Download Assignment 3 - UBC Physics

Assignment 3
Given January 16, Due January 23, 2015.
A tutorial on the applications of discrete groups.
Consider the group C3v which is the cyclic group with three elements,
C3 , augmented by a reflection symmetry through a “vertical” plane, thus
the “v”. It can be thought of as the symmetries of a pyramid which has
an equilateral triangle base, as displayed in figure 1. It has two generators,
c which implements a rotation of the triangle at the base of the pyramid
by angle 2π/3 radians, and thus obeys c3 = e and b which implements a
reflection through a vertical plane which is perpendicular to the equilateral
triangle base of the pyramid and bisects one of the angles. Being a reflection
it obeys b2 = e. The b and c transformations are clearly symmetries of the
object. Note that the net effect of the b and c transformations are identical
to those of the generators of D3 , the proper isometries of an equilateral
triangle. Their net effect is to permute the positions A, B and C. They
must therefore obey the relations of D3 , that is, (bc)2 = e, and in fact, then,
C3v is isomorphic to D3 which is isomorphic to S3 , the symmetric group of
permutations of three objects. This means that, for example, they have the
same conjugacy classes, the same irreducible representations and the same
character table.
However, unlike D3 whose operations can be thought of as proper rotations (without reflections) and can be embedded in the rotation group,
SO(3), C3v contains a reflection and there are no reflections in SO(3). It
therefore cannot be embedded in SO(3). What is needed is O(3). The matrix corresponding to the reflection generator b has determinant −1. Like the
embedding of D3 in SO(3), the embedding of C3v in O(3) forms a reducible
representation of C3v .
1. What irreducible representations of C3v are contained in the
three dimensional representation which are constructed by
embedding C3v as O(3) matrices?
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Figure 1: The pyramid has a base which is an equilateral triangle with vertices A, B, C placed in the x-y plane.
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The tetrahedral group, denoted T , is the symmetry group of a proper
(without reflection) transformations of a regular tetrahedron. It is of
order 12 and it consists of four C3 subgroups, corresponding to rotations by angles 2π/3 and 4π/3 about the centres of the faces of the
tetrahedron. The group has two generators c and b with the three
relations c3 = e, b2 = e, (bc)3 = e. The latter relation can be seen
by tracing the path of one of the vertices of the tetrahedron under
subsequent transformations. The group elements are
{e, c, c2 , b, bc, bc2 , cb, c2 b, cbc, cbc2 , c2 bc, c2 bc2 }
2. Show that the group elements which are listed in the set above
are all of the independent words that can be written given
the two generators c and b and the relations b2 = e, c3 = e
and (bc)3 = e. ( Hint:Observe that the relation (bc)3 = e allows us to
reduce any sequence of generators with two b’s to one with one b, that
is beb = e, bcb = c2 bc2 , bc2 b = cbc.)
3. The group transformations permute the vertices of the tetrahedron.
Thus, we expect that T is isomorphic to a subgroup of the symmetric
group S4 of permutations of four objects. The latter group has order
24, so T , being of order 12, cannot be isomorphic to S4 itself. S4 does
have an order 12 subgroup, the alternating group A4 . A4 is the set of all
“even permutations”, that is, permutations which can be implemented
by an even number of exchanges of neighbouring objects (recall the
generators σi in the presentation of Sn where σi exchanged the i’th
objet to the i + 1’th object). Also, recall the cycle decomposition of
S4 . A cycle of even length corresponds to an odd permutation and a
cycle of odd length corresponds to an even permutation. For the cycle
decomposition of a given permutation, the degree of the permutation
is even if there is an even number of even order cycles and it is odd of
there is an odd number of even order cycles. Use this fact to show
that the alternating group, A4 has order 12.
As for conjugacy classes, remember that, all rotations by the same angle
about any axis are in the same conjugacy class of SO(3). Thus, since T
is a subgroup of SO(3), a necessary condition for two group elements to
be in the same conjugacy class of T is that they must first be in the same
conjugacy class of SO(3). This sufficient condition divides the group
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Figure 2: The regular tetrahedron, whose four corners are the points a, b, c, d
is inscribed in a cube. The axis of 2-fold rotation symmetries are denoted
by X, Y and Z. The body diagonals of the cube are four three-fold rotation
axes.
elements of T into three potential classes, all rotations by 0, that is {e},
rotations by 2π/3, that is {c, c2 , bc, bc2 , cb, c2 b, cbc, c2 bc2 } and rotations
by π radians, that is {b, cbc2 , c2 bc}. Thus, candidates for the classes
are {e}, {c, c2 , bc, bc2 , cb, c2 b, cbc, c2 bc2 } and {b, cbc2 , c2 bc}. However,
even though the elements of these sets are linked by conjugation with
rotations, that is, with elements of SO(3), they are not necessarily
linked by conjugation by rotations which are themselves elements of
the group T , which is a subgroup of SO(3). For example, whereas there
are rotations in SO(3) which invert the direction of a body diagonal of
the cube in figure 2, there is no element of the group T which reverses
the direction of a body diagonal of the cube. We therefore expect that
c and c2 are actually in different classes, even though both are rotations
by 2π/3 radians.
4. Show that the classes are
{e}, {c, bc, cb, c2 bc2 }, {c2 , bc2 , c2 b, cbc}{b, cbc2 , c2 bc}
Since there are four classes, there are four irreducible representations.
The dimensions squared of the irreducible representations must add up
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to the order of the group, 12. Thus 1 + d22 + d23 + d24 = 12, where we have
taken into account that there always must be a trivial one-dimensional
representation.
5. Show that the only solution of the remaining integers is that
there are two more one-dimensional representations and one
three dimensional representation so that in total, there are
three one-dimensional representations and one three dimensional representation.
We can begin constructing the character table as
T
[e] 4[c] 4[c2 ] 3[b]
A1
A2
A3
T
1
1
1
3
1
1
1
6. Show that the only possible one-dimensional representations
have the generators identified as
c=ω , b=1
and
c = ω2 , b = 1
where ω = exp(2πi/3). This allows us to fill in two more rows of the
character table,
T
[e] 4[c] 4[c2 ] 3[b]
A1
A2
A3
T
1
1
1
3
1
ω
ω2
1
ω2
ω
1
1
1
7. Show that the remainder of the character table can be filled
in using orthogonality of the rows or the columns, so that
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T
[e] 4[c] 4[c2 ] 3[b]
A1
A2
A3
T
1
1
1
3
1
ω
ω2
0
1
ω2
ω
0
1
1
1
−1
8. Consider the 27-dimensional reducible representation which is
formed by taking the direct product of three three-dimensional
representations, T ⊗ T ⊗ T . Show that it can be decomposed
as the direct sum
T ⊗ T ⊗ T = 2A1 ⊕ 2A2 ⊕ 2A3 ⊕ 7T
Magnetic Moment
A material has a magnetic moment if it responds to the magnetic field
by experiencing a torque. It has a polarization, or an electric dipole
moment if it experiences a torque when exposed to an electric field. The
magnetic moment and electric polarization transform like a pseudovector and a vector under rotations,
Mi → M̃i = det(R)
3
X
Rij Mj , Pi → P̃i =
j=1
3
X
Rij Pj
j=1
If R ∈ SO(3), these transformation laws are identical. However, if the
transformation contains a reflection, that if it is necessarily in O(3),
these transformations differ by an overall sign. The magnetic moment
is said to be a pseudo-vector and the electric polarization is a vector.
In an isotropic material, all quantities should be invariant under rotations. This tells us that the electric or magnetic dipole moments should
obey
3
3
X
X
Mi = det(R)
Rij Mj , Pi =
Rij Pj
j=1
j=1
9. Show that the above equations imply Pi = 0 and Mi = 0 for an
SO(3) invariant physical system.
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10. Can Mi or Pi be nonzero if the symmetry of the physical system is reduced to T , D3 or C3v ?
Conductivity Tensor
Now, let us ask the question as to whether the conductivity of a material
with various symmetries can be zero or non-zero. The conductivity
~ which
tensor is a relationship between the strength of an electric field E
is applied to a material and the electric current ~j which results when
the electric field is applied. The current is a linear function of the
applied field which has the form
3
X
ji =
σij Ej
j=1
Time reversal symmetry implies that σij is an real, symmetric matrix,
σij = σji . Under a spatial rotation, it transforms as
3
X
σij → σ̃ij =
Rim Rjn σmn
m,n=1
We could ask what the form of the conductivity tensor is in a material
which is completely isotropic. Complete isotropy implies that the conductivity tensor does not change under a spatial rotation. That is, in
an isotropic environment,
σij =
3
X
Rim Rjn σmn
(1)
m,n=1
for any rotation, that is, for any matrix R ∈ SO(3).
11. Show that the equation (1) implies that the only possibility
for σij is that it is proportional to the unit matrix,
σij = σδij
(Hint: you are allowed to use Schur’s Lemma for SO(3) and the fact
that the R’s are an irreducible representation of SO(3), the representation with ` = 1.)
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Now, let us assume that the symmetry of the environment of a quantum
system are reduced form SO(3) to one of the groups D3 , C3v or T . In
general, the lower degree of symmetry should pose fewer constraints
than the full rotation group. We want to know whether a conductivity
tensor which is more general than σij = σδij is allowed by considering
an environment with less symmery. Let us begin with the tetragonal
group T .
To answer the question about the form of σ, we need to find the reducible representation under which the conductivity tensor transforms.
Since it is a symmetric matrix, and, under a rotation, it is easy to argue
that it transforms like
3
i
X
1 h (`=1)
(`=1)
(`=1)
(`=1)
Dim (g)Djn (g) + Din (g)Djm (g) σmn
σij →
2
m,n=1
(`=1)
where Dij (g) is the three-dimensional representation of SO(3), which
we were calling the orthogonal matrices Rij in the above, and g is an
element of T (which is also an element of SO(3), where we regard T
as a subgroup of SO(3)). Let us call this representation (V ⊗ V )+ .
Generally, we only need the character of the above representation. We
can find the character from the character formula for SO(3) group
elements. To begin, we take the “trace” of the transformation matrix
in the above formula by setting i = m and j = n and then summing
over m and n. We get
2
1h
TrD(`=1) (g) + TrD(`=1) (g 2 )
χ(V ⊗V )+ (g) =
2
12. Show that the character of the representation (V ⊗ V )+ above
has the characters which are summarized in the last row of
the following character table:
T
[e] 4[c] 4[c2 ] 3[b]
A1
A2
A3
T
(V ⊗ V )+
1
1
1
3
6
8
1
ω
ω2
0
0
1
ω2
ω
0
0
1
1
1
−1
2
13. Show that, studying the above character table implies
(V ⊗ V )+ = A1 ⊕ A2 ⊕ A3 ⊕ T
Since the identity representation only appears once on the right-handside of the equation above, this implies that there is only one way for
the conductivity tensor to be invariant under the tetragonal group, if
σij = σδij , as before.
Now, let us consider the group D3 . Recall that the character table of
D3 is
D3
[e] 2[c] 3[b]
A1
A2
E
(V ⊗ V )+
1
1
2
6
1
1
−1
0
1
−1
0
2
(2)
where we have added the character of the reducible representation (V ⊗
V )+ .
14. How does the reducible representation (V ⊗ V )+ decompose
into irreducible representations of D3 ?
If we define the representation (V ⊗ V )− as one which transforms an
anti-symmetric tensor
3
i
X
1 h (`=1)
(`=1)
(`=1)
(`=1)
Dim (g)Djn (g) − Din (g)Djm (g) amn
aij →
2
m,n=1
with characters defined by
χ(V ⊗V )− (g) =
2
1h
TrD(`=1) (g) − TrD(`=1) (g 2 )
2
15. Find the decomposition of (V ⊗V )− in terms of representations
of T and D3 .
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