Solutions to Exercises
479
C. One has demonstrated that ˆ
H is diagonal if n i > n max or n f > n max , so that
n i = n f = n, therein. As one assumes that | i j i τ i = | f j f τ f = |j τ as well,
H if reads
H if =
n i =n f
u n i (r i ) u n f (r f ) ˆ
A {{Ψ
J T f
T f
| ⊗ ⊗u n f j τ |}
J A
M A
| ˆ
H | ˆ
A {|Ψ
J T i
T i
⊗ |u n i j τ
J A
M A
+
n≤nmax
u n (r i ) u n (r f ) ˆ
A {{Ψ
J T f
T f
| ⊗ ⊗u n j τ |}
J A
M A
| ˆ
H | ˆ
A {|Ψ
J T i
T i
⊗ |u n j τ
J A
M A
+
n≥0
u n (r i ) u n (r f ) [[Ψ
J T f
T f
| ⊗ ⊗u n j τ |]
J A
M A
| ˆ
H T + ˆ
H P |[|Ψ
J T i
T i
⊗ |u n j τ
J A
M A
−
n≤nmax
u n (r i ) u n (r f ) [[Ψ
J T f
T f
| ⊗ ⊗u n j τ |]
J A
M A
| ˆ
H T + ˆ
H P |[|Ψ
J T i
T i
⊗ |u n j τ
J A
M A
.
(9.180)
The infinite sum in the H if matrix element has been separated in two parts: one
where n ≥ 0 and the other one where n ≤ n max . It is indeed more convenient to
deal with a sum starting at n = 0 to have completeness relations appear. Let us
then calculate the infinite sum of Eq. (9.180):
n≥0
u n (r i )u n (r f ) {{Ψ
J T f
T f
| ⊗ ⊗u n j τ |}
J A
M A
| ˆ
H T + ˆ
H P |{|Ψ
J T i
T i
⊗ |u n j τ
J A
M A
=
n≥0
u n (r i )u n (r f )
J T f
T f
| ˆ
T + ˆ
U core + ˆ
V res |Ψ
J T i
T i
n j τ |u n j τ
+ +Ψ
J T f
T f
|Ψ
J T f
T i
n j τ |ˆ t + ˆ
U basis |u n j τ
= =Ψ
J T f
T f
| ˆ
H |Ψ
J T i
T i
n≥0
u n (r i )u n (r f ) + +Ψ
J T f
T f
|Ψ
J T i
T i
n≥0
e n u n (r i )u n (r f ) ,
(9.181)
where e n is the single-particle energy of |u n j τ , and where Eq. (9.13) has been
used.
As |Ψ
J T i
T i
and |Ψ
J T f
T f
are eigenvectors of ˆ
H , the matrix element of Eq. (9.181)
is nonzero only if |Ψ
J T i
T i
= |Ψ
J T f
T f
= |Ψ
J T
T . Hence, in this case, the matrix
element of Eq. (9.181) is equal to
¯
h 2
2m
−
d 2
dr 2
i
+
+ 1)
r 2
i
+ E T
δ(r i − r f ) + U basis (r i , r f ) ,
(9.182)