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so that
cos
c 2
2 r
H 0 exp ð cr Þ r cos
½
u
¼ exp ð cr Þð 2c 1 Þ
u
ð 10
:
129 Þ
We then obtain for the integral:
"
# exp ð 2cr Þ
ð
dxx 2 1
1
c 5
p
c 2
2 r 2
h 2p z j H 0 j 2p z
drr 2
2
p
ð 2c 1 Þ r
1
0
2
4
3
5 ¼
c 2
2
c 2
2
4
3 2
ð 2c Þ
4 3 2
ð 2c Þ
c
2
3 c 5
¼
ð 2c 1 Þ
4
5
ð 10
:
130 Þ
giving for the excitation energy
1
« ¼h 2p z j H 0 E 0 j 2p z
2 ð c 2
c þ 1 Þ
ð 10
:
131 Þ
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