Physics 745 - Group Theory
Solution to Final, Spring 2009
You may use (1) class notes, (2) the text, (3) former homeworks and solutions (available
online), or (4) any math references, such as integral
Physics 745 - Group Theory
Final, Spring 2009
You may use (1) class notes, (2) the text, (3) former homeworks and solutions (available
online), or (4) any math references, such as integral tables, Map
Physics 745 - Group Theory
Solutions to Midterm Exam
1. Let us define three more matrices, F, G, and H, defined by
0 1
1 0
1 0
F =
, G =
, H =
1 0
0 1
0 1
The original set is not a group, s
Homework Set 3
ei 2
1. The eigenvectors are v = i 2 , with eigenvalues = a bei( + ) 2 . If you want
e
them normalized, just divide by 2 .
a
2. N v = i
be
i 2
be i ei 2 aei 2 be i 2
i + 2 e
*
=
Homework Set 4
1a. Obviously, E is the identity matrix, so EX = XE = X, and we dont need to check this.
For the others, we have 25 multiplications still to check. The work will be skipped as
much as p
Solution Set 5 Tinkham 3-1
a) There are five character classes, and hence five inequivalent irreducible
representations. The sums of the squares of the dimensions must equal eight, which
can only be a
Homework Set 6
Trivially, there are two character classes, and there are two
iEi
irreps, each of which is one-dimensional. It isnt that tough to find the
other irreps for this group, as illustrated at
Solution Set 7
We first need to figure out if the rotation given is a proper one, which means it
can actually be performed or not. This can be done by checking if the determinant is +1
(proper rotatio
Solution Set 8
Lets first figure out which of the ones listed correspond to which ones in the
table for the group O. Obviously, E is item (1) on Natalies list. C4 means a rotation
around the principle
Solution Set 9
First we are supposed to demonstrate that D3 is a subgroup of D6, where the
former is the symmetry group of proper rotations of an equilateral triangle, and the latter
is the group for
Solution Set 10
First we are supposed to show that the listed wave functions have the proper
boundary conditions; that is, that they vanish at the boundaries. Plugging in the boundary
values, we see t
Solution Set 11
JE
First we need the character table for Oh, which is easily determined
+ 1
from the character table for O and that for the inversion group J. They are
- 1
in the table below.
Next we
Physics 745 - Group Theory
Solution Set 12
Due Monday, February 16
1. Diamond is a version of carbon. The position of the carbon atoms takes the form
r = d ( n1 + x ) x + d ( n2 + y ) y + d ( n3 + z )
Physics 745 - Group Theory
Solution Set 13
The original basis is:
a = ax
b = b ( cos x + sin y )
c = cz
An arbitrary element of the translation group is then n1a + n2b + n3c , where ( n1 , n2 , n3 ) a
Physics 745 - Group Theory
Solution Set 14
We use the general formula given by Dr. Holzwarth to find the scattering
amplitude
S ( k )
( 2 )
=
3
( k G ) F ( k ) e
3
iG a
a
G
a
a
As inspired in class,