2
2 - 35y35
0 1 8
4
0
-6
0 1
’ 51q5
. 14112/7
0 1 1
FOUR-BODY
2
4
2
2
193
PROBLEM
2 * 15q5
0 2 1
4 * 1051/2/105
0 -1 -1 -4 -13 -6
2
0
2 * 421j2/21
0 0 3 8
0
4
-2
* 141y7
1
0
2
2 * 21j2/7
1 12
5
1
3
3
5y5
3
1
35y35
1
3
2101/2/105
1
1
301y105
-3
* 7q7
1 -1 -6 -4 3 6 28 1 22 6 -3 -24 -18 -42 -92 -32
194
W.
ZICKENDRAHT
Li=L,=5,G2=G$: Ki
Ke
P
4
4
215
A=5
A=6 5
0 -1 4
4
2
4 * 31/z/15
-5
0 1 -2
4
0
2
2
--I
2 * 7oy35 s/15
5 5 10
0 -1
-1 1 -8 4
2
0
4 . 2101f2/315
0
0
4163
5
5
1
5
3
51/2/l 5
5
1
21O1/2/1O5
3
3
l/45
3
I
42112/315
1
I
21105
-15 -40
3 36
45 30 160 64
-9 -6 -144 -192
1 -5
1 -3
5
1
25 30 160
9 -66 96
-25 -10 -240 25 20 30 360 240 320
-3 114 -48 1 -172 -114 24 -496 64
FOUR-BODY
195
PROBLEM
Li = 69L, = 0: &
K?
P
4
0
661/2/l 1
X=6
-1 2
2
0
5sy55
5 6 -16 16
0
0
2 . 231112/231
-5 -3 18 12 -24 16
Li = 6, L, = 1: no solution for X < 6; Li = Ci,L, = 2: Ki
Kt?
P
6
2
1
6
0
6112/3
-1
4
2
661j2/33
-3 -2
X=6 1
4 4
0
2 * 11’/2/11
1 0
2
2
551j2/165
15 20 26 -32 -32 16
196
W.
2
0
ZICKENDRAHT
330112/165
-5 -6 0 16
0
2
2 * 23111a/231
-5 -6 12 -8
0
0
2 - 15W2/23 1
5 3 0 0 -24 32
3
1
2 * 551J2/55
-5 -2 8
1
1
2 - 221Ja/33
5 4 6 -12 -8 8
Li = 6, L, = 3: no solution for A < 6; Li = 6, L, = 4: Ki
Ke
P
6
4
1
6
2
2 * 71/2/7
6
0
3 * 7oy35
A=6 1 -1 1
FOUR-BODY
197
PROBLEM
Ki
Ke
P
4
4
66112/33
X=6 -3 -4 2
4
2
2 - 462112/231
4
0
2 * 1 155112/385
2
4
55112/165
3 2 10 -3 -14 15 10
-16 2
2
2 . 3851/2/1155
-15 -20 -26 -80 -80 96
2
0
1541y1155
45 54 336 -256
0
4
2 . 2311/2/231
-5 6
0
2
4 . 33112/231
5 6 30 -48
0
-0
2 * 3301q1155
-15 -9 -126 -84 128 128
5
3
2 +6112/3
1
198
W.
Ki
K,
ZICKENDRAHT
P
X=6
5
1
2 . 421J2/7
--I
3
3
2 * 1101y55
-5 -6 4
3
1
2 * 770112J385
15 6 4
1
3
4 * 11112/33
5 2 -6
1
1
4 . 771J2/231
-15 -12 -18 -6 -4 32
L, =:6, La = 5: no solution for h < 6; L,=Li = 6,G2 = G$: Ki
Ke
6
6
6
4
P
66112/l 1
X=6
-1
551/2/l 1 6
0
10 . 231112/231
4
4
2/11
4
2
301y33
-1 3 4 20 -3 -2 -32
FOUR-BODY
199
PROBLEM
Ki
Ke
4
0
2
2
l/33
2
0
1051J2/231
0
0
l/231
5
5
2
5
3
2 - 16W2/1 I
5
1
10 . 661J2/33
1
3
3
l/11
30 36 64
3
1
101q11
1
1
4133
P 10 . 14y77
X=6 1 12 15 20 26 256 256 256 -10 -12 -192 -256 100 60 2160 1440 3840 1024 1 -1
-10 -4 -32 25 20 30 120 80 64
200
W.
ZICKENDRAHT REFERENCES
Ann. Phys. (N.Y.) 35 (1965), 18. Phys. 3 (1966), 461. 3. W. ZICKENDRAHT, J. Math. Phys. 10 (1968), 30. 4. E. L. SURKOV, Sov. J. Nucl. Phys. 5 (1967), 644. 5. H. W. GALBRAITH, J. Math. Phys. 12 (1971), 782. 6. E. -CON AND A. AMAYA, Ann. Phys. (N.Y.) 97 (1976), 266. 7. J. M. L&Y-LEBLOND, J. Math. Phys. 7 (1966), 2217. 8. W. ZICKENDRAHT, Z. Phys. 251 (1972), 365. 9. W. ZICKENDRAHT AND H. STENSCHKE, Phys. Lett. 17 (1965), 243. 10. L. W. BRUCH AND H. STENSCHKE, J. Chem. Phys. 57 (1972), 1019. 11. W. ZICKENDRAHT, J. Math. Phys. 12 (1971), 1663. 1. W.
2. Yu.
ZICKENDRAHT,
A. SIMONOV,
Sov. J. Nucl.
12. A. Y. DZYUBLIK, V. I. OVCHARENKO, A. I. STESHENKO, AND G. F. FILIPPOV, Sov. J. Nucl. Phys. 15 (1972), 487. 13. P. GUI~HANI AND D. J. ROWE, Canad. J. Phys. 54 (1976), 970. 14. A. BOHR, Kgl. Dan. Vidensk. Selsk., Mat. Fys. Medd. 26 (1952), 14. 15. W. ZICKENDRAHT, Z. Phys. 253 (1972), 356. 16. V. V. VANAGAS, Sov. J. Nucl. Phys. 23 (1976), 505. 17. J. D. LOUCK AND H. W. GALBRAITH, Rev. Mod. Phys. 44 (1972), 540. 18. M. E. ROSE, “Elementary Theory of Angular Momentum,” Wiley, New York, 1957. 19. K. WILDERMUTH AND W. MCCLURE, “Cluster Representations of Nuclei,” p. 1, Springer Tracts in Modern Physics 41, Springer, New York, 1966.