The vibrational modes associated with the
NH bonds in NH_{3} is found as follows:
The ammonia
molecule belongs to the C_{3v}
point group.
The
number of bonds that remain unshifted by the operations are represented
by the following:

E

2C_{3}

3s_{v}

G_{3H}

3

0

1

There are three types of irrepresentations in this point group: A_{1},
A_{2} and E.
The total
number of symmetry operations is 6. There is one E element, two C_{3}
axes and three s_{v}
planes: h = 1 + 2 + 3 = 6.
The irrepresentations
are calculated as follows using the formula (the colours indicate where
the numbers came from)
The number of times the A_{1} occurs in the
representation is:
n_{A1}

=

1/6

[

1*1*3

+

2*1*0

+

3*1*1

]

=

1/6 [3 + 0 + 3]

=

1





1E


2C_{3}


3s_{v}










g_{E}
= 1


g_{C3}
= 2


g_{s}_{v}
= 3










c_{E}(A_{1})
= 1


c_{C3}(A_{1})
= 1


c_{s}_{v}(A_{1})
= 1










c_{E}(3H)
= 3


c_{C3}(3H)
= 0


c_{s}_{v}(3H)
= 1






The number of times the A_{2} occurs in the representation
is:
n_{A2}

=

1/6

[

1*1*3

+

2*1*0

+

3*1*1

]

=

1/6 [3 + 0 3]

=

0





1E


2C_{3}


3s_{v}










g_{E}
= 1


g_{C3}
= 2


g_{s}_{v}
= 3










c_{E}(A_{2})
= 1


c_{C3}(A_{2})
= 1


c_{s}_{v}(A_{2})
= 1










c_{E}(3H)
= 3


c_{C3}(3H)
= 0


c_{s}_{v}(3H)
= 1






The number of times the E occurs in the representation is:
n_{E}

=

1/6

[

1*2*3

+

2*1*0

+

3*0*1

]

=

1/6 [6 + 0 + 0]

=

1





1E


2C_{3}


3s_{v}










g_{E}
= 1


g_{C3}
= 2


g_{s}_{v}
= 3










c_{E}(E)
= 2


c_{C3}(E)
= 1


c_{s}_{v}(E)
= 0










c_{E}(3H)
= 3


c_{C3}(3H)
= 0


c_{s}_{v}(3H)
= 1






The representations reduces to A_{1} and E:
G_{3H}
= A_{1} + E
This can be readily checked by adding up the characters for these two
rows: if correct, the representation should be regenerated:
