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Decomposition of group representations

The decomposition of a reducible representation of a group reduces it to the direct sum of irreducible representations of the group. We have shown in an earlier article that this involves decomposing a block diagonal matrix into the direct sum of its constituent square matrices. In this article, we shall derive some useful equations involving the characters of the representations of a group (see eq25, eq27a, eq29 and eq30 below).

Consider an element of a reducible representation that has undergone a similarity transformation to the following block diagonal matrix:

where , ,  and .

Clearly, . Since the trace of a matrix belonging to the -th class of the -th irreducible representation is called the character of a class , we have . However, a particular constituent irreducible representation may appear more than once in the decomposition. Therefore,

where  is the number of times  appear in the direct sum.

Multiplying eq25 by , where  is the number of elements in the -th class, and sum over ,

Substituting eq23 in eq26,

Since ,

Eq27 is sometimes written as a sum over the individual symmetry operations  rather than over the various classes:

Question

Show that .

Answer

The complex conjugate of eq25, with a change of the dummy index from to , is . Multiply this by eq25 and by and sum over ,

Swap the dummy indices of eq23 to , which when substituted in the above equation, gives

Since ,

 

With reference to eq25 and eq28, if one of the is equal to 1, with the rest of the  equal to zero, then  is an irreducible representation. This implies that

 

Question

Show that if two reducible representations and of a group are equivalent, they decompose into the same direct sum of irreducible representations of .

Answer

Since and  are related by a similarity transformation,  for every . Using eq27a,

 

 

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Regular representation

The regular representation of a group is a reducible representation that is generated from a rearranged multiplication table of the group. It is used to derive an important property (see eq40 below) for constructing character tables.

Consider the re-arranged multiplication table for the  point group such that all the identity elements are along the diagonal:

An element of the regular representation of the group  is derived from table II in the form of a  matrix, whose entries are 1 when the element of  occurs in table II and zero otherwise. For example,

In other words, we have

where  is the -th row and -th column matrix entry of the -th element of the  representation of .

Question

Show that each matrix of the regular representation of the point group has an inverse.

Answer

The entry 1 appears only once in every row (or column) of a matrix, e.g. , of the regular representation. We can therefore swap the rows (or columns) of  to form . According to determinant property 1,  and consequently, according to determinant property 8, . Therefore,  is non-singular, according to determinant property 11. The same logic applies to the rest of the matrices.

 

If these derived matrices are truly elements of a representation of , then they must satisfy the closure property of , i.e.

where

Therefore,  in eq35 needs to be equal to  to satisfy eq32. To prove this, we multiply the condition  in eq31 on the left by  to give . Similarly, we multiply the condition  in eq34 on the left by  and then on the right by  to give . Combining the two results, we have , which when multiplied on the right by  and then on the left by , gives , which completes the proof.

By inspecting table II, the character of the regular representation is

Question

Show that the regular representation is reducible.

Answer

If the regular representation is reducible, the LHS of eq28 must be greater than . Applying the LHS of eq28 to the regular representation and using eq36,

For non-single element groups, and hence .

 

Finally, we shall prove that . From eq25 and eq27, we have

where  and are the characters for the -th class of the regular representation and irreducible representation respectively.

Expanding eq38 and using eq36,

Question

Why is ?

Answer

The regular representation for  is a reducible representation that is the direct sum of irreducible representations under the class . Each of these constituent irreducible representations is in general a  matrix with trace of .

 

Eq39 therefore states that each constituent irreducible representation occurs in the regular representation a number of times that is equal to the dimension of the corresponding irreducible representation. For example, with reference to the  table below, , where ,  and . In other words, each of the irreducible representations  and  appears once in the  regular representation matrix element , while the irreducible representation  appears twice.

Since the matrix dimension for each element of the regular representation is ,

where we have used eq39 and where  refers to the dimension of the -th irreducible representation of a group.

 

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Character table

A character table is a square matrix whose rows and columns are irreducible representations  of a point group  and classes  of , respectively, with the entries being the characters  of elements of the corresponding irreducible representation.

It has the general form:

where  is the number of elements in the -th class.

The construction of a character table is based on the following properties, some of which are related to the little orthogonality theorem:

    1. There is always a one-dimensional irreducible representation, called a totally symmetric representation, in which all the characters are 1. It is conventionally denoted by . This is because a representation of a group is a collection of matrices that multiply according to the multiplication table of the group, and the collection of matrices, each with entry 1, always multiply according to the multiplication table of that group.
    2. The number of irreducible representations is equal to the number of classes of the group. This is a consequence of the 1st little orthogonality relation.
    3. The sum of the squares of the dimensions of all the irreducible representations of a group is equal to the order of the group, i.e. , which is eq40.
    4. Each irreducible representation is regarded as a basis vector in a -dimensional vector space, with the weighted characters of an irreducible representation being the components of the corresponding basis vector. Therefore, irreducible representations have orthonormal relations with one another. These relations are characterised by either
      a)     (see eq24); or
      b)     (see eq24b)
    5. The weighted sum of the squares of the characters of an irreducible representation is equal to the order of the group, i.e. , which is eq29.

For example, the character table for the  point group is constructed by first noting that there are six symmetry operations () that belong to three classes. Using properties 1 and 2 mentioned above, there are three irreducible representations, with the first being the totally symmetric representation:

Using the 3rd property, . Since , let  and . Furthermore, the characters of a  identity matrix and a  identity matrix are 1 and 2 respectively. Therefore, we have

For orthogonal vectors and , property 4a states that

with the 5th property requiring

Since a representation of a group is a collection of matrices that multiply according to the multiplication table of the group, the  matrices of  (where the character itself is a matrix) must multiply according to the  multiplication table as follows:

In other words,  and , where the elements (in matrix form) of the same class in a group have the same trace. Substituting  and  in eq42 gives  ( is rejected, see Q&A below for explanation), which when substituted in eq41 gives . We now have:

Question
    1. How about the solution set of and ?
    2. Can we use the same logic in determining the characters of to derive the characters of ?
Answer
    1. This solution set is rejected, as it is not consistent with property 5.
    2. No, because we will end up with more variables than we can solve, e.g. for , we have , where and . Therefore, the method used to determine the characters of only works if  is one-dimensional.

 

The remaining characters are obtained using property 4b, where  and . Therefore,  and , with


The character table of a point group can also be generated using basis functions, which are defined in the next article. Some point groups, e.g.  and , have a two-dimensional irreducible representation that contains two rows of characters that are complex conjugates of each other (). Each of these rows of character is an irreducible representation in its own right, with the total number of irreducible representations equal to the number of classes of the group (property 2).

Finally, the generic irreducible representation symbols of  are replaced by Mulliken symbols in character tables for clarity, where

    1. : one-dimensional representation that is symmetric with respect to rotation about the principal axis, i.e. .
    2. : one-dimensional representation that is antisymmetric with respect to rotation about the principal axis, i.e. .
    3. : two-dimensional representation (not to be confused with the identity element of the group).
    4. : three-dimensional representation.
    5. If there are more than one -dimensional representation, subscripts are used to differentiate them, e.g.  and .
    6. Molecular term symbols (, , and ), in addition to the Mulliken and , are used to label the irreducible representations of the axial point groups and .  The molecular term symbol for an irreducible representation corresponds to the magnitude of the quantum number  of a hydrogenic basis wavefunction that is used to generate the representation, e.g. (see this article for derivation), where . We have,
      The use of molecular term symbols in character tables have useful applications in spectroscopy.
    7. The subscript (from the German gerade, meaning even) refers to representations that are symmetric with respect to inversion, while the subscript  (from the German ungerade, meaning uneven) refers to representations that are antisymmetric with respect to inversion.
    8. The superscript ‘ refers to representations that are symmetric with respect to , while the superscript ” refers to representations that are antisymmetric with respect to .
    9. The superscript + refers to representations that are symmetric with respect to , while the superscript – refers to representations that are antisymmetric with respect to .

So, we have

The last part of a character table lists basis functions that transform according to the irreducible representations of the group. We shall elaborate on this in the next article.

 

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Basis

A basis of a group is a set of objects that transforms according to representations of the group. The objects may be vectors, pseudovectors, functions, bond angles, etc. They are not necessarily orthogonal but are usually chosen to be orthogonal for certain applications.

To explain the definition of a basis, we consider the  point group with the following character table:

Let’s examine how symmetry operators of  transform  -orbital wavefunctions.

, by inspection, is invariant when acted upon by each of the four symmetry operators of  (see diagram below). Mathematically, we obtain the same eigenvalue of +1 after each operation, e.g. . We say that  transforms according to the totally symmetric representation . Repeating the logic,  and  transform according to and  respectively.

From this article,

where  and  is the distance from the origin.

Hence, the function  is invariant to all symmetry operations of a point group. This implies that the -orbital wavefunctions transform in the same way as the linear functions ,  and . We call these functions basis functions and include them in the character table as follows:

The basis functions , and in the character table of a point group also represent the independent translational motions of a molecule of that point group. To elaborate further, let’s define the independent translational motions of a molecule as the displacement of the molecule in the , and directions. For example, the diagram below describes the displacement of in the -direction in terms of unit instantaneous displacements vectors that are centred on the atoms, or equivalently, a single instantaneous displacement vector on the centre of mass.

Similarly, the translational motions of in the -direction and -direction are described by corresponding displacement vectors on the centre of mass. These three linearly independent displacement vectors form a basis set of a representation  of  because any one of the vectors is transformed by the symmetry operations of into a linear combination of the vectors in the set. Symmetry operations acting on the basis set produce the following results:

The corresponding matrix transformation equations are:

The transformation matrices form , which is reducible and in block-diagonal form, with , and transforming according to ,  and respectively. Therefore, the bases , and in the character table represent the independent translational motions of .

We can also use instantaneous displacement vectors to describe rotational motions of a molecule. The rotation of  about the -axis is shown in the diagram below, with three instantaneous displacement vectors tangent to the two circular paths of motion.

Just like the way we reduce the three translational displacement vectors to a single vector centred on the centre of mass of , we represent the three rotational vectors with , which is a pseudovector characterised by two quantities: a rotation direction that is defined by the right-hand rule and a magnitude that represents the rotation angle . Therefore, the transformation of the three instantaneous vectors by the symmetry operation is equivalent to the reflection of the rotation direction of (curved grey arrow) about the  plane. The symmetry operation leaves the instantaneous vectors, and hence the rotation direction of , unchanged.

Symmetry operations acting on the basis set produce the following results:

The corresponding matrix transformation equations are:

The transformation matrices form , which is again reducible and in block-diagonal form, with , and  transforming according to , and respectively. Therefore, the bases , and  in the character table represent the independent rotational motions of .

As such, we can revise the character table as follows:

-orbitals wavefunctions, which are quadratic polynomials, also form a basis. For example, the orbital wavefunction transforms according to (see diagram below).

Consequently, we have

Question

Can -orbital wavefunctions form a basis for a group?

Answer

Yes, they are classified as a basis of cubic functions.

 

Lastly, character tables for some point groups have certain elements of a basis in parentheses. An example is the  point group:

which in the matrix representation form is

The notation of  means that  and  transform together but not independently under a symmetry operation. To elaborate on this, let’s consider the wavefunctions and . Separately, each wavefunction does not transform according to any of the three irreducible representations of . This can be verified by letting any point on any of the wavefunctions undergo the  symmetry operation. The eigenvalue obtained does not match any element of any irreducible representation of .

However, if we subject both wavefunctions together to the symmetry operation , e.g. by choosing the points  and  on and respectively (see diagram below), we obtain, after some simple geometry, the transformed points  and .

In matrix form, we have

or, in a new notation,

Repeating the same logic for the rest of the symmetry operators and comparing the results with the table depicting the matrix representation of , we find that the wavefunctions and , which are equivalent to the basis functions and , transform together according to the irreducible representation . We also find that a basis can be used to give rise to matrix representations of a group, since the matrix form of  is generated in the process.

Using the same logic for the rotation vectors and the -orbitals, we find that the pairs ,  and transform according to the irreducible representation .

An important implication of the basis functions and  transforming together under the symmetry operations of a particular point group is that there is no difference in symmetry between them with respect to that point group. In other words, the two basis functions and certain properties they possess may be treated as equivalent and indistinguishable. An example is the common energy state that corresponds to and , i.e. a degenerate energy state. Therefore, group theory plays a useful role in spectroscopic analysis.

 

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Symmetry of the Hamiltonian

The Hamiltonian is invariant under a symmetry operation if its expression in different bases related by the symmetry operation is the same.

Our objective is to prove the above statement and that the Hamiltonian is invariant under a symmetry operation  if it commutes with .

Let’s consider the non-relativistic multi-electron Hamiltonian :

where , and .

The first term on the RHS of eq43 is the kinetic energy operator. To prove that it is invariant under any symmetry operation, we need to show that it is rotation-invariant and reflection-invariant. As mentioned in this article, we can analyse a symmetry operation in terms of the change of basis of a coordinate system. The following matrix equation expresses the change of basis of the operator by a rotation about the -axis:

where  are components of a vector in the old basis, while  are components of the same vector in the new basis.

Since , we have . That leaves us to show that . From eq44,

Using the multivariable chain rule ,

Substituting eq47 in  and , we have  and  respectively. We then substitute these two equalities in eq48 to give:

Repeating the above logic for , we have

Adding eq49 and eq50 completes the proof.

The change of basis by a reflection, e.g. along the -plane, is given by

Since  and , we only need to show that , which is achieved using the above logic for proving .

The second and third terms on the RHS of eq43 (potential energy operator) are dependent on the electron-nuclear distance  and the inter-electron distance  respectively. Since electron-nuclear distances and inter-electron distances do not change under symmetry operations,  and .

Hence,  is invariant to a change of basis related by , and therefore, invariant under a symmetry operation . Since  is invariant to a change of basis, it undergoes the following similarity transformation:

or equivalently,

Eq53 states that commutes with , which is a consequence of  being invariant under the symmetry operation .

 

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Symmetry and degeneracy

Symmetry and degeneracy are related as a result of the non-relativistic Hamiltonian being invariant under symmetry operations.

Consider the stationary state of a quantum mechanical system that is described by a set of linearly independent wavefunctions , which satisfies the time-independent Schrodinger equation . Subjecting both sides of the equation to the symmetry operation , where  are elements of a group , we have . Using eq53 and noting that  is a scalar, we obtain

which implies that  is an eigenfunction of , i.e. .

We say that the set of symmetry operations form the group of the Hamiltonian. If  is -fold degenerate, any linear combination of the subset of wavefunctions associated with the degenerate eigenvalue will be a solution of eq54. We can then write:

According to the closure property of ,

So,

Since  is a linear operator,

Subtracting eq59 from eq57, we have . If the wavefunctions are linearly independent, all coefficients of  must be zero, i.e.

With reference to eq55 through eq57,  and range from to . Thus, eq60 implies that  and are entries of the square matrices  and respectively, i.e.

Comparing eq61 with eq58, the matrices  and  multiply the same way as the symmetry operations  and  and therefore form a representation of . Such a representation has a dimension equal to  and may be reducible or irreducible. However, the following analysis shows that it is irreducible, assuming that there is no accidental degeneracy, which happens when two eigenvalues are the same even though their corresponding eigenfunctions transform differently under the symmetry operations of .

 

Question

What is an example of accidental degeneracy?

Answer

The hydrogenic wavefunctions and , which describe the -orbital and -orbitals respectively, obviously have different symmetry and may transform according to different irreducible representations, e.g. the -orbital transforms according to the totally symmetric irreducible representation, while the -orbitals may not. Therefore, they may have different energies. However, and  are degenerate wavefunctions for a particular value of . In fact, the number of degenerate energy levels of the hydrogen atom is  instead of , which is expected from a non-relativistic Hamiltonian with a central potential. This implies that spin-orbit coupling is not considered when deriving the hydrogenic wavefunctions. The degeneracy is lifted once spin-orbit coupling is included.

 

To show that  is irreducible, we express eq55 as

where .

If  is reducible, then it may be in block-diagonal form. For example, if ,

Substituting eq61b in eq61a, and  are linear combinations of the subset with elements  and , while  and are linear combinations of the subset with elements  and . If so, the eigenvalues corresponding to the different subsets could be different (e.g. when one subset is one-dimensional and the other is two-dimensional), which would contradict our original assumption that  is -fold degenerate.

If is not initially in block-diagonal form, it can be converted into a block-diagonal matrix  via a similarity transformation, where . We can then rewrite eq61a as

where  is the basis for the transformed representation.

Since is in block-diagonal form, we arrive at the same conclusion as before. Therefore, assuming that there is no accidental degeneracy, a representation that is generated by a set of orthogonal degenerate wavefunctions is irreducible and we called the set of wavefunctions , basis wavefunctions of .

Question

1) Why is the number of basis functions needed to generate  equal to the dimension of ?

2) Can any set of functions, other than eigenfunctions of the Hamiltonian, be a set of basis functions for a representation of ?

3) Show that if  is a basis of a representation of a group, then any linear combination of  is a basis of a representation that is equivalent to .

Answer

1) Each  in of eq55 creates a column of matrix entries (see Q&A below for an illustration). So, the  matrix  is generated by  number of . The same logic applies to matrices  and . This implies that the number of linearly independent basis functions of a representation corresponds to the dimension of the representation. This set of linearly independent basis functions can be made orthogonal to one another using the Gram-Schmidt process. Therefore, the number of orthogonal basis functions of a representation corresponds to the dimension of the representation.

2) By inspecting eq55 through eq61, any set of linearly independent functions , not necessary eigenfunctions of the Hamiltonian, is a set of basis functions for a representation of  if  is transformed by a symmetry operation into a linear combination of . The converse is obviously also true, i.e. if  is a set of basis functions for a representation of , then . In this case, there is no requirement that the functions are degenerate and hence the representation generated can be either reducible or irreducible.

3) Since matrix multiplication is distributive,


Let’s rewrite eq55 as . Substituting this expression in the above equation and rearranging, gives


where .
Using eq61d and repeating the steps from eq55 through eq61, is also basis of a representation  of the group.
To Show that is equivalent to , let , which can be written as the following matrix equation:


where .

Using eq55 with in place of , we have , whose matrix equation is


where .

Multiplying eq61e by on the right, we have , which when substituted in eq61f gives


The matrix equation of eq55 with in place of is


Comparing eq61g with eq61h,  is related to by a similarity transformation. This implies that is equivalent to .

 

If is non-degenerate, the only eigenfunctions satisfying eq54 are , where is a constant. Therefore, . Normalising , we get , which means that . Consequently, the wavefunctions associated with non-degenerate eigenvalues transform according to one-dimensional irreducible representations with characters of +1 or -1.

 

Question

If , show how eq55 is related to the matrix representation .

Answer

The two degenerate wavefunctions are  and . We have

 

We can conclude from the above analysis that irreducible representations with dimension of one are called non-degenerate representations, while those with dimensions of more than one are known as degenerate representations.

 

Question

With regard to the  symmetry operation of the  point group, show that  and  transform together according to the two-dimensional irreducible representation .

Answer

With respect to the green line in the below diagram depicting the spherical coordinate system,  and are invariant to the symmetry operation of the point group.

The  and  wavefunctions in spherical coordinates are  and (see this article for derivation).

In other words, the symmetry operation  transforms  into a linear combination of  and (c.f. eq55). Similarly,

Therefore,  and  transform together according to the two-dimensional irreducible representation , with the coefficients of the basis functions  and  in eq62 and eq63 being the entries of the matrix .

 

Repeating the same procedure shown in the Q&A above for the rest of the symmetry operations of the  point group, we have

The transformation of the function  is determined by summing the separate transformations of the -orbitals  and .

 

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Direct product representation

A direct product representation of a group is a representation with each element being the Kronecker product of the elements of two other irreducible representations  and  of  corresponding to the same symmetry operation .

For example, if the dimensions of  and  are 2 and 3 respectively, i.e.  and , then

In general, if the dimensions of  and  are  and  respectively, then the dimension of  is :

 

Question

Show that

Answer

With reference to eq75, let’s denote the block  by . So,

 

If  is also a representation of , then it must be consistent with the closure property of , i.e. . The proof is as follows:

where the 2nd equality uses the identity proven in the Q&A above.

With reference to eq75, the trace of  is

It follows that the direct product of three representations is

which can be extended to direct products of more than three representations.

 

Question

With reference to the  point group, show that , while  can be decomposed into the direct sum of .

Answer

Using eq76, we include  and  in the character table of the  point group as follows:

Clearly, , which implies that  is an irreducible representation of . Since the number of irreducible representations of a point group is equal to the number of classes of the group,  is a reducible representation because its characters are not equivalent to the characters of any of the 3 irreducible representations. We then use eq27 for the decomposition of , where

Therefore, .

 

Finally, consider the sets of linearly independent functions

i) , where
ii) , where
iii) , where

that transform according to ,  and  respectively.

Since , and are sets of linearly independent functions, we can write

According to a previous article, we can also write

where is a symmetry operation of and , and  are the matrix entries of , and respectively.

 

Question

What is the relation between the matrix entries of , and in ?

Answer

Let the matrix entries of , and  be , and respectively, where

Using the ordering convention called dictionary order, where the order of or  is given by

or in terms of the notation or ,

In other words, is determined by and , and is determined by and . For example, if and ,

We can then express the matrix entries of as

 

Multiplying eq77a by eq77b,

From eq77g, we have  and we can rewrite the RHS of eq77h as , where . Hence, eq77h is equivalent to eq77c. Combining eq77f and eq77h, we have

This implies that if the functions and are bases for the irreducible representations of and of a point group respectively, then is the basis for , which is the direct product of  and . In other words,

If the functions  and  transform according to the irreducible representations of  and  of a point group respectively, then the function , which is the product of the functions  and  must transform according to the direct product of .

 

Question

Show that the reducible representation of the direct product of two irreducible representations  and  contains the totally symmetric representation if and only if is the complex-conjugate representation of .

Answer

Using eq27a, the number of times the totally symmetric representation appears in the decomposition of is

Substitute eq76 in eq78

Comparing eq79 with eq22,  if and only if  and vanishes otherwise.

 

 

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Vanishing integrals

A vanishing integral is a continuous sum of a function that evaluates to zero. Group theory allows us to identify a vanishing integral as a result of the symmetry of the system.

Consider a set of basis functions  that transforms according to a representation  of a point group . The integral of an element of the function  is a scalar, which is invariant to any symmetry operation  of the coordinate system, i.e.

Since there are  number of symmetry operations of , we can write  equations of eq82 and sum them to give

Let , where  and  belong to the sets of linearly independent functions  and respectively, and where  and  are bases for the irreducible representations  and respectively of . Substituting  and eq77i in eq83,

Rewrite eq20 as  and substitute it in eq84 to give

We conclude that:

Rule 1

If  and  transform according to two different non-equivalent irreducible representations  and  respectively, the integral of their product over all space is necessarily zero

This implies that basis functions belonging to different non-equivalent irreducible representations are orthogonal. With respect to eq76 and eq77i, if  transforms according to the totally symmetric representation of and  transforms according to a different irreducible representation  of , then  must transform according to , which is obviously not the total symmetric representation of . Therefore,

Rule 2 If a function transforms according to an irreducible representation that is not the totally symmetric representation of a group, its integral over all space is necessarily zero

Next, we look at the quantum-mechanical integral , where  and  are wavefunctions and  is a quantum-mechanical operator, which is also a function.

 

Question

Why is an operator a function?

Answer

A function is a mapping of each element of the set  in one vector space of functions  to one element of the set in another vector space of functions , which may be the same space as . Similarly, an operator maps each element (e.g. a wavefunction) of a vector space of functions  to another element in another vector space of functions , which may be the same space as .

 

Due to rule 1,  is necessarily zero if  and  transform according to two different non-equivalent irreducible representations. Let’s consider the integral , where ,  and  transform according to the irreducible representations ,  and respectively of a group . Since the non-relativistic Hamiltonian  is invariant under any symmetry operation, it transforms according to the totally symmetric irreducible representation of . With reference to eq76 and eq77i,  consequently transforms according to . Therefore, if .

 

Question

Do the integrals and vanish in a molecule?

Answer

is totally symmetric, while  and  transform according to two different non-equivalent irreducible representations and respectively. Therefore, . Using eq76,  transforms according to . Hence,  is not necessarily zero.

 

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Projection operator (group theory)

A projection operator is used to construct a linear combination of a set of basis functions that spans an irreducible representation of a point group .

Using eq55, we can write

where  is the -th symmetry operation of ,

 is a set of  basis functions of the -th irreducible representation  of , and  is the -th row and -column matrix entry of .

Multiplying eq90 by  and sum over ,

Substituting eq20a in the above equation,

where .

is a linear combination of the symmetry operators  with coefficients that are entries of the matrix representations of . If  and ,

We call  the projection operator, which generates a basis  of the irreducible representation  from another basis  of . The significance of this is that if we know one member of a set of basis functions of an irreducible representation, then we can project all the other members of the set.

Question

Use eq93 and the general function  to show that the basis functions  and  belong to the degenerate irreducible representation  of the point group .

Answer

For ,

Substituting the values of this table into the above equation and simplifying, we have . If we repeat the procedure for , we have . Both projections eliminate . It is obvious by inspecting the matrix entries of the matrices of  that and also eliminate . Since  is not projected out of , it does not belong to . This implies that  transforms according to . In other words, any linear combination of a set of functions that transforms according to an irreducible representation of a point group is also a basis of the irreducible representation.

 

We can also define another projection operator, , where  and is the dimension of . This projection operator employs the characters of an irreducible representation instead of matrix entries of every matrix representation, i.e.

In contrast with  (c.f. eq93), which projects or  from  for , projects out the same basis function from (easily shown using eq94). Since any linear combination of a set of basis functions of an irreducible representation of a point group is also a function that transforms according to the irreducible representation, we can express eq96 and eq97 as

where .

Finally, let’s examine the effect of a projection operator on a set of linearly independent basis functions  of an -dimensional reducible representation of a point group . Since  is a basis for , we have  or equivalently,

where .

Let’s analyse the example for , where :

can undergo a similarity transformation to , which has the block-diagonal form of:

where each block is an irreducible representation.

It is obvious that  is a direct sum of irreducible representations of . The direct sum can also be obtained using eq27a. Let’s denote the basis functions of the transformed reducible representation by

where  refers to the dimension of the -th irreducible representation of .

Therefore, we have

Since a similarity transformation involves a change of basis, the old basis functions  can be expressed as a linear combination of the new basis functions :

where , i.e.  also transforms according to the -th irreducible representation of .

Applying the projection operator to eq102 and using eq98 and eq99,

With reference to the LHS of eq103,  results in a linear combination of  because  is itself a linear combination of symmetry operators of  (c.f. eq94). This linear combination of , according to the RHS of eq103, is equal to a function that transforms according to an irreducible representation  of . Such a consequence is used to construct symmetry-adapted linear combination (SALC) of orbitals, which shall be discussed in the next article.

 

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Symmetry-adapted linear combination (SALC)

A symmetry-adapted linear combination (SALC) is a function that consists of a sum of basis functions that are related by symmetry transformations. SALC wave functions of a molecule are building blocks of a molecular orbital (MO) of the molecule. Such SALCs of a certain symmetry combine with other atomic orbitals of the same symmetry to form MO wave functions, which can be used as trial functions in the Hartree-Fock-Roothaan procedure to calculate energy states of a molecule. These energy states are eventually presented in an MO diagram.

Consider the molecule . The first step is to choose a basis set to generate a representation of the  point group, which the molecule belongs to. The choice of a basis set is related to the way bonds are formed in . The elements of the set may be vectors representing the bonds or wave functions of atoms forming the bonds. For this example, we shall use the basis set of valence -orbitals of the four atoms.

Question

How do we know if the basis set that we have chosen transform according to a representation of the point group?

Answer

Since a symmetry operation transforms into an indistinguishable copy of itself, each of the element of the basis set transforms into another element of the set. We can therefore express the transformation as eq55, which implies that the set generates a representation of the point group.

 

The diagram above shows the four orbitals, with the principal axis of rotation perpendicular to the plane of the screen.  Symmetry operations acting on the basis set produce the following results:

The corresponding matrix transformation equations are:

The transformation matrices form a reducible representation of the  point group:

 

Question
    1. How do we verify that the matrices form a representation of the  point group?
    2. Why is a reducible representation?
Answer
    1. Check that they satisfy the closure property of the group.
    2. The original character table before the addition of has no 4-dimensional irreducible representation. Therefore, must be reducible.

 

Using eq27a, we have

which implies that the decomposition of the reducible representation is .

Applying the projection operator on for ,

This means that , which is an SALC that normalises to , is a basis for . Similarly, the projections of  and for result in , while the projection of  for  returns .

 

Question

How do we normalise ?

Answer

Since , we have .

 

Applying the projection operator on , ,  and for , we have

Since the number of orthogonal basis functions of a representation corresponds to the dimension of the representation (see this article for explanation), there are four orthogonal basis functions for . If the first two are and , the remaining two must transform according to (see Q&A below for explanation).

 

Question

Explain the last sentence in the above paragraph.

Answer

A symmetry operation sends each orthogonal basis into another linear combination, whose coefficients form a column of the matrix representation under that symmetry operation (see this article for explanation). is related to by a similarity transformation. This implies that they are equivalent and have the same trace. can also undergo a similarity transformation to block-diagonal form  because it is a reducible representation. The blocks correspond to the direct sum of . Therefore, we can select a set of four orthogonal basis functions that transform according to , with the coefficients of each basis function forming a column entry for . Since the first two basis functions generate two columns corresponding to and , the last two basis functions must produce the columns corresponding to .

 

However, the normalised functions , , and  are not orthogonal to each other. To find the two remaining orthogonal basis functions for , we arbitrarily select as one the two functions. As any linear combination of a set of functions that transforms according to is a basis of that is equivalent to (see this article for explanation), the other basis SALC function is determined by taking a linear combination of ,  and , e.g. , which normalises to .

 

Question

Show that is orthogonal to  but not orthogonal to .

Answer


 

Therefore, the three orthogonal SALC functions describing the orbitals of the three hydrogen atoms (see above diagram) are ,  and . They combine with the atomic orbitals of the central nitrogen atom to form MO wave functions, which will be further elaborated in the next article.

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