Solutions to Exercises
481
where
V c i c f (r i , r f ) =
n max
n i n f α i α f
u n i (r i )u n f (r f )c
(i)
α i
c
(f)
α f
× ×SD α f | ˆ
H |SD α i − δ c i c f
n≤n max
(E T + e n ) u n (r i )u n (r f )
is the finite-range coupling potential induced by the nucleon–nucleon interaction
between c i and c f .
Exercise II.
A. ˆ
Q is a Hermitian operator because all the one-body states entering ˆ
Q are bound
states. As these one-body states are orthogonal, ˆ
Q 2 reads
ˆ
Q
2
=
s∈core
|s
s ∈core
|s
| =
s,s ∈core
|s
|
=
s,s ∈core
δ ss |s
| =
s∈core
|s = ˆ
Q .
ˆ
Q is a projector because ˆ
Q is Hermitian and ˆ
Q 2 = ˆ
Q. As ˆ
P = ˆ
1 − ˆ
Q, therefore
ˆ
P is Hermitian and commutes with ˆ
Q. One directly obtains
ˆ
P
2
= ( ˆ
1 − ˆ
Q)
2
= ˆ
1 + ˆ
Q
2
− 2 ˆ
Q = ˆ
1 − ˆ
Q = ˆ
P
ˆ
Q ˆ
P = ˆ
P ˆ
Q = ˆ
Q − ˆ
Q
2
= 0 .
Consequently, ˆ
P is a projector orthogonal to ˆ
Q.
If |s ∈ core, then ˆ
Q |s = |s, and one has ˆ
Q |s = 0 if |s /
∈ core.
Conversely, ˆ
P |s = |s if |s /
∈ core, so that ˆ
P |s = 0 if |s ∈ core. The
one-body states inside and outside of the core form a complete basis, so that one
cannot generate other eigenvectors as those stated. Thus, the eigenvectors of ˆ
Q
and ˆ
P are the states inside and outside the core. Their eigenvalues are 0 or 1, as
is the case for all projectors.
B. One considers an arbitrary operator ˆ
A acting on the Hilbert space. If one wants
ˆ
A to act on the P space, one multiplies ˆ
A on the right by ˆ
P to eliminate all the
core components of input states.
Output states should not have any component in the Q space either. Hence,
one must also multiply ˆ
A by ˆ
P from the left. Therefore, the ˆ
A operator restricted
to the P space is ˆ
P ˆ
A ˆ
P .
If ˆ
A is Hermitian, one has
( ˆ
P ˆ
A ˆ
P )
†
= ˆ
P
† ( ˆ
P ˆ
A)
†
= ˆ
P
† ˆ
A
† ˆ
P
†
= ˆ
P ˆ
A ˆ
P .
Consequently, if ˆ
A is Hermitian, ˆ
P ˆ
A ˆ
P is also Hermitian.