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Matrix Elements of the Finite-Range Interaction

  • Walid Younes
  • Daniel Marc Gogny
  • Jean-François Berger
Chapter
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Part of the Lecture Notes in Physics book series (LNP, volume 950)

Abstract

In this chapter we give explicit expression for the matrix elements of the finite-range interaction used in the second part of the book in the deformed harmonic-oscillator basis. The components of the interaction explicitly given are: the central (Gaussian) part, the spin-orbit part, a density-dependent contribution, and both the exact and Slater approximation to the Coulomb interaction.

Microscopic approaches using phenomenological effective interactions of the Skyrme [1, 2] and finite-range [3, 4] type have a long history of successful applications to a variety of nuclear-physics phenomena [5, 6, 7, 8]. These interactions are parameterizations which usually include a density-dependent part, as suggested by Brueckner’s G-matrix theory [9, 10], and are based on the hypothesis that nuclear properties can be well reproduced at the mean-field approximation with these forces. This is essentially the philosophy of the Brueckner-Hartree-Fock approach [11], The parameters of the interaction can then be adjusted within this framework using a judicious choice of nuclear data, such as the properties of nuclear matter and of select finite nuclei [4, 12, 13]. The calculations in the second part of this book will rely on a finite-range effective interaction [3, 4], which we describe in this chapter.

2.1 General Form of the Interaction

The finite-range effective interaction potential used in this work takes the form [3, 4] where \(\hat {P}_{\sigma }\) and \(\hat {P}_{\tau }\) are spin- and isospin-exchange operators, ρ is the total nuclear density, σ1 and σ2 are vectors whose components are the Pauli spin matrices, and the remaining parameters are listed in Table 2.1. Throughout this book, we have used the D1S parameterization given in Table 2.1.
Table 2.1

Different parameterization for the finite-range effective interaction in Eq. (2.1). (The values are adapted from [14, 15, 16])

 

i

\(\begin {array}{@{\hspace{-2pt}}l} \mu _{i}\\ \mathrm{(fm)} \end {array}\)

\(\begin {array}{@{\hspace{-2pt}}l} W_{i}\\ \left (\mathrm{MeV}\right )\end {array}\)

\(\begin {array}{@{\hspace{-2pt}}l} B_{i}\\ \mathrm{(MeV)} \end {array}\)

\(\begin {array}{@{\hspace{-2pt}}l} H_{i}\\ \mathrm{(MeV)} \end {array}\)

\(\begin {array}{@{\hspace{-2pt}}l} M_{i}\\ \mathrm{(MeV)} \end {array}\)

\(\begin {array}{@{\hspace{-2pt}}l} t_{0}\left (\mathrm{MeV}\right .\\ \left .\cdot \mathrm{fm}^{3\left (\alpha +1\right )}\right ) \end {array}\)

x 0

α

\(\begin {array}{@{\hspace{-2pt}}l} W_{LS}\\ \left (\mathrm{MeV}\cdot \mathrm{fm}^{5}\right ) \end {array}\)

D1

1

0.7

−402.40

−100.00

−496.17

−23.561

1350.0

1

1/3

115

 

2

1.2

−21.297

−11.772

37.270

−68.810

    

D1S

1

0.7

1720.30

1300.0

−1813.53

1397.6

1390.6

1

1/3

130

 

2

1.2

103.639

−163.483

162.812

−223.933

    

D1N

1

0.8

−2047.6

1700.0

−2414.9

1519.4

1609.5

1

1/3

115

 

2

1.2

293.02

−300.78

414.59

−316.84

    

D1M

1

0.50

−12797.57

14048.85

−15144.43

11963.89

1562.22

1

1/3

115.36

 

2

1.00

490.95

−752.27

675.12

−693.57

    

2.2 The Deformed Harmonic-Oscillator Basis

In applications to fission presented subsequently, the HFB equations have been solved by expanding the qp states onto finite bases constituted of truncated sets of axially deformed harmonic oscillator (HO) states. This choice is very convenient since the matrix elements of the different terms of the finite-range interaction (2.1) then assumes a closed analytical form (see, e.g., Refs. [17, 18] for a review of the numerical methods that are used in this context). 1 Such a deformed HO basis can be written in cylindrical coordinates \(\left (\rho ,\varphi ,z\right )\) as
$$\displaystyle \begin{aligned} \begin{aligned}\varPhi_{n_{r},\varLambda,n_{z}}\left(\mathbf{r};b_{\perp},b_{z}\right) & =\varPhi_{n_{r},\varLambda}\left(\rho,\varphi;b_{\perp}\right)\varPhi_{n_{z}}\left(z;b_{z}\right)\\ & =\varPhi_{n_{r},\left|\varLambda\right|}\left(\rho;b_{\perp}\right)\frac{e^{\mathrm{i}\varLambda\varphi}}{\sqrt{2\pi}}\varPhi_{n_{z}}\left(z;b_{z}\right) \end{aligned} {} \end{aligned} $$
(2.2)
with the radial-component function
$$\displaystyle \begin{aligned} \begin{aligned}\varPhi_{n_{r},\left|\varLambda\right|}\left(\rho;b_{\perp}\right) & =\mathcal{N}_{n_{r}}^{\left|\varLambda\right|}\eta^{\left|\varLambda\right|/2}e^{-\eta/2}L_{n_{r}}^{\left|\varLambda\right|}\left(\eta\right)\end{aligned} {} \end{aligned} $$
(2.3)
defined in terms of associated Laguerre polynomials \(L_{n_{r}}^{\left |\varLambda \right |}\left (\eta \right )\) as a function of
$$\displaystyle \begin{aligned} \begin{array}{rcl} \eta &\displaystyle \equiv &\displaystyle \rho^{2}/b_{\perp}^{2} \end{array} \end{aligned} $$
The quantity Λ corresponds to the projection of the orbital angular momentum onto the symmetry axis Oz, whereas nr and nz count the number of nodes of the wave function along the radial and z directions, respectively. In Eq. (2.3) the normalization constant is given by
$$\displaystyle \begin{aligned} \begin{aligned}\mathcal{N}_{n_{r},\left|\varLambda\right|} & \equiv\frac{1}{b_{\perp}}\left[\frac{2n_{r}!}{\left(n_{r}+\left|\varLambda\right|\right)!}\right]^{1/2}\end{aligned} {} \end{aligned} $$
(2.4)
The Cartesian, z-axis-component function in Eq. (2.2),
$$\displaystyle \begin{aligned} \begin{aligned}\varPhi_{n_{z}}\left(z;b_{z}\right) & =\mathcal{N}_{n_{z}}e^{-\xi^{2}/2}H_{n_{z}}\left(\xi\right)\end{aligned} {} \end{aligned} $$
(2.5)
is expressed in terms of Hermite polynomials \(H_{n_{z}}\left (\xi \right )\) with
$$\displaystyle \begin{aligned} \begin{array}{rcl} \xi &\displaystyle \equiv &\displaystyle z/b_{z} \end{array} \end{aligned} $$
and normalization constant
$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{N}_{n_{z}} &\displaystyle \equiv &\displaystyle \frac{1}{\left(b_{z}\sqrt{\pi}2^{n_{z}}n_{z}!\right)^{1/2}} \end{array} \end{aligned} $$
The harmonic-oscillator functions defined in Eqs. (2.2) and (2.5) satisfy the orthonormalization conditions
$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{0}^{\infty}\rho d\rho\int_{0}^{2\pi}d\varphi\,\varPhi_{n_{r},\varLambda}^{*}\left(\rho,\varphi;b_{\perp}\right)\varPhi_{n_{r}^{\prime},\varLambda^{\prime}}\left(\rho,\varphi;b_{\perp}\right) &\displaystyle = &\displaystyle \delta_{n_{r},n_{r}^{\prime}}\delta_{\varLambda,\varLambda^{\prime}} \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{-\infty}^{\infty}dz\,\varPhi_{n_{z}}\left(z;b_{z}\right)\varPhi_{n_{z}^{\prime}}\left(z;b_{z}\right) &\displaystyle = &\displaystyle \delta_{n_{z},n_{z}^{\prime}} \end{array} \end{aligned} $$
The parameters b and bz appearing in the harmonic-oscillator function definitions are usually treated as variational parameters in HFB calculations, and chosen to minimize the energy. These parameters can also be written in terms of harmonic-oscillator energies
$$\displaystyle \begin{aligned} \hbar\omega_{\bot}=\frac{\hbar^{2}}{mb_{\bot}^{2}},\quad \hbar\omega_{z}=\frac{\hbar^{2}}{mb_{z}^{2}},\quad \omega_{0}\equiv\omega_{\bot}^{2}\omega_{z} \end{aligned}$$
where m is the nucleon mass. The quantum numbers of the basis, nr = 0, 1, 2, …, Λ = 0, ±1, ±2, …, nz = 0, 1, 2, …, are truncated for a given maximum shell number N according to the relation [19]
$$\displaystyle \begin{aligned} \begin{array}{rcl} \hbar\omega_{\bot}\left(2n_{r}+\left|\varLambda\right|+1\right)+\hbar\omega_{z}\left(n_{z}+\frac{1}{2}\right) &\displaystyle \leq &\displaystyle \hbar\omega_{0}\left(N+2\right) \end{array} \end{aligned} $$
In addition, a maximum value \(n_{z}^{\left (\mathrm{max}\right )}\) can be imposed on the nz quantum number. Through a systematic study of the optimal values of ħω and ħωz for 240Pu as a function of quadrupole (Q20) and octupole (Q30) constraints, we have obtained the following relations for N = 13 and \(n_{z}^{\left (\mathrm{max}\right )}=25\), by fitting the individual ħω0 and q ≡ ωωz with the functional form
$$\displaystyle \begin{aligned} \begin{aligned}f\left(Q_{30},Q_{20}\right) & =\sum_{i,j}c_{ij}Q_{30}^{i}Q_{20}^{j}\end{aligned} {} \end{aligned} $$
(2.6)
The numerical values of the cij coefficients for ħω0 and the ratio q are given in Tables 2.2 and 2.3, respectively. From this, we can then calculate
$$\displaystyle \begin{aligned} \begin{array}{rcl} \hbar\omega_{\bot} &\displaystyle = &\displaystyle q^{1/3}\hbar\omega_{0}\\ \hbar\omega_{z} &\displaystyle = &\displaystyle q^{-2/3}\hbar\omega_{0} \end{array} \end{aligned} $$
Table 2.2

Coefficients for ħω0 using the functional form of Eq. (2.6)

i

j

c ij

0

0

7.916212603959e+00

0

1

4.416886583910e-03

0

2

−8.119414361741e-06

1

0

1.401836208522e-01

1

1

−1.046648642584e-03

1

2

1.618642016886e-06

2

0

−5.828609820447e-03

2

1

4.468841617704e-05

2

2

−7.011156064446e-08

3

0

6.498801501801e-05

3

1

−4.805951376011e-07

3

2

6.592668106802e-10

4

0

−2.438595317082e-07

4

1

1.551342759362e-09

4

2

−1.401951420300e-12

Table 2.3

Coefficients for q ≡ ωωz using the functional form of Eq. (2.6)

i

j

c ij

0

0

9.574894959751e-01

0

1

8.261269094176e-03

0

2

−6.909503821655e-06

1

0

2.005584042966e-02

1

1

−1.854775781290e-04

1

2

3.043733634606e-07

2

0

−5.096586816081e-05

2

1

1.227377881974e-06

2

2

−2.708065016489e-09

Including the spin quantum number σ = ±1∕2, the basis kets can be written as
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\alpha\right\rangle &\displaystyle = &\displaystyle \left|n_{r}^{\alpha},\varLambda_{\alpha},n_{z}^{\alpha},\sigma_{\alpha}\right\rangle \end{array} \end{aligned} $$
and the total angular-momentum projection on the symmetry axis of the nucleus is Ωα = Λα + σα. If axial symmetry is preserved, then the projection Ω is a good quantum number for the system, and the mean and pairing field matrices discussed in Sects. 2.3.2 and 2.3.3, respectively, will adopt a block-diagonal structure, with the blocks labeled by Ω. Following the notation in [20], we write the time-reversed basis states as
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\bar{\alpha}\right\rangle &\displaystyle = &\displaystyle 2\sigma_{\alpha}\left|\underline{\alpha}\right\rangle \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left|\underline{\alpha}\right\rangle &\displaystyle = &\displaystyle \left|n_{r}^{\alpha},-\varLambda_{\alpha},n_{z}^{\alpha},-\sigma_{\alpha}\right\rangle \end{array} \end{aligned} $$

2.3 Form of the HFB Equations

2.3.1 Total Energy of the System

Compact forms of the HFB equations have previously been given for various symmetries [4, 20]. For convenience, we present here forms of those equations appropriate for axially symmetric solutions.

The total energy of the system, including constraints on the nucleon numbers, can be written as [4]
$$\displaystyle \begin{aligned} \begin{array}{rcl} E &\displaystyle \equiv &\displaystyle \left\langle \tilde{0}\left|\hat{H}-\lambda_{n}\hat{N}_{n}-\lambda_{p}\hat{N}_{p}\right|\tilde{0}\right\rangle \end{array} \end{aligned} $$
or, more explicitly,
$$\displaystyle \begin{aligned} \begin{array}{rcl} E &\displaystyle = &\displaystyle \sum_{q,\alpha\beta}\left(T_{\alpha\beta}-\lambda_{q}\delta_{\alpha\beta}\right)\rho_{\beta\alpha}^{q}+\frac{1}{2}\sum_{qq^{\prime},\alpha\beta\gamma\delta}\bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}}\rho_{\delta\beta}^{q^{\prime}}\rho_{\gamma\alpha}^{q}\\ &\displaystyle &\displaystyle +\frac{1}{4}\sum_{qq^{\prime},\alpha\beta\gamma\delta}\bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}}\kappa_{\delta\gamma}^{q^{\prime}}\left(\kappa_{\beta\alpha}^{q}\right)^{*} \end{array} \end{aligned} $$
where the indices α, β, γ, δ span the basis states, q, q = 0, 1 designate neutron and proton states, respectively, the \(\rho _{\beta \alpha }^{q}\equiv \left \langle \tilde {0}\left |a_{\alpha }^{q\dagger }a_{\beta }^{q}\right |\tilde {0}\right \rangle \) are the matrix elements of the one-body density, and the \(\kappa _{\beta \alpha }^{q}\equiv \left \langle \tilde {0}\left |a_{\beta }^{q}a_{\alpha }^{q}\right |\tilde {0}\right \rangle \) are the matrix elements of the pairing tensor. Let us note that, implicit in the above expression of the energy, is the assumption that the HFB vacuum is the direct product of a proton and neutron vacuum; otherwise ρ and κ would not necessarily be diagonal in q. The coefficients Tαβ are matrix elements of the kinetic-energy operator, and the anti-symmetrized matrix elements of the density-dependent effective interaction are given by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}} &\displaystyle \equiv &\displaystyle \left\langle \alpha\beta\left|V\right|\widetilde{\gamma\delta}\right\rangle \\ {} &\displaystyle = &\displaystyle \left\langle \alpha\beta\left|V\right|\gamma\delta\right\rangle -\left\langle \alpha\beta\left|V\right|\delta\gamma\right\rangle \end{array} \end{aligned} $$
with the properties
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{V}_{\beta\alpha\gamma\delta} &\displaystyle = &\displaystyle -\bar{V}_{\alpha\beta\gamma\delta}\\ {} \bar{V}_{\alpha\beta\delta\gamma} &\displaystyle = &\displaystyle -\bar{V}_{\alpha\beta\gamma\delta} \end{array} \end{aligned} $$
For convenience, we define
$$\displaystyle \begin{aligned} \begin{aligned}E_{kin} & \equiv\sum_{q,\alpha\beta}\left(T_{\alpha\beta}-\lambda_{q}\delta_{\alpha\beta}\right)\rho_{\beta\alpha}^{q}\\ E_{pot} & \equiv\frac{1}{2}\sum_{qq^{\prime},\alpha\beta\gamma\delta}\bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}}\rho_{\delta\beta}^{q^{\prime}}\rho_{\gamma\alpha}^{q}\\ E_{pair} & \equiv\frac{1}{4}\sum_{qq^{\prime},\alpha\beta\gamma\delta}\bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}}\kappa_{\delta\gamma}^{q^{\prime}}\left(\kappa_{\beta\alpha}^{q}\right)^{*} \end{aligned} {} \end{aligned} $$
(2.7)
so that
$$\displaystyle \begin{aligned}\begin{array}{rcl} E &\displaystyle = &\displaystyle E_{kin}+E_{pot}+E_{pair} \end{array} \end{aligned} $$
The matrix elements of the density and pairing tensor operators satisfy the properties
$$\displaystyle \begin{aligned}\begin{array}{rcl} \rho_{\alpha\beta}^{q*} &\displaystyle = &\displaystyle \rho_{\beta\alpha}^{q}\\ \rho_{\bar{\alpha}\bar{\beta}}^{q} &\displaystyle = &\displaystyle \rho_{\alpha\beta}^{q*}\\ \rho_{\bar{\alpha}\beta}^{q} &\displaystyle = &\displaystyle \rho_{\alpha\bar{\beta}}^{q}=0\quad \mathrm{for} \ \alpha,\beta>0 \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned}\begin{array}{rcl} \kappa_{\beta\alpha} &\displaystyle = &\displaystyle -\kappa_{\alpha\beta}\\ \kappa_{\bar{\alpha}\beta} &\displaystyle = &\displaystyle -\kappa_{\alpha\bar{\beta}}^{*}\\ \kappa_{\alpha\beta} &\displaystyle = &\displaystyle \kappa_{\bar{\alpha}\bar{\beta}}=0\quad \mathrm{for} \ \alpha,\beta>0 \end{array} \end{aligned} $$
Then, using the hermiticity of the matrix elements, we can simplify the expression above to get
$$\displaystyle \begin{aligned} \begin{array}{rcl} E_{kin} &\displaystyle = &\displaystyle 2\sum_{q;\alpha,\beta>0}\left(T_{\alpha\beta}-\lambda_{q}\delta_{\alpha\beta}\right)\rho_{\beta\alpha}^{q}\\ &\displaystyle = &\displaystyle 4\sum_{q;\alpha>\beta>0}\left(T_{\alpha\beta}-\lambda_{q}\delta_{\alpha\beta}\right)\rho_{\beta\alpha}^{q}+2\sum_{q;\alpha}\left(T_{\alpha\alpha}-\lambda_{q}\delta_{\alpha\alpha}\right)\rho_{\alpha\alpha}^{q} \end{array} \end{aligned} $$
Or, in a more compact form, and similarly, where we have defined the Hartree-Fock field (with no re-arrangement terms) as
$$\displaystyle \begin{aligned} \begin{aligned}\varGamma_{\alpha\gamma}^{q,nr}\left(pot\right) & \equiv\sum_{q^{\prime},\beta\delta}\bar{V}_{\alpha\beta\gamma\delta}^{qq^{\prime}}\rho_{\delta\beta}^{q^{\prime}}\end{aligned} {} \end{aligned} $$
(2.8)
and which satisfies the properties
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{\bar{\alpha}\bar{\gamma}}^{q,nr}\left(pot\right) &\displaystyle = &\displaystyle \left[\varGamma_{\alpha\gamma}^{q,nr}\left(pot\right)\right]^{*}\\ {} \varGamma_{\gamma\alpha}^{q,nr}\left(pot\right) &\displaystyle = &\displaystyle \left[\varGamma_{\alpha\gamma}^{q,nr}\left(pot\right)\right]^{*} \end{array} \end{aligned} $$
Next, we simplify the pairing contribution to the total energy, where we have defined the pairing field for α, β > 0
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{\alpha\bar{\beta}}^{q} &\displaystyle \equiv &\displaystyle \frac{1}{2}\sum_{q^{\prime};\gamma,\delta>0}\left[\bar{V}_{\alpha\bar{\beta}\gamma\bar{\delta}}^{qq^{\prime}}\kappa_{\bar{\delta}\gamma}^{q^{\prime}}+\bar{V}_{\alpha\bar{\beta}\bar{\gamma}\delta}^{qq^{\prime}}\kappa_{\delta\bar{\gamma}}^{q^{\prime}}\right]\\ {} &\displaystyle = &\displaystyle \frac{1}{2}\sum_{q^{\prime};\gamma,\delta>0}\left[-\bar{V}_{\alpha\bar{\beta}\gamma\bar{\delta}}^{qq^{\prime}}\kappa_{\delta\bar{\gamma}}^{q^{\prime}*}+\bar{V}_{\alpha\bar{\beta}\bar{\gamma}\delta}^{qq^{\prime}}\kappa_{\delta\bar{\gamma}}^{q^{\prime}}\right] \end{array} \end{aligned} $$
and which satisfies the properties
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{\bar{\alpha}\beta}^{q} &\displaystyle = &\displaystyle -\left(\varDelta_{\alpha\bar{\beta}}^{q}\right)^{*}\\ {} \varDelta_{\bar{\beta}\alpha}^{q} &\displaystyle = &\displaystyle -\varDelta_{\alpha\bar{\beta}}^{q} \end{array} \end{aligned} $$

2.3.2 The Hartree-Fock Field

The Hartree-Fock field defined in Eq. (2.8) did not include the so-called “re-arrangement” terms. The more general definition of the field, which includes re-arrangement terms, is
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{\alpha\gamma}^{q} &\displaystyle = &\displaystyle \frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}E_{pot}+\frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}E_{pair}\\ {} &\displaystyle \equiv &\displaystyle \varGamma_{\alpha\gamma}^{q}\left(pot\right)+\varGamma_{\alpha\gamma}^{q}\left(pair\right) \end{array} \end{aligned} $$
The second term, \(\varGamma _{\alpha \gamma }^{q}\left (pair\right )\), arises if the effective interaction contains an explicit density dependence (which is the case of the interaction (2.1)). It is expressed as
$$\displaystyle \begin{aligned} \begin{aligned}\varGamma_{\alpha\gamma}^{q}\left(pair\right) & \equiv\frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}\frac{1}{4}\sum_{q_{1}q_{2},\alpha^{\prime}\beta^{\prime}\gamma^{\prime}\delta^{\prime}}V_{\alpha^{\prime}\beta^{\prime}\gamma^{\prime}\delta^{\prime}}^{q_{1}q_{2}}\kappa_{\delta^{\prime}\gamma^{\prime}}^{q_{2}}\left(\kappa_{\beta^{\prime}\alpha^{\prime}}^{q_{1}}\right)^{*}\\ {} & =\frac{1}{4}\sum_{q_{1}q_{2},\alpha^{\prime}\beta^{\prime}\gamma^{\prime}\delta^{\prime}}\left(\frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}V_{\alpha^{\prime}\beta^{\prime}\gamma^{\prime}\delta^{\prime}}^{q_{1}q_{2}}\right)\kappa_{\delta^{\prime}\gamma^{\prime}}^{q_{2}}\left(\kappa_{\beta^{\prime}\alpha^{\prime}}^{q_{1}}\right)^{*} \end{aligned} {} \end{aligned} $$
(2.9)
As we will see in Sect. 2.4.6, the density-dependent term in the effective interaction of Eq. (2.1) will contribute matrix elements in Eq. (2.9) of the form
$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{\alpha\beta\gamma\delta}^{qq^{\prime}} &\displaystyle = &\displaystyle t_{0}\left(\delta_{\sigma_{\alpha},\sigma_{\gamma}}\delta_{\sigma_{\beta},\sigma_{\delta}}-\delta_{\sigma_{\alpha},\sigma_{\delta}}\delta_{\sigma_{\beta},\sigma_{\gamma}}\right)\left(1-x_{0}\right)\delta_{qq^{\prime}}\\ &\displaystyle &\displaystyle \times\int d^{3}r\,\varPhi_{\alpha}^{*}\left(\mathbf{r}\right)\varPhi_{\beta}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma}\left(\mathbf{r}\right)\varPhi_{\delta}\left(\mathbf{r}\right)\rho^{\lambda}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
which vanish for x0 = 1. Therefore, the Hartree-Fock field reduces to
$$\displaystyle \begin{aligned} \begin{aligned}\varGamma_{\alpha\gamma}^{q} & =\frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}E_{pot}\end{aligned} {} \end{aligned} $$
(2.10)
At this point, it is useful to separate the contributions to \(\varGamma _{\alpha \gamma }^{q}\) into density-dependent and density-independent terms. For the density-independent part of the field, we have where we have defined
$$\displaystyle \begin{aligned} \begin{aligned}F_{\alpha\beta\gamma\delta} & \equiv\left\langle \alpha\beta\left|V\right|\gamma\delta\right\rangle -\left\langle \alpha\beta\left|V\right|\delta\gamma\right\rangle +\left\langle \alpha\bar{\delta}\left|V\right|\gamma\bar{\beta}\right\rangle -\left\langle \alpha\bar{\delta}\left|V\right|\bar{\beta}\gamma\right\rangle \\ & =\left\langle \alpha\beta\left|V\right|\gamma\delta\right\rangle -\left\langle \alpha\beta\left|V\right|\delta\gamma\right\rangle +4\sigma_{\beta}\sigma_{\delta}\left(\left\langle \alpha\underline{\delta}\left|V\right|\gamma\underline{\beta}\right\rangle -\left\langle \alpha\underline{\delta}\left|V\right|\underline{\beta}\gamma\right\rangle \right) \end{aligned} {} \end{aligned} $$
(2.11)
For real-valued density matrix elements, the field reduces to
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{\alpha\gamma}^{q,di} &\displaystyle = &\displaystyle \sum_{q^{\prime};\beta\geq\delta>0}\frac{1}{1+\delta_{\beta\delta}}\left(F_{\alpha\beta\gamma\delta}+F_{\alpha\delta\gamma\beta}\right)\rho_{\delta\beta}^{q^{\prime}} \end{array} \end{aligned} $$
The density-dependent field (\(\varGamma _{\alpha \gamma }^{q,dd}\)) can be obtained directly from the differentiation in Eq. (2.10). The spatial dependence of the one-body density has the form
$$\displaystyle \begin{aligned} \begin{aligned}\rho^{q}\left(\mathbf{r}\right) & =\sum_{\alpha^{\prime}\gamma^{\prime}}\delta_{\sigma_{\alpha^{\prime}},\sigma_{\gamma^{\prime}}}\varPhi_{\alpha^{\prime}}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma^{\prime}}\left(\mathbf{r}\right)\rho_{\gamma^{\prime}\alpha^{\prime}}^{q}\end{aligned} {}\end{aligned} $$
(2.12)
Imposing the symmetries of the density matrix elements, this expression reduces to
$$\displaystyle \begin{aligned} \begin{aligned}\rho^{q}\left(\mathbf{r}\right) & =4\sum_{\alpha^{\prime}\geq\gamma^{\prime}>0}\frac{\delta_{\sigma_{\alpha^{\prime}},\sigma_{\gamma^{\prime}}}}{1+\delta_{\alpha^{\prime}\gamma^{\prime}}}\mathrm{Re}\left[\varPhi_{\alpha^{\prime}}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma^{\prime}}\left(\mathbf{r}\right)\rho_{\gamma^{\prime}\alpha^{\prime}}^{q}\right]\end{aligned} {} \end{aligned} $$
(2.13)
Thus, for a generic function of the density, we write the derivative with respect \(\rho _{\gamma \alpha }^{q}\) using Eq. (2.12),
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial}{\partial\rho_{\gamma\alpha}^{q}}f\left(\rho^{q}\left(\mathbf{r}\right)\right) &\displaystyle = &\displaystyle \delta_{\sigma_{\alpha},\sigma_{\gamma}}\varPhi_{\alpha}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma}\left(\mathbf{r}\right)f^{\prime}\left(\rho^{q}\left(\mathbf{r}\right)\right) \end{array} \end{aligned} $$
where f is the derivative of the function f.

2.3.3 The Pairing Field

The pairing field is given in general by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{\alpha\beta}^{q} &\displaystyle \equiv &\displaystyle \frac{\partial}{\partial\left(\kappa_{\beta\alpha}^{q}\right)^{*}}E_{pair}\\ &\displaystyle = &\displaystyle \frac{1}{2}\sum_{q^{\prime},\gamma\delta}V_{\alpha\beta\gamma\delta}^{qq^{\prime}}\kappa_{\delta\gamma}^{q^{\prime}} \end{array} \end{aligned} $$
Taking into account the symmetry properties of the matrix elements, where we have defined
$$\displaystyle \begin{aligned} \begin{array}{rcl} G_{\alpha\beta\gamma\delta} &\displaystyle \equiv &\displaystyle V_{\alpha\bar{\beta}\bar{\gamma}\delta}-V_{\alpha\bar{\beta}\delta\bar{\gamma}}-V_{\alpha\bar{\beta}\gamma\bar{\delta}}+V_{\alpha\bar{\beta}\bar{\delta}\gamma} \end{array} \end{aligned} $$

2.4 Matrix Elements in an Axially Deformed Harmonic-Oscillator Basis

2.4.1 Matrix Element of the Kinetic-Energy Operator

The contribution from the kinetic-energy operator follows from Eq. (2.7)
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial}{\partial\rho_{ki}^{q}}E_{kin} &\displaystyle = &\displaystyle T_{ik}-\lambda_{q}\delta_{ik} \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} T_{ik} &\displaystyle = &\displaystyle \left\langle i\left|-\frac{\hbar^{2}}{2m}\nabla^{2}\right|k\right\rangle \end{array} \end{aligned} $$
We can use the fact that the harmonic-oscillator basis functions satisfy the harmonic-oscillator Schrödinger equation, which in cylindrical coordinates takes the form
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left(-\frac{\hbar^{2}}{2m}\nabla^{2}+\frac{1}{2}m\omega_{\bot}^{2}r^{2}+\frac{1}{2}m\omega_{z}^{2}z^{2}\right)\varPhi_{n_{r},\varLambda}\left(\rho,\varLambda;b_{\perp}\right)\varPhi_{n_{z}}\left(z;b_{z}\right)\\ {} &\displaystyle = &\displaystyle E\varPhi_{n_{r},\varLambda}\left(\rho,\varLambda;b_{\perp}\right)\varPhi_{n_{z}}\left(z;b_{z}\right) \end{array} \end{aligned} $$
with
$$\displaystyle \begin{aligned} \begin{array}{rcl} E &\displaystyle = &\displaystyle \hbar\omega_{\bot}\left(2n_{r}+\left|\varLambda\right|+1\right)+\hbar\omega_{z}\left(n_{z}+\frac{1}{2}\right) \end{array} \end{aligned} $$
along with recurrence relations for the Laguerre and Hermite polynomials to calculate
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_{r}^{\prime},\varLambda^{\prime},n_{z}^{\prime},\sigma^{\prime}\left|-\frac{\hbar^{2}}{2m}\nabla^{2}\right|n_{r},\varLambda,n_{z},\sigma\right\rangle &\displaystyle = &\displaystyle \delta_{\varLambda^{\prime}\varLambda}\delta_{\sigma^{\prime}\sigma}\\ &\displaystyle &\displaystyle \times\left\{ \delta_{n_{z}^{\prime}n_{z}}\left\langle n_{r}^{\prime},\varLambda^{\prime}\left|-\frac{\hbar^{2}}{2m}\nabla_{r}^{2}\right|n_{r},\varLambda\right\rangle \right.\\ &\displaystyle &\displaystyle \left.+\delta_{n_{r}^{\prime}n_{r}}\left\langle n_{z}^{\prime}\left|-\frac{\hbar^{2}}{2m}\nabla_{z}^{2}\right|n_{z}\right\rangle \right\} \end{array} \end{aligned} $$
with
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_{r}^{\prime},\varLambda^{\prime}\left|-\frac{\hbar^{2}}{2m}\nabla_{r}^{2}\right|n_{r},\varLambda\right\rangle &\displaystyle = &\displaystyle \frac{\hbar\omega_{\bot}}{2}\left\{ \delta_{n_{r}^{\prime}n_{r}}\left(2n_{r}+\left|\varLambda\right|+1\right)\right.\\ &\displaystyle &\displaystyle +\delta_{n_{r}^{\prime},n_{r}-1}\sqrt{n_{r}\left(n_{r}+\left|\varLambda\right|\right)}\\ &\displaystyle &\displaystyle \left.+\delta_{n_{r}^{\prime},n_{r}+1}\sqrt{\left(n_{r}+1\right)\left(n_{r}+\left|\varLambda\right|+1\right)}\right\} \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_{z}^{\prime}\left|-\frac{\hbar^{2}}{2m}\nabla_{z}^{2}\right|n_{z}\right\rangle &\displaystyle = &\displaystyle \frac{\hbar\omega_{z}}{2}\left\{ \delta_{n_{z}^{\prime}n_{z}}\left(n_{z}+\frac{1}{2}\right)\right.\\ &\displaystyle &\displaystyle -\frac{1}{2}\delta_{n_{z}^{\prime},n_{z}-2}\sqrt{n_{z}\left(n_{z}-1\right)}\\ &\displaystyle &\displaystyle \left.-\frac{1}{2}\delta_{n_{z}^{\prime},n_{z}+2}\sqrt{\left(n_{z}+1\right)\left(n_{z}+2\right)}\right\} \end{array} \end{aligned} $$
See also the discussion in Sect. 2.4.5.

2.4.2 Matrix Elements of the Central Contribution

For the central terms in Eq. (2.1),
$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{cent}\left({\mathbf{r}}_{1},{\mathbf{r}}_{2}\right) &\displaystyle = &\displaystyle \left(W+B\hat{P}_{\sigma}-H\hat{P}_{\tau}-M\hat{P}_{\sigma}\hat{P}_{\tau}\right)e^{-\left({\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right)^{2}/\mu^{2}} \end{array} \end{aligned} $$
To calculate the matrix elements needed in Eq. (2.11) it is convenient to separate the spin, isospin, and spatial parts,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle ij\left|V\right|kl\right\rangle &\displaystyle = &\displaystyle \left\langle ij\left|V_{\sigma}\right|kl\right\rangle \left\langle ij\left|V_{\tau}\right|kl\right\rangle \left\langle ij\left|V_{r}\right|kl\right\rangle \end{array} \end{aligned} $$
After some simplifications, we can write for σi = σk and σj = σl:
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle F_{ijkl}+F_{ilkj}\\ {} &\displaystyle = &\displaystyle \left[2\left(W-H\delta_{qq^{\prime}}\right)+\left(B-M\delta_{qq^{\prime}}\right)\right]\left(\left\langle ij\left|V_{r}\right|kl\right\rangle +\left\langle il\left|V_{r}\right|kj\right\rangle \right)\\ {} &\displaystyle &\displaystyle -\left[\left(B\delta_{qq^{\prime}}-M\right)+\left(W\delta_{qq^{\prime}}-H\right)\delta_{\sigma_{i}\sigma_{j}}\right]\left(\left\langle ij\left|V_{r}\right|lk\right\rangle +\left\langle il\left|V_{r}\right|jk\right\rangle \right)\\ {} &\displaystyle &\displaystyle -\left[\left(B\delta_{qq^{\prime}}-M\right)+\left(W\delta_{qq^{\prime}}-H\right)\delta_{\sigma_{i},-\sigma_{j}}\right]\left(\left\langle i\underline{l}\left|V_{r}\right|\underline{j}k\right\rangle +\left\langle i\underline{j}\left|V_{r}\right|\underline{l}k\right\rangle \right) \end{array} \end{aligned} $$
and for σi = −σk and σj = −σk:
$$\displaystyle \begin{aligned} \begin{array}{rcl} F_{ijkl}+F_{ilkj} &\displaystyle = &\displaystyle \left(W\delta_{qq^{\prime}}-H\right)\left[\delta_{\sigma_{i},-\sigma_{j}}\left(\left\langle i\underline{l}\left|V_{r}\right|\underline{j}k\right\rangle -\left\langle ij\left|V_{r}\right|lk\right\rangle \right)\right.\\ {} &\displaystyle &\displaystyle \left.-\delta_{\sigma_{i},\sigma_{j}}\left(\left\langle il\left|V_{r}\right|jk\right\rangle +\left\langle i\underline{j}\left|V_{r}\right|\underline{l}k\right\rangle \right)\right] \end{array} \end{aligned} $$
while the remaining cases give
$$\displaystyle \begin{aligned} \begin{array}{rcl} F_{ijkl}+F_{ilkj} &\displaystyle = &\displaystyle 0 \end{array} \end{aligned} $$
For the spatial part of the matrix elements, we recall the results in [17]. A matrix element of the gaussian function in a cylindrical harmonic-oscillator basis can be separated into radial and Cartesian components,
$$\displaystyle \begin{aligned} \begin{aligned}\left\langle ij\left|V_{r}\right|kl\right\rangle & =V_{ijkl}^{\left(r\right)}V_{ijkl}^{\left(z\right)}\end{aligned} {} \end{aligned} $$
(2.14)
where, in cylindrical coordinates \(\left (r,\varphi ,z\right )\),
$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{ijkl}^{\left(r\right)} &\displaystyle \equiv &\displaystyle \int_{0}^{\infty}r_{1}dr_{1}\int_{0}^{2\pi}d\varphi_{1}\int_{0}^{\infty}r_{2}dr_{2}\int_{0}^{2\pi}d\varphi_{2}\\ {} &\displaystyle &\displaystyle \varPhi_{n_{r}^{\left(i\right)},\varLambda_{i}}^{*}\left(r_{1},\varphi_{1};b_{\perp}\right)\varPhi_{n_{r}^{\left(j\right)},\varLambda_{j}}^{*}\left(r_{2},\varphi_{2};b_{\perp}\right)\\ {} &\displaystyle &\displaystyle V\left(r_{1},\varphi_{1};r_{2},\varphi_{2}\right)\varPhi_{n_{r}^{\left(k\right)},\varLambda_{k}}\left(r_{1},\varphi_{1};b_{\perp}\right)\varPhi_{n_{r}^{\left(l\right)},\varLambda_{l}}\left(r_{2},\varphi_{2};b_{\perp}\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{ijkl}^{\left(z\right)} &\displaystyle \equiv &\displaystyle \int_{-\infty}^{\infty}dz_{1}\int_{-\infty}^{\infty}dz_{2}\,\varPhi_{n_{z}^{\left(i\right)}}\left(z_{1};b_{z}\right)\varPhi_{n_{z}^{\left(j\right)}}\left(z_{2};b_{z}\right)\\ {} &\displaystyle &\displaystyle V\left(z_{1};z_{2}\right)\varPhi_{n_{z}^{\left(k\right)}}\left(z_{1};b_{z}\right)\varPhi_{n_{z}^{\left(l\right)}}\left(z_{2};b_{z}\right) \end{array} \end{aligned} $$
In [17], explicit forms were derived for \(V_{ijkl}^{\left (r\right )}\) and \(V_{ijkl}^{\left (z\right )}\) with the gaussian interaction. We have
$$\displaystyle \begin{aligned} \begin{array}{rcl} V_{ijkl}^{\left(r\right)} &\displaystyle = &\displaystyle \frac{G_{\perp}-1}{G_{\perp}+1}\sum_{n_{r}=0}^{n_{\bar{j},l}}\sum_{n=0}^{n_{\bar{i},k}}T_{n_{r}^{(i)},-\varLambda_{i};n_{r}^{(k)},\varLambda_{k}}^{n,-\varLambda_{i}+\varLambda_{k}}T_{n_{r}^{(j)},-\varLambda_{j};n_{r}^{(l)},\varLambda_{l}}^{n_{r},-\varLambda_{j}+\varLambda_{l}}\\ {} &\displaystyle &\displaystyle \times\bar{I}\left(n_{r},-\varLambda_{j}+\varLambda_{l};n,-\varLambda_{i}+\varLambda_{k}\right) \end{array} \end{aligned} $$
with
$$\displaystyle \begin{aligned} \begin{aligned}G_{\bot} & \equiv1+\frac{\mu^{2}}{b_{\bot}^{2}}\end{aligned} {} \end{aligned} $$
(2.15)
and
$$\displaystyle \begin{aligned} \begin{aligned}n_{\mu,\nu} & \equiv n_{r}^{(\mu)}+n_{r}^{(\nu)}+\frac{\left|\varLambda_{\mu}\right|+\left|\varLambda_{\nu}\right|-\left|\varLambda_{\mu}+\varLambda_{\nu}\right|}{2}\\ {} n_{\bar{\mu},\nu} & \equiv n_{r}^{(\mu)}+n_{r}^{(\nu)}+\frac{\left|\varLambda_{\mu}\right|+\left|\varLambda_{\nu}\right|-\left|-\varLambda_{\mu}+\varLambda_{\nu}\right|}{2} \end{aligned} {} \end{aligned} $$
(2.16)
The T coefficients are given by
$$\displaystyle \begin{aligned} \begin{array}{rcl} T_{n_{1},k_{1};n_{2},k_{2}}^{n,k_{1}+k_{2}} &\displaystyle = &\displaystyle \left(-1\right)^{n_{1}+n_{2}-n}\sqrt{\frac{n!\left(n_{1}+\left|k_{1}\right|\right)!\left(n_{2}+\left|k_{2}\right|\right)!}{n_{1}!n_{2}!\left(n+\left|k_{1}+k_{2}\right|\right)!}}\\ {} &\displaystyle &\displaystyle \times\sum_{m_{1}=0}^{n_{1}}\sum_{m_{2}=0}^{n_{2}}\delta_{n\leq n_{1,2}-m_{1}-m_{2}}C_{n_{1},k_{1},m_{1};n_{2},k_{2},m_{2}}^{n,k_{1}+k_{2}} \end{array} \end{aligned} $$
where the Kronecker-delta function \(\delta _{n\leq n_{1,2}-m_{1}-m_{2}}\) ensures that we always have n ≤ n1,2,
$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{n_{1},k_{1},m_{1};n_{2},k_{2},m_{2}}^{n,k_{1}+k_{2}} &\displaystyle \equiv &\displaystyle \left(-1\right)^{m_{1}+m_{2}}\left(\begin{array}{c} n_{1}\\ m_{1} \end{array}\right)\left(\begin{array}{c} n_{2}\\ {} m_{2} \end{array}\right)\left(\begin{array}{c} n_{1,2}-m_{1}-m_{2}\\ {} n \end{array}\right)\\ &\displaystyle &\displaystyle \times\frac{\left(n_{1,2}+\left|k_{1}+k_{2}\right|-m_{1}-m_{2}\right)!}{\left(n_{1}+\left|k_{1}\right|-m_{1}\right)!\left(n_{2}+\left|k_{2}\right|-m_{2}\right)!} \end{array} \end{aligned} $$
The coefficient \(\bar {I}\) is given by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{I}\left(n_{1},k_{1};n_{2},k_{2}\right) &\displaystyle \equiv &\displaystyle \delta_{k_{1}+k_{2},0}\frac{\sqrt{n_{1}!\left(n_{1}+\left|k\right|\right)!n_{2}!\left(n_{2}+\left|k\right|\right)!}}{\left(G_{\perp}+1\right)^{n_{1}+n_{2}+\left|k\right|}}\varXi\left(n_{1},n_{2},\left|k\right|\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varXi\left(n_{1},n_{2},\left|k\right|\right) &\displaystyle = &\displaystyle \frac{1}{\left(n_{1}+n_{2}+\left|k\right|\right)!}\left(\begin{array}{c} n_{1}+n_{2}+\left|k\right|\\ n_{1} \end{array}\right)\left(\begin{array}{c} n_{1}+n_{2}+\left|k\right|\\ n_{2} \end{array}\right) \end{array} \end{aligned} $$
Next, for the term \(V_{ijkl}^{\left (z\right )}\) in Eq. (2.14), we have
$$\displaystyle \begin{aligned} \begin{aligned}V_{ijkl}^{\left(z\right)} & =\sqrt{\frac{G_{z}-1}{G_{z}+1}}\sum_{m_{z}=\left|n_{z}^{(i)}-n_{z}^{(k)}\right|,2}^{n_{z}^{(i)}+n_{z}^{(k)}}T_{n_{z}^{(i)},n_{z}^{(k)}}^{m_{z}}\\ & \times\sum_{n_{z}=\left|n_{z}^{(j)}-n_{z}^{(l)}\right|,2}^{n_{z}^{(j)}+n_{z}^{(l)}}T_{n_{z}^{(j)},n_{z}^{(l)}}^{n_{z}}\bar{I}\left(m_{z},n_{z}\right) \end{aligned} {} \end{aligned} $$
(2.17)
where the “,2” notation indicates that the summations are incremented in steps of 2, and with
$$\displaystyle \begin{aligned} \begin{aligned}G_{z} & \equiv1+\frac{\mu^{2}}{b_{z}^{2}}\end{aligned} {} \end{aligned} $$
(2.18)
The T coefficients for the Cartesian term are given by
$$\displaystyle \begin{aligned} \begin{array}{rcl} T_{k_{1},k_{2}}^{k} &\displaystyle = &\displaystyle \frac{\sqrt{k_{1}!k_{2}!k!}}{\left(\frac{k_{1}-k_{2}+k}{2}\right)!\left(\frac{k_{2}-k_{1}+k}{2}\right)!\left(\frac{k_{1}+k_{2}-k}{2}\right)!} \end{array} \end{aligned} $$
and the \(\bar {I}\) coefficient by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{I}\left(m,n\right) &\displaystyle = &\displaystyle \sqrt{\frac{m!n!}{2^{m+n}}}\frac{\left(-1\right)^{\left(m-n\right)/2}}{\left(\frac{m+n}{2}\right)!\left(1+G_{z}\right)^{\left(m+n\right)/2}}\left(\begin{array}{c} m+n\\ n \end{array}\right) \end{array} \end{aligned} $$

2.4.2.1 Modification for Large Oscillator Number in z Direction

In the case where the nz quantum numbers are large, Eq. (2.17) requires the evaluation of sums of products of large (T) and small (\(\bar {I}\)) coefficients, which can be numerically unstable. As shown in [17, 21], we can replace Eq. (2.17) with
$$\displaystyle \begin{aligned} \begin{aligned}V_{ijkl}^{\left(z\right)} & =\frac{\mu}{\sqrt{2\pi^{3}}b_{z}}\sum_{n_{z}=\left|n_{z}^{(j)}-n_{z}^{(l)}\right|,2}^{n_{z}^{(j)}+n_{z}^{(l)}}T_{n_{z}^{(j)},n_{z}^{(l)}}^{n_{z}}\bar{F}_{n_{z}^{(i)},n_{z}^{(k)}}^{n_{z}}\end{aligned} {} \end{aligned} $$
(2.19)
where where \(\,_{2}F_{1}\left (a,b;c;z\right )\) is a hypergeometric function, and
$$\displaystyle \begin{aligned} \begin{aligned}\xi & \equiv\frac{n_{z}^{(i)}+n_{z}^{(k)}+n_{z}+1}{2}\end{aligned} {} \end{aligned} $$
(2.21)
and
$$\displaystyle \begin{aligned} \begin{aligned}z & \equiv1+\frac{\mu^{2}}{2b^{2}}\end{aligned} {}\end{aligned} $$
(2.22)

2.4.3 Matrix Elements of the Coulomb Contribution

The Coulomb matrix elements can be expressed as
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle ij\left|V_{C}^{\left(D\right)}\right|kl\right\rangle &\displaystyle = &\displaystyle e^{2}\delta_{\sigma_{i},\sigma_{k}}\delta_{\sigma_{j},\sigma_{l}}\delta_{\sigma_{j},\sigma_{l}}\delta_{-\varLambda_{i}+\varLambda_{k}-\varLambda_{j}+\varLambda_{l},0}\delta_{q,p}\delta_{q^{\prime},p}\times\mathcal{I}_{ijkl} \end{array} \end{aligned} $$
where we have defined the integral
$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{I}_{ijkl} &\displaystyle \equiv &\displaystyle \int d^{3}r_{1}\int d^{3}r_{2}\varPhi_{i}^{*}\left({\mathbf{r}}_{1}\right)\varPhi_{j}^{*}\left({\mathbf{r}}_{2}\right)\frac{1}{\left|{\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right|}\varPhi_{k}\left({\mathbf{r}}_{1}\right)\varPhi_{l}\left({\mathbf{r}}_{2}\right) \end{array} \end{aligned} $$
Using the integral identity
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{1}{\left|{\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right|} &\displaystyle = &\displaystyle \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}du\,\frac{e^{-\left({\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right)^{2}/u^{2}}}{u^{2}} \end{array} \end{aligned} $$
we can write
$$\displaystyle \begin{aligned} \begin{aligned}\mathcal{I}_{ijkl} & =\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\frac{du}{u^{2}}C_{ijkl}^{\left(r\right)}\left(u\right)\times C_{ijkl}^{\left(z\right)}\left(u\right)\end{aligned} {} \end{aligned} $$
(2.23)
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{ijkl}^{\left(r\right)}\left(u\right) &\displaystyle \equiv &\displaystyle \int_{0}^{2\pi}d\varphi_{1}\int_{0}^{\infty}\rho_{1}d\rho_{1}\int_{0}^{2\pi}d\varphi_{2}\int_{0}^{\infty}\rho_{2}d\rho_{2}\varPhi_{n_{r}^{\left(i\right)},\varLambda_{i}}^{*}\left(\rho_{1},\varphi_{1}\right)\\ &\displaystyle &\displaystyle \varPhi_{n_{r}^{\left(j\right)},\varLambda_{j}}^{*}\left(\rho_{2},\varphi_{2}\right)e^{-\left(\mathbf{\rho}_{1}-\mathbf{\rho}_{2}\right)^{2}/u^{2}}\varPhi_{n_{r}^{\left(k\right)},\varLambda_{k}}\left(\rho_{1},\varphi_{1}\right)\varPhi_{n_{r}^{\left(l\right)},\varLambda_{l}}\left(\rho_{2},\varphi_{2}\right) \end{array} \end{aligned} $$
and Using the techniques described in [17], we can write where \(G_{z}\left (u\right )\) is given by Eq. (2.18) with the integration variable u replacing μ, and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{J}_{z}\left(m,n\right) &\displaystyle \equiv &\displaystyle \frac{\left(-1\right)^{\left(m-n\right)/2}\sqrt{m!n!}}{2^{\left(m+n\right)/2}\left(\frac{m+n}{2}\right)!}\left(\begin{array}{c} m+n\\ m \end{array}\right) \end{array} \end{aligned} $$
with the implicit requirement that m + n is even. For the radial coefficients, we can show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{ijkl}^{\left(r\right)}\left(u\right) &\displaystyle = &\displaystyle \sum_{n_{r}=0}^{n_{\bar{j}l}}T_{n_{r}^{\left(j\right)},-\varLambda_{j};n_{r}^{\left(l\right)},\varLambda_{l}}^{n_{r}}\sum_{n=0}^{n_{\bar{i}k}}T_{n_{r}^{\left(i\right)},-\varLambda_{i};n_{r}^{\left(k\right)},\varLambda_{k}}^{n}\frac{G_{\bot}\left(u\right)-1}{G_{\bot}\left(u\right)+1}\\ &\displaystyle &\displaystyle \times\frac{\bar{J}_{r}\left(n,n_{r},-\varLambda_{i}+\varLambda_{k}\right)}{\left(G_{\bot}\left(u\right)+1\right)^{n+n_{r}+\left|-\varLambda_{i}+\varLambda_{k}\right|}} \end{array} \end{aligned} $$
where \(G_{\bot }\left (u\right )\) is given by Eq. (2.15) with the integration variable u replacing μ, and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{J}_{r}\left(n_{1},n_{2},\varLambda\right) &\displaystyle = &\displaystyle \sqrt{n_{1}!\left(n_{1}+\left|\varLambda\right|\right)n_{2}!\left(n_{2}+\left|\varLambda\right|\right)}\varXi\left(n_{1},n_{2},\left|\varLambda\right|\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varXi\left(n_{2},n_{1},\left|\varLambda\right|\right) &\displaystyle = &\displaystyle \frac{1}{\left(n_{1}+n_{2}+\left|\varLambda\right|\right)!}\left(\begin{array}{c} n_{1}+n_{2}+\left|\varLambda\right|\\ n_{1} \end{array}\right)\left(\begin{array}{c} n_{1}+n_{2}+\left|\varLambda\right|\\ n_{2} \end{array}\right) \end{array} \end{aligned} $$
We have thus re-written Eq. (2.23) in the form We define the integral and introduce the coefficients and
$$\displaystyle \begin{aligned} \begin{array}{rcl} D_{ijkl}^{\left(z\right)}\left(q\right) &\displaystyle \equiv &\displaystyle \sum_{n_{z}=\left|n_{z}^{\left(j\right)}-n_{z}^{\left(l\right)}\right|,2}^{n_{z}^{\left(j\right)}+n_{z}^{\left(l\right)}}\sum_{m_{z}=\left|n_{z}^{\left(i\right)}-n_{z}^{\left(k\right)}\right|,2}^{n_{z}^{\left(i\right)}+n_{z}^{\left(k\right)}}\delta_{m_{z}+n_{z},q}T_{n_{z}^{\left(j\right)},n_{z}^{\left(l\right)}}^{n_{z}}T_{n_{z}^{\left(i\right)},n_{z}^{\left(k\right)}}^{m_{z}}\\ {} &\displaystyle &\displaystyle \times\bar{J}_{z}\left(m_{z},n_{z}\right) \end{array} \end{aligned} $$
Note that Kronecker delta symbols were introduced to group the terms in the summations by the sum of the indices (i.e., mr + nr for \(D_{ijkl}^{\left (r\right )}\left (p\right )\) and mz + nz for \(D_{ijkl}^{\left (z\right )}\left (q\right )\)) because the remaining integral over u, defined in Eq. (2.27), only depends on the sum of these indices. Then Eq. (2.26) can be cast in the form All that remains now is to further evaluate the integral in Eq. (2.27). We can re-write this integral as
$$\displaystyle \begin{aligned} \begin{array}{rcl} M\left(p,2r\right) &\displaystyle = &\displaystyle \frac{1}{\sqrt{\pi}b_{z}}\int_{0}^{\infty}dt\frac{1}{\left(2+t\right)^{p+1}\left(2+\frac{b_{\bot}^{2}}{b_{z}^{2}}t\right)^{r+1/2}} \end{array} \end{aligned} $$
and note that where \(\,_{2}F_{1}\left (a,b;c;z\right )\) is a hypergeometric function. This last result can be obtained by first making the variable substitution 2 + αt = 2∕y in the integral, which then becomes
$$\displaystyle \begin{aligned} \begin{aligned}\int_{0}^{\infty}dt\frac{1}{\left(2+\alpha t\right)^{k}\left(2+t\right)^{l}} & =\frac{\alpha^{l-1}}{2^{k+l-1}}\int_{0}^{1}dy\,\frac{y^{k+l-2}}{\left[1-\left(1-\alpha\right)y\right]^{l}}\end{aligned} {} \end{aligned} $$
(2.31)
Equation (2.30) then follows from identifying the right-hand side of Eq. (2.31) with the integral definition of the hypergeometric function (see, e.g., (15.3.1) in [22]),
$$\displaystyle \begin{aligned} \begin{aligned}\,_{2}F_{1}\left(a,b,c,z\right) & \equiv\frac{\varGamma\left(c\right)}{\varGamma\left(b\right)\varGamma\left(c-b\right)}\int_{0}^{1}dt\end{aligned} \,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-tz\right)^{a}} \end{aligned}$$
Note that for deformed systems bz > b, and the hypergeometric function of 1 − α above converges.

2.4.3.1 Modification for Large Oscillator Number in z Direction

In the case where the nz quantum numbers are large, we use the results in [17] to replace Eq. (2.25) with
$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{ijkl}^{\left(z\right)}\left(u\right) &\displaystyle = &\displaystyle \frac{u}{\sqrt{2\pi^{3}}b_{z}}\sum_{n_{z}=\left|n_{z}^{(j)}-n_{z}^{(l)}\right|,2}^{n_{z}^{(j)}+n_{z}^{(l)}}T_{n_{z}^{(j)},n_{z}^{(l)}}^{n_{z}}\bar{F}_{n_{z}^{(i)},n_{z}^{(k)}}^{n_{z}}\left(u\right) \end{array} \end{aligned} $$
where \(\bar {F}_{n_{z}^{(i)},n_{z}^{(k)}}^{n_{z}}\) is given by Eq. (2.20). Equation (2.29) can then be re-written as where \(D_{ijkl}^{\left (r\right )}\left (p\right )\) is the same coefficient defined in Eq. (2.28). Next, we have defined
$$\displaystyle \begin{aligned} \begin{aligned}\bar{D}_{ijkl}^{\left(z\right)}\left(n_{z}\right) & \equiv\frac{1}{\sqrt{2\pi^{3}}}T_{n_{z}^{(j)},n_{z}^{(l)}}^{n_{z}}\frac{\varGamma\left(\xi-n_{z}^{(i)}\right)\varGamma\left(\xi-n_{z}^{(k)}\right)\varGamma\left(\xi-n_{z}\right)}{\sqrt{n_{z}!n_{z}^{(i)}!n_{z}^{(k)}!}}\end{aligned} {} \end{aligned} $$
(2.33)
with ξ given by Eq. (2.21) and The coefficient in Eq. (2.33) can be re-written as and the integral in Eq. (2.34) can be re-written as
$$\displaystyle \begin{aligned} \begin{aligned}\bar{M}\left(p,n_{z}^{(i)},n_{z}^{(k)},n_{z}\right) & =\frac{2^{\xi}}{\sqrt{\pi}b_{z}}\int_{0}^{\infty}dt\,\frac{\,_{2}F_{1}\left(-n_{z}^{(i)},-n_{z}^{(k)};-\xi+n_{z}+1;-\frac{b_{\bot}^{2}}{2b_{z}^{2}}t\right)}{\left(2+t\right)^{p+1}\left(2+\frac{b_{\bot}^{2}}{b_{z}^{2}}t\right)^{\xi}}\end{aligned} {} \end{aligned} $$
(2.36)
Next, we expand the hypergeometric function in a power series, where the notation \(\left (z\right )_{r}\) indicates a rising Pochhammer symbol, and where
$$\displaystyle \begin{aligned} \begin{array}{rcl} \beta &\displaystyle \equiv &\displaystyle \frac{b_{\bot}^{2}}{b_{z}^{2}} \end{array} \end{aligned} $$
Note that since both \(n_{z}^{(i)}\) and \(n_{z}^{(k)}\) are integers in Eq. (2.37), the series terminates after a finite number of terms, and since ξ is always a half-integer number, there are no singularities caused by the terms in the denominator. Then the integral in Eq. (2.36) becomes
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{M}\left(p,n_{z}^{(i)},n_{z}^{(k)},n_{z}\right) &\displaystyle = &\displaystyle \frac{2^{\xi}}{\sqrt{\pi}b_{z}}\sum_{r=0}^{\min\left(n_{z}^{(i)},n_{z}^{(k)}\right)}\frac{\left(-n_{z}^{(i)}\right)_{r}\left(-n_{z}^{(k)}\right)_{r}}{\left(-\xi+n_{z}+1\right)_{r}r!}\left(-\frac{\beta}{2}\right)^{r}\\ &\displaystyle &\displaystyle \times\int_{0}^{\infty}dt\,\frac{t^{r}}{\left(2+t\right)^{p+1}\left(2+\beta t\right)^{\xi}} \end{array} \end{aligned} $$
We can evaluate the remaining integral by making the substitution
$$\displaystyle \begin{aligned} \begin{array}{rcl} 2+\beta t &\displaystyle \equiv &\displaystyle \frac{2}{y}\vspace{-4pt} \end{array} \end{aligned} $$
Then, we can show that the exponents satisfy the required relations for the integral to converge and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_{0}^{\infty}dt\,\frac{t^{r}}{\left(2+t\right)^{p+1}\left(2+\beta t\right)^{\xi}} &\displaystyle = &\displaystyle \frac{\beta^{p-r}}{2^{p+\xi-r}}\int_{0}^{1}dy\,\frac{y^{p+\xi-r-1}\left(1-y\right)^{r}}{\left[1-\left(1-\beta\right)y\right]^{p+1}}\\ {} &\displaystyle = &\displaystyle \frac{\beta^{p-r}}{2^{p+\xi-r}}B\left(p+\xi-r,r+1\right)\\ {} &\displaystyle &\displaystyle \times\,_{2}F_{1}\left(p+1,p+\xi-r;p+\xi+1;1-\beta\right)\\ {} &\displaystyle = &\displaystyle \frac{B\left(p+\xi-r,r+1\right)}{2^{p+\xi-r}}\\ {} &\displaystyle &\displaystyle \times\,_{2}F_{1}\left(\xi,r+1;p+\xi+1;1-\beta\right) \end{array} \end{aligned} $$
where \(B\left (x,y\right )\) is the Beta function. Using this result, and after some simplifications, we obtain
$$\displaystyle \begin{aligned} \begin{array}{rcl} \bar{M}\left(p,n_{z}^{(i)},n_{z}^{(k)},n_{z}\right) &\displaystyle = &\displaystyle \frac{1}{2^{p}\sqrt{\pi}\left(p+\xi\right)b_{z}}\sum_{r=0}^{\min\left(n_{z}^{(i)},n_{z}^{(k)}\right)}C_{r}\beta^{r}\\ {} &\displaystyle &\displaystyle \times\,_{2}F_{1}\left(\xi,r+1;p+\xi+1;1-\beta\right) \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{r} &\displaystyle \equiv &\displaystyle \begin{cases} 1 &\displaystyle r=0\\ {} \prod_{q=0}^{r-1}\frac{\left(q-n_{z}^{(i)}\right)\left(q-n_{z}^{(k)}\right)}{\left(q-\xi+n_{z}+1\right)\left(q-p-\xi+1\right)} &\displaystyle r>0 \end{cases} \end{array} \end{aligned} $$
With this result we have all the ingredients needed to calculate the Coulomb integral in Eq. (2.32).

2.4.4 Matrix Elements of the Spin-Orbit Contribution

The spin-orbit energy is given by [23]
$$\displaystyle \begin{aligned} \begin{array}{rcl} E_{so} &\displaystyle = &\displaystyle -\frac{W_{0}}{2}\int d^{3}r\left\{ \rho\left(\mathbf{r}\right)\mathbf{\nabla}\cdot\mathbf{J}\left(\mathbf{r}\right)+\sum_{q}\rho_{q}\left(\mathbf{r}\right)\mathbf{\nabla}\cdot{\mathbf{J}}_{q}\left(\mathbf{r}\right)\right\} \end{array} \end{aligned} $$
where the density for charge q is
$$\displaystyle \begin{aligned} \begin{aligned}\rho_{q}\left(\mathbf{r}\right) & \equiv\sum_{\alpha,\beta,\sigma}\rho_{\beta\alpha}^{\left(q\right)}\varPhi_{\alpha}^{*}\left(\mathbf{r},\sigma\right)\varPhi_{\beta}\left(\mathbf{r},\sigma\right)\\ {} & \equiv\sum_{\alpha,\beta,\sigma}\rho_{\beta\alpha}^{\left(q\right)}R_{\alpha\beta}\left(\mathbf{r},\sigma\right) \end{aligned} {} \end{aligned} $$
(2.38)
and the current is
$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathbf{J}}_{q}\left(\mathbf{r}\right) &\displaystyle \equiv &\displaystyle -\mathrm{i}\sum_{\alpha,\beta,\sigma,\sigma^{\prime}}\rho_{\beta\alpha}^{\left(q\right)}\varPhi_{\alpha}^{*}\left(\mathbf{r},\sigma\right)\mathbf{\nabla}\varPhi_{\beta}\left(\mathbf{r},\sigma^{\prime}\right)\times\left\langle \sigma\left|\mathbf{\sigma}\right|\sigma^{\prime}\right\rangle \end{array} \end{aligned} $$
where the components of σ are the Pauli spin matrices. We calculate the divergence
$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathbf{\nabla}\cdot{\mathbf{J}}_{q}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle -\mathrm{i}\sum_{\alpha,\beta,\sigma,\sigma^{\prime}}\rho_{\beta\alpha}^{\left(q\right)}\mathbf{\nabla}\cdot\left[\varPhi_{\alpha}^{*}\left(\mathbf{r},\sigma\right)\mathbf{\nabla}\varPhi_{\beta}\left(\mathbf{r},\sigma^{\prime}\right)\times\left\langle \sigma\left|\mathbf{\sigma}\right|\sigma^{\prime}\right\rangle \right]\\ {} &\displaystyle = &\displaystyle -\mathrm{i}\sum_{\alpha,\beta,\sigma,\sigma^{\prime}}\rho_{\beta\alpha}^{\left(q\right)}\left\langle \sigma\left|\mathbf{\sigma}\right|\sigma^{\prime}\right\rangle \cdot\mathbf{\nabla}\varPhi_{\alpha}^{*}\left(\mathbf{r},\sigma\right)\times\mathbf{\nabla}\varPhi_{\beta}\left(\mathbf{r},\sigma^{\prime}\right)\\ {} &\displaystyle \equiv &\displaystyle \sum_{\alpha,\beta,\sigma,\sigma^{\prime}}\rho_{\beta\alpha}^{\left(q\right)}\varDelta_{\alpha\beta}\left(\mathbf{r},\sigma,\sigma^{\prime}\right) \end{array} \end{aligned} $$
and define
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{ik}\left(\mathbf{r}\right) &\displaystyle \equiv &\displaystyle \sum_{\sigma}R_{ik}\left(\mathbf{r},\sigma\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ik}\left(\mathbf{r}\right) &\displaystyle \equiv &\displaystyle \sum_{\sigma,\sigma^{\prime}}\varDelta_{ik}\left(\mathbf{r},\sigma,\sigma^{\prime}\right) \end{array} \end{aligned} $$
From this, it follows that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial}{\partial\rho_{ki}^{\left(q^{\prime}\right)}}\rho_{q}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \delta_{q,q^{\prime}}R_{ik}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial}{\partial\rho_{ki}^{\left(q^{\prime}\right)}}\mathbf{\nabla}\cdot{\mathbf{J}}_{q}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \delta_{q,q^{\prime}}\varDelta_{ik}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
and therefore,
$$\displaystyle \begin{aligned} \begin{aligned}\left\langle ij\left|V_{so}\right|kl\right\rangle & =\frac{\partial^{2}}{\partial\rho_{lj}^{\left(q_{jl}\right)}\partial\rho_{ki}^{\left(q_{ik}\right)}}E_{so}\\ & =-\frac{W_{0}}{2}\left(1+\delta_{q_{ik},q_{jl}}\right)\int d^{3}r\left[R_{ik}\left(\mathbf{r}\right)\varDelta_{jl}\left(\mathbf{r}\right)+R_{jl}\left(\mathbf{r}\right)\varDelta_{ik}\left(\mathbf{r}\right)\right] \end{aligned} {} \end{aligned} $$
(2.39)
Note that this result contains the exchange part of the matrix element, which appears as the \(\delta _{q_{ik},q_{jl}}\) term.

2.4.4.1 Explicit Form of \(R_{ik}\left (\mathbf {r}\right )\) and \(\varDelta _{jl}\left (\mathbf {r}\right )\)

We write
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{ik}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \sum_{\sigma}\varPhi_{i}^{*}\left(\mathbf{r},\sigma\right)\varPhi_{k}\left(\mathbf{r},\sigma\right)\\ &\displaystyle = &\displaystyle \frac{e^{i\left(\varLambda_{k}-\varLambda_{i}\right)\varphi}}{2\pi}\delta_{\sigma_{i},\sigma_{k}}\tilde{\varPhi}_{i}\left(\rho,z\right)\tilde{\varPhi}_{k}\left(\rho,z\right) \end{array} \end{aligned} $$
where we have written
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varPhi_{n}\left(\mathbf{r},\sigma\right) &\displaystyle \equiv &\displaystyle \frac{e^{i\varLambda_{n}\varphi}}{\sqrt{2\pi}}\tilde{\varPhi}_{n}\left(\rho,z\right)\chi_{\sigma} \end{array} \end{aligned} $$
to separate the dependence of the basis states on \(\left (\rho ,z\right )\) alone. Next we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ik}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \sum_{\sigma,\sigma^{\prime}}-\mathrm{i}\left\langle \sigma\left|\mathbf{\sigma}\right|\sigma^{\prime}\right\rangle \cdot\mathbf{\nabla}\varPhi_{i}^{*}\left(\mathbf{r},\sigma\right)\times\mathbf{\nabla}\varPhi_{k}\left(\mathbf{r},\sigma^{\prime}\right)\\ &\displaystyle = &\displaystyle -\mathrm{i}\left\langle \sigma_{i}\left|\mathbf{\sigma}\right|\sigma_{k}\right\rangle \cdot\mathbf{\nabla}\varPhi_{i}^{*}\left(\mathbf{r}\right)\times\mathbf{\nabla}\varPhi_{k}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
We further write
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle \sigma_{i}\left|\mathbf{\sigma}\right|\sigma_{k}\right\rangle &\displaystyle = &\displaystyle \delta_{\sigma_{i},-\sigma_{k}}\hat{x}+2\mathrm{i}\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}\hat{y}+2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\hat{z}\\ &\displaystyle = &\displaystyle \delta_{\sigma_{i},-\sigma_{k}}\left(\cos\varphi\hat{\rho}-\sin\varphi\hat{\varphi}\right)+2\mathrm{i}\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}\left(\sin\varphi\hat{\rho}+\cos\varphi\hat{\varphi}\right)\\ &\displaystyle &\displaystyle +2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\hat{z}\\ &\displaystyle = &\displaystyle \delta_{\sigma_{i},-\sigma_{k}}e^{2\mathrm{i}\sigma_{k}\varphi}\hat{\rho}+2\mathrm{i}\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}e^{2\mathrm{i}\sigma_{k}\varphi}\hat{\varphi}+2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\hat{z} \end{array} \end{aligned} $$
Now, since i and k are from the same symmetry block then Ωi = Ωk, and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varLambda_{k}-\varLambda_{i}+2\sigma_{k}-2\sigma_{i} &\displaystyle = &\displaystyle 0 \end{array} \end{aligned} $$
Therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle \sigma_{i}\left|\mathbf{\sigma}\right|\sigma_{k}\right\rangle e^{\mathrm{i}\left(\varLambda_{k}-\varLambda_{i}\right)\varphi} &\displaystyle = &\displaystyle \delta_{\sigma_{i},-\sigma_{k}}\hat{\rho}+2\mathrm{i}\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}\hat{\varphi}+2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\hat{z} \end{array} \end{aligned} $$
Then, carrying out the vector operations,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ik}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \frac{1}{2\pi}\left\{ -\delta_{\sigma_{i},-\sigma_{k}}\left[\frac{\varLambda_{i}}{\rho}\tilde{\varPhi}_{i}\partial_{z}\tilde{\varPhi}_{k}+\partial_{z}\tilde{\varPhi}_{i}\frac{\varLambda_{k}}{\rho}\tilde{\varPhi}_{k}\right]\right.\\ &\displaystyle &\displaystyle -2\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}\left[\partial_{\rho}\tilde{\varPhi}_{i}\partial_{z}\tilde{\varPhi}_{k}-\partial_{z}\tilde{\varPhi}_{i}\partial_{\rho}\tilde{\varPhi}_{k}\right]\\ &\displaystyle &\displaystyle \left.+2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\left[\partial_{\rho}\tilde{\varPhi}_{i}\frac{\varLambda_{k}}{\rho}\tilde{\varPhi}_{k}+\frac{\varLambda_{i}}{\rho}\tilde{\varPhi}_{i}\partial_{\rho}\tilde{\varPhi}_{k}\right]\right\} \end{array} \end{aligned} $$

2.4.4.2 Separation of Coordinates

Returning to Eq. (2.39), we can write
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle ij\left|V_{so}\right|kl\right\rangle &\displaystyle = &\displaystyle -\frac{W_{0}}{2}\left(1+\delta_{q_{ik},q_{jl}}\right)\left(I_{ik;jl}+I_{jl;ik}\right) \end{array} \end{aligned} $$
where we have defined the integral
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{ik;jl} &\displaystyle \equiv &\displaystyle \int d^{3}r\: R_{ik}\left(\mathbf{r}\right)\varDelta_{jl}\left(\mathbf{r}\right)\\ &\displaystyle = &\displaystyle \int d^{3}r_{1}\int d^{3}r_{2}\:\delta^{3}\left({\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right)R_{ik}\left({\mathbf{r}}_{1}\right)\varDelta_{jl}\left({\mathbf{r}}_{2}\right) \end{array} \end{aligned} $$
Using the techniques outlined in [17], we can show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \delta^{3}\left({\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right) &\displaystyle = &\displaystyle \sum_{n_{r},\varLambda,n_{z}}f_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{1}\right)\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{2}\right) \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} f_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{1}\right) &\displaystyle \equiv &\displaystyle e^{-\eta_{1}/2}e^{-\xi_{1}^{2}/2}\varPhi_{n_{r},\varLambda,n_{z}}^{*}\left({\mathbf{r}}_{1}\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{2}\right) &\displaystyle \equiv &\displaystyle e^{\eta_{2}/2}e^{\xi_{2}^{2}/2}\varPhi_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{2}\right) \end{array} \end{aligned} $$
with the variables η and ξ defined as in Sect. 2.2. Thus, we need to calculate the integrals
$$\displaystyle \begin{aligned} \begin{aligned}R_{ik}^{\left(n_{r},\varLambda,n_{z}\right)} & \equiv\int d^{3}r_{1}R_{ik}\left({\mathbf{r}}_{1}\right)f_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{1}\right)\end{aligned} {} \end{aligned} $$
(2.40)
and
$$\displaystyle \begin{aligned} \begin{aligned}\varDelta_{jl}^{\left(n_{r},\varLambda,n_{z}\right)} & \equiv\int d^{3}r_{2}\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left({\mathbf{r}}_{2}\right)\varDelta_{jl}\left({\mathbf{r}}_{2}\right)\end{aligned} {} \end{aligned} $$
(2.41)
from which we will obtain
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{ik;jl} &\displaystyle = &\displaystyle \sum_{n_{r},\varLambda,n_{z}}R_{ik}^{\left(n_{r},\varLambda,n_{z}\right)}\varDelta_{jl}^{\left(n_{r},\varLambda,n_{z}\right)} \end{array} \end{aligned} $$

2.4.4.3 Calculation of the Integral \(R_{ik}^{\left (n_{r},\varLambda ,n_{z}\right )}\) in Eq. (2.40)

Expanding the products of harmonic-oscillator functions as in [17], we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{ik}^{\left(n_{r},\varLambda,n_{z}\right)} &\displaystyle = &\displaystyle \frac{1}{\sqrt{\pi}b_{\bot}}\frac{1}{\left(\pi b_{z}^{2}\right)^{1/4}}\delta_{\sigma_{i},\sigma_{k}}\delta_{\varLambda-\varLambda_{i}+\varLambda_{k},0}\\ &\displaystyle &\displaystyle \times\left[\sum_{m=0}^{n_{i,k}}T_{n_{r}^{(i)},-\varLambda_{i};n_{r}^{(k)},\varLambda_{k}}^{m}I\left(n_{i,k}-m,-\varLambda_{i}+\varLambda_{k},n_{r},\varLambda;b\right)\right]\\ &\displaystyle &\displaystyle \times\left[\sum_{n=\left|n_{z}^{(i)}-n_{z}^{(k)}\right|,2}^{n_{z}^{(i)}+n_{z}^{(k)}}T_{n_{z}^{(i)},n_{z}^{(k)}}^{n}I\left(n,n_{z};b\right)\right] \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} I\left(n_{1},\varLambda_{1},n_{2},\varLambda_{2};b\right) &\displaystyle \equiv &\displaystyle \int_{0}^{\infty}\rho d\rho\int_{0}^{2\pi}d\varphi\, e^{-\rho^{2}/b^{2}}\varPhi_{n_{1},\varLambda_{1}}\left(\rho,\varphi;b\right)\varPhi_{n_{2},\varLambda_{2}}\left(\rho,\varphi;b\right) \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} I\left(n_{1},n_{2};b\right) &\displaystyle \equiv &\displaystyle \int_{-\infty}^{+\infty}dz\, e^{-z^{2}/b^{2}}\varPhi_{n_{1}}\left(z;b\right)\varPhi_{n_{2}}\left(z;b\right) \end{array} \end{aligned} $$
Note that this implies we always have Λ = 0 (since i and k are from the same block, Ωi = Ωk, and since σi = σk we must therefore have Λi = Λk). We can separate this further as
$$\displaystyle \begin{aligned} \begin{aligned}R_{ik}^{\left(n_{r},\varLambda,n_{z}\right)} & =R_{ik}^{\left(n_{r},\varLambda\right)}R_{ik}^{\left(n_{z}\right)}\end{aligned} {} \end{aligned} $$
(2.42)
with
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{ik}^{\left(n_{r},\varLambda\right)} &\displaystyle \equiv &\displaystyle \frac{1}{\sqrt{\pi}b_{\bot}}\delta_{\sigma_{i},\sigma_{k}}\delta_{\varLambda-\varLambda_{i}+\varLambda_{k},0}\\ &\displaystyle &\displaystyle \times\left[\sum_{m=0}^{n_{i,k}}T_{n_{r}^{(i)},-\varLambda_{i};n_{r}^{(k)},\varLambda_{k}}^{m}I\left(n_{i,k}-m,-\varLambda_{i}+\varLambda_{k},n_{r},\varLambda;b\right)\right] \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{ik}^{\left(n_{z}\right)} &\displaystyle \equiv &\displaystyle \frac{1}{\left(\pi b_{z}^{2}\right)^{1/4}}\left[\sum_{n=\left|n_{z}^{(i)}-n_{z}^{(k)}\right|,2}^{n_{z}^{(i)}+n_{z}^{(k)}}T_{n_{z}^{(i)},n_{z}^{(k)}}^{n}I\left(n,n_{z};b\right)\right] \end{array} \end{aligned} $$
Using the generating-function methods in [17] we calculate the remaining integrals explicitly,
$$\displaystyle \begin{aligned} \begin{array}{rcl} I\left(n_{1},\varLambda_{1},n_{2},\varLambda_{2};b\right) &\displaystyle = &\displaystyle \frac{\left(-1\right)^{n_{1}+n_{2}}}{2^{n_{1,2}+1}n_{1,2}!}\sqrt{n_{1}!\left(n_{1}+\left|\varLambda_{1}\right|\right)!}\sqrt{n_{2}!\left(n_{2}+\left|\varLambda_{2}\right|\right)!}\delta_{\varLambda_{1}+\varLambda_{2},0}\\ &\displaystyle &\displaystyle \times\sum_{p=0}^{n_{1,2}}\left(-1\right)^{p}\left(\begin{array}{c} n_{1,2}\\ p \end{array}\right)\sum_{q=0}^{p}\left(\begin{array}{c} p\\ q \end{array}\right)\left(\begin{array}{c} n_{1,2}-p\\ \frac{n_{1,2}-p-\varLambda_{1}}{2} \end{array}\right)\\ &\displaystyle &\displaystyle \times\delta_{2q-p,2n_{1}+\left|\varLambda_{1}\right|-n_{1,2}}\delta_{n_{1,2}-p-\varLambda_{1},\mathrm{even}}\delta_{p-n_{1,2}\leq\varLambda_{1}\leq n_{1,2}-p} \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} I\left(n_{1},n_{2};b\right) &\displaystyle = &\displaystyle \frac{\left(n_{1}+n_{2}-1\right)!!}{2^{\left(n_{1}+n_{2}+1\right)/2}\sqrt{n_{1}!n_{2}!}}\left(-1\right)^{\left(n_{1}+n_{2}\right)/2}\frac{\left(-1\right)^{n_{1}}+\left(-1\right)^{n_{2}}}{2} \end{array} \end{aligned} $$

2.4.4.4 Calculation of the Integral \(\varDelta _{jl}^{\left (n_{r},\varLambda ,n_{z}\right )}\) in Eq. (2.41)

For the integral \(\varDelta _{ik}^{\left (n_{r},\varLambda ,n_{z}\right )}\), it will be convenient to re-introduce the φ dependence explicitly. We then write
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ik}^{\left(n_{r},\varLambda,n_{z}\right)} &\displaystyle = &\displaystyle -\delta_{\sigma_{i},-\sigma_{k}}\left(I_{\varphi z}-I_{z\varphi}\right)-2\sigma_{k}\delta_{\sigma_{i},-\sigma_{k}}\left(I_{\rho z}-I_{z\rho}\right)\\ &\displaystyle &\displaystyle +2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\left(I_{\rho\varphi}-I_{\varphi\rho}\right)\\ &\displaystyle = &\displaystyle -\delta_{\sigma_{i},-\sigma_{k}}\left(I_{\varphi z}+2\sigma_{k}I_{\rho z}\right)+\delta_{\sigma_{i},-\sigma_{k}}\left(I_{z\varphi}+2\sigma_{k}I_{z\rho}\right)\\ &\displaystyle &\displaystyle +2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\left(I_{\rho\varphi}-I_{\varphi\rho}\right)\\ &\displaystyle \equiv &\displaystyle A_{ik}^{\left(n_{r},\varLambda\right)}A_{ik}^{\left(n_{z}\right)}+B_{ik}^{\left(n_{r},\varLambda\right)}B_{ik}^{\left(n_{z}\right)}+Z_{ik}^{\left(n_{r},\varLambda\right)}Z_{ik}^{\left(n_{z}\right)} \end{array} \end{aligned} $$
with
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{\varphi z} &\displaystyle \equiv &\displaystyle \int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)e^{2\mathrm{i}\sigma_{k}\varphi}\mathrm{i}\partial_{\varphi}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{z}\varPhi_{k}\left(\mathbf{r}\right)\\ I_{z\varphi} &\displaystyle \equiv &\displaystyle \int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)e^{2\mathrm{i}\sigma_{k}\varphi}\mathrm{i}\partial_{z}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{\varphi}\varPhi_{k}\left(\mathbf{r}\right)\\ I_{\rho z} &\displaystyle \equiv &\displaystyle \int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)e^{2\mathrm{i}\sigma_{k}\varphi}\partial_{\rho}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{z}\varPhi_{k}\left(\mathbf{r}\right)\\ I_{z\rho} &\displaystyle \equiv &\displaystyle \int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)e^{2\mathrm{i}\sigma_{k}\varphi}\partial_{z}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{\rho}\varPhi_{k}\left(\mathbf{r}\right)\\ I_{\rho\varphi} &\displaystyle \equiv &\displaystyle -\mathrm{i}\int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)\partial_{\rho}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{\varphi}\varPhi_{k}\left(\mathbf{r}\right)\\ I_{\varphi\rho} &\displaystyle \equiv &\displaystyle \mathrm{i}\int d^{3}r\,\hat{\varPhi}_{n_{r},\varLambda,n_{z}}\left(\mathbf{r}\right)\partial_{\varphi}\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{\rho}\varPhi_{k}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
To proceed further, we separate the radial (ρ, φ) and z dependence of each term in \(\varDelta _{ik}^{\left (n_{r},\varLambda ,n_{z}\right )}\),
$$\displaystyle \begin{aligned} \begin{array}{rcl} -\delta_{\sigma_{i},-\sigma_{k}}\left(I_{\varphi z}+2\sigma_{k}I_{\rho z}\right) &\displaystyle \equiv &\displaystyle A_{ik}^{\left(n_{r},\varLambda\right)}A_{ik}^{\left(n_{z}\right)}\\ \delta_{\sigma_{i},-\sigma_{k}}\left(I_{z\varphi}+2\sigma_{k}I_{z\rho}\right) &\displaystyle \equiv &\displaystyle B_{ik}^{\left(n_{r},\varLambda\right)}B_{ik}^{\left(n_{z}\right)}\\ 2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}\left(I_{\rho\varphi}-I_{\varphi\rho}\right) &\displaystyle \equiv &\displaystyle Z_{ik}^{\left(n_{r},\varLambda\right)}Z_{ik}^{\left(n_{z}\right)} \end{array} \end{aligned} $$
Next, we introduce the general integrals of the form and
$$\displaystyle \begin{aligned} \begin{aligned}I^{\left(n_{z}\right)}\left(n_{1};n_{2}\right) & \equiv\int_{-\infty}^{\infty}dz\,\hat{\varPhi}_{n_{z}}\left(z\right)\varPhi_{n_{1}}\left(z\right)\varPhi_{n_{2}}\left(z\right)\end{aligned} {} \end{aligned} $$
(2.44)
Using recurrence relations for the derivatives of harmonic-oscillator functions, we can then write
$$\displaystyle \begin{aligned} \begin{array}{rcl} A_{ik}^{\left(n_{r},\varLambda\right)} &\displaystyle = &\displaystyle -\frac{\delta_{\sigma_{i},-\sigma_{k}}}{b_{\bot}}\left\{ \sqrt{n_{r}^{\left(i\right)}+\left|\varLambda_{i}\right|}\left(1-\delta_{\varLambda_{i},0}\right)\left(-s_{-\varLambda_{i}}+2\sigma_{k}\right)\right.\\ &\displaystyle &\displaystyle \times I_{0,2\sigma_{k}+s_{-\varLambda_{i}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i}+s_{-\varLambda_{i}};n_{r}^{\left(k\right)},\varLambda_{k}\right)\\ &\displaystyle &\displaystyle +\sqrt{n_{r}^{\left(i\right)}}\left[-s_{-\varLambda_{i}}\left(1-\delta_{\varLambda_{i},0}\right)-2\sigma_{k}\left(1+\delta_{\varLambda_{i},0}\right)\right]\\ &\displaystyle &\displaystyle \times I_{0,2\sigma_{k}-s_{-\varLambda_{i}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)}-1,\varLambda_{i}-s_{-\varLambda_{i}};n_{r}^{\left(k\right)},\varLambda_{k}\right)\\ &\displaystyle &\displaystyle \left.-\frac{2\sigma_{k}}{b_{\perp}}I_{1,2\sigma_{k}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)},\varLambda_{k}\right)\right\} \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} A_{ik}^{\left(n_{z}\right)} &\displaystyle = &\displaystyle \frac{1}{b_{z}}\left[\sqrt{\frac{n_{z}^{\left(k\right)}}{2}}I^{\left(n_{z}\right)}\left(n_{z}^{\left(i\right)};n_{z}^{\left(k\right)}-1\right)-\sqrt{\frac{n_{z}^{\left(k\right)}+1}{2}}I^{\left(n_{z}\right)}\left(n_{z}^{\left(i\right)};n_{z}^{\left(k\right)}+1\right)\right] \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} B_{ik}^{\left(n_{r},\varLambda\right)} &\displaystyle = &\displaystyle \frac{\delta_{\sigma_{i},-\sigma_{k}}}{b_{\perp}}\left\{ \sqrt{n_{r}^{\left(k\right)}+\left|\varLambda_{k}\right|}\left(1-\delta_{\varLambda_{k},0}\right)\left(-s_{\varLambda_{k}}+2\sigma_{k}\right)\right.\\ &\displaystyle &\displaystyle \times I_{0,2\sigma_{k}+s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)},\varLambda_{k}-s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle +\sqrt{n_{r}^{\left(k\right)}}\left[-s_{\varLambda_{k}}\left(1-\delta_{\varLambda_{k},0}\right)-2\sigma_{k}\left(1+\delta_{\varLambda_{k},0}\right)\right]\\ &\displaystyle &\displaystyle \times I_{0,2\sigma_{k}-s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)}-1,\varLambda_{k}+s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle \left.-\frac{2\sigma_{k}}{b_{\perp}}I_{1,2\sigma_{k}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)},\varLambda_{k}\right)\right\} \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} B_{ik}^{\left(n_{z}\right)} &\displaystyle = &\displaystyle \frac{1}{b_{z}}\left[\sqrt{\frac{n_{z}^{\left(i\right)}}{2}}I^{\left(n_{z}\right)}\left(n_{z}^{\left(i\right)}-1;n_{z}^{\left(k\right)}\right)-\sqrt{\frac{n_{z}^{\left(i\right)}+1}{2}}I^{\left(n_{z}\right)}\left(n_{z}^{\left(i\right)}+1;n_{z}^{\left(k\right)}\right)\right] \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} Z_{ik}^{\left(n_{r},\varLambda\right)} &\displaystyle = &\displaystyle \frac{2\sigma_{k}\delta_{\sigma_{i},\sigma_{k}}}{b_{\perp}^{2}}s_{-\varLambda_{i}}\left(1-\delta_{\varLambda_{i},0}\right)\left\{ -2\sqrt{n_{r}^{\left(i\right)}+\left|\varLambda_{i}\right|}\sqrt{n_{r}^{\left(k\right)}}\right.\\ &\displaystyle &\displaystyle \times I_{0,s_{-\varLambda_{i}}-s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i}+s_{-\varLambda_{i}};n_{r}^{\left(k\right)}-1,\varLambda_{k}+s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle +2\sqrt{n_{r}^{\left(i\right)}}\sqrt{n_{r}^{\left(k\right)}+\left|\varLambda_{k}\right|}\\ &\displaystyle &\displaystyle \times I_{0,-s_{-\varLambda_{i}}+s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)}-1,\varLambda_{i}-s_{-\varLambda_{i}};n_{r}^{\left(k\right)},\varLambda_{k}-s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle +\frac{1}{b_{\perp}}\sqrt{n_{r}^{\left(k\right)}+\left|\varLambda_{k}\right|}I_{1,s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)},\varLambda_{k}-s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle +\frac{1}{b_{\perp}}\sqrt{n_{r}^{\left(k\right)}}I_{1,-s_{\varLambda_{k}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(k\right)}-1,\varLambda_{k}+s_{\varLambda_{k}}\right)\\ &\displaystyle &\displaystyle -\frac{1}{b_{\perp}}\sqrt{n_{r}^{\left(i\right)}+\left|\varLambda_{i}\right|}I_{1,s_{-\varLambda_{i}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i}+s_{-\varLambda_{i}};n_{r}^{\left(k\right)},\varLambda_{k}\right)\\ &\displaystyle &\displaystyle \left.-\frac{1}{b_{\perp}}\sqrt{n_{r}^{\left(i\right)}}I_{1,-s_{-\varLambda_{i}}}^{\left(n_{r},\varLambda\right)}\left(n_{r}^{\left(i\right)}-1,\varLambda_{i}-s_{-\varLambda_{i}};n_{r}^{\left(k\right)},\varLambda_{k}\right)\right\}\vspace{-5pt} \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} Z_{ik}^{\left(n_{z}\right)} &\displaystyle \equiv &\displaystyle I^{\left(n_{z}\right)}\left(n_{z}^{\left(i\right)};n_{z}^{\left(k\right)}\right) \end{array} \end{aligned} $$
and where
$$\displaystyle \begin{aligned} \begin{aligned}s_{k} & \equiv\begin{cases} 1 & k\geq0\\ -1 & k<0 \end{cases}\end{aligned} {} \end{aligned} $$
(2.45)
The remaining integrals can be evaluated using the formula for products of harmonic-oscillator functions in [17]. For the integral in Eq. (2.43) with α = 0 and μ = 0, For α = 0 and μ = ±1, Next, for α = 0 and μ = ±2 we have two cases of interest, first for \(\left |-\varLambda _{1}+\varLambda _{2}\right |=0\), and for \(\left |-\varLambda _{1}+\varLambda _{2}\right |=2\), Next, for α = 1 and μ = 0 we find and for α = 1 and μ = ±1
For the \(I^{\left (n_{z}\right )}\) integral in Eq. (2.44), we have

2.4.4.5 Modification for Large Oscillator Number in z Direction

In the case where the nz quantum numbers are large, we need to re-write the integral \(R_{ik}^{\left (n_{z}\right )}\) which appears in Eq. (2.42) in a form that is more stable numerically. Writing this integral explicitly, we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{n_{1},n_{2}}^{n} &\displaystyle \equiv &\displaystyle \int_{-\infty}^{\infty}dz\,\phi_{n_{1}}^{*}\left(z;b\right)\left[e^{-\frac{z^{2}}{2b^{2}}}\phi_{n}^{*}\left(z;b\right)\right]\phi_{n_{2}}\left(z;b\right) \end{array} \end{aligned} $$
In appendix D of [17] the integral
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_{1}\left|f_{n}\right|n_{2}\right\rangle &\displaystyle = &\displaystyle K_{z}^{1/2}\lambda_{n}\int_{-\infty}^{\infty}dz\,\varPhi_{n_{1}}\left(z;b\right)e^{-z^{2}/\left(2Gb^{2}\right)}\varPhi_{n}\left(z;G^{1/2}b\right)\varPhi_{n_{2}}\left(z;b\right) \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} K_{z} &\displaystyle \equiv &\displaystyle \frac{\pi\mu^{2}}{G^{1/2}}\\ \lambda_{n} &\displaystyle \equiv &\displaystyle G^{-n/2}\\ G &\displaystyle \equiv &\displaystyle 1+\frac{\mu^{2}}{b^{2}} \end{array} \end{aligned} $$
is evaluated to
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_{1}\left|f_{n}\right|n_{2}\right\rangle &\displaystyle = &\displaystyle \frac{\mu b^{-1/2}}{\sqrt{2\pi^{5/2}}}\frac{\varGamma\left(\xi-n_{1}\right)\varGamma\left(\xi-n_{2}\right)\varGamma\left(\xi-n\right)}{z^{\xi}\sqrt{n!n_{1}!n_{2}!}}\\ &\displaystyle &\displaystyle \times\,_{2}F_{1}\left(-n_{1},-n_{2};-\xi+n+1;1-z\right) \end{array} \end{aligned} $$
Taking the limit as μ → 0, we can then deduce
$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{n_{1},n_{2}}^{n} &\displaystyle = &\displaystyle \lim_{\mu\rightarrow0}\frac{\left\langle n_{1}\left|f_{n}\right|n_{2}\right\rangle }{K^{1/2}\lambda_{n}}\\ &\displaystyle = &\displaystyle \frac{b^{-1/2}}{\sqrt{2\pi^{7/2}}}\frac{\varGamma\left(\xi-n_{1}\right)\varGamma\left(\xi-n_{2}\right)\varGamma\left(\xi-n\right)}{\sqrt{n!n_{1}!n_{2}!}} \end{array} \end{aligned} $$
which can also be written as

2.4.5 Two-Body Center-of-Mass Correction

The kinetic energy of the center of mass is written
$$\displaystyle \begin{aligned} \begin{array}{rcl} K &\displaystyle = &\displaystyle \frac{{\mathbf{P}}^{2}}{2mA} \end{array} \end{aligned} $$
where m is the nucleon mass, and A is the number of nucleons. This energy must be subtracted from the Hartree-Fock Hamiltonian. The total momentum P can be decomposed into the individual nucleon momenta,
$$\displaystyle \begin{aligned} \begin{array}{rcl} K &\displaystyle = &\displaystyle \frac{1}{2mA}\left(\sum_{i=1}^{A}{\mathbf{p}}_{i}\right)^{2}\\ &\displaystyle = &\displaystyle \frac{1}{2mA}\left(\sum_{i}{\mathbf{p}}_{i}^{2}+2\sum_{i>j}{\mathbf{p}}_{i}\cdot{\mathbf{p}}_{j}\right) \end{array} \end{aligned} $$
where the individual momentum operator can be written in coordinate space
$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathbf{p} &\displaystyle = &\displaystyle \frac{\hbar}{i}\mathbf{\nabla} \end{array} \end{aligned} $$
The operator P2 can therefore be written in second-quantized form as
$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathbf{P}}^{2} &\displaystyle = &\displaystyle \sum_{ij}\left\langle i\left|{\mathbf{p}}^{2}\right|j\right\rangle a_{i}^{\dagger}a_{j}+\frac{1}{2}\sum_{ijkl}\left\langle ij\left|2{\mathbf{p}}_{1}\cdot{\mathbf{p}}_{2}\right|kl\right\rangle a_{i}^{\dagger}a_{j}^{\dagger}a_{l}a_{k}\\ {} &\displaystyle = &\displaystyle \sum_{ij}\left\langle i\left|{\mathbf{p}}^{2}\right|j\right\rangle a_{i}^{\dagger}a_{j}+\sum_{ijkl}\left\langle ij\left|{\mathbf{p}}_{1}\cdot{\mathbf{p}}_{2}\right|kl\right\rangle a_{i}^{\dagger}a_{j}^{\dagger}a_{l}a_{k} \end{array} \end{aligned} $$
The expectation value of this operator in the Hartree-Fock ground state is
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle \varPhi\left|{\mathbf{P}}^{2}\right|\varPhi\right\rangle &\displaystyle = &\displaystyle \sum_{ij}\left\langle i\left|{\mathbf{p}}^{2}\right|j\right\rangle \rho_{ji}+\sum_{ijkl}\left\langle ij\left|{\mathbf{p}}_{1}\cdot{\mathbf{p}}_{2}\right|\widetilde{kl}\right\rangle \rho_{ki}\rho_{lj} \end{array} \end{aligned} $$
where \(\left |\widetilde {kl}\right \rangle =\left |kl\right \rangle -\left |lk\right \rangle \) enforces antisymmetrization. Therefore, we can separate the kinetic energy into one- and two-body contributions
$$\displaystyle \begin{aligned} \begin{array}{rcl} K &\displaystyle = &\displaystyle K_{1}+K_{2} \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} K_{1} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{2mA}\sum_{ij}\left\langle i\left|\mathbf{\nabla}^{2}\right|j\right\rangle \rho_{ji} \end{array} \end{aligned} $$
and
$$\displaystyle \begin{aligned} \begin{array}{rcl} K_{2} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{2mA}\sum_{ijkl}\left\langle ij\left|\mathbf{\nabla}_{1}\cdot\mathbf{\nabla}_{2}\right|\widetilde{kl}\right\rangle \rho_{ki}\rho_{lj} \end{array} \end{aligned} $$
The one-body contribution, K1, simply introduces a well-known \(\left (1-1/A\right )\) multiplication factor to the matrix elements of the kinetic-energy operator derived in Sect. 2.4.1. For the remainder of this section, we focus on the two-body contribution, K2. The field that corresponds to this energy is
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(K\right)} &\displaystyle = &\displaystyle \frac{\partial}{\partial\rho_{ki}}K_{2}\\ {} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\sum_{jl}\left(\left\langle i\left|\mathbf{\nabla}\right|k\right\rangle \cdot\left\langle j\left|\mathbf{\nabla}\right|l\right\rangle -\left\langle i\left|\mathbf{\nabla}\right|l\right\rangle \cdot\left\langle j\left|\mathbf{\nabla}\right|k\right\rangle \right)\rho_{lj} \end{array} \end{aligned} $$
It will be useful to express the derivatives in terms of ladder operators,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \partial_{+} &\displaystyle \equiv &\displaystyle \partial_{x}+i\partial_{y}\\ \partial_{-} &\displaystyle \equiv &\displaystyle \partial_{x}-i\partial_{y} \end{array} \end{aligned} $$
so that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathbf{\nabla}_{1}\cdot\mathbf{\nabla}_{2} &\displaystyle = &\displaystyle \frac{1}{2}\left(\partial_{+}^{\left(1\right)}\partial_{-}^{\left(2\right)}+\partial_{-}^{\left(1\right)}\partial_{+}^{\left(2\right)}\right)+\partial_{z_{1}}\partial_{z_{2}} \end{array} \end{aligned} $$
From this point, we can calculate the direct and exchange contributions explicitly.

2.4.5.1 The Direct Contribution

we calculate the direct contribution to the field defined by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,D\right)} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\sum_{jl}\sum_{q^{\prime}}\left\langle ij\left|\mathbf{\nabla}_{1}\cdot\mathbf{\nabla}_{2}\right|kl\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\\ &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\sum_{q^{\prime}}\left(\frac{1}{2}\left\langle i\left|\partial_{+}\right|k\right\rangle \sum_{jl}\left\langle j\left|\partial_{-}\right|l\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\right.\\ &\displaystyle &\displaystyle +\frac{1}{2}\left\langle i\left|\partial_{-}\right|k\right\rangle \sum_{jl}\left\langle j\left|\partial_{+}\right|l\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\\ &\displaystyle &\displaystyle \left.+\left\langle i\left|\partial_{z}\right|k\right\rangle \sum_{jl}\left\langle j\left|\partial_{z}\right|l\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\right) \end{array} \end{aligned} $$
Since Ωi = Ωk and Ωj = Ωl, the conditions σi = σk and σj = σl imposed by the matrix elements imply that Λi = Λk and Λj = Λl, respectively. Then by using the recurrence relations for derivatives of harmonic-oscillator functions it is easy to show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle i\left|\partial_{\pm}\right|k\right\rangle &\displaystyle = &\displaystyle \left\langle j\left|\partial_{\pm}\right|l\right\rangle =0 \end{array} \end{aligned} $$
Thus we are left with
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,D\right)} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\left\langle i\left|\partial_{z}\right|k\right\rangle \sum_{q^{\prime}}\sum_{jl}\left\langle j\left|\partial_{z}\right|l\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\\ &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\delta_{\sigma_{i},\sigma_{k}}I_{ik}^{\left(z\right)}\sum_{jl}\delta_{\sigma_{j},\sigma_{l}}I_{jl}^{\left(z\right)}\rho_{lj} \end{array} \end{aligned} $$
where we have defined
$$\displaystyle \begin{aligned} \begin{aligned}I_{ij}^{\left(z\right)} & \equiv\int d^{3}r\,\varPhi_{i}^{*}\left(\mathbf{r}\right)\frac{\partial}{\partial z}\varPhi_{j}\left(\mathbf{r}\right)\end{aligned} {} \end{aligned} $$
(2.46)
We can explicitly take into account time-reversed states and order the indices to get
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,D\right)} &\displaystyle = &\displaystyle \frac{-\hbar^{2}}{mA}\delta_{\sigma_{i},\sigma_{k}}I_{ik}^{\left(z\right)}\left[\sum_{j>l>0}2\delta_{\sigma_{j},\sigma_{l}}\left(I_{jl}^{\left(z\right)}+I_{lj}^{\left(z\right)}\right)\rho_{lj}+\sum_{j>0}2I_{jj}^{\left(z\right)}\rho_{jj}\right] \end{array} \end{aligned} $$
Now, using recurrence relations, we can show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{jj}^{\left(z\right)} &\displaystyle = &\displaystyle 0 \end{array} \end{aligned} $$
and integrating by parts, we can show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{lj}^{\left(z\right)} &\displaystyle = &\displaystyle -I_{jl}^{\left(z\right)} \end{array} \end{aligned} $$
and therefore the direct contribution vanishes,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,D\right)} &\displaystyle = &\displaystyle 0 \end{array} \end{aligned} $$

2.4.5.2 The Exchange Contribution

We calculate
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,E\right)} &\displaystyle = &\displaystyle \frac{\hbar^{2}}{mA}\sum_{q^{\prime}}\sum_{jl}\left\langle i\left|\mathbf{\nabla}\right|l\right\rangle \cdot\left\langle j\left|\mathbf{\nabla}\right|k\right\rangle \rho_{lj}^{\left(q^{\prime}\right)}\\ &\displaystyle = &\displaystyle \frac{\hbar^{2}}{mA}\sum_{jl}\delta_{\sigma_{i},\sigma_{l}}\delta_{\sigma_{j},\sigma_{k}}\left[\frac{1}{2}\left(I_{il}^{\left(+\right)}I_{jk}^{\left(-\right)}+I_{il}^{\left(-\right)}I_{jk}^{\left(+\right)}\right)+I_{il}^{\left(z\right)}I_{jk}^{\left(z\right)}\right]\rho_{lj}^{\left(q\right)} \end{array} \end{aligned} $$
where we have defined
$$\displaystyle \begin{aligned} \begin{aligned}I_{ij}^{\left(\pm\right)} & \equiv\int d^{3}r\,\varPhi_{i}^{*}\left(\mathbf{r}\right)\partial_{\pm}\varPhi_{j}\left(\mathbf{r}\right)\end{aligned} {} \end{aligned} $$
(2.47)
For convenience, we define
$$\displaystyle \begin{aligned} \begin{aligned}I_{il;jk} & \equiv\frac{1}{2}\left(I_{il}^{\left(+\right)}I_{jk}^{\left(-\right)}+I_{il}^{\left(-\right)}I_{jk}^{\left(+\right)}\right)+I_{il}^{\left(z\right)}I_{jk}^{\left(z\right)}\end{aligned} {} \end{aligned} $$
(2.48)
so that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,E\right)} &\displaystyle = &\displaystyle \frac{\hbar^{2}}{mA}\sum_{jl}\delta_{\sigma_{i},\sigma_{l}}\delta_{\sigma_{j},\sigma_{k}}I_{il;jk}\rho_{lj}^{\left(q\right)} \end{array} \end{aligned} $$
Taking explicitly into account time-reversed states and ordering of indices, we can write this as
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,E\right)} &\displaystyle = &\displaystyle \frac{\hbar^{2}}{mA}\sum_{j\geq l>0}\frac{\rho_{lj}^{\left(q\right)}}{1+\delta_{jl}}\left\{ \delta_{\sigma_{i},\sigma_{l}}\delta_{\sigma_{j},\sigma_{k}}I_{il;jk}+4\sigma_{j}\sigma_{l}\delta_{\sigma_{i},-\sigma_{l}}\delta_{-\sigma_{j},\sigma_{k}}I_{i\underline{l};\underline{j}k}\right.\\ &\displaystyle &\displaystyle \left.+\delta_{\sigma_{i},\sigma_{j}}\delta_{\sigma_{l},\sigma_{k}}I_{ij;lk}+4\sigma_{j}\sigma_{l}\delta_{\sigma_{i},-\sigma_{j}}\delta_{-\sigma_{l},\sigma_{k}}I_{i\underline{j};\underline{l}k}\right\} \end{array} \end{aligned} $$
and with the help of the identity
$$\displaystyle \begin{aligned} \begin{array}{rcl} \delta_{\sigma_{i},\sigma_{l}}\delta_{\sigma_{j},\sigma_{k}} &\displaystyle = &\displaystyle \delta_{\sigma_{i},\sigma_{k}}\delta_{\sigma_{j},\sigma_{l}}\delta_{\sigma_{i},\sigma_{j}}+\delta_{\sigma_{i},-\sigma_{k}}\delta_{\sigma_{j},-\sigma_{l}}\delta_{\sigma_{i},-\sigma_{j}} \end{array} \end{aligned} $$
we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(q,K,E\right)} &\displaystyle = &\displaystyle \frac{\hbar^{2}}{mA}\sum_{j\geq l>0}\frac{\rho_{lj}^{\left(q\right)}}{1+\delta_{jl}}\left\{ \delta_{\sigma_{i},\sigma_{k}}\delta_{\sigma_{j},\sigma_{l}}\left[\delta_{\sigma_{i},\sigma_{j}}\left(I_{il;jk}+I_{ij;lk}\right)\right.\right.\\ &\displaystyle &\displaystyle \left.+\delta_{\sigma_{i},-\sigma_{j}}\left(I_{i\underline{l};\underline{j}k}+I_{i\underline{j};\underline{l}k}\right)\right]\\ &\displaystyle &\displaystyle +\delta_{\sigma_{i},-\sigma_{k}}\delta_{\sigma_{j},-\sigma_{l}}\left[\delta_{\sigma_{i},\sigma_{j}}\left(I_{ij;lk}-I_{i\underline{l};\underline{j}k}\right)\right.\\ &\displaystyle &\displaystyle \left.\left.+\delta_{\sigma_{i},-\sigma_{j}}\left(I_{il;jk}-I_{i\underline{j};\underline{l}k}\right)\right]\right\} \end{array} \end{aligned} $$
This is the field that must be subtracted from the Hartree-Fock field to correct for the two-body contributions of the center-of-mass motion. Next, we will give the explicit form of the integrals Iil;jk defined in Eq. (2.48).
First we note that in cylindrical coordinates,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \partial_{\pm} &\displaystyle = &\displaystyle e^{\pm\mathrm{i}\varphi}\left(\frac{\partial}{\partial\rho}\pm\mathrm{i}\frac{1}{\rho}\frac{\partial}{\partial\varphi}\right) \end{array} \end{aligned} $$
Using recurrence relations for the derivatives of harmonic-oscillator functions, we can then show that the integral in Eq. (2.47) can be written as
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{ij}^{\left(\pm\right)} &\displaystyle = &\displaystyle \sqrt{\pi}\delta_{n_{z}^{\left(i\right)},n_{z}^{\left(j\right)}}\left\{ \left(1\mp s_{\varLambda_{j}}\right)\left(1-\delta_{\varLambda_{j},0}\right)\sqrt{n_{r}^{\left(j\right)}+\left|\varLambda_{j}\right|}\right.\\ &\displaystyle &\displaystyle \times I_{0,s_{\varLambda_{j}}\pm1}^{\left(0,0\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(j\right)},\varLambda_{j}-s_{\varLambda_{j}}\right)\\ &\displaystyle &\displaystyle -\left[\left(1\pm s_{\varLambda_{j}}\right)+\left(1\mp s_{\varLambda_{j}}\right)\delta_{\varLambda_{j},0}\right]\sqrt{n_{r}^{\left(j\right)}}\\ &\displaystyle &\displaystyle \times I_{0,-s_{\varLambda_{j}}\pm1}^{\left(0,0\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(j\right)}-1,\varLambda_{j}+s_{\varLambda_{j}}\right)\\ &\displaystyle &\displaystyle \left.-\frac{1}{b_{\perp}}I_{1,\pm1}^{\left(0,0\right)}\left(n_{r}^{\left(i\right)},\varLambda_{i};n_{r}^{\left(j\right)},\varLambda_{j}\right)\right\} \end{array} \end{aligned} $$
where sΛ was defined in Eq. (2.45), and the remaining integrals \(I_{\alpha ,\mu }^{\left (n_{r},\varLambda \right )}\) were calculated in Sect. 2.4.4. Next, for the integrals \(I_{\alpha \beta }^{\left (z\right )}\) defined in Eq. (2.46), we can similarly show that
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{\alpha\beta}^{\left(z\right)} &\displaystyle = &\displaystyle \frac{1}{b_{z}}\delta_{n_{r}^{\left(\alpha\right)},n_{r}^{\left(\beta\right)}}\delta_{\varLambda_{\alpha},\varLambda_{\beta}}\left[\sqrt{\frac{n_{z}^{\left(\beta\right)}}{2}}\delta_{n_{z}^{\left(\alpha\right)},n_{z}^{\left(\beta\right)}-1}-\sqrt{\frac{n_{z}^{\left(\beta\right)}+1}{2}}\delta_{n_{z}^{\left(\alpha\right)},n_{z}^{\left(\beta\right)}+1}\right]\vspace{-4pt} \end{array} \end{aligned} $$

2.4.6 The Density-Dependent Contribution

We start with the potential function
$$\displaystyle \begin{aligned} \begin{aligned}V\left({\mathbf{r}}_{1},{\mathbf{r}}_{2}\right) & \equiv t_{0}\left(1+x_{0}P_{\sigma}\right)\rho^{\lambda}\left(\frac{{\mathbf{r}}_{1}+{\mathbf{r}}_{2}}{2}\right)\delta^{3}\left({\mathbf{r}}_{1}-{\mathbf{r}}_{2}\right)\end{aligned} {} \end{aligned} $$
(2.49)
where \(\rho \left (\mathbf {r}\right )\) is the spatial one-body nucleon density given by the sum of neutron and proton densities,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \rho\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \rho_{n}\left(\mathbf{r}\right)+\rho_{p}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
with
$$\displaystyle \begin{aligned}\begin{array}{rcl} \rho_{q}\left(\mathbf{r}\right) &\displaystyle = &\displaystyle \sum_{i^{\prime}\gamma^{\prime}}\delta_{\sigma_{i^{\prime}},\sigma_{\gamma^{\prime}}}\varPhi_{i^{\prime}}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma^{\prime}}\left(\mathbf{r}\right)\rho_{\gamma^{\prime}i^{\prime}}^{q} \end{array} \end{aligned} $$
The total energy corresponding to the density-dependent potential is
$$\displaystyle \begin{aligned} \begin{aligned}E_{DD} & \equiv\frac{1}{4}\sum_{qq^{\prime},ijkl}\left[\left\langle ij\left|V\right|\widetilde{kl}\right\rangle \rho_{lj}^{q^{\prime}}\rho_{ki}^{q}+\left\langle ij\left|V\right|\widetilde{kl}\right\rangle ^{*}\left(\rho_{lj}^{q^{\prime}}\right)^{*}\left(\rho_{ki}^{q}\right)^{*}\right]\end{aligned} {} \end{aligned} $$
(2.50)
After some simplifications, this can be written in the form
$$\displaystyle \begin{aligned} \begin{array}{rcl} E_{DD} &\displaystyle = &\displaystyle \frac{1}{4}t_{0}\left(1-x_{0}\right)\int d^{3}r\,\rho^{\lambda+2}\left(\mathbf{r}\right)\\ &\displaystyle &\displaystyle +t_{0}\left(\frac{1}{2}+x_{0}\right)\int d^{3}r\,\rho^{\lambda}\left(\mathbf{r}\right)\rho_{n}\left(\mathbf{r}\right)\rho_{p}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
from which we deduce the corresponding Hartree-Fock field [4], which includes the so-called “re-arrangement” terms [24]
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{q} &\displaystyle = &\displaystyle \frac{\partial E_{DD}}{\partial\rho_{ki}^{q}}\\ &\displaystyle = &\displaystyle t_{0}\delta_{\sigma_{i},\sigma_{k}}\int d^{3}r\,\varPhi_{i}^{*}\left(\mathbf{r}\right)\varPhi_{k}\left(\mathbf{r}\right)\left\{ \left[1+\frac{x_{0}}{2}+\frac{1}{4}\lambda\left(1-x_{0}\right)\right]\rho^{\lambda+1}\left(\mathbf{r}\right)\right.\\ &\displaystyle &\displaystyle \left.-\left(\frac{1}{2}+x_{0}\right)\rho^{\lambda}\left(\mathbf{r}\right)\rho_{q}\left(\mathbf{r}\right)+\left(\frac{1}{2}+x_{0}\right)\lambda\rho^{\lambda-1}\left(\mathbf{r}\right)\rho_{n}\left(\mathbf{r}\right)\rho_{p}\left(\mathbf{r}\right)\right\} \end{array} \end{aligned} $$
If we define the integral
$$\displaystyle \begin{aligned} \begin{aligned}D_{\alpha\gamma}^{\left(qq^{\prime}\right)} & \equiv\int d^{3}r\,\varPhi_{\alpha}^{*}\left(\mathbf{r}\right)\varPhi_{\gamma}\left(\mathbf{r}\right)\rho^{\lambda-1}\left(\mathbf{r}\right)\rho_{q}\left(\mathbf{r}\right)\rho_{q^{\prime}}\left(\mathbf{r}\right)\end{aligned} {} \end{aligned} $$
(2.51)
we can then write
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{q} &\displaystyle = &\displaystyle t_{0}\delta_{\sigma_{i},\sigma_{k}}\left\{ \left[1+\frac{x_{0}}{2}+\frac{1}{4}\lambda\left(1-x_{0}\right)\right]D_{ik}^{\left(tt\right)}-\left(\frac{1}{2}+x_{0}\right)D_{ik}^{\left(tq\right)}\right.\\ &\displaystyle &\displaystyle \left.+\left(\frac{1}{2}+x_{0}\right)\lambda D_{ik}^{\left(np\right)}\right\} \end{array} \end{aligned} $$
where the superscript t refers to the total density: \(\rho _{t}\left (\mathbf {r}\right )=\rho _{n}\left (\mathbf {r}\right )+\rho _{p}\left (\mathbf {r}\right )\). For a non-integer exponent λ, the integrals in Eq. (2.51) cannot in general be performed analytically and must instead be evaluated numerically, e.g. by a quadrature method.
The contribution to the total energy of the system from the density-dependent part of the interaction is given in Eq. (2.50) and be written in terms of a Hartree-Fock field with no re-arrangement terms,
$$\displaystyle \begin{aligned} \begin{array}{rcl} E_{DD} &\displaystyle = &\displaystyle \frac{1}{4}\sum_{q,ik}\left[\varGamma_{ik}^{q,\mathrm{nr}}\rho_{ki}^{q}+\left(\varGamma_{ik}^{q,\mathrm{nr}}\right)^{*}\left(\rho_{ki}^{q}\right)^{*}\right] \end{array} \end{aligned} $$
where
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{q,\mathrm{nr}} &\displaystyle \equiv &\displaystyle \sum_{q^{\prime},jl}\left\langle ij\left|V\right|\widetilde{kl}\right\rangle \rho_{lj}^{q^{\prime}} \end{array} \end{aligned} $$
Starting from Eq. (2.49), and using the integrals in Eq. (2.51) we can show that this is
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{q,\mathrm{nr}} &\displaystyle = &\displaystyle t_{0}\delta_{\sigma_{i},\sigma_{k}}\left[\left(1+\frac{x_{0}}{2}\right)D_{ik}^{\left(tt\right)}-\left(\frac{1}{2}+x_{0}\right)D_{ik}^{\left(tq\right)}\right] \end{array} \end{aligned} $$
Finally, we reiterate the discussion at the end of [20] concerning the contribution of the density-dependent potential in Eq. (2.49) to the pairing field. This contribution is given by
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ij}^{q} &\displaystyle = &\displaystyle \frac{1}{2}\sum_{q^{\prime},kl}\left\langle ij\left|V\right|\widetilde{kl}\right\rangle \kappa_{lk}^{q^{\prime}} \end{array} \end{aligned} $$
which we can write in terms of spin-exchange (Pσ), isospin-exchange (Pτ), and spatial-exchange (Pr) operators,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varDelta_{ij}^{q} &\displaystyle = &\displaystyle \frac{1}{2}\sum_{q^{\prime},kl}\left\langle ij\left|V\left(1-P_{\sigma}P_{\tau}P_{r}\right)\right|kl\right\rangle \kappa_{lk}^{q^{\prime}} \end{array} \end{aligned} $$
Note however that because V  contains \(\delta ^{3}\left ({\mathbf {r}}_{1}-{\mathbf {r}}_{2}\right )\), the operator Pr has no effect. Furthermore, the states k and l are forced to correspond to the same particle type by the pairing tensor \(\kappa _{lk}^{q^{\prime }}\), therefore the operator Pτ also has no effect. Thus we are left with
$$\displaystyle \begin{aligned} \begin{aligned}\varDelta_{ij}^{q} & =\frac{1}{2}\sum_{q^{\prime},kl}\left\langle ij\left|V\left(1-P_{\sigma}\right)\right|kl\right\rangle \kappa_{lk}^{q^{\prime}}\end{aligned} {} \end{aligned} $$
(2.52)
If we assign x0 = 1 in Eq. (2.49), then the density-dependent matrix element in Eq. (2.52) is proportional to the term
$$\displaystyle \begin{aligned} \begin{array}{rcl} \left(1+P_{\sigma}\right)\left(1-P_{\sigma}\right) &\displaystyle = &\displaystyle 1-P_{\sigma}P_{\sigma}=0 \end{array} \end{aligned} $$
In other words, by choosing x0 = 1, we eliminate the contribution of the density-dependent potential in Eq. (2.49) to the pairing field.

2.4.7 The Slater Approximation to the Coulomb Exchange Term

The matrix elements derived in Sect. 2.4.3 can be used to calculate both the direct and exchange Coulomb terms, however these calculations can be computationally demanding, especially for the exchange term. As an alternate approach, an approximation due to Slater [25] can be used for the exchange energy,
$$\displaystyle \begin{aligned} \begin{array}{rcl} E_{C,\mathrm{Slater}}^{\left(E\right)} &\displaystyle = &\displaystyle -\frac{3}{4}e^{2}\left(\frac{3}{\pi}\right)^{1/3}\int d^{3}r\rho_{p}^{4/3}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
and the corresponding contribution to the Hartree-Fock field is
$$\displaystyle \begin{aligned} \begin{array}{rcl} \varGamma_{ik}^{\left(C,E\right)} &\displaystyle = &\displaystyle \frac{\partial}{\partial\rho_{ki}}E_{C,\mathrm{Slater}}^{\left(E\right)}\\ &\displaystyle = &\displaystyle -e^{2}\left(\frac{3}{\pi}\right)^{1/3}\delta_{\sigma_{i},\sigma_{k}}\delta_{q_{i},q_{k}}\delta_{q_{i},p}\int d^{3}r\varPhi_{i}^{*}\left(\mathbf{r}\right)\varPhi_{k}\left(\mathbf{r}\right)\rho_{p}^{1/3}\left(\mathbf{r}\right) \end{array} \end{aligned} $$
As in the case of the integrals in Eq. (2.51), these remaining integrals must be carried out numerically. The Slater approximation can be used for all HFB iterations to accelerate the calculations. The last HFB iteration could be performed using the exact form of the Coulomb field if additional accuracy is required at a minimal additional computational cost.

Footnotes

  1. 1.

    Let us mention that using an axially deformed HO basis does not necessarily imply that nuclear shapes have to be restricted to axially symmetric ones. In fact, triaxial shapes can be described using such a basis by allowing a mixing of the orbital momentum projection quantum number Λ in the expansion of the HFB qp states, although such a mixing is convenient in practice only for relatively small triaxiality – which is generally the case along fission paths

Notes

Acknowledgements

This chapter was prepared by a contractor of the U.S. Government under contract number DE-AC52-06NA27344. Accordingly, the U.S. Government retains a nonexclusive, royalty-free license to publish or reproduce the published form of this contribution, or allow others to do so, for U.S. Government purposes.

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Copyright information

© Springer Nature Switzerland AG 2019

Authors and Affiliations

  • Walid Younes
    • 1
  • Daniel Marc Gogny
    • 2
  • Jean-François Berger
    • 3
  1. 1.Department of PhysicsLawrence Livermore National LaboratoryLivermoreUSA
  2. 2.Lawrence Livermore National LaboratLivermoreUSA
  3. 3.Centre DAM-Ile de FranceRetired from Commissariat à l’Energie AtomiqueBruyères-le-Châtel, ArpajonFrance

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