There are many ways to "verify" a character table. One way to do this is to calculate the character table starting with the given basis functions. This is what we shall do here. This will solve both parts (a) and (b).
We note that the character of a class can be obtained by applying one of the symmetry operations in the class to the basis function (or functions) and then summing up the diagonal elements. To do this for all the classes it is convenient to first create a table listing the effect of the symmetry operations on the spatial coordinates x, y and z.
For the Td group we obtain immediately the following list:
{E}: identity ; xyz-->xyz
{C2}: 2-fold rotation about the x-axis; xyz-->x,-y,-z
{C3}: 3-fold clockwise rotation about the [111]-axis; xyz-->yzx
{S4}: 4-fold clockwise rotation about the x-axis followed by a reflection onto the yz-plane ; xyz-->-x,z,-y
{s}: reflection onto the [110]-plane; xyz-->yxz
The effects of these operations on the different basis functions are summarized below:
A1 Irreducible Representation with basis function: xyz
Since all the symmetry operations either do not change the signs or change the signs in pairs it is clear that all the them leaves the function xyz unchanged and therefore all the characters are equal to unity as expected on the identity representaion.
Combining these results we obtain the characters for A1 as:
| Classes | {E} | {C2} | {S4}: | {s} | {C3} |
| Characters of A1 | 1 | 1 | 1 | 1 | 1 |
A2 Irreducible Representation with basis function: x4(y2-z2)+ y4(z2-x2)+ z4(x2-y2)
If a symmetry operation does not change the ordering of x, y and z then it will not change the basis function and the corresponding character should be unity. This is the case for the operation {E}and the two proper rotations: {C2}and {C3}.
In the case of the other two improper rotations {S4}and {s}, the cyclic order of x, y and z is reversed. This changes the sign of the basis function and therefore the characters are both -1.
Combining these results we obtain the characters for A2 as:
| Classes | {E} | {C2} | {S4}: | {s} | {C3} |
| Characters of A2 | 1 | 1 | -1 | -1 | 1 |
The operation {C2} changes only the signs of y and z and therefore leaves both f1 and f2 unchanged i.e. its character is 2.
The operation {C3} changes the orders of x, y and z. It changes f1 into (y2-z2)=
(-½)(f1+2f2) and f2 into x2-½(y2+z2) =(¾)f1-(f2/2). Thus the two diagonal elements are both -½ and the character is -1.
The operation {S4} interchanges y2 and z2. It changes f1 into (x2-z2)=
(½)(f1-2f2) and f2 into y2-½(x2+z2) =(-¾)f1-(f2/2). Thus the two diagonal elements are ½ and -½, respectively and the character is 0.
The operation {s} interchanges x2 and y2. It changes f1 into (y2-x2)=-f1 and f2 into z2-½(x2+y2) =f2. Thus the two diagonal elements are -1 and 1, respectively and the character is 0.
Combining these results we obtain the characters for E as:
| Classes | {E} | {C2} | {S4}: | {s} | {C3} |
| Characters of E | 2 | 2 | 0 | 0 | -1 |