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Physical applications of group theory

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Physical applications of group theory
29.11 PHYSICAL APPLICATIONS OF GROUP THEORY
It follows that
χprod (X) =
nλ nµ
prod
D (X) kk
k=1
nµ
nλ (λ)
D (X) ii D(µ) (X) jj
=
i=1 j=1
#n
. # nµ
.
λ
(λ)
D (X) ii
D(µ) (X) jj
=
i=1
j=1
= χ(λ) (X) χ(µ) (X).
(29.23)
This proves the theorem, and a similar argument leads to the corresponding result
for integrands in the form of a product of three or more factors.
An immediate corollary is that an integral whose integrand is the product of
two functions transforming according to two different irreps is necessarily zero. To
see this, we use (29.18) to determine whether irrep A1 appears in the product
character set χprod (X):
mA1 =
∗
1 (A1 )
1 prod
1 (λ)
χ (X) χprod (X) =
χ (X) =
χ (X)χ(µ) (X).
g X
g X
g X
We have used the fact that χ(A1 ) (X) = 1 for all X but now note that, by virtue of
(29.14), the expression on the right of this equation is equal to zero unless λ = µ.
Any complications due to non-real characters have been ignored – in practice,
they are handled automatically as it is usually Ψ∗ φ, rather than Ψφ, that appears
in integrands, though many functions are real in any case, and nearly all characters
are.
Equation (29.23) is a general result for integrands but, specifically in the context
of chemical bonding, it implies that for the possibility of bonding to exist, the
two quantum wavefunctions must transform according to the same irrep. This is
discussed further in the next section.
29.11 Physical applications of group theory
As we indicated at the start of chapter 28 and discussed in a little more detail at
the beginning of the present chapter, some physical systems possess symmetries
that allow the results of the present chapter to be used in their analysis. We
consider now some of the more common sorts of problem in which these results
find ready application.
1105
REPRESENTATION THEORY
y
1
4
2
x
3
Figure 29.4 A molecule consisting of four atoms of iodine and one of
manganese.
29.11.1 Bonding in molecules
We have just seen that whether chemical bonding can take place in a molecule
is strongly dependent upon whether the wavefunctions of the two atoms forming
a bond transform according to the same irrep. Thus it is sometimes useful to be
able to find a wavefunction that does transform according to a particular irrep
of a group of transformations. This can be done if the characters of the irrep are
known and a sensible starting point can be guessed. We state without proof that
starting from any n-dimensional basis vector Ψ ≡ (Ψ1 Ψ2 · · · Ψn )T , where {Ψi }
· · · Ψn(λ) )T generated
is a set of wavefunctions, the new vector Ψ(λ) ≡ (Ψ1(λ) Ψ(λ)
2
by
∗
χ(λ) (X)XΨi
(29.24)
Ψi(λ) =
X
will transform according to the λth irrep. If the randomly chosen Ψ happens not
to contain any component that transforms in the desired way then the Ψ(λ) so
generated is found to be a zero vector and it is necessary to select a new starting
vector. An illustration of the use of this ‘projection operator’ is given in the next
example.
Consider a molecule made up of four iodine atoms lying at the corners of a square in the
xy-plane, with a manganese atom at its centre, as shown in figure 29.4. Investigate whether
the molecular orbital given by the superposition of p-state (angular momentum l = 1)
atomic orbitals
Ψ1 = Ψy (r − R1 ) + Ψx (r − R2 ) − Ψy (r − R3 ) − Ψx (r − R4 )
can bond to the d-state atomic orbitals of the manganese atom described by either (i) φ1 =
(3z 2 − r2 )f(r) or (ii) φ2 = (x2 − y 2 )f(r), where f(r) is a function of r and so is unchanged by
any of the symmetry operations of the molecule. Such linear combinations of atomic orbitals
are known as ring orbitals.
We have eight basis functions, the atomic orbitals Ψx (N) and Ψy (N), where N = 1, 2, 3, 4
and indicates the position of an iodine atom. Since the wavefunctions are those of p-states
they have the forms xf(r) or yf(r) and lie in the directions of the x- and y-axes shown in
the figure. Since r is not changed by any of the symmetry operations, f(r) can be treated as
a constant. The symmetry group of the system is 4mm, whose character table is table 29.4.
1106
29.11 PHYSICAL APPLICATIONS OF GROUP THEORY
Case (i). The manganese atomic orbital φ1 = (3z 2 − r2 )f(r), lying at the centre of the
molecule, is not affected by any of the symmetry operations since z and r are unchanged
by them. It clearly transforms according to the identity irrep A1 . We therefore need to
know which combination of the iodine orbitals Ψx (N) and Ψy (N), if any, also transforms
according to A1 .
We use the projection operator (29.24). If we choose Ψx (1) as the arbitrary onedimensional starting vector, we unfortunately obtain zero (as the reader may wish to
verify), but Ψy (1) is found to generate a new non-zero one-dimensional vector transforming
according to A1 . The results of acting on Ψy (1) with the various symmetry elements X
can be written down by inspection (see the discussion in section 29.2). So, for example, the
Ψy (1) orbital centred on iodine atom 1 and aligned along the positive y-axis is changed
by the anticlockwise rotation of π/2 produced by R into an orbital centred on atom 4
and aligned along the negative x-axis; thus R Ψy (1) = −Ψx (4). The complete set of group
actions on Ψy (1) is:
I, Ψy (1);
Q, −Ψy (3);
R, Ψx (2);
mx , Ψy (1);
my , −Ψy (3);
md , Ψx (2);
R , −Ψx (4);
md , −Ψx (4).
Now χ(A1 ) (X) = 1 for all X, so (29.24) states that the sum of the above results for XΨy (1),
all with weight 1, gives a vector (here, since the irrep is one-dimensional, just a wavefunction) that transforms according to A1 and is therefore capable of forming a chemical
bond with the manganese wavefunction φ1 . It is
Ψ(A1 ) = 2[Ψy (1) − Ψy (3) + Ψx (2) − Ψx (4)],
though, of course, the factor 2 is irrelevant. This is precisely the ring orbital Ψ1 given in
the problem, but here it is generated rather than guessed beforehand.
Case (ii). The atomic orbital φ2 = (x2 − y 2 )f(r) behaves as follows under the action of
typical conjugacy class members:
I, φ2 ;
Q, φ2 ;
R, (y 2 − x2 )f(r) = −φ2 ;
mx , φ2 ;
md , −φ2 .
From this we see that φ2 transforms as a one-dimensional irrep, but, from table 29.4, that
irrep is B1 not A1 (the irrep according to which Ψ1 transforms, as already shown). Thus
φ2 and Ψ1 cannot form a bond. The original question did not ask for the the ring orbital to which φ2 may
bond, but it can be generated easily by using the values of XΨy (1) calculated in
case (i) and now weighting them according to the characters of B1 :
Ψ(B1 ) = Ψy (1) − Ψy (3) + (−1)Ψx (2) − (−1)Ψx (4)
+ Ψy (1) − Ψy (3) + (−1)Ψx (2) − (−1)Ψx (4)
= 2[Ψy (1) − Ψx (2) − Ψy (3) + Ψx (4)].
Now we will find the other irreps of 4mm present in the space spanned by
the basis functions Ψx (N) and Ψy (N); at the same time this will illustrate the
important point that since we are working with characters we are only interested
in the diagonal elements of the representative matrices. This means (section 29.2)
that if we work in the natural representation Dnat we need consider only those
functions that transform, wholly or partially, into themselves. Since we have no
need to write out the matrices explicitly, their size (8 × 8) is no drawback. All the
irreps spanned by the basis functions Ψx (N) and Ψy (N) can be determined by
considering the actions of the group elements upon them, as follows.
1107
REPRESENTATION THEORY
(i) Under I all eight basis functions are unchanged, and χ(I) = 8.
(ii) The rotations R, R and Q change the value of N in every case and so
all diagonal elements of the natural representation are zero and χ(R) =
χ(Q) = 0.
(iii) mx takes x into −x and y into y and, for N = 1 and 3, leaves N unchanged,
with the consequences (remember the forms of Ψx (N) and Ψy (N)) that
Ψx (1) → −Ψx (1),
Ψy (1) → Ψy (1),
Ψx (3) → −Ψx (3),
Ψy (3) → Ψy (3).
Thus χ(mx ) has four non-zero contributions, −1, −1, 1 and 1, together
with four zero contributions. The total is thus zero.
(iv) md and md leave no atom unchanged and so χ(md ) = 0.
The character set of the natural representation is thus 8, 0, 0, 0, 0, which, either
by inspection or by applying formula (29.18), shows that
Dnat = A1 ⊕ A2 ⊕ B1 ⊕ B2 ⊕ 2E,
i.e. that all possible irreps are present. We have constructed previously the
combinations of Ψx (N) and Ψy (N) that transform according to A1 and B1 .
The others can be found in the same way.
29.11.2 Matrix elements in quantum mechanics
In section 29.10 we outlined the procedure for determining whether a matrix
element that involves the product of three factors as an integrand is necessarily
zero. We now illustrate this with a specific worked example.
Determine whether a ‘dipole’ matrix element of the form
J = Ψd1 xΨd2 dτ,
where Ψd1 and Ψd2 are d-state wavefunctions of the forms xyf(r) and (x2 − y 2 )g(r) respectively, can be non-zero (i) in a molecule with symmetry C3v (or 3m), such as ammonia, and
(ii) in a molecule with symmetry C4v (or 4mm), such as the MnI4 molecule considered in
the previous example.
We will need to make reference to the character tables of the two groups. The table for
C3v is table 29.1 (section 29.6); that for C4v is reproduced as table 29.5 from table 29.4 but
with the addition of another column showing how some common functions transform.
We make use of (29.23), extended to the product of three functions. No attention need
be paid to f(r) and g(r) as they are unaffected by the group operations.
Case (i). From the character table 29.1 for C3v , we see that each of xy, x and x2 − y 2
forms part of a basis set transforming according to the two-dimensional irrep E. Thus we
may fill in the array of characters (using chemical notation for the classes, except that
we continue to use I rather than E) as shown in table 29.6. The last line is obtained by
1108
29.11 PHYSICAL APPLICATIONS OF GROUP THEORY
4mm
I
Q
R, R mx , my
md , md
A1
A2
B1
B2
E
1
1
1
1
2
1
1
1
1
−2
1
1
−1
−1
0
1
−1
1
−1
0
1
−1
−1
1
0
z; z 2 ; x2 + y 2
Rz
x2 − y 2
xy
(x, y); (xz, yz); (Rx , Ry )
Table 29.5 The character table for the irreps of group 4mm (or C4v ). The
right-hand column lists some common functions, or, for the two-dimensional
irrep E, pairs of functions, that transform according to the irrep against which
they are shown.
Function
Irrep
I
xy
x
x2 − y 2
E
E
E
product
Classes
2C3 3σv
2
2
2
−1
−1
−1
0
0
0
8
−1
0
Table 29.6 The character sets, for the group C3v (or 3mm), of three functions
and of their product x2 y(x2 − y 2 ).
Function
xy
x
x2 − y 2
product
Irrep
B2
E
B1
Classes
2C6 2σv
I
C2
1
2
1
1
−2
1
−1
0
−1
−1
0
1
1
0
−1
2
−2
0
0
0
2σd
Table 29.7 The character sets, for the group C4v (or 4mm), of three functions,
and of their product x2 y(x2 − y 2 ).
multiplying together the corresponding characters for each of the three elements. Now, by
inspection, or by applying (29.18), i.e.
mA1 = 16 [1(1)(8) + 2(1)(−1) + 3(1)(0)] = 1,
we see that irrep A1 does appear in the reduced representation of the product, and so J
is not necessarily zero.
Case (ii). From table 29.5 we find that, under the group C4v , xy and x2 − y 2 transform
as irreps B2 and B1 respectively and that x is part of a basis set transforming as E. Thus
the calculation table takes the form of table 29.7 (again, chemical notation for the classes
has been used).
Here inspection is sufficient, as the product is exactly that of irrep E and irrep A1 is
certainly not present. Thus J is necessarily zero and the dipole matrix element vanishes. 1109
REPRESENTATION THEORY
y3
x3
y1
y2
x2
x1
Figure 29.5 An equilateral array of masses and springs.
29.11.3 Degeneracy of normal modes
As our final area for illustrating the usefulness of group theoretical results we
consider the normal modes of a vibrating system (see chapter 9). This analysis
has far-reaching applications in physics, chemistry and engineering. For a given
system, normal modes that are related by some symmetry operation have the same
frequency of vibration; the modes are said to be degenerate. It can be shown that
such modes span a vector space that transforms according to some irrep of the
group G of symmetry operations of the system. Moreover, the degeneracy of
the modes equals the dimension of the irrep. As an illustration, we consider the
following example.
Investigate the possible vibrational modes of the equilateral triangular arrangement of
equal masses and springs shown in figure 29.5. Demonstrate that two are degenerate.
Clearly the symmetry group is that of the symmetry operations on an equilateral triangle,
namely 3m (or C3v ), whose character table is table 29.1. As on a previous occasion, it is
most convenient to use the natural representation Dnat of this group (it almost always
saves having to write out matrices explicitly) acting on the six-dimensional vector space
(x1 , y1 , x2 , y2 , x3 , y3 ). In this example the natural and regular representations coincide, but
this is not usually the case.
We note that in table 29.1 the second class contains the rotations A (by π/3) and B (by
2π/3), also known as R and R . This class is known as 3z in crystallographic notation, or
C3 in chemical notation, as explained in section 29.9. The third class contains C, D, E, the
three mirror reflections.
Clearly χ(I) = 6. Since all position labels are changed by a rotation, χ(3z ) = 0. For the
mirror reflections the simplest representative class member to choose is the reflection my in
the plane containing the y3 -axis, since then only label 3 is unchanged; under my , x3 → −x3
and y3 → y3 , leading to the conclusion that χ(my ) = 0. Thus the character set is 6, 0, 0.
Using (29.18) and the character table 29.1 shows that
Dnat = A1 ⊕ A2 ⊕ 2E.
1110
29.11 PHYSICAL APPLICATIONS OF GROUP THEORY
However, we have so far allowed xi , yi to be completely general, and we must now identify
and remove those irreps that do not correspond to vibrations. These will be the irreps
corresponding to bodily translations of the triangle and to its rotation without relative
motion of the three masses.
Bodily translations are linear motions of the centre of mass, which has coordinates
x = (x1 + x2 + x3 )/3 and
y = (y1 + y2 + y3 )/3).
Table 29.1 shows that such a coordinate pair (x, y) transforms according to the twodimensional irrep E; this accounts for one of the two such irreps found in the natural
representation.
It can be shown that, as stated in table 29.1, planar bodily rotations of the triangle –
rotations about the z-axis, denoted by Rz – transform as irrep A2 . Thus, when the linear
motions of the centre of mass, and pure rotation about it, are removed from our reduced
representation, we are left with E⊕A1 . So, E and A1 must be the irreps corresponding to the
internal vibrations of the triangle – one doubly degenerate mode and one non-degenerate
mode.
The physical interpretation of this is that two of the normal modes of the system have
the same frequency and one normal mode has a different frequency (barring accidental
coincidences for other reasons). It may be noted that in quantum mechanics the energy
quantum of a normal mode is proportional to its frequency. In general, group theory does not tell us what the frequencies are, since it is
entirely concerned with the symmetry of the system and not with the values of
masses and spring constants. However, using this type of reasoning, the results
from representation theory can be used to predict the degeneracies of atomic
energy levels and, given a perturbation whose Hamiltonian (energy operator) has
some degree of symmetry, the extent to which the perturbation will resolve the
degeneracy. Some of these ideas are explored a little further in the next section
and in the exercises.
29.11.4 Breaking of degeneracies
If a physical system has a high degree of symmetry, invariant under a group G of
reflections and rotations, say, then, as implied above, it will normally be the case
that some of its eigenvalues (of energy, frequency, angular momentum etc.) are
degenerate. However, if a perturbation that is invariant only under the operations
of the elements of a smaller symmetry group (a subgroup of G) is added, some of
the original degeneracies may be broken. The results derived from representation
theory can be used to decide the extent of the degeneracy-breaking.
The normal procedure is to use an N-dimensional basis vector, consisting of the
N degenerate eigenfunctions, to generate an N-dimensional representation of the
symmetry group of the perturbation. This representation is then decomposed into
irreps. In general, eigenfunctions that transform according to different irreps no
longer share the same frequency of vibration. We illustrate this with the following
example.
1111
REPRESENTATION THEORY
M
M
M
Figure 29.6 A circular drumskin loaded with three symmetrically placed
masses.
A circular drumskin has three equal masses placed on it at the vertices of an equilateral
triangle, as shown in figure 29.6. Determine which degenerate normal modes of the drumskin
can be split in frequency by this perturbation.
When no masses are present the normal modes of the drum-skin are either non-degenerate
or two-fold degenerate (see chapter 21). The degenerate eigenfunctions Ψ of the nth normal
mode have the forms
Jn (kr)(cos nθ)e±iωt
or
Jn (kr)(sin nθ)e±iωt .
Therefore, as explained above, we need to consider the two-dimensional vector space
spanned by Ψ1 = sin nθ and Ψ2 = cos nθ. This will generate a two-dimensional representation of the group 3m (or C3v ), the symmetry group of the perturbation. Taking the easiest
element from each of the three classes (identity, rotations, and reflections) of group 3m,
we have
IΨ1 = Ψ1 ,
IΨ2 = Ψ2 ,
AΨ1 = sin n θ − 23 π = cos 23 nπ Ψ1 − sin 23 nπ Ψ2 ,
AΨ2 = cos n θ − 23 π = cos 23 nπ Ψ2 + sin 23 nπ Ψ1 ,
CΨ1 = sin[n(π − θ)] = −(cos nπ)Ψ1 ,
CΨ2 = cos[n(π − θ)] = (cos nπ)Ψ2 .
The three representative matrices are therefore
cos 23 nπ − sin 23 nπ
D(I) = I2 , D(A) =
,
cos 23 nπ
sin 23 nπ
D(C) =
− cos nπ
0
0
cos nπ
.
The characters of this representation are χ(I) = 2, χ(A) = 2 cos(2nπ/3) and χ(C) = 0.
Using (29.18) and table 29.1, we find that
mA1 = 16 2 + 4 cos 23 nπ = mA2
mE = 16 4 − 4 cos 23 nπ .
Thus
#
D=
A1 ⊕ A2
E
if n = 3, 6, 9, . . . ,
otherwise.
Hence the normal modes n = 3, 6, 9, . . . each transform under the operations of 3m
1112
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