H
h i ¼
0jH 0 j0
h
iÀ 2eFa 0jxj1
h
iþ a
2 1jH 0 j1
h
i
1 þ a 2
:
ð5:113Þ
Here, defining the ground state energy and first excited energy as E 0 and E 1 ,
respectively, we put
H 0 0i ¼ E 0
j
j 0i and H 0 j 1i ¼ E 1 j 1i,
ð5:114Þ
where E 0 6 ¼ E 1 . As already shown in Chaps. 1 and 2, both the systems have
nondegenerate energy eigenstates. Then, we have
H
h i ¼
E 0 À 2eFa 0jxj1
h
iþ a
2 E 1
1 þ a 2
:
ð5:115Þ
Now we wish to determine the condition where hHi has an extremum. That is, we
are seeking a that satisfies
∂ H
h i
∂a
¼ 0:
ð5:116Þ
Calculating LHS of (5.116) by use of (5.115), we have
∂ H
h i
∂a
¼
2eF 0jxj1
h
ia
2
þ 2 E 1 À E 0
ð
Þ a À 2eF 0jxj1
h
i
1 þ a 2
ð
Þ
2
:
ð5:117Þ
For
∂ H
h i
∂a
to be zero, we must have
eF 0jxj1
h
ia
2
þ E 1 À E 0
ð
Þ a À eF 0jxj1
h
i¼ 0:
ð5:118Þ
Then, solving the quadratic equation (5.118), we obtain
a ¼
E 0 À E 1 Æ E 1 À E 0
ð
Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 þ
4 eF 0jxj1
h
i
½
Š
2
E 1 ÀE 0
ð
Þ
2
r
2eF 0jxj1
h
i
:
ð5:119Þ
Taking the plus sign in the numerator of (5.119) and using
ffiffiffiffiffiffiffiffiffiffiffiffi
1 þ Δ
p
% 1 þ
Δ
2
ð5:120Þ
for small Δ, we obtain
176
5 Approximation Methods of Quantum Mechanics