Spin-orbit and tensor energy densities

Within the energy-density approach, one does not relate the EDF to an average of the two-body force, but one postulates the EDF based on symmetry conditions only. Then, the most general EDF, depending on the spin-current densities, reads[4]

$$\begin{array}{rcl} {\mathcal H}_{SO} &=& \sum_{t=0,1} C_{t}... ... + C_{t}^{J2} \underline{\mathsf J}_{t}^{2}\Big) , \end{array}$$ (5)

where the spherical-symmetry condition has been released. The standard pseudoscalar $\underline{{J}}_{t}$, vector $\mbox{\boldmath${J}$\unboldmath }_{t}$, and pseudotensor $\underline{\mathsf J}_{t}$ parts of the spin-current density,[16]

$$\begin{array}{rcl} J_{abt} &=&{\textstyle{\frac{1}{2i}}}\Big... ...boldmath }=\mbox{\boldmath${r}$\unboldmath }'}, \\ \end{array}$$ (6)

are defined as

$$\begin{array}{rcl} \underline{{J}}_{t}&=& \sum_{a=x,y,z}{J}_... ...style{\frac{1}{3}}} \underline{{J}}_{t}\delta_{ab}. \end{array}$$ (7)

The tensor energy density ${\mathcal H}_T$ now depends on six coupling constants, $C_{t}^{J0}$, $C_{t}^{J1}$, and $C_{t}^{J2}$, for $t$=0,1, and not on two coupling constants, $t_e$ and $t_o$, as in Eq. 3. Similarly, the SO energy density ${\mathcal H}_{SO}$ depends now on two coupling constants $C_{t}^{\nabla {J}}$ for $t$=0,1, and not on one, $W_0$, (the latter generalization has been introduced and studied in Ref.[17]).

From the symmetry conditions imposed by the spherical, axial, and reflection symmetries one obtains[18] that:

1$^\circ$

The pseudoscalar densities $\underline{{J}}_{t}$ vanish unless the axial or reflection symmetries are broken.

2$^\circ$

The pseudotensor densities $\underline{\mathsf J}_{t}$ vanish unless the spherical symmetry is broken.

Finally, the gauge-invariance symmetry conditions[4] require that there are only two gauge-invariant combinations of the pseudoscalar, vector, and pseudotensor terms, namely,

$$\begin{array}{rcl} G^{T}_t&=& {\textstyle{\frac{1}{3}}} \und... ...tstyle{\frac{1}{2}}}\underline{{\mathsf J}}_t^{2} . \end{array}$$ (8)

In such a case, only four out of the six tensor coupling constants are linearly independent, i.e.,

$$\begin{array}{rcl} C_t^{J0 } &=& {\textstyle{\frac{1}{3}}} A... ...2 } &=& ~ A_t + {\textstyle{\frac{1}{2}}} B_t . \\ \end{array}$$ (9)

On the other hand, the averaging of the tensor forces (1) implies that only two out of the six tensor coupling constants remain linearly independent, i.e.,

$$\begin{array}{rcl} B_0 &=& -3 A_0 = - {\textstyle{\frac{3}{8... ...textstyle{\frac{3}{8}}} \left(t_e- t_o\right) . \\ \end{array}$$ (10)