Next: Mean-field calculations with tensor
Up: Tensor densities in the
Previous: Spin-orbit and tensor forces
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{displaymath}
\begin{array}{rcl}
{\mathcal H}_{SO} &=&
\sum_{t=0,1}
C_{t}...
...
+ C_{t}^{J2} \underline{\mathsf J}_{t}^{2}\Big) ,
\end{array}\end{displaymath}](img30.png) |
(5) |
where the spherical-symmetry condition has been released.
The
standard pseudoscalar
, vector
,
and pseudotensor
parts of the
spin-current density,[16]
![\begin{displaymath}
\begin{array}{rcl}
J_{abt}
&=&{\textstyle{\frac{1}{2i}}}\Big...
...boldmath }=\mbox{\boldmath${r}$\unboldmath }'}, \\
\end{array}\end{displaymath}](img34.png) |
(6) |
are defined as
![\begin{displaymath}
\begin{array}{rcl}
\underline{{J}}_{t}&=&
\sum_{a=x,y,z}{J}_...
...style{\frac{1}{3}}} \underline{{J}}_{t}\delta_{ab}.
\end{array}\end{displaymath}](img35.png) |
(7) |
The tensor energy density
now depends
on six coupling constants,
,
, and
,
for
=0,1, and not on two coupling constants,
and
, as in Eq. 3.
Similarly, the SO energy density
depends now on two coupling constants
for
=0,1,
and not on one,
, (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
![$^\circ$](img44.png)
- The pseudoscalar densities
vanish unless the axial or reflection symmetries are broken.
- 2
![$^\circ$](img44.png)
- The pseudotensor densities
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{displaymath}
\begin{array}{rcl}
G^{T}_t&=& {\textstyle{\frac{1}{3}}} \und...
...tstyle{\frac{1}{2}}}\underline{{\mathsf J}}_t^{2} .
\end{array}\end{displaymath}](img45.png) |
(8) |
In such a case, only four out of the six tensor coupling constants are linearly
independent, i.e.,
![\begin{displaymath}
\begin{array}{rcl}
C_t^{J0 } &=& {\textstyle{\frac{1}{3}}} A...
...2 } &=& ~ A_t + {\textstyle{\frac{1}{2}}} B_t . \\
\end{array}\end{displaymath}](img46.png) |
(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{displaymath}
\begin{array}{rcl}
B_0 &=& -3 A_0 = - {\textstyle{\frac{3}{8...
...textstyle{\frac{3}{8}}} \left(t_e- t_o\right) . \\
\end{array}\end{displaymath}](img47.png) |
(10) |
Next: Mean-field calculations with tensor
Up: Tensor densities in the
Previous: Spin-orbit and tensor forces
Jacek Dobaczewski
2006-04-22