出版社: Oxford University Press, USA
出版年: 19880107
页数: 548
定价: USD 77.95
装帧: Paperback
ISBN: 9780198519614
内容简介 · · · · · ·
This is a practical introduction to the principal ideas in gauge theory and their applications to elementary particle physics. It explains technique and methodology with simple exposition backed up by many illustrative examples. Derivations, some of well known results, are presented in sufficient detail to make the text accessible to readers entering the field for the first tim...
This is a practical introduction to the principal ideas in gauge theory and their applications to elementary particle physics. It explains technique and methodology with simple exposition backed up by many illustrative examples. Derivations, some of well known results, are presented in sufficient detail to make the text accessible to readers entering the field for the first time. The book focuses on the strong interaction theory of quantum chromodynamics and the electroweak interaction theory of Glashow, Weinberg, and Salam, as well as the grand unification theory, exemplified by the simplest SU(5) model. Not intended as an exhaustive survey, the book nevertheless provides the general background necessary for a serious student who wishes to specialize in the field of elementary particle theory. Physicists with an interest in general aspects of gauge theory will also find the book highly useful.
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驰生武🇨🇳 (lithromantic)
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets  isospin invariance  SU(2) symmetry the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons p and n form an isospin doublet isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k strong interraction does not dis...20190118 14:58
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets

isospin invariance  SU(2) symmetry
the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons
p and n form an isospin doublet
isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k
strong interraction does not distinguish n from p means [T_i, H_s]=0 where H_s is the strong interaction Hamiltonlian
and the isospin symmetry is not an exact symmetry because n and p are not strictly degnerate in mass. H=H_0+H_I, where H_I is the symmetry breaking part H_0>>H_I

SU(3) symmtrey and the quark model
Λ and K decay with a long life time> postulated these particles possessed a new additive quantum number, strangeness S, conserved in strong interaction, violated via weak interaction
Q=T_3+Y/2 with Y=B+S
SU(2)_T * U(1)_Y as a subgroup
SU(3) symmetry ? 0 1 and 1/2+ baryons (p=q=1) octet representation, 3/2+ (p=3,q=0)decuplete representation
but the eight fold way scheme is not identified with any known particles
quark model
all hadrons are built out of spin 1/2 quarks which transform as members of the fundamental representation(p=1,q=0) of SU(3)
(1) 3 flavous of quarks up down and strange
(2)mesons (B=0) are q q^ bound states 3 * 3*=1+8
(2)baryons(B=1) are qqq bound states 3 * 3 * 3=1+8+8+10, the octet parts have the same quantum number(T_3,Y) as the octet mesons, because T_3and Y are generators of SU(3) group and their eigenvalues for a given representation are uniquely defined

The GellmannOkubo mass formulas
回应 20190118 14:58 
驰生武🇨🇳 (lithromantic)
transformation law of tenors 简单的张量变换  irreducible representation and Young Tableaux first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with finint...20190116 10:44
transformation law of tenors
简单的张量变换

irreducible representation and Young Tableaux
first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with fininte permutation symmetry among its indices.
ψ^ij, P_12ψ^ij=ψ^ji; S^ij=1/2(ψ^ij+ψ^ji), A^ij=1/2(ψ^ijψ^ji); P_12 S^ij=S^ij, P_12 A^ij= A^ij
so we link the Young tableau with representation of SU(n)
rank f, we have f boxes to arrange, and length of rows should not increase from top to bottom f_1>=f_2>=...., and f_1+f_2+...=f, each box has an index i_k. ψ_{i_1,i_2,...,i_f1,i_f1+1,....,i_f1+f2....}
simplest case, symmetric and antisymmetric
dimensionality of irreducible representations
λ_1=f_1f_2, λ_2=f_2f_3,..., λ_n1=f_n1
d(λ_1,λ_2,...,λ_n1)=(1+λ_1)(1+λ_2)...(1+λ_n1)(1+(λ_1+λ_2)/2)(1+(λ_2+λ_3)/2)...(1+(λ_n2+λ_n1)/2) (1+(λ_1+λ_2+λ_3)/3)... (1+(λ_n3+λ_n2+λ_n1)/3)...(1+(λ_1+λ_2+...+λ_n1)/(n1))
example: SU(2), we have only 1 row d(λ_1)=1+λ_1, λ_1=2j
SU(3),2 row, d(λ_1,λ_2)=(1+λ_1)(1+λ_2))(1+(λ_1+λ_2)/2), λ_1=p,λ_2=q
for large n d=Π (n+D_i)/h_i, where h_i is the hook number and D_i is defined to be the number of sateps going from the box in the upper lefthand corner of the tableau to the ith box with each step toeards the right counted as +1 and each downward step as 1.

reduction of the product representation
1. in the tableau for the first factor, assign the same symbol, say a to all the boxes in the first row, the symbol b to all the boxes in the second row, etc.
2. attach boxes labelled by the symbol a to the tableau of the second factor in all possible ways, subject to the rules that no two a's appear in the same colume and that the resultant graph is still a Young tableau
3. after all symbols have been added to the tableau, these added symbols are then read from right to left in the first row
example SU(3)

group generators in tensor notation
(1)Hermitian and real generator matrices
any n*n unitary matrix U can be written in the form U=exp(iH), where H is hermitian and traceless. we can choose the group parameter to be real if ε_a=ε_a*, the H=ε_a λ_a, where λ_a are n*n hermitian matrices, their commutation relation [λ_a/2,λ_b/2]=i f_{abc} λ_c/2, f_{abc}is the structure const, and we can have λ_a/2=F_a for simplification.
for tensor method we can write the hermitian matrix H as H^j_i=ε^α_β(W^α_β)^j_i , indices from 1 to n
we can choose to have real generator matrices (W^β_α)^j_i=δ_αi δ^βj 1/n δ^β_α δ^j_i
and we can expand an arbitsry hermitian matrix M, M=m0 I_n+\sum ^{n^21} m_a λ^a
(2)real generators in vector representation
for the defining vector representation, W^β_α has a nonzero element only at α row and β column
infinitesimal SU(n) transformation on basis ψ^i > ψ'^i=ψ^i + δψ^i with δψ^i=ε^α_β(W^β_α)^ι_j ψ^j
(F^β_α ψ)^i= δ^i_α ψ^β, which means F^β_α takes the αth component to ψ and turn it into the βth component, zero for other components
(3) real generators on higher rank tensors
F ψ^..._...= \sum W ψ^..._. m .\sumW ψ^. n ._...
应该是可以简化特征值的计算
example
real generator for SU(2)
giving eigenvalues d(ψ_1)=1/2 and d(ψ_2)=1/2 for two states in ψ_i
and we can read off the quantum numbers for the triplet states as
d(ψ_11)=1/2+1/2=1, d(ψ_2)=1/21/2=0, d(ψ_22)=1/21/2=1 W^1_2=(0,0;1,0)
回应 20190116 10:44

驰生武🇨🇳 (lithromantic)
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets  isospin invariance  SU(2) symmetry the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons p and n form an isospin doublet isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k strong interraction does not dis...20190118 14:58
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets

isospin invariance  SU(2) symmetry
the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons
p and n form an isospin doublet
isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k
strong interraction does not distinguish n from p means [T_i, H_s]=0 where H_s is the strong interaction Hamiltonlian
and the isospin symmetry is not an exact symmetry because n and p are not strictly degnerate in mass. H=H_0+H_I, where H_I is the symmetry breaking part H_0>>H_I

SU(3) symmtrey and the quark model
Λ and K decay with a long life time> postulated these particles possessed a new additive quantum number, strangeness S, conserved in strong interaction, violated via weak interaction
Q=T_3+Y/2 with Y=B+S
SU(2)_T * U(1)_Y as a subgroup
SU(3) symmetry ? 0 1 and 1/2+ baryons (p=q=1) octet representation, 3/2+ (p=3,q=0)decuplete representation
but the eight fold way scheme is not identified with any known particles
quark model
all hadrons are built out of spin 1/2 quarks which transform as members of the fundamental representation(p=1,q=0) of SU(3)
(1) 3 flavous of quarks up down and strange
(2)mesons (B=0) are q q^ bound states 3 * 3*=1+8
(2)baryons(B=1) are qqq bound states 3 * 3 * 3=1+8+8+10, the octet parts have the same quantum number(T_3,Y) as the octet mesons, because T_3and Y are generators of SU(3) group and their eigenvalues for a given representation are uniquely defined

The GellmannOkubo mass formulas
回应 20190118 14:58 
驰生武🇨🇳 (lithromantic)
transformation law of tenors 简单的张量变换  irreducible representation and Young Tableaux first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with finint...20190116 10:44
transformation law of tenors
简单的张量变换

irreducible representation and Young Tableaux
first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with fininte permutation symmetry among its indices.
ψ^ij, P_12ψ^ij=ψ^ji; S^ij=1/2(ψ^ij+ψ^ji), A^ij=1/2(ψ^ijψ^ji); P_12 S^ij=S^ij, P_12 A^ij= A^ij
so we link the Young tableau with representation of SU(n)
rank f, we have f boxes to arrange, and length of rows should not increase from top to bottom f_1>=f_2>=...., and f_1+f_2+...=f, each box has an index i_k. ψ_{i_1,i_2,...,i_f1,i_f1+1,....,i_f1+f2....}
simplest case, symmetric and antisymmetric
dimensionality of irreducible representations
λ_1=f_1f_2, λ_2=f_2f_3,..., λ_n1=f_n1
d(λ_1,λ_2,...,λ_n1)=(1+λ_1)(1+λ_2)...(1+λ_n1)(1+(λ_1+λ_2)/2)(1+(λ_2+λ_3)/2)...(1+(λ_n2+λ_n1)/2) (1+(λ_1+λ_2+λ_3)/3)... (1+(λ_n3+λ_n2+λ_n1)/3)...(1+(λ_1+λ_2+...+λ_n1)/(n1))
example: SU(2), we have only 1 row d(λ_1)=1+λ_1, λ_1=2j
SU(3),2 row, d(λ_1,λ_2)=(1+λ_1)(1+λ_2))(1+(λ_1+λ_2)/2), λ_1=p,λ_2=q
for large n d=Π (n+D_i)/h_i, where h_i is the hook number and D_i is defined to be the number of sateps going from the box in the upper lefthand corner of the tableau to the ith box with each step toeards the right counted as +1 and each downward step as 1.

reduction of the product representation
1. in the tableau for the first factor, assign the same symbol, say a to all the boxes in the first row, the symbol b to all the boxes in the second row, etc.
2. attach boxes labelled by the symbol a to the tableau of the second factor in all possible ways, subject to the rules that no two a's appear in the same colume and that the resultant graph is still a Young tableau
3. after all symbols have been added to the tableau, these added symbols are then read from right to left in the first row
example SU(3)

group generators in tensor notation
(1)Hermitian and real generator matrices
any n*n unitary matrix U can be written in the form U=exp(iH), where H is hermitian and traceless. we can choose the group parameter to be real if ε_a=ε_a*, the H=ε_a λ_a, where λ_a are n*n hermitian matrices, their commutation relation [λ_a/2,λ_b/2]=i f_{abc} λ_c/2, f_{abc}is the structure const, and we can have λ_a/2=F_a for simplification.
for tensor method we can write the hermitian matrix H as H^j_i=ε^α_β(W^α_β)^j_i , indices from 1 to n
we can choose to have real generator matrices (W^β_α)^j_i=δ_αi δ^βj 1/n δ^β_α δ^j_i
and we can expand an arbitsry hermitian matrix M, M=m0 I_n+\sum ^{n^21} m_a λ^a
(2)real generators in vector representation
for the defining vector representation, W^β_α has a nonzero element only at α row and β column
infinitesimal SU(n) transformation on basis ψ^i > ψ'^i=ψ^i + δψ^i with δψ^i=ε^α_β(W^β_α)^ι_j ψ^j
(F^β_α ψ)^i= δ^i_α ψ^β, which means F^β_α takes the αth component to ψ and turn it into the βth component, zero for other components
(3) real generators on higher rank tensors
F ψ^..._...= \sum W ψ^..._. m .\sumW ψ^. n ._...
应该是可以简化特征值的计算
example
real generator for SU(2)
giving eigenvalues d(ψ_1)=1/2 and d(ψ_2)=1/2 for two states in ψ_i
and we can read off the quantum numbers for the triplet states as
d(ψ_11)=1/2+1/2=1, d(ψ_2)=1/21/2=0, d(ψ_22)=1/21/2=1 W^1_2=(0,0;1,0)
回应 20190116 10:44

驰生武🇨🇳 (lithromantic)
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets  isospin invariance  SU(2) symmetry the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons p and n form an isospin doublet isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k strong interraction does not dis...20190118 14:58
在某个变换下不变（具有某种对称性，该变换和体系的哈密顿量对易）>简并度相关，multiplets

isospin invariance  SU(2) symmetry
the nuclear forces(strong interactions) are independent of the electric charge (U(1) symmetry?) carried by nucleons
p and n form an isospin doublet
isospin generator T_i, satisfy the Lie algebra of SU(2) [T_i,T_j]=i ε_{ijk}Τ_k
strong interraction does not distinguish n from p means [T_i, H_s]=0 where H_s is the strong interaction Hamiltonlian
and the isospin symmetry is not an exact symmetry because n and p are not strictly degnerate in mass. H=H_0+H_I, where H_I is the symmetry breaking part H_0>>H_I

SU(3) symmtrey and the quark model
Λ and K decay with a long life time> postulated these particles possessed a new additive quantum number, strangeness S, conserved in strong interaction, violated via weak interaction
Q=T_3+Y/2 with Y=B+S
SU(2)_T * U(1)_Y as a subgroup
SU(3) symmetry ? 0 1 and 1/2+ baryons (p=q=1) octet representation, 3/2+ (p=3,q=0)decuplete representation
but the eight fold way scheme is not identified with any known particles
quark model
all hadrons are built out of spin 1/2 quarks which transform as members of the fundamental representation(p=1,q=0) of SU(3)
(1) 3 flavous of quarks up down and strange
(2)mesons (B=0) are q q^ bound states 3 * 3*=1+8
(2)baryons(B=1) are qqq bound states 3 * 3 * 3=1+8+8+10, the octet parts have the same quantum number(T_3,Y) as the octet mesons, because T_3and Y are generators of SU(3) group and their eigenvalues for a given representation are uniquely defined

The GellmannOkubo mass formulas
回应 20190118 14:58 
驰生武🇨🇳 (lithromantic)
transformation law of tenors 简单的张量变换  irreducible representation and Young Tableaux first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with finint...20190116 10:44
transformation law of tenors
简单的张量变换

irreducible representation and Young Tableaux
first take tank2 tensor as an example, that it can be decomposed in to symmetric and antismmtric part, and they do not mix with each other. => it can be generalized to tensors of higher rank with the result that the basis for irreducible representations of SU(n) correspond to tensors with fininte permutation symmetry among its indices.
ψ^ij, P_12ψ^ij=ψ^ji; S^ij=1/2(ψ^ij+ψ^ji), A^ij=1/2(ψ^ijψ^ji); P_12 S^ij=S^ij, P_12 A^ij= A^ij
so we link the Young tableau with representation of SU(n)
rank f, we have f boxes to arrange, and length of rows should not increase from top to bottom f_1>=f_2>=...., and f_1+f_2+...=f, each box has an index i_k. ψ_{i_1,i_2,...,i_f1,i_f1+1,....,i_f1+f2....}
simplest case, symmetric and antisymmetric
dimensionality of irreducible representations
λ_1=f_1f_2, λ_2=f_2f_3,..., λ_n1=f_n1
d(λ_1,λ_2,...,λ_n1)=(1+λ_1)(1+λ_2)...(1+λ_n1)(1+(λ_1+λ_2)/2)(1+(λ_2+λ_3)/2)...(1+(λ_n2+λ_n1)/2) (1+(λ_1+λ_2+λ_3)/3)... (1+(λ_n3+λ_n2+λ_n1)/3)...(1+(λ_1+λ_2+...+λ_n1)/(n1))
example: SU(2), we have only 1 row d(λ_1)=1+λ_1, λ_1=2j
SU(3),2 row, d(λ_1,λ_2)=(1+λ_1)(1+λ_2))(1+(λ_1+λ_2)/2), λ_1=p,λ_2=q
for large n d=Π (n+D_i)/h_i, where h_i is the hook number and D_i is defined to be the number of sateps going from the box in the upper lefthand corner of the tableau to the ith box with each step toeards the right counted as +1 and each downward step as 1.

reduction of the product representation
1. in the tableau for the first factor, assign the same symbol, say a to all the boxes in the first row, the symbol b to all the boxes in the second row, etc.
2. attach boxes labelled by the symbol a to the tableau of the second factor in all possible ways, subject to the rules that no two a's appear in the same colume and that the resultant graph is still a Young tableau
3. after all symbols have been added to the tableau, these added symbols are then read from right to left in the first row
example SU(3)

group generators in tensor notation
(1)Hermitian and real generator matrices
any n*n unitary matrix U can be written in the form U=exp(iH), where H is hermitian and traceless. we can choose the group parameter to be real if ε_a=ε_a*, the H=ε_a λ_a, where λ_a are n*n hermitian matrices, their commutation relation [λ_a/2,λ_b/2]=i f_{abc} λ_c/2, f_{abc}is the structure const, and we can have λ_a/2=F_a for simplification.
for tensor method we can write the hermitian matrix H as H^j_i=ε^α_β(W^α_β)^j_i , indices from 1 to n
we can choose to have real generator matrices (W^β_α)^j_i=δ_αi δ^βj 1/n δ^β_α δ^j_i
and we can expand an arbitsry hermitian matrix M, M=m0 I_n+\sum ^{n^21} m_a λ^a
(2)real generators in vector representation
for the defining vector representation, W^β_α has a nonzero element only at α row and β column
infinitesimal SU(n) transformation on basis ψ^i > ψ'^i=ψ^i + δψ^i with δψ^i=ε^α_β(W^β_α)^ι_j ψ^j
(F^β_α ψ)^i= δ^i_α ψ^β, which means F^β_α takes the αth component to ψ and turn it into the βth component, zero for other components
(3) real generators on higher rank tensors
F ψ^..._...= \sum W ψ^..._. m .\sumW ψ^. n ._...
应该是可以简化特征值的计算
example
real generator for SU(2)
giving eigenvalues d(ψ_1)=1/2 and d(ψ_2)=1/2 for two states in ψ_i
and we can read off the quantum numbers for the triplet states as
d(ψ_11)=1/2+1/2=1, d(ψ_2)=1/21/2=0, d(ψ_22)=1/21/2=1 W^1_2=(0,0;1,0)
回应 20190116 10:44
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订阅关于Gauge Theory of elementary particle physics的评论:
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0 有用 [已注销] 20161101
至少要學過QED和重整化，可以整理思路。
0 有用 Varagnac 20200601
是本出名的并且被广泛使用的书，但是实在太丑陋了，尤其磁单极和瞬子都弄的什么东西。
0 有用 Varagnac 20200601
是本出名的并且被广泛使用的书，但是实在太丑陋了，尤其磁单极和瞬子都弄的什么东西。
0 有用 [已注销] 20161101
至少要學過QED和重整化，可以整理思路。