Solutions to Exercises
235
C. In this situation, the natural orbitals have been calculated from the scalar density
matrix of an approximate ground state of 7 He. Thus, nucleon occupation is not
negligible even on high lying natural orbitals. Consequently, the ground state of
7 He is only approximately described using a few natural orbitals, even without
the truncation of model space. Indeed, the 3p-3h configurations present in a nontruncated model space cannot compensate for the missing high lying natural
orbitals.
Exercise V.
A. One can clearly identify bound states in the negative energy region. On the
contrary, unbound resonant states are embedded in the discretized continuum of
many-body scattering states, making them impossible to identify. This situation
is qualitatively the same as that of Fig. 5.2.
B. As continuum coupling is small for 18 O, the overlap between the eigenstates
obtained in pole approximation and in full space is close to one. As a consequence, the identification of resonances can be made without ambiguity. As
the Berggren basis is complete, resonant eigenstates are stationary by slightly
changing the Berggren basis contour, in full analogy with Fig. 5.2.
Exercise VI.
A. The ground state and the first excited state of 6 He and 6 Be are either weakly
bound or resonance states. As one has either two valence neutrons or two valence
protons, the two-body interaction is in T = 1 channel. Hence, the nuclear
interaction does not provide with much binding energy. Consequently, the use of
Berggren basis generated by the Woods-Saxon potential of the core can provide
with numerically stable results. This explains why the bases generated by the
multi-Slater determinant coupled Hartree-Fock and Woods-Saxon potentials are
equivalent.
B. In the case of 6 Li, one has a valence proton and a valence neutron, so that both
T = 0 and T = 1 components of the nuclear interaction are present. The T = 0
part of the nuclear interaction is responsible for the well-bound character of
6 Li. Indeed, the ground state of 5 Li is a broad resonance, whereas that of 6 Li is
bound by 3.7 MeV. Moreover, the spectrum of 6 Li has unbound states built from
the 0p 1/2 one-body resonance states, whose widths are small compared to that
of 5 Li. 0p 1/2 one-body state has an extremely large width of 5–7 MeV when it is
generated by the Woods-Saxon potential of a core. Consequently, the Berggren
basis produced by the Woods-Saxon potential of the core has very large twobody matrix elements. This makes the calculation imprecise.
On the contrary, the multi-Slater determinant coupled Hartree-Fock potential
removes strong continuum coupling, so that calculations are precise using its
Berggren basis. Indeed, the 0p 3/2 and 0p 1/2 one-body states are either bound or
narrow resonances in the multi-Slater determinant coupled Hartree-Fock basis,
so that Hamiltonian matrix elements involving unbound states are smaller. Let
235
C. In this situation, the natural orbitals have been calculated from the scalar density
matrix of an approximate ground state of 7 He. Thus, nucleon occupation is not
negligible even on high lying natural orbitals. Consequently, the ground state of
7 He is only approximately described using a few natural orbitals, even without
the truncation of model space. Indeed, the 3p-3h configurations present in a nontruncated model space cannot compensate for the missing high lying natural
orbitals.
Exercise V.
A. One can clearly identify bound states in the negative energy region. On the
contrary, unbound resonant states are embedded in the discretized continuum of
many-body scattering states, making them impossible to identify. This situation
is qualitatively the same as that of Fig. 5.2.
B. As continuum coupling is small for 18 O, the overlap between the eigenstates
obtained in pole approximation and in full space is close to one. As a consequence, the identification of resonances can be made without ambiguity. As
the Berggren basis is complete, resonant eigenstates are stationary by slightly
changing the Berggren basis contour, in full analogy with Fig. 5.2.
Exercise VI.
A. The ground state and the first excited state of 6 He and 6 Be are either weakly
bound or resonance states. As one has either two valence neutrons or two valence
protons, the two-body interaction is in T = 1 channel. Hence, the nuclear
interaction does not provide with much binding energy. Consequently, the use of
Berggren basis generated by the Woods-Saxon potential of the core can provide
with numerically stable results. This explains why the bases generated by the
multi-Slater determinant coupled Hartree-Fock and Woods-Saxon potentials are
equivalent.
B. In the case of 6 Li, one has a valence proton and a valence neutron, so that both
T = 0 and T = 1 components of the nuclear interaction are present. The T = 0
part of the nuclear interaction is responsible for the well-bound character of
6 Li. Indeed, the ground state of 5 Li is a broad resonance, whereas that of 6 Li is
bound by 3.7 MeV. Moreover, the spectrum of 6 Li has unbound states built from
the 0p 1/2 one-body resonance states, whose widths are small compared to that
of 5 Li. 0p 1/2 one-body state has an extremely large width of 5–7 MeV when it is
generated by the Woods-Saxon potential of a core. Consequently, the Berggren
basis produced by the Woods-Saxon potential of the core has very large twobody matrix elements. This makes the calculation imprecise.
On the contrary, the multi-Slater determinant coupled Hartree-Fock potential
removes strong continuum coupling, so that calculations are precise using its
Berggren basis. Indeed, the 0p 3/2 and 0p 1/2 one-body states are either bound or
narrow resonances in the multi-Slater determinant coupled Hartree-Fock basis,
so that Hamiltonian matrix elements involving unbound states are smaller. Let