# жже0 tI tI tI ¤ жже1

```Tutorial #5: Characters and Orthogonality Relations
April 12, 2016
1
Schur's lemma in quantum mechanics
1.1
Particle on a pyramid
Consider a particle hopping between the verticies of a pyramid whose base is an equilateral triangle.
Figure 1: Pyramid
The particle hops between the base verticies with amplitude
amplitude
t1 ;
that is, the Hamiltonian is
0
t1
H
t1
t1
where
on site
1
2
0
0
0
T
t1
t
0
t
and from the base to the top of the pyramid with
t1
t
t
0
is the state corresponding to the particle on site
and so on. The Hamiltonian is obviously
basis of the form
t1
0
t
t
t,
C3v
(1)
1 (head of the pyramid), 0
1
0
0
T
is the state
symmetric; as you saw in class, we can construct an orthonormal
|a, l, xy, with a denoting irreducible representation, l the state within the representation, and x is some
extra index. In order to do that, we need to know what are the relevant irreducible representations. Our Hilbert space
is four-dimensional, so it must transform according to some reducible representation (because the irreducible ones are
only one- and two-dimensional). First, we notice that action of the group only exchange the base verticies, and leave the
1
|1, 1, 1y |1y 0
0
(2)
0
is an invariant state - that is, it transform according to the trivial representation.
to denote the rst irreducible representation of
C3v ,
Mind the notation: the rst 1 is
the trivial one; the second 1 is for the state itself (but the trivial
representation is one-dimensional, so this index cannot obtain any other value); the third 1 is for dierent copies of the
trivial representation. And indeed, there is another copy: if we consider the representation of the group that act on the
Hilbert space:
1
0
D pRq 0
0
0
0
0
1
0
1
0
0
0
0
,
1
0
1
0
D pσ1 q 0
0
1
0
1
0
0
0
0
0
1
0
0
,
1
0
etc.
(3)
Tutorial 5: Characters and Orthogonality Relations
3
CHARACTERS
we can notice that
|1, 1, 2y |2y 0
1 1
? 3 1
1
(4)
is also invariant under the action of the group. Finally, since the Hilbert space representation is faithful, we must have
the (only) faithful representation of
orthogonal to both
C3v ,
the two-dimensional one. This means that the remaining two dimensional space
|1y and |2y transform according the the representation Dp3q , and we are free to choose a basis, for
example:
0
1 ? 11 ,
6
2
|3, 1, 1y |3y |3, 2, 1y |4y 0
1 ? 11
2
0
(5)
Now, recall that according to Schur's lemma the action of the Hamiltonian in this basis is in the form
H |a, l, xy &cedil;
paq |a, l, yy
fxy
(6)
y
we can see this explicitly:
?
3t1
?2 t ?
H |2y ?23 3t1 |1y
3t ?23 t
0
t1 ? 1
H |1y t1 3t |2y ,
t1
that is,
2t |2y
(7)
p1q f p1q ?3t1 , f p1q 2t, so we now only have to diagonalize a 2 2 matrix, rather then a 4 4:
22
21
? 1
0
3t
f p1q ? 1
3t
2t
f12
(8)
for the other two states the situation is even better, since they are already eigenstates:
p3q
where
2
p3q
f11
H |3, l, 1y f11 |3, l, 1y
is just a number (that happens to be
t), because the representation Dp3q only appear once.
Schur orthogonality relation
Reminder: Schur orthogonality relation is the equation
&cedil;
P
Dpaq g 1
•
note that there is a small caveat: if
P Dpaq pg q P 1
for some xed
P,
Dpaq
mn
g G
and
Dpbq
|dG| δab δni δmj
Dpbq pg q
ij
(10)
a
are equivalent representations, but not identical, namely
Dpbq pg q the orthogonality relation reads
&cedil;
P
Dpaq g 1
mn
g G
3
(9)
Dpbq pg q
|G| P P 1
in mj
d
ij
(11)
a
Characters
Reminder: The (rst) orthogonality relation for characters is
&cedil;
kα χa pCα q χb pCα q |G| δab
(12)
Cα
(where
a, b
are indecies for irreducible representations, and
relation (or completeness relation) is
&cedil;
α, β
for conjugacy classes). The seconed orthogonality
χa pCα q χa pCβ q a
|G| δ
kα
αβ
(13)
As you saw in the lecture, we can construct the following vectors:
d
vaα
using dirac notation, the rst relation is
Group Theory in Physics
|kGα| χa pCα q
(14)
xva |vb y δab , while the seconed one is &deg;a |va y xva | 1.
page 2 of 3
Spring 2016
Tutorial 5: Characters and Orthogonality Relations
3.1
Alternating group
A4
Recall the alternating group
Construct the character table of
A4
1
- the group of even permutations on four letters. It consists of four conjugacy classes :
A4
d21
is a normal subgroup of
d1
A4
d22
d23
d24
|A4 | 12
G
H of a group
from the irreducible representations of G H :
D pg q D pgH q
{
In our case, A4 H is a group of order 3, so it must be isomorphic to
Dp1q
Dp2q
Dp3q
e
1
1
1
R
1
ω
ω2
2πi
3
{
G,
you can immediately construct
(18)
C3 ,
for which the character table is
R̄
1
ω2
ω
Table 1: Characters of
e
(17)
(you can verify directly that it is closed under multiplication, and thus a subgroup, and it
irreducible representations of
ω
(16)
d2 d3 1, d4 3. Now, notice that
H tid, p12q p34q , p13q p24q , p14q p23qu
is normal it is a union of conjugacy classes). Given a normal subgroup,
where
(15)
has four conjugacy classes, it also has four irreducible representations. The dimensionality
of the representations must satisfy
The only possible choice id
CHARACTERS
A4
C1 tidu , Cπ tp12q p34q , p13q p24q , p14q p23qu
tp123q , p142q , p134q , p243qu , Cα1 tp132q , p124q , p143q , p234qu
Cα
Solution: First, Since
3
C3
. This gives us all the one-dimensional representations of
A4 ,
and we can use orthogonality to nd the
character for the three-dimensional representation.
C1
1
1
1
3
Dp1q
Dp2q
Dp3q
Dp4q
3Cπ
1
1
1
1
4Cα
1
ω
ω2
0
Table 2: Characters of
3.2
4Cα
1
ω2
ω
0
1
A4
Quaternion group
Construct the character table for the Quaternion group.
Solution: As you recall,
Q
has 5 conjugacy classes:
teu , Cw twu , Cx tx, x̄u , Cy ty, ȳu , Cz tz, z̄u
(19)
&deg; 2
So it also has ve irreducible representations. From the constraint
a da |Q| 8 we obtain d1 d2 d3 d4 1,
d5 2. Consider now the Subgroup X te, w, x, x̄u. The quotient group Q{X is isomorphic to Z2 , so it induces
a sign representation: 1 for the elements contained in X , and 1 for elements of QzX (elements contained in Q
C1
but not in
X ).
The same argument can be also applied to the subgroups
Y
and
Z,
and thus we obtain the three
non-trivial one-dimensional representations. The characters for the two-dimensional representation can be obtained
using orthogonality.
Dp1q
Dp2q
Dp3q
Dp4q
Dp5q
C1
1
1
1
1
2
Cw
1
1
1
1
2
2Cx
1
1
1
1
0
2Cy
1
1
1
1
0
Table 3: Characters of
1
2Cz
1
1
1
1
0
Q
The derivation of this partition appears in tutorial #3, even though we haven't done it in class
Group Theory in Physics
page 3 of 3
Spring 2016
```