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Chiral symmetry breaking and the Cabibbo angle
Nuclear Physics BS0 (1972) 87-92. North-Holland Publishing Company
CHIRAL SYMMETRY BREAKING AND THE CABIBBO ANGLE P. DITTNER and S. ELIEZER Department of Physics, Imperial College, Prince Consort Road, London P.H. DONDI Department of Applied Mathematics and TheoreticalPhysics, Cambridge
Received 27 September 1972
Abstract: Following the suggestion of Oakes we consider the Cabibbo rotation in models of chiral-symmetry breaking. The m 2--* 0 limit implies exact SU(2) X SU(2) symmetry only. in the (3, ~) + (3, 3) model. Thus with other models we proceed either from Hamiland octet parts) giving tertians (with only smglet " . • m,r 2 = 0, or from Hamiltonians displaying exact SU(2) × SU(2) symmetry. We show that in the first case the results are independent of the chiral properties of the symmetry breaking and that the second method is inconsistent in the (8, 8) and (6, 6) + (6, 6) models.
It has been suggested by Oakes [1] that the SU(2) X SU(2) breaking of strong interactions arises from a Cabibbo rotation on an SU(2) X SU(2) invariant Hamiltonian, which relates m,r/m K to the Cabibbo angle. Oakes has worked within the model [2, 3] where the hadronic symmetry-breaking Hamiltonian transforms according to the (3, 3-) + (5, 3) representation of SU(3) X SU(3); starting from the SU(2) X SU(2) symmetric Hamiltonian H, -~ = H~ - e ( u o -
C-2u8),
(1)
where/-/s is the SU(3) X SU(3) symmetric part of the Hamiltonian and where u 0 and ui(i = 1 , . . . , 8) are the scalar components of the (3, 3-) + (5, 3) representation, and by applying a Cabibbo rotation he obtained
"= "s + "~ = ( e-~°°¥-~ e~°°Y)~s=-o = "s - ~(Uo - v~ sin20u3
(2) -,,/f(1
- {- sin E 0) Us),
Qv and QA being the vector and axial-vector generators of the group and 0 the Cabibbo angle. From this Hamfltonian the divergences of the axial vector currents can be calculated using
P. Dittner et aL, Chiralsymmetry breaking
88
(3)
~uA~ = -i[QiA, H] , which, together with the PCAC hypothesis
4.m2$ij = (Ol a~A~ lq~j ),
(4)
gives 2fm 2 sin20 =
+ O(e) =
m I2t
+ O(e).
(5)
Numerically this gives sin 0 = 0.27 (for fK = f~r) in reasonable agreement with experiment. (At this point we would like to mention [4] that if the vacuum is an SU(3) singlet in the chiral limit thenfK/f ~ = 1 + O(e) and thus in eq. (5) it is inconsistent to take fK =/=fTr). More recently, Genz et al. [5], have considered the connection between mTr and the Cabibbo angle assuming that the symmetry breaking Hamiltonian transforms as the (8, 8) representation [6-8] of SU(3) X SU(3). In this model the divergences of the axial currents contain 10 and 1-0-pieces as well as an octet piece so that m 2 = 0 does not imply exact SU(2) X SU(2) (i.e. ~uA~U#= 0 for i = 1, 2, 3). Indeed, for symmetry breaking from any (X, Y0 + (-~, X) representation [9] with X 4= 3, SU(2) X SU(2) does not become exact when m 2 ~ 0. Moreover, with only an SU(3) singlet and octet part in the breaking Hamiltonian it is possible to choose the parameters to make the pion massless, which corresponds to the octet part of the axial divergence vanishing ((auA~U)8 = 0 for i = 1, 2, 3) but in order to have the whole axial divergence zero one must include other SU(3) representations (for example the twenty-seven) in the Hamiltonian. Thus for models other than (3, 3) + (3, 3) one can apply Oakes' analysis in two ways, by applying the Cabibbo rotation to two different Hamiltonians: (i) A symmetry breaking Hamiltonian, with singlet and octet parts with parameters chosen to give m 2 = 0 or equivalently (auA~) 8 = 0 but with auA ~ =/=0 (for i =1,2,3). (ii) A Hamiltonian with singlet, octet and (say) twenty-seven parts with a A u = 0 for i = 1, 2, 3. In this case, of course, m2rrwill also be zero. Genz et al. [5] c~oose the first possibility and arrive at the same result as Oakes [1]. In this paper we shall show that the first method (i) is (to first order in the symmetry breaking) independent of the chiral representation chosen to describe the symmetry violation, so that the resulti of refs. [1, 5] are not peculiar to the models assumed there. Further investigation excludes possibility (ii) in both the (8, 8) and (6, 6) + (6, 6) [10] models of chiral symmetry breaking. (i) Let = S - co 0 8 ,
(6)
P. Dittner et al., Chiralsymmetry breaking
89
where S is a singlet and O a is the eighth component of an octet under SU(3). (Here we have absorbed the formal parameter e, which denotes the order of the symmetry breaking, into S and 08). Putting m 2 = 0 gives
(~rlSlrr> Co - ( ~r1081 ~r>"
(7)
Applying the Cabibbo rotation (which is an SU(3) rotation about the seventh axis) gives
H = (e -2iOQ~ He2i°Q~)zxs=_o=S-½v~c 0 sin2003-co(l -~sin20) O8 ,
(8)
which yields 2m 2 = (rr[ Hlrt> = (rr ISIrr> - Co(1 - ] - sin2 O) (triO8 I~r>, 2m 2 = m2o + m2+ = ( K I S I K ) - e0(1 - ~ sin 2 0)(KIO8IK).
(9) (10)
The Wigner-Eckart theorem gives (to first order in the symmetry breaking) OrlSlTr> = (KISIK>,
(~riO817r>= _2(KIO81K >.
(I I)
Eqs. (7), (9), (10) and (11) give m2 7r
2 sin20
m2
2 - sin20 '
i.e. sin20 _
2m 2 ~r + O(e), 2m 2 + m2r
(12)
which is Oakes' result eq. (5). (Note that this also substantiates our earlier statement that fK/fn = 1 + O(e)). We also remark that the sum rule 2 m 2K ° - - m K2 + = m ~r'
(13)
obtained in refs. [1, 5], is also independent of the chiral properties of the Hamiltonian. (ii) The (8, 8) and (6, 6) + (6, 6) representations of SU(3) X SU(3) can be written as an SU(3) 8 X 8 tensor Si/and two symmetric 8 X 8 tensors T~ respectively [6-8, 10]. Thus their vector transformations are
[Qv, SiTc]= if¢.pSpk + if&pSjp ,
(14)
[QV, Tlk] - " ± +i - %, r;,k
(15)
+ r:,, .
90
P. Dittner et al., Chiral symmetry breaking
Their commutations with the axial generators are of course different and are given by [QiA, s,k ],'= --if iip S pk + if ~pSjp ,
(16)
± = - { d iipr;k -v + d ~ v T~ v} . [~iA, r~l lp +Siflk J pq T:~ pq +S&d.lpq T,q
(17)
Each tensor can be decomposed into its SU(3) representations as follows:
%= s ,1 s
(sij + 5,)+
(sij-sji)=s
,l
+
,
= S$1) + Si/(8S) + Sij(27),
S eA = Si](8A) + Si](10)
+ Si] ( 1 0 ) ,
(18) (19) (2 0)
where
sij(1) = ~ 8ijSkk,
(21)
S~¢(8s) = ~ di1~dk,.mSlm,
(22)
_1
S#(8 A) - TfilkfklmSlm ,
(23)
and T~ are decomposed in the same way as S S. Using eq. (3) and requiring avA ~ = 0 for i/--" 1, 2, 3, the SU(2) X SU(2) symmetric Hamiltonians are obtained:
Hsb{(8, 8)}
H-sb {(6, 6) + (6, 6)}
= e($88(1) + ~ $88(8S) + $88(27))
(24)
= e(~ Skk -- ~ x/~astmSlm + $88(27)),
(24)'
=-g(rffs(1 ) + 2T~+8(8)+½ Tff8(27))
(25)
= ~ (',a L T kk + _ 2_ s "v/3dsva T ~+ + 1 T~8(27)) "
(25)'
The Cabibbo rotation on the "8" component of an octet and "88" component of a symmetric second rank SU(3) tensor gives e
_2/oov ~ 2/oov U81m~ime • = (1 _ a sin 2 O)dsimSi m +x/r~ sin 0 cos 0 d61mSlm
+ ½ X/~ sin 2 0 d3lmStm,
(26)
P. Dittner et al., Chiralsymmetry breaking
91
e-2i°QVs88e2i°QV = (1 - { sin 2 0)2S88 + 3 sin20 cos20 $66 + ¼ sin40S33 + 2x/~'(1 - ~ sin20) sin 0 cos 0 $86 + x/r3"sin20 (1 - ~ sin20)S83 + 3 sin30 cos 0S63.
(27)
Thus a Cabibbo rotation of the Hamiltonians in eqs. (24)' and (25)' yields Hsb((8, 8))
= ~ eSkk -- ~x/3-e (1 - ~ Sin2 O)dalmSlm+e(1-2a-sin20)2S88(27)
+...,
(28)
Hsb.{(6,6-) +(6,6)}= ale-T+ " 2O)d81mT~+ +~-g(1 -~sin20)2Tff8(27 ) kk --sZX/~'e'(1--~ sm
+ ....
(29)
If in general the SU(2) invariant (8, 8) and (6, 6-) + (6, 6) Hamiltonians are Hsb {(8, 8)}
= relSkk--X/r~Te8d81mSlm ~ + e27S88 (27)
Hsb ((6, 6) + (6, 6)} = ge I T~k -v~re8ds~T~ + e27 T88(27) --
--
1--
+
1
--
+
--
+
(30) (31)
we can use eqs. (3) and (4) to relate the symmetry breaking parameters to the masses of the pseudoscalar mesons: e8 - -
6 ( - 3 m 2 + 2m 2 + m 2) =
el
5(3m 2 + 4m 2K + m 2n)
e27
(m 2 _ 4m 2 + 3m 2)
el
(3m 2 +4m 2 + m 2)
e-8 el
=
0.658,
(32)
-- --0.046,
(33)
2 + m2~ 20(-3m2 + zmK n) 7(3m 2 +4m 2 + m 2)
e27
5(m2-4m2 + 3m2)-
Se
9(3m 2 + 4m 2 + m 2)
1.567
0.026.
(34)
(35)
92
P. Dittner et al., Chiral symmetry breaking
If the Hamiltonians (30) and (31) arise from a Cabibbo rotation as given in eqs. (28) and (29) then this implies, in each case, two equations for sin20 which in both models are not only inconsistent equations but lead to imaginary values for sin0 ! To summarise, in models other than (3, 3) + (5, 3) breaking, the starting point for the Cabibbo rotation is not unique, and one might proceed either from Hamiltonians giving m27r= 0 (i.e. the octet part of auA~ = 0 for i = 1, 2, 3) or from SU(2) X SU(2) symmetric Hamiltonians (i.e. the whole of auA~= 0 for i = 1, 2, 3). We have shown that in the first case the results are independent of the chiral properties of the symmetry breaking and that the second method is inconsistent in the (8, 8) and (6, ~ + (6, 6) models. This last result comes as no great surprise; the smallness of the pion mass does not indicate approximate SU(2) X SU(2) symmetry in these models and thus to a large extent one loses the motivation to relate the SU(2) X SU(2) symmetry breaking to the Cabibbo angle. Two of us (P. D. and P.H.D.) are grateful to the Science Research Council for financial support.
REFERENCES [1] [2] [3] [4]
R.J. Oakes, Phys. Letters, 29B (1969) 683. M. GeU-Mann,R.J. Oakes and B. Renner, Phys. Rev. 175 (1968) 2195. S. Glashow and S. Weinberg, Phys. Rev. Letters 20 (1968) 224. R. Dashen, Lectures on chiral symmetry breaking, presented at the 1971 Vaxenna Summer School. [5] H. Genz, J. Katz, L.R. Ram Mohan and S. Tatur, Chiral symmetry breaking and the Cabibbo angle, Berlin preprint. [6] K.J. Barnes and C.J. Isham, Nucl. Phys. B15 (1970) 333. [7] H. Genz and J. Katz, Nucl. Phys. B21 (1970) 333. [8] J.J. Brehm, Nucl. Phys. B34 (1971) 269. [9] R. Dashen, Phys. Rev. D3 (1971) 1879. [10] P. Dittner, S. Eliezer and P.H. Dondi, Nucl. Phys. B49 (1972) 242; P. Awil, SLAC-PUB-1078 preprint.
CHIRAL SYMMETRY BREAKING AND THE CABIBBO ANGLE P. DITTNER and S. ELIEZER Department of Physics, Imperial College, Prince Consort Road, London P.H. DONDI Department of Applied Mathematics and TheoreticalPhysics, Cambridge
Received 27 September 1972
Abstract: Following the suggestion of Oakes we consider the Cabibbo rotation in models of chiral-symmetry breaking. The m 2--* 0 limit implies exact SU(2) X SU(2) symmetry only. in the (3, ~) + (3, 3) model. Thus with other models we proceed either from Hamiland octet parts) giving tertians (with only smglet " . • m,r 2 = 0, or from Hamiltonians displaying exact SU(2) × SU(2) symmetry. We show that in the first case the results are independent of the chiral properties of the symmetry breaking and that the second method is inconsistent in the (8, 8) and (6, 6) + (6, 6) models.
It has been suggested by Oakes [1] that the SU(2) X SU(2) breaking of strong interactions arises from a Cabibbo rotation on an SU(2) X SU(2) invariant Hamiltonian, which relates m,r/m K to the Cabibbo angle. Oakes has worked within the model [2, 3] where the hadronic symmetry-breaking Hamiltonian transforms according to the (3, 3-) + (5, 3) representation of SU(3) X SU(3); starting from the SU(2) X SU(2) symmetric Hamiltonian H, -~ = H~ - e ( u o -
C-2u8),
(1)
where/-/s is the SU(3) X SU(3) symmetric part of the Hamiltonian and where u 0 and ui(i = 1 , . . . , 8) are the scalar components of the (3, 3-) + (5, 3) representation, and by applying a Cabibbo rotation he obtained
"= "s + "~ = ( e-~°°¥-~ e~°°Y)~s=-o = "s - ~(Uo - v~ sin20u3
(2) -,,/f(1
- {- sin E 0) Us),
Qv and QA being the vector and axial-vector generators of the group and 0 the Cabibbo angle. From this Hamfltonian the divergences of the axial vector currents can be calculated using
P. Dittner et aL, Chiralsymmetry breaking
88
(3)
~uA~ = -i[QiA, H] , which, together with the PCAC hypothesis
4.m2$ij = (Ol a~A~ lq~j ),
(4)
gives 2fm 2 sin20 =
+ O(e) =
m I2t
+ O(e).
(5)
Numerically this gives sin 0 = 0.27 (for fK = f~r) in reasonable agreement with experiment. (At this point we would like to mention [4] that if the vacuum is an SU(3) singlet in the chiral limit thenfK/f ~ = 1 + O(e) and thus in eq. (5) it is inconsistent to take fK =/=fTr). More recently, Genz et al. [5], have considered the connection between mTr and the Cabibbo angle assuming that the symmetry breaking Hamiltonian transforms as the (8, 8) representation [6-8] of SU(3) X SU(3). In this model the divergences of the axial currents contain 10 and 1-0-pieces as well as an octet piece so that m 2 = 0 does not imply exact SU(2) X SU(2) (i.e. ~uA~U#= 0 for i = 1, 2, 3). Indeed, for symmetry breaking from any (X, Y0 + (-~, X) representation [9] with X 4= 3, SU(2) X SU(2) does not become exact when m 2 ~ 0. Moreover, with only an SU(3) singlet and octet part in the breaking Hamiltonian it is possible to choose the parameters to make the pion massless, which corresponds to the octet part of the axial divergence vanishing ((auA~U)8 = 0 for i = 1, 2, 3) but in order to have the whole axial divergence zero one must include other SU(3) representations (for example the twenty-seven) in the Hamiltonian. Thus for models other than (3, 3) + (3, 3) one can apply Oakes' analysis in two ways, by applying the Cabibbo rotation to two different Hamiltonians: (i) A symmetry breaking Hamiltonian, with singlet and octet parts with parameters chosen to give m 2 = 0 or equivalently (auA~) 8 = 0 but with auA ~ =/=0 (for i =1,2,3). (ii) A Hamiltonian with singlet, octet and (say) twenty-seven parts with a A u = 0 for i = 1, 2, 3. In this case, of course, m2rrwill also be zero. Genz et al. [5] c~oose the first possibility and arrive at the same result as Oakes [1]. In this paper we shall show that the first method (i) is (to first order in the symmetry breaking) independent of the chiral representation chosen to describe the symmetry violation, so that the resulti of refs. [1, 5] are not peculiar to the models assumed there. Further investigation excludes possibility (ii) in both the (8, 8) and (6, 6) + (6, 6) [10] models of chiral symmetry breaking. (i) Let = S - co 0 8 ,
(6)
P. Dittner et al., Chiralsymmetry breaking
89
where S is a singlet and O a is the eighth component of an octet under SU(3). (Here we have absorbed the formal parameter e, which denotes the order of the symmetry breaking, into S and 08). Putting m 2 = 0 gives
(~rlSlrr> Co - ( ~r1081 ~r>"
(7)
Applying the Cabibbo rotation (which is an SU(3) rotation about the seventh axis) gives
H = (e -2iOQ~ He2i°Q~)zxs=_o=S-½v~c 0 sin2003-co(l -~sin20) O8 ,
(8)
which yields 2m 2 = (rr[ Hlrt> = (rr ISIrr> - Co(1 - ] - sin2 O) (triO8 I~r>, 2m 2 = m2o + m2+ = ( K I S I K ) - e0(1 - ~ sin 2 0)(KIO8IK).
(9) (10)
The Wigner-Eckart theorem gives (to first order in the symmetry breaking) OrlSlTr> = (KISIK>,
(~riO817r>= _2(KIO81K >.
(I I)
Eqs. (7), (9), (10) and (11) give m2 7r
2 sin20
m2
2 - sin20 '
i.e. sin20 _
2m 2 ~r + O(e), 2m 2 + m2r
(12)
which is Oakes' result eq. (5). (Note that this also substantiates our earlier statement that fK/fn = 1 + O(e)). We also remark that the sum rule 2 m 2K ° - - m K2 + = m ~r'
(13)
obtained in refs. [1, 5], is also independent of the chiral properties of the Hamiltonian. (ii) The (8, 8) and (6, 6) + (6, 6) representations of SU(3) X SU(3) can be written as an SU(3) 8 X 8 tensor Si/and two symmetric 8 X 8 tensors T~ respectively [6-8, 10]. Thus their vector transformations are
[Qv, SiTc]= if¢.pSpk + if&pSjp ,
(14)
[QV, Tlk] - " ± +i - %, r;,k
(15)
+ r:,, .
90
P. Dittner et al., Chiral symmetry breaking
Their commutations with the axial generators are of course different and are given by [QiA, s,k ],'= --if iip S pk + if ~pSjp ,
(16)
± = - { d iipr;k -v + d ~ v T~ v} . [~iA, r~l lp +Siflk J pq T:~ pq +S&d.lpq T,q
(17)
Each tensor can be decomposed into its SU(3) representations as follows:
%= s ,1 s
(sij + 5,)+
(sij-sji)=s
,l
+
,
= S$1) + Si/(8S) + Sij(27),
S eA = Si](8A) + Si](10)
+ Si] ( 1 0 ) ,
(18) (19) (2 0)
where
sij(1) = ~ 8ijSkk,
(21)
S~¢(8s) = ~ di1~dk,.mSlm,
(22)
_1
S#(8 A) - TfilkfklmSlm ,
(23)
and T~ are decomposed in the same way as S S. Using eq. (3) and requiring avA ~ = 0 for i/--" 1, 2, 3, the SU(2) X SU(2) symmetric Hamiltonians are obtained:
Hsb{(8, 8)}
H-sb {(6, 6) + (6, 6)}
= e($88(1) + ~ $88(8S) + $88(27))
(24)
= e(~ Skk -- ~ x/~astmSlm + $88(27)),
(24)'
=-g(rffs(1 ) + 2T~+8(8)+½ Tff8(27))
(25)
= ~ (',a L T kk + _ 2_ s "v/3dsva T ~+ + 1 T~8(27)) "
(25)'
The Cabibbo rotation on the "8" component of an octet and "88" component of a symmetric second rank SU(3) tensor gives e
_2/oov ~ 2/oov U81m~ime • = (1 _ a sin 2 O)dsimSi m +x/r~ sin 0 cos 0 d61mSlm
+ ½ X/~ sin 2 0 d3lmStm,
(26)
P. Dittner et al., Chiralsymmetry breaking
91
e-2i°QVs88e2i°QV = (1 - { sin 2 0)2S88 + 3 sin20 cos20 $66 + ¼ sin40S33 + 2x/~'(1 - ~ sin20) sin 0 cos 0 $86 + x/r3"sin20 (1 - ~ sin20)S83 + 3 sin30 cos 0S63.
(27)
Thus a Cabibbo rotation of the Hamiltonians in eqs. (24)' and (25)' yields Hsb((8, 8))
= ~ eSkk -- ~x/3-e (1 - ~ Sin2 O)dalmSlm+e(1-2a-sin20)2S88(27)
+...,
(28)
Hsb.{(6,6-) +(6,6)}= ale-T+ " 2O)d81mT~+ +~-g(1 -~sin20)2Tff8(27 ) kk --sZX/~'e'(1--~ sm
+ ....
(29)
If in general the SU(2) invariant (8, 8) and (6, 6-) + (6, 6) Hamiltonians are Hsb {(8, 8)}
= relSkk--X/r~Te8d81mSlm ~ + e27S88 (27)
Hsb ((6, 6) + (6, 6)} = ge I T~k -v~re8ds~T~ + e27 T88(27) --
--
1--
+
1
--
+
--
+
(30) (31)
we can use eqs. (3) and (4) to relate the symmetry breaking parameters to the masses of the pseudoscalar mesons: e8 - -
6 ( - 3 m 2 + 2m 2 + m 2) =
el
5(3m 2 + 4m 2K + m 2n)
e27
(m 2 _ 4m 2 + 3m 2)
el
(3m 2 +4m 2 + m 2)
e-8 el
=
0.658,
(32)
-- --0.046,
(33)
2 + m2~ 20(-3m2 + zmK n) 7(3m 2 +4m 2 + m 2)
e27
5(m2-4m2 + 3m2)-
Se
9(3m 2 + 4m 2 + m 2)
1.567
0.026.
(34)
(35)
92
P. Dittner et al., Chiral symmetry breaking
If the Hamiltonians (30) and (31) arise from a Cabibbo rotation as given in eqs. (28) and (29) then this implies, in each case, two equations for sin20 which in both models are not only inconsistent equations but lead to imaginary values for sin0 ! To summarise, in models other than (3, 3) + (5, 3) breaking, the starting point for the Cabibbo rotation is not unique, and one might proceed either from Hamiltonians giving m27r= 0 (i.e. the octet part of auA~ = 0 for i = 1, 2, 3) or from SU(2) X SU(2) symmetric Hamiltonians (i.e. the whole of auA~= 0 for i = 1, 2, 3). We have shown that in the first case the results are independent of the chiral properties of the symmetry breaking and that the second method is inconsistent in the (8, 8) and (6, ~ + (6, 6) models. This last result comes as no great surprise; the smallness of the pion mass does not indicate approximate SU(2) X SU(2) symmetry in these models and thus to a large extent one loses the motivation to relate the SU(2) X SU(2) symmetry breaking to the Cabibbo angle. Two of us (P. D. and P.H.D.) are grateful to the Science Research Council for financial support.
REFERENCES [1] [2] [3] [4]
R.J. Oakes, Phys. Letters, 29B (1969) 683. M. GeU-Mann,R.J. Oakes and B. Renner, Phys. Rev. 175 (1968) 2195. S. Glashow and S. Weinberg, Phys. Rev. Letters 20 (1968) 224. R. Dashen, Lectures on chiral symmetry breaking, presented at the 1971 Vaxenna Summer School. [5] H. Genz, J. Katz, L.R. Ram Mohan and S. Tatur, Chiral symmetry breaking and the Cabibbo angle, Berlin preprint. [6] K.J. Barnes and C.J. Isham, Nucl. Phys. B15 (1970) 333. [7] H. Genz and J. Katz, Nucl. Phys. B21 (1970) 333. [8] J.J. Brehm, Nucl. Phys. B34 (1971) 269. [9] R. Dashen, Phys. Rev. D3 (1971) 1879. [10] P. Dittner, S. Eliezer and P.H. Dondi, Nucl. Phys. B49 (1972) 242; P. Awil, SLAC-PUB-1078 preprint.