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Axial vector renormalization constant from algebra of currents. II
ANNALS
OF PHYSICS:
47, 103-115 (1968)
Axial
Vector Renormalization Constant from Algebra of Currents. II MAMATA PATTNAIK
Saha Institute
of Nuclear
Physics,
Calcutta,
India
A general sum rule connecting the matrix elements of the SU(4) current operators between zero momentum states, characterized by arbitrary spin and isospin quantum numbers is derived using SLr(4) current commutation relation and Racah algebra. The sum rule thus obtained gives more insight into the problem and reproduces the results obtained by other authors for calculating the axial vector renormalization constant.
I. INTRODUCTION
In recent years, the implications,of the algebra of currents approach to higher symmetry groups is being widely studied, because it reproduces the grouptheoretical results and at the same time maintains a control over the corrections due to the symmetry-breaking effects. Since Gellmann (1) suggested the possibility, that the experimentally observed weak vector and axial vector currents satisfy the commutation relations of many symmetry groups, several attempts (2) have been made in this line to obtain some fruitful results. Following this line, Adler (3) and Weisberger (4) have derived a very interesting sum rule, from chiral SU(3)(X) SU(3) algebra and dispersion technique, which relates the axial vector renormalization constant (GA/Gy) to pion-nucleon scattering cross section. This sum rule has a spectacular success in calculating the value of (GA/G,) = 1.2, in remarkable agreement with the experimental value (GA/GY) = 1.18. On the other hand, Ryan (5) has studied the implications of such algebra of currents to SU(4) symmetry, which is probably the most straightforward and economic generalization of SU(3) symmetry. In that paper he showed that, a value of (GA/G,) = Q previously deduced from SU(6) symmetry (6) can be obtained by an appropriate choice of the states. He also concluded that by the inclusion of higher states, the algebra of currents approach can be made to yield an arbitrary high value of (GA/GY). The purpose of the present paper is to study in what manner the inclusion of higher states affects the value of (GA/GY). For this, we obtain sum rules for the reduced matrix elements of the current operators between states characterized by 103
104
PATTNAIK
arbitrary spin and isospin quantum numbers, by taking matrix elements of both sides of the commutation relation
between these states, and using Racah algebra (7). These sum rules are a set of infinite number of coupled equations involving reduced matrix elements which can be solved by truncation and the value of (GA/GV) is obtained from these solutions. We have considered various cases of truncations and studied how the value of (GAG”) varies as higher states are included. This has been demonstrated in Sections II and III of this paper. Finally in Section IV we discuss the results obtained by our procedure.
II. DERIVATION
OF THE SUM RULE
The derivation of the sum rules for the reduced matrix elements of the current operators will be done with the aid of the commutation relation [Aiw, Aja] = &e,,,V,y
+ i&,c,~~A~~,
(2-l)
where
VdU =scPX#+(X, t)+ $qx, t),
(2.3)
Aio = [ d%#+(x, t) g
(2.4)
#(x, t),
(TVand T, are, respectively, the 2 x 2 Pauli spin and isospin matrices. VJa are the generators of the isospin group and are constant with time, where as all others vary with time. However, for simplicity, we will neglect the time dependence. The next step is to find the matrix elements of the commutation relation, Eq. (2.1) between zero-momentum physical states, which are characterized by two quantum numbers 1I, J), where Z is the isospin and J is the spin quantum numbers of the state. We note here that the operator Aimis a triplet tensor in both spin and isospin space. The subscript i, j and superscript 01and ,3 take the values 1,2, 3. This means that Eq. (2.1) is given in the Hermitian basis. For convenience we will convert this equation into the spherical basis (8). In this basis a state is specified by 1I J I3 Js),
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where Z3 and J3 are, respectively, the third components Equation (2.1) in spherical basis is now converted to
of isospin Z and spin J.
I Vi,
(2.5)
where the new (Y’, /3’, i’ and j’ are the magnetic quantum numbers, each of them having values 1, 0 and - 1. j2 j3 m2 m3
Jl
1
[ ml
represents the Clebsch-Gordan coefficient. We define the following quantities: (ZJr,J,
1A;: 1 Z’J’Z;J;)
= [;,
1,
;I[;:
3
i’, ;]
3
3
G(ZJ, Z’J’),
(2.6)
3
(ZJZ3J3 1A;, 1Z’J’Z;J;)
= ~3~1,6,,;[J(J
+ I)]‘/” g(J, J’),
(ZJZ3J3 1 V,y’ 1Z’J’Z;J;)
= I&/ 6,aJi[Z(Z + l)]‘/” 6,,, ,
(2.7) (2.8)
where G(ZJ, Z’J’) is the actual reduced matrix element, usually denoted by (ZJll A 11Z’J’) and g(J, J’) is proportional to the reduced matrix element
I’J’)
=
(+#j”’
(-)Z’+J’-Z-J
G(Z’, J’; ZJ),
G(ZJ, Z’J’) = G(JZ, J’Z’).
(2.9) (2.10)
Our convention is (A;:)+ = (-)s’+i’
A:;: ’
We now take the matrix element of the commutation relation (2.1) between ) Z J Z3 J3) and I I” J” 1: J,“) states. Introducing the intermediate states 1I’ J’ Z; Jj), we find &
[(ZJZ3J3 1A$ I Z’J’Z;J;)(Z’J’Z;J;
I A$ 1ZnJ”Z;J;)
I’J’ 35
-
(ZJZ3J3 I A$ I Z’J’Z;J;)
1A$ 1Z”J”Z;J;)]
106
PATTNAIK
The left-hand side of the equation can be written
Then using the properties
of Clebsch-Gordan 12
5, (-)l’+J’-l-J ($##I
1 I’
’ IL
Z
as
coefficients (8) we get 1 I”
I’
1 J’
J
1 J”
z; J L/Y z; ZJ L’ J; JJ?
J
Ji; J,]
G(ZJ, Z’J’) G(Z”J”, Z’J’)
= - d/2; S,,,, SII- S,,; C [ !, t, ;,] c: ;, k’ 1 J 3
We now multiply
;:] [J(J +
111””g(J, J”>
3
Eq. (2.12) by 2 L1 6’1’
;,
and then sum over ~4, p’, manipulations f’+J’---J
;,I[;,
;,
:,l[;,
;,
+
(-)I"--l+l
;j
U(1, lZZ”, IZ’) U(llJJ”,
[$&f)l”
(-)J”-J+l
;,
i’, j’, Ii, and Ji to obtain after some algebraical
x G(ZJ, Z’J’) G(Z”J”, Z’J’)(l =
;]f
JJ’)
- (-)-I-j)
SI,, S.r,I[J(J + 1)l”” dJ, J”> SW
ii,“”
(5,""
61~
&,,[z(z
+
l)]“”
b
sJJ-
*
(2.13)
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OF CURRENTS
The U-coefficients in Eq. (2.13) are converted to W-coefficients by the relation
Wlj2jj2 ; i’i”) = Wl[~l>1~2 WjIi2ii3 ; ii”),
(2.14)
where [j] = 2j + 1. Using Eq. (2.14) in (2.13) we obtain finally “+J’-‘-J([zqJ”][i][~])l~2
W(1 IZZ”; izy W(1 IJJ”; V)
x G(ZJ, Z’J’) G(Z”J”, Z’S)(l =
(-y--J+1
+
sr,, sr.1bW
($'"
(-)r'--1+1
- (-)-‘-J>
($1'"
+ l)P2 gv, J”>
SI>, sr,, b[Z(Z
+ l)]“” &I” .
(2.15)
Eq. (2.15) expresses the relation between the various matrix elements of the axial vector current operators. Depending upon the initial, final, and intermediate states considered, a set of equations satisfied by the reduced matrix element of the operator A are obtained. The solution of these equations gives the values of the axial vector renormalization constant (GA/GY). Before deriving these set of equations from Eq. (2.15) which we will reserve for the next section, we note that the right-hand side of Eq. (2.15) vanishes identically unless I + .I is odd. We recall here that Z and .I are introduced through the expression after Eq. (2.12) and that it is this plus the fact that (Z.Z) and (Z”J”) must be states that can be connected by the operator on the left-hand side of the commutation relation (2.1) that restricts 1 and J to the values 2, 1, 0. In order that the Eq. (15) gives nonvanishing matrix elements, we have the following four possibilities to have (1 + J) odd. The possibilities are 1. f = 1, 2. i = 0, 3. i = 2, 4.i= 1, III. CALCULATION
J=O 3=1 J=l J=2
OF THE AXIAL
VECTOR
so I+J=1. so I+J=1. SO i+J=3. SO i+,i=
3.
RENORMALIZATION
CONSTANT
In the previous section we have derived the sum rule for the reduced matrix elements of the current operators between states characterized by arbitrary spin and isospin quantum numbers, by taking matrix elements of both sides of the commutation relation (2.1) between the states (ZJ) and (Z”J”). In this section we will give some definite examples of these sum rules at various truncations and study how (GA/GV) varies as higher states are included.
108
PATTNAIK
We know, that if the matrix element of the commutation relation (2.1) is taken between zero-momentum nucleon states, the allowed intermediate states are (3, &); ($, 3), ($, 4) and (#,$). For convenience let us denote these states by numbers 1. 2, 3, and 4, respectively. As for example we will denote the reduced matrix elements G(+ 4, Q #) by G(l, 4). We will consider now the following few cases of truncation and find the value for the axial vector renormalisation constant (GA/G”). Case I:
(I’, J’) = (i, is>.
.
Let us take the example where the first of the above set of four states contribute. From Eq. (2.15) we obtain, taking (IJ) = (I”J”) = ($ $), jG2(1, 1) = 1
(3.1)
g(Q, +.I = 1.
(3.2)
and
Equation (3.1) is the definition of the axial vector renormalisation (GA/G”). So we find (GA 1 Gy)2 = $G2(1, 1) = 1, or (GA I GY) = ztl. Case 2:
(I’, 0
constant
(3.3)
= (4, !d + (3, $1.
We will now consider the case where the first two of the above set of states contribute. Then taking the matrix element of Eq. (2.1) between (4 +) states and using Eq. (2.15) we get the following sum rules, for different combinations of I and J to have the right-hand side of Eq. (2.15) nonvanishing. These are (a) I = 1, J = 0; W2(1, 1) + +G2(1, 2) = 1,
(3.4)
+G2(1, 1) - QG2(1, 2) = g(+, 4).
(3.5)
and (b)I=O,J=
1;
We see that there are two equations-(3.4) and (3.5)-having three unknowns. So to get a solution for (GA/G,) we must consider some more sum rules in this case. We, therefore, now take the matrix element of the commutation relation between
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CURRENTS
(4 4) and (4 $) states with the above set of intermediate we obtain the sum rules
states. Then from Eq. (2.15)
(a) I = 0,J = 1; W’~>
G(l, 2) W,
1) + (l/2/3) ~‘5 G(1,2) G(2,2) = g($, 8).
(3.6)
(b) I = 1, J = 2; 1.6 G(1, 1) G(l, 2) + G(2,2) G&2)
= 0.
(3.7)
Equations (3.4)-(3.7) together do not give solutions because these are four equations having five unknowns. So we now proceed to take the case where (I .Z) = (Z”.Z”) = (Q Q). Equation (2.15) then yields the following sum rules (a) I = 1, J = 0;
(b)f=O,J=
$G2(1, 2) + +G2(2, 2) = 1.
(3.8)
+G2(1, 2) + &G2(2, 2) = g($, #).
(3.9)
1;
(c) 1 = 1, .7 = 2; 5G2(1, 2) - 8G2(2, 2) = 0.
(3.10)
From Eqs. (3.4)-(3.10) we find (GA 1Gv)2 = +G2(1, 1) = + ;
(3.11)
so
(3.12) and G2(1, 2) = $,
(3.13a)
G2(2, 2) = +g ,
(3.13b)
g(3, 3) = - 5 ,
(3.13c)
Ed% 4) = + s ,
(3.13d)
g(& ;> = -(gy2.
(3.13e)
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PATTNAIK
Case 3:
(I’J’) = &I,*) + (8, 4).
On taking (N) = (Z”.Z”) = (4 4) an d inserting the set of intermediate (4 $) and (3 +), Eq. (2.15) gives the sum rules given below.
states
(a) I = 1, J = 0; $G2(1, 1) - $G2(1, 3) = 1.
(3.14)
(b) f = 0, .I = 1; this leads to (3.15)
+G2(1, 1) + +G2(1, 3) = g($, 3). While taking (Z.Z) = (4 4) and (Z”J”) = (8 Q) and introducing states we get from Eq. (2.15) the sum rules
the same intermediate
(a) I = 1, J = 0; G(1, 1) G(1, 3) - 6
G(1, 3) G(3, 3) = 0,
(3.16)
and (b) f = 2,J = 1; lOG(1, 1) G(l, 3) + 245 G(1, 3) G(3, 3) = 0.
(3.17)
Unlike the previous cases we also find, as in Ref. (5), examining Eqs. (3.14)-(3.17) that this set of states give no solution. Case 4:
(I’J’) = ($, 4) + (8, Q).
If we take (ZJ) = (Z”J”) = (4 4) and use the intermediate states (i 4) and (Q 8) then we get from Eq. (2.15) the sum rules satisfied by the following reduced matrix elements. These are (a) I = I,3
= 0; $G2(1, 1) - $G2(1, 4) = 1.
(3.18)
$G2(1, 1) - $G2(1, 4) = g(&, 4).
(3.19)
(b) I = 0, J = 1;
Upon taking the matrix element of Eq. (2.1) between (8 4) and (3 $) states (i.e., ZJ = 4 4 and Z”.Z” = $3) and introducing the same intermediate states (* +) and (Q 9) we obtain from Eq. (2.15) i = 1, i = 2,
J=O J = 1 the sum rule I
(3.20)
G(l, 4) G(l, 1) - G(l, 4) G(4,4) = 0.
(3.21)
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CURRENTS
Then we consider (ZJ) = (Z”J”) = ($ Q). Using the above two intermediate from Eq. (2.15) we obtain the sum rules
111 states
(a) I = 1, J = 0; $G2( 1, 4) + &,G”(4, 4) = 1.
(3.22)
&G2(1 7 4) + &G2(4, 4) = g(g, 8).
(3.23)
(b) 1 = 0, J = 1;
(c) I = 1, J = 2; 25G2( I, 4) - 32G2(4, 4) solving Eqs. (3.18)-(3.24)
(3.24)
we get (GA 1 Gv)2 = $G2(1, 1) := 25 9 7
or fQ,
(3.25)
G2(1, 4) = Q ,
(3.26a)
G2(4, 4) = IQ ,
(3.26b)
(GAIGY)= and
s(Q, 4) = g(% $1 = 1.
(3.26~)
Equation (3.25) is the SU(6) result (6) and has also been obtained in Ref. (5). Of course neither the SU(6) nor our procedure can calculate the sign of GA/G”. Case 5:
(rl.7) = (4, c$) + (+, 8) + ($, 4) + (3, 8).
We now consider the case where the above set of four states contribute. This case has been considered by the present author in a previous paper (9) following Ryan’s procedure and has given a value of GA/G” = 1.22. We have shown in the above four cases that the present approach has reproduced the results obtained in Ref. (5). So from previous considerations one must also expect here to get the same value for GA/GY (9). Let us examine this. As usual we must first take (N) = (Z”J”) = (+ i) and using the above set of four intermediate states we get from Eq. (2. IS) (a) I = 1, J = 0; $G2(1, 1) + $G2(1, 2) -
$G2(1, 3) - -G2(1, 4) = 1.
(3.27)
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PATTNAIK
(b) i: = 0, J = 1; QG2(1, 1) - QG2(1, 2) + +G2(1, 3) - $G2(1, 4) = g(&, 4). Next we take (ZJ) = (4 Q) and (Z”J”) = (Q, 3). With the same intermediate as above, Eq. (2.15) yields (a)I=2,J= 5G(1, 1) G&4)
(3.28) states
1; - 545 G&2)
G(2,4) + d/s G(l, 3) G(3,4) - 5G(1, 4) G(4,4) = 0. (3.29)
(b) I = 1, J = 2; 5G(l, 1) G(l, 4) + d/s G(l, 2) G(2,4) - 5d/5 G(l, 3) G(3,4) - 5G(l, 4) G(4,4) = 0. (3.30) On taking now (ZJ) = (4 8) and (Z”J”) = ($ &) and the same intermediate we obtain the following sum rules from Eq. (2.15):
states
(a) I = 2, J = 1; 5G(1, 2) G(l, 3) + 5(101j2) G(2, 2) G(2, 3) - (10’12) G(2, 3) G(3, 3) - 5G(2, 4) G(3, 4) = 0.
(3.31)
(b) I = 1,3 = 2; 5G(1, 2) G(l, 3) - ( 101/2) G(2,2) G(2, 3) + 5(10112)G(2, 3) G(3, 3) - 5G(2,4) G(3,4) = 0.
(3.32)
Further, we take (ZJ) = (4 $) and (Z”J”) = ($8) and truncate the intermediate states just after (& &), (& $), (8 a), and ($ Q) states. From Eq. (2.15), we find for (a)I= y.0
l,J=O
the sum rule
G(l, 2) G(l, 4) - 22/2 G(2,2) G(2,4) + 22/5 G(2,3) G(3,4) + 2(10112)G(2,4) G(4,4) = 0.
(3.33)
(b) I = 2, J = 1 252/2 G( 1,2) G( 1,4) + 202/z G(2,2) G(2,4) - 102/5 G(2,3) G(3,4) + 4(10112)G(2,4) G(4,4) = 0. (C)f=
l,J=2
(3.34)
the sum rule
- 52/2 G(1,2) G( 1,4) + 82/2 G(2,2) G(2,4) - 101/s G(2,3) G(3,4) - 8(101/2) G(2,4) G(4,4) = 0. (3.35)
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113
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Finally, to get a solution we have to take (IJ) = (I”J”) = (8 4). Truncating at the above set of intermediate states (Z’J’) we obtain from Eq. (2.15) the sum rules (a)I=
l,J=O +G’(lp 4) + $G2(2, 4) + &G”(3, 4) + &G”(4, 4) = 1.
(b)j=O,J= WV,
(3.36)
I; 4) + &G2(2, 4) + +G2(3, 4) + &G2(4, 4) = g($, g).
(3.37)
(c) I = 2, J = 1; 25G2(1, 4) + 20G2(2, 4) - 40G2(3, 4) - 32G2(4, 4) = 0.
(3.38)
(d) 1 = 1, J = 2; 25G2(1, 4) - 40G2(2, 4) + 20G2(3, 4) - 32G2(4, 4) = 0. There are now thirteen equations-(3.27) we solve them and find
(3.39)
to (3.39)-and
twelve unknowns. So
(GA 1 Gy)2 = +G2(1, 1) = -& ;
(3.40)
(GA 1GY) FE i ++ s 1.22
(3.41)
so
and G2(1, 2) = G”(1, 3) = $+,
(3.42a)
G2(1 3 4) = @,
(3.42b)
G2(2, 4) = G”(3, 4) = #,
(3.42~)
G2(4, 4) = 2% ,
(3.42d)
G2(2, 2) = G2(3, 3) = #$,
(3.42e)
G”(2, 3) = 2% )
(3.42f)
g(3, $1 = g(#, 2) = 1.
w&J
In the present approach we also find for this case the value 1.22 for (G,,/GV). 595/47/I-g
114
PATTNAIK
We have also calculated two further cases and are going to give the value of (GA/GV) for these cases without giving the detailed calculations. These are Case 6:
(Z’J’) = ($, $) + (Q, 8) + (& $).
(GA I G,) = i $. Case 7:
(Z’J’) = (B, it) + (+, 8) + (%, 4) + (2,s) + (I, #) + (2,9. (G, / G,) = f L93.
(3.43) + (9, $1 + (S, S) + (4, Q) (3.44)
The calculation is very similar to the previous cases. One has to first find a set of coupled equations from (2.15) by suitable choice of the initial and final states, which when solved gives the above values for (GA/G,). The values of other reduced matrix elements are also obtained from these equations but they are not so important here. IV.
DISCUSSIONS
In our analysis we have studied in what manner (G,/G,) varies with truncation. We find that, as one goes from one representation to another, certain regularities in the value of (GJG,) appear in different directions. For example, as one goes from (TJ’) = (4 &) to (+ 4) + (% 3) to (B Q) + ($3) + (+ 3) and so on, the value of (GA/GY) changes from 1 = $ to Q to z, and so on, respectively. This regularity has been observed also in Ref. (5). Another regularity which has not been observed before but appears here is that, as one goes from (I’J’) = ($ 4) to (+ &) + (3 $) + (8 4) + (# +) which are allowed when the matrix element of the commutator [&, A,01 is taken between (4 3) states) to ($4) + (8 8) + (3 3) + (4 3) + (4 9 + (3 4) + (3 Q> + (8 $9 + @ $1 ( w h ic h are allowed when the matrix element of the commutator is taken between (8 Q) states) and likewise, the value of (GA/Gy) changes from (GA/G,) = 1 = $ to J& to 9, and so on, respectively. Thus we find that (GA/GV) depends very much on truncation. However, what one can say in its favor is that our approach is very general and straightforward and reproduces the results of complicated higher symmetry groups. In addition to this, it is seen that by choosing the states (Z’J’) = (4 4) + (4 8) + (# Q) + (Q g), our method yields a value (GA/GY) = 1.22 which is about 4 % above the experimental value (GA/G,) = 1.18). It may be worth remarking here that in static Chew-Low theory with one pion approximation these are the states allowed, and higher states are ignored. It may be due to this fact that with this set of states we get a value of (G,/GV) close to the experimental value.
ALGEBRA OF CURRENTS
115
ACKNOWLEDGMENT
The author is extremely grateful to Professor T. Pradhan for suggesting this investigation and for many stimulating and helpful discussions. RECEIVED: September 6, 1967
REFERENCES 1. M. GELL-MANN, Phys. Rev. 125, 1067 (1962). 2. B. W. LEE, Phys. Rev. Letters 14,673 (1965); C. RYAN, Phys. Rev. 140, B480 (1965); S. OKUBO, Phys. Letters 17, 172 (1965); R. F. DASHEN AND M. GELL-MANN, ibid. 17, 145 (1965): E. C. G. SUDERSHAN, ibid. 14, 1057 (1965). 3. S. L. ADLER, Phys. Rev. Letters 14, 1051 (1965); Phys. Rev. 140, B736 (1965). 4. W. I. WEISBERGER, Phys. Rev. Letters 14, 1047 (1965); Phys. Rev. 143, 1302 (1966). 5. C. RYAN, Ann. Phys. (N. Y.) 38, 1 (1966). 6. F. GURSEY, A. PAIS, AND L. RADICATI, Phys. Rev. Letters 13, 299 (1964). 7. G. RACAH, Phys. Rev. 62, 438 (1942). 8. M. E. ROSE, “Elementary theory of Angular Momentum.” Wiley, London, 1957. 9. MAMATA PAITNAIK, Ann. Phys. (N. Y.) 46, 317 (1968).
OF PHYSICS:
47, 103-115 (1968)
Axial
Vector Renormalization Constant from Algebra of Currents. II MAMATA PATTNAIK
Saha Institute
of Nuclear
Physics,
Calcutta,
India
A general sum rule connecting the matrix elements of the SU(4) current operators between zero momentum states, characterized by arbitrary spin and isospin quantum numbers is derived using SLr(4) current commutation relation and Racah algebra. The sum rule thus obtained gives more insight into the problem and reproduces the results obtained by other authors for calculating the axial vector renormalization constant.
I. INTRODUCTION
In recent years, the implications,of the algebra of currents approach to higher symmetry groups is being widely studied, because it reproduces the grouptheoretical results and at the same time maintains a control over the corrections due to the symmetry-breaking effects. Since Gellmann (1) suggested the possibility, that the experimentally observed weak vector and axial vector currents satisfy the commutation relations of many symmetry groups, several attempts (2) have been made in this line to obtain some fruitful results. Following this line, Adler (3) and Weisberger (4) have derived a very interesting sum rule, from chiral SU(3)(X) SU(3) algebra and dispersion technique, which relates the axial vector renormalization constant (GA/Gy) to pion-nucleon scattering cross section. This sum rule has a spectacular success in calculating the value of (GA/G,) = 1.2, in remarkable agreement with the experimental value (GA/GY) = 1.18. On the other hand, Ryan (5) has studied the implications of such algebra of currents to SU(4) symmetry, which is probably the most straightforward and economic generalization of SU(3) symmetry. In that paper he showed that, a value of (GA/G,) = Q previously deduced from SU(6) symmetry (6) can be obtained by an appropriate choice of the states. He also concluded that by the inclusion of higher states, the algebra of currents approach can be made to yield an arbitrary high value of (GA/GY). The purpose of the present paper is to study in what manner the inclusion of higher states affects the value of (GA/GY). For this, we obtain sum rules for the reduced matrix elements of the current operators between states characterized by 103
104
PATTNAIK
arbitrary spin and isospin quantum numbers, by taking matrix elements of both sides of the commutation relation
between these states, and using Racah algebra (7). These sum rules are a set of infinite number of coupled equations involving reduced matrix elements which can be solved by truncation and the value of (GA/GV) is obtained from these solutions. We have considered various cases of truncations and studied how the value of (GAG”) varies as higher states are included. This has been demonstrated in Sections II and III of this paper. Finally in Section IV we discuss the results obtained by our procedure.
II. DERIVATION
OF THE SUM RULE
The derivation of the sum rules for the reduced matrix elements of the current operators will be done with the aid of the commutation relation [Aiw, Aja] = &e,,,V,y
+ i&,c,~~A~~,
(2-l)
where
VdU =scPX#+(X, t)+ $qx, t),
(2.3)
Aio = [ d%#+(x, t) g
(2.4)
#(x, t),
(TVand T, are, respectively, the 2 x 2 Pauli spin and isospin matrices. VJa are the generators of the isospin group and are constant with time, where as all others vary with time. However, for simplicity, we will neglect the time dependence. The next step is to find the matrix elements of the commutation relation, Eq. (2.1) between zero-momentum physical states, which are characterized by two quantum numbers 1I, J), where Z is the isospin and J is the spin quantum numbers of the state. We note here that the operator Aimis a triplet tensor in both spin and isospin space. The subscript i, j and superscript 01and ,3 take the values 1,2, 3. This means that Eq. (2.1) is given in the Hermitian basis. For convenience we will convert this equation into the spherical basis (8). In this basis a state is specified by 1I J I3 Js),
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where Z3 and J3 are, respectively, the third components Equation (2.1) in spherical basis is now converted to
of isospin Z and spin J.
I Vi,
(2.5)
where the new (Y’, /3’, i’ and j’ are the magnetic quantum numbers, each of them having values 1, 0 and - 1. j2 j3 m2 m3
Jl
1
[ ml
represents the Clebsch-Gordan coefficient. We define the following quantities: (ZJr,J,
1A;: 1 Z’J’Z;J;)
= [;,
1,
;I[;:
3
i’, ;]
3
3
G(ZJ, Z’J’),
(2.6)
3
(ZJZ3J3 1A;, 1Z’J’Z;J;)
= ~3~1,6,,;[J(J
+ I)]‘/” g(J, J’),
(ZJZ3J3 1 V,y’ 1Z’J’Z;J;)
= I&/ 6,aJi[Z(Z + l)]‘/” 6,,, ,
(2.7) (2.8)
where G(ZJ, Z’J’) is the actual reduced matrix element, usually denoted by (ZJll A 11Z’J’) and g(J, J’) is proportional to the reduced matrix element
I’J’)
=
(+#j”’
(-)Z’+J’-Z-J
G(Z’, J’; ZJ),
G(ZJ, Z’J’) = G(JZ, J’Z’).
(2.9) (2.10)
Our convention is (A;:)+ = (-)s’+i’
A:;: ’
We now take the matrix element of the commutation relation (2.1) between ) Z J Z3 J3) and I I” J” 1: J,“) states. Introducing the intermediate states 1I’ J’ Z; Jj), we find &
[(ZJZ3J3 1A$ I Z’J’Z;J;)(Z’J’Z;J;
I A$ 1ZnJ”Z;J;)
I’J’ 35
-
(ZJZ3J3 I A$ I Z’J’Z;J;)
1A$ 1Z”J”Z;J;)]
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The left-hand side of the equation can be written
Then using the properties
of Clebsch-Gordan 12
5, (-)l’+J’-l-J ($##I
1 I’
’ IL
Z
as
coefficients (8) we get 1 I”
I’
1 J’
J
1 J”
z; J L/Y z; ZJ L’ J; JJ?
J
Ji; J,]
G(ZJ, Z’J’) G(Z”J”, Z’J’)
= - d/2; S,,,, SII- S,,; C [ !, t, ;,] c: ;, k’ 1 J 3
We now multiply
;:] [J(J +
111””g(J, J”>
3
Eq. (2.12) by 2 L1 6’1’
;,
and then sum over ~4, p’, manipulations f’+J’---J
;,I[;,
;,
:,l[;,
;,
+
(-)I"--l+l
;j
U(1, lZZ”, IZ’) U(llJJ”,
[$&f)l”
(-)J”-J+l
;,
i’, j’, Ii, and Ji to obtain after some algebraical
x G(ZJ, Z’J’) G(Z”J”, Z’J’)(l =
;]f
JJ’)
- (-)-I-j)
SI,, S.r,I[J(J + 1)l”” dJ, J”> SW
ii,“”
(5,""
61~
&,,[z(z
+
l)]“”
b
sJJ-
*
(2.13)
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OF CURRENTS
The U-coefficients in Eq. (2.13) are converted to W-coefficients by the relation
Wlj2jj2 ; i’i”) = Wl[~l>1~2 WjIi2ii3 ; ii”),
(2.14)
where [j] = 2j + 1. Using Eq. (2.14) in (2.13) we obtain finally “+J’-‘-J([zqJ”][i][~])l~2
W(1 IZZ”; izy W(1 IJJ”; V)
x G(ZJ, Z’J’) G(Z”J”, Z’S)(l =
(-y--J+1
+
sr,, sr.1bW
($'"
(-)r'--1+1
- (-)-‘-J>
($1'"
+ l)P2 gv, J”>
SI>, sr,, b[Z(Z
+ l)]“” &I” .
(2.15)
Eq. (2.15) expresses the relation between the various matrix elements of the axial vector current operators. Depending upon the initial, final, and intermediate states considered, a set of equations satisfied by the reduced matrix element of the operator A are obtained. The solution of these equations gives the values of the axial vector renormalization constant (GA/GY). Before deriving these set of equations from Eq. (2.15) which we will reserve for the next section, we note that the right-hand side of Eq. (2.15) vanishes identically unless I + .I is odd. We recall here that Z and .I are introduced through the expression after Eq. (2.12) and that it is this plus the fact that (Z.Z) and (Z”J”) must be states that can be connected by the operator on the left-hand side of the commutation relation (2.1) that restricts 1 and J to the values 2, 1, 0. In order that the Eq. (15) gives nonvanishing matrix elements, we have the following four possibilities to have (1 + J) odd. The possibilities are 1. f = 1, 2. i = 0, 3. i = 2, 4.i= 1, III. CALCULATION
J=O 3=1 J=l J=2
OF THE AXIAL
VECTOR
so I+J=1. so I+J=1. SO i+J=3. SO i+,i=
3.
RENORMALIZATION
CONSTANT
In the previous section we have derived the sum rule for the reduced matrix elements of the current operators between states characterized by arbitrary spin and isospin quantum numbers, by taking matrix elements of both sides of the commutation relation (2.1) between the states (ZJ) and (Z”J”). In this section we will give some definite examples of these sum rules at various truncations and study how (GA/GV) varies as higher states are included.
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We know, that if the matrix element of the commutation relation (2.1) is taken between zero-momentum nucleon states, the allowed intermediate states are (3, &); ($, 3), ($, 4) and (#,$). For convenience let us denote these states by numbers 1. 2, 3, and 4, respectively. As for example we will denote the reduced matrix elements G(+ 4, Q #) by G(l, 4). We will consider now the following few cases of truncation and find the value for the axial vector renormalisation constant (GA/G”). Case I:
(I’, J’) = (i, is>.
.
Let us take the example where the first of the above set of four states contribute. From Eq. (2.15) we obtain, taking (IJ) = (I”J”) = ($ $), jG2(1, 1) = 1
(3.1)
g(Q, +.I = 1.
(3.2)
and
Equation (3.1) is the definition of the axial vector renormalisation (GA/G”). So we find (GA 1 Gy)2 = $G2(1, 1) = 1, or (GA I GY) = ztl. Case 2:
(I’, 0
constant
(3.3)
= (4, !d + (3, $1.
We will now consider the case where the first two of the above set of states contribute. Then taking the matrix element of Eq. (2.1) between (4 +) states and using Eq. (2.15) we get the following sum rules, for different combinations of I and J to have the right-hand side of Eq. (2.15) nonvanishing. These are (a) I = 1, J = 0; W2(1, 1) + +G2(1, 2) = 1,
(3.4)
+G2(1, 1) - QG2(1, 2) = g(+, 4).
(3.5)
and (b)I=O,J=
1;
We see that there are two equations-(3.4) and (3.5)-having three unknowns. So to get a solution for (GA/G,) we must consider some more sum rules in this case. We, therefore, now take the matrix element of the commutation relation between
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(4 4) and (4 $) states with the above set of intermediate we obtain the sum rules
states. Then from Eq. (2.15)
(a) I = 0,J = 1; W’~>
G(l, 2) W,
1) + (l/2/3) ~‘5 G(1,2) G(2,2) = g($, 8).
(3.6)
(b) I = 1, J = 2; 1.6 G(1, 1) G(l, 2) + G(2,2) G&2)
= 0.
(3.7)
Equations (3.4)-(3.7) together do not give solutions because these are four equations having five unknowns. So we now proceed to take the case where (I .Z) = (Z”.Z”) = (Q Q). Equation (2.15) then yields the following sum rules (a) I = 1, J = 0;
(b)f=O,J=
$G2(1, 2) + +G2(2, 2) = 1.
(3.8)
+G2(1, 2) + &G2(2, 2) = g($, #).
(3.9)
1;
(c) 1 = 1, .7 = 2; 5G2(1, 2) - 8G2(2, 2) = 0.
(3.10)
From Eqs. (3.4)-(3.10) we find (GA 1Gv)2 = +G2(1, 1) = + ;
(3.11)
so
(3.12) and G2(1, 2) = $,
(3.13a)
G2(2, 2) = +g ,
(3.13b)
g(3, 3) = - 5 ,
(3.13c)
Ed% 4) = + s ,
(3.13d)
g(& ;> = -(gy2.
(3.13e)
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PATTNAIK
Case 3:
(I’J’) = &I,*) + (8, 4).
On taking (N) = (Z”.Z”) = (4 4) an d inserting the set of intermediate (4 $) and (3 +), Eq. (2.15) gives the sum rules given below.
states
(a) I = 1, J = 0; $G2(1, 1) - $G2(1, 3) = 1.
(3.14)
(b) f = 0, .I = 1; this leads to (3.15)
+G2(1, 1) + +G2(1, 3) = g($, 3). While taking (Z.Z) = (4 4) and (Z”J”) = (8 Q) and introducing states we get from Eq. (2.15) the sum rules
the same intermediate
(a) I = 1, J = 0; G(1, 1) G(1, 3) - 6
G(1, 3) G(3, 3) = 0,
(3.16)
and (b) f = 2,J = 1; lOG(1, 1) G(l, 3) + 245 G(1, 3) G(3, 3) = 0.
(3.17)
Unlike the previous cases we also find, as in Ref. (5), examining Eqs. (3.14)-(3.17) that this set of states give no solution. Case 4:
(I’J’) = ($, 4) + (8, Q).
If we take (ZJ) = (Z”J”) = (4 4) and use the intermediate states (i 4) and (Q 8) then we get from Eq. (2.15) the sum rules satisfied by the following reduced matrix elements. These are (a) I = I,3
= 0; $G2(1, 1) - $G2(1, 4) = 1.
(3.18)
$G2(1, 1) - $G2(1, 4) = g(&, 4).
(3.19)
(b) I = 0, J = 1;
Upon taking the matrix element of Eq. (2.1) between (8 4) and (3 $) states (i.e., ZJ = 4 4 and Z”.Z” = $3) and introducing the same intermediate states (* +) and (Q 9) we obtain from Eq. (2.15) i = 1, i = 2,
J=O J = 1 the sum rule I
(3.20)
G(l, 4) G(l, 1) - G(l, 4) G(4,4) = 0.
(3.21)
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Then we consider (ZJ) = (Z”J”) = ($ Q). Using the above two intermediate from Eq. (2.15) we obtain the sum rules
111 states
(a) I = 1, J = 0; $G2( 1, 4) + &,G”(4, 4) = 1.
(3.22)
&G2(1 7 4) + &G2(4, 4) = g(g, 8).
(3.23)
(b) 1 = 0, J = 1;
(c) I = 1, J = 2; 25G2( I, 4) - 32G2(4, 4) solving Eqs. (3.18)-(3.24)
(3.24)
we get (GA 1 Gv)2 = $G2(1, 1) := 25 9 7
or fQ,
(3.25)
G2(1, 4) = Q ,
(3.26a)
G2(4, 4) = IQ ,
(3.26b)
(GAIGY)= and
s(Q, 4) = g(% $1 = 1.
(3.26~)
Equation (3.25) is the SU(6) result (6) and has also been obtained in Ref. (5). Of course neither the SU(6) nor our procedure can calculate the sign of GA/G”. Case 5:
(rl.7) = (4, c$) + (+, 8) + ($, 4) + (3, 8).
We now consider the case where the above set of four states contribute. This case has been considered by the present author in a previous paper (9) following Ryan’s procedure and has given a value of GA/G” = 1.22. We have shown in the above four cases that the present approach has reproduced the results obtained in Ref. (5). So from previous considerations one must also expect here to get the same value for GA/GY (9). Let us examine this. As usual we must first take (N) = (Z”J”) = (+ i) and using the above set of four intermediate states we get from Eq. (2. IS) (a) I = 1, J = 0; $G2(1, 1) + $G2(1, 2) -
$G2(1, 3) - -G2(1, 4) = 1.
(3.27)
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PATTNAIK
(b) i: = 0, J = 1; QG2(1, 1) - QG2(1, 2) + +G2(1, 3) - $G2(1, 4) = g(&, 4). Next we take (ZJ) = (4 Q) and (Z”J”) = (Q, 3). With the same intermediate as above, Eq. (2.15) yields (a)I=2,J= 5G(1, 1) G&4)
(3.28) states
1; - 545 G&2)
G(2,4) + d/s G(l, 3) G(3,4) - 5G(1, 4) G(4,4) = 0. (3.29)
(b) I = 1, J = 2; 5G(l, 1) G(l, 4) + d/s G(l, 2) G(2,4) - 5d/5 G(l, 3) G(3,4) - 5G(l, 4) G(4,4) = 0. (3.30) On taking now (ZJ) = (4 8) and (Z”J”) = ($ &) and the same intermediate we obtain the following sum rules from Eq. (2.15):
states
(a) I = 2, J = 1; 5G(1, 2) G(l, 3) + 5(101j2) G(2, 2) G(2, 3) - (10’12) G(2, 3) G(3, 3) - 5G(2, 4) G(3, 4) = 0.
(3.31)
(b) I = 1,3 = 2; 5G(1, 2) G(l, 3) - ( 101/2) G(2,2) G(2, 3) + 5(10112)G(2, 3) G(3, 3) - 5G(2,4) G(3,4) = 0.
(3.32)
Further, we take (ZJ) = (4 $) and (Z”J”) = ($8) and truncate the intermediate states just after (& &), (& $), (8 a), and ($ Q) states. From Eq. (2.15), we find for (a)I= y.0
l,J=O
the sum rule
G(l, 2) G(l, 4) - 22/2 G(2,2) G(2,4) + 22/5 G(2,3) G(3,4) + 2(10112)G(2,4) G(4,4) = 0.
(3.33)
(b) I = 2, J = 1 252/2 G( 1,2) G( 1,4) + 202/z G(2,2) G(2,4) - 102/5 G(2,3) G(3,4) + 4(10112)G(2,4) G(4,4) = 0. (C)f=
l,J=2
(3.34)
the sum rule
- 52/2 G(1,2) G( 1,4) + 82/2 G(2,2) G(2,4) - 101/s G(2,3) G(3,4) - 8(101/2) G(2,4) G(4,4) = 0. (3.35)
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Finally, to get a solution we have to take (IJ) = (I”J”) = (8 4). Truncating at the above set of intermediate states (Z’J’) we obtain from Eq. (2.15) the sum rules (a)I=
l,J=O +G’(lp 4) + $G2(2, 4) + &G”(3, 4) + &G”(4, 4) = 1.
(b)j=O,J= WV,
(3.36)
I; 4) + &G2(2, 4) + +G2(3, 4) + &G2(4, 4) = g($, g).
(3.37)
(c) I = 2, J = 1; 25G2(1, 4) + 20G2(2, 4) - 40G2(3, 4) - 32G2(4, 4) = 0.
(3.38)
(d) 1 = 1, J = 2; 25G2(1, 4) - 40G2(2, 4) + 20G2(3, 4) - 32G2(4, 4) = 0. There are now thirteen equations-(3.27) we solve them and find
(3.39)
to (3.39)-and
twelve unknowns. So
(GA 1 Gy)2 = +G2(1, 1) = -& ;
(3.40)
(GA 1GY) FE i ++ s 1.22
(3.41)
so
and G2(1, 2) = G”(1, 3) = $+,
(3.42a)
G2(1 3 4) = @,
(3.42b)
G2(2, 4) = G”(3, 4) = #,
(3.42~)
G2(4, 4) = 2% ,
(3.42d)
G2(2, 2) = G2(3, 3) = #$,
(3.42e)
G”(2, 3) = 2% )
(3.42f)
g(3, $1 = g(#, 2) = 1.
w&J
In the present approach we also find for this case the value 1.22 for (G,,/GV). 595/47/I-g
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PATTNAIK
We have also calculated two further cases and are going to give the value of (GA/GV) for these cases without giving the detailed calculations. These are Case 6:
(Z’J’) = ($, $) + (Q, 8) + (& $).
(GA I G,) = i $. Case 7:
(Z’J’) = (B, it) + (+, 8) + (%, 4) + (2,s) + (I, #) + (2,9. (G, / G,) = f L93.
(3.43) + (9, $1 + (S, S) + (4, Q) (3.44)
The calculation is very similar to the previous cases. One has to first find a set of coupled equations from (2.15) by suitable choice of the initial and final states, which when solved gives the above values for (GA/G,). The values of other reduced matrix elements are also obtained from these equations but they are not so important here. IV.
DISCUSSIONS
In our analysis we have studied in what manner (G,/G,) varies with truncation. We find that, as one goes from one representation to another, certain regularities in the value of (GJG,) appear in different directions. For example, as one goes from (TJ’) = (4 &) to (+ 4) + (% 3) to (B Q) + ($3) + (+ 3) and so on, the value of (GA/GY) changes from 1 = $ to Q to z, and so on, respectively. This regularity has been observed also in Ref. (5). Another regularity which has not been observed before but appears here is that, as one goes from (I’J’) = ($ 4) to (+ &) + (3 $) + (8 4) + (# +) which are allowed when the matrix element of the commutator [&, A,01 is taken between (4 3) states) to ($4) + (8 8) + (3 3) + (4 3) + (4 9 + (3 4) + (3 Q> + (8 $9 + @ $1 ( w h ic h are allowed when the matrix element of the commutator is taken between (8 Q) states) and likewise, the value of (GA/Gy) changes from (GA/G,) = 1 = $ to J& to 9, and so on, respectively. Thus we find that (GA/GV) depends very much on truncation. However, what one can say in its favor is that our approach is very general and straightforward and reproduces the results of complicated higher symmetry groups. In addition to this, it is seen that by choosing the states (Z’J’) = (4 4) + (4 8) + (# Q) + (Q g), our method yields a value (GA/GY) = 1.22 which is about 4 % above the experimental value (GA/G,) = 1.18). It may be worth remarking here that in static Chew-Low theory with one pion approximation these are the states allowed, and higher states are ignored. It may be due to this fact that with this set of states we get a value of (G,/GV) close to the experimental value.
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115
ACKNOWLEDGMENT
The author is extremely grateful to Professor T. Pradhan for suggesting this investigation and for many stimulating and helpful discussions. RECEIVED: September 6, 1967
REFERENCES 1. M. GELL-MANN, Phys. Rev. 125, 1067 (1962). 2. B. W. LEE, Phys. Rev. Letters 14,673 (1965); C. RYAN, Phys. Rev. 140, B480 (1965); S. OKUBO, Phys. Letters 17, 172 (1965); R. F. DASHEN AND M. GELL-MANN, ibid. 17, 145 (1965): E. C. G. SUDERSHAN, ibid. 14, 1057 (1965). 3. S. L. ADLER, Phys. Rev. Letters 14, 1051 (1965); Phys. Rev. 140, B736 (1965). 4. W. I. WEISBERGER, Phys. Rev. Letters 14, 1047 (1965); Phys. Rev. 143, 1302 (1966). 5. C. RYAN, Ann. Phys. (N. Y.) 38, 1 (1966). 6. F. GURSEY, A. PAIS, AND L. RADICATI, Phys. Rev. Letters 13, 299 (1964). 7. G. RACAH, Phys. Rev. 62, 438 (1942). 8. M. E. ROSE, “Elementary theory of Angular Momentum.” Wiley, London, 1957. 9. MAMATA PAITNAIK, Ann. Phys. (N. Y.) 46, 317 (1968).