Instructor Solutions Manual
Symmetry, Broken Symmetry, and Topology
in Modern Physics
Mike Guidry and Yang Sun
This document gives the solutions for all 344 problems found at the ends of
chapters for the first edition of Symmetry, Broken Symmetry, and Topology
in Modern Physics: A First Course by Mike Guidry and Yang Sun (Cam-
bridge University Press, 2022). Unless otherwise indicated, literature refer-
ences, equation numbers, figure references, table references, and chapter or
section numbers refer to the print version of that book. The occasional refer-
ence to figures or tables contained specifically in this document (rather than
the book) will be indicated explicitly by a tag “[this document]”.
,
,1 Introduction
No problems for this introductory chapter.
1
, 2 Some Properties of Groups
2.1 Suppose that there are two group identities, e and e′ . Then e = e · e′ = e′ . Thus there is
only one group identity. Suppose that a−1 and ā are both inverses of the group element a.
Then using that a · a−1 and a · ā must both be equal to the group identity,
ā = ā · (a · a−1) = (ā · a) · a−1 = ea−1 = a−1 .
Thus each group element has a unique inverse. From the properties of the inverse and
identity, (a · b)−1 · (a · b) = e. Multiply this expression from the right by b−1 and use b ·
b−1 = e to give
(a · b)−1 · a = b−1 .
Multiply from the right by a−1 and use a · a−1 = e to give
(a · b)−1 = b−1 · a−1.
Check this result against entries in Table 2.2: Let a = (12) and b = (23). Then from the
table a−1 = (12) and b−1 = (23), and also from the table
a · b = (12) · (23) = (321).
Thus from the table (a · b)−1 = (321)−1 = (123). From the formula derived above and the
table
(a · b)−1 = b−1 a−1 = (23) · (12) = (123).
Hence for this example (a · b)−1 = b−1 a−1 , as asserted.
2.2 Define the permutations
123 123 123
()= (12) = (23) =
123 213 132
123 123 123
(13) = (123) = (321) =
321 231 312
so that, for example, (12){abc} → {bac}. Carrying out all possible combinations, the mul-
tiplication table for A · B is
A\B e (12) (23) (13) (123) (321)
e e (12) (23) (13) (123) (321)
(12) (12) e (321) (123) (13) (23)
(23) (23) (123) e (321) (12) (13)
(13) (13) (321) (123) e (23) (12)
(123) (123) (23) (13) (12) (321) e
(321) (321) (13) (12) (23) e (123)
2
Symmetry, Broken Symmetry, and Topology
in Modern Physics
Mike Guidry and Yang Sun
This document gives the solutions for all 344 problems found at the ends of
chapters for the first edition of Symmetry, Broken Symmetry, and Topology
in Modern Physics: A First Course by Mike Guidry and Yang Sun (Cam-
bridge University Press, 2022). Unless otherwise indicated, literature refer-
ences, equation numbers, figure references, table references, and chapter or
section numbers refer to the print version of that book. The occasional refer-
ence to figures or tables contained specifically in this document (rather than
the book) will be indicated explicitly by a tag “[this document]”.
,
,1 Introduction
No problems for this introductory chapter.
1
, 2 Some Properties of Groups
2.1 Suppose that there are two group identities, e and e′ . Then e = e · e′ = e′ . Thus there is
only one group identity. Suppose that a−1 and ā are both inverses of the group element a.
Then using that a · a−1 and a · ā must both be equal to the group identity,
ā = ā · (a · a−1) = (ā · a) · a−1 = ea−1 = a−1 .
Thus each group element has a unique inverse. From the properties of the inverse and
identity, (a · b)−1 · (a · b) = e. Multiply this expression from the right by b−1 and use b ·
b−1 = e to give
(a · b)−1 · a = b−1 .
Multiply from the right by a−1 and use a · a−1 = e to give
(a · b)−1 = b−1 · a−1.
Check this result against entries in Table 2.2: Let a = (12) and b = (23). Then from the
table a−1 = (12) and b−1 = (23), and also from the table
a · b = (12) · (23) = (321).
Thus from the table (a · b)−1 = (321)−1 = (123). From the formula derived above and the
table
(a · b)−1 = b−1 a−1 = (23) · (12) = (123).
Hence for this example (a · b)−1 = b−1 a−1 , as asserted.
2.2 Define the permutations
123 123 123
()= (12) = (23) =
123 213 132
123 123 123
(13) = (123) = (321) =
321 231 312
so that, for example, (12){abc} → {bac}. Carrying out all possible combinations, the mul-
tiplication table for A · B is
A\B e (12) (23) (13) (123) (321)
e e (12) (23) (13) (123) (321)
(12) (12) e (321) (123) (13) (23)
(23) (23) (123) e (321) (12) (13)
(13) (13) (321) (123) e (23) (12)
(123) (123) (23) (13) (12) (321) e
(321) (321) (13) (12) (23) e (123)
2
