Symmetries

Within the HF approximation, symmetry properties of the interaction carry over to symmetry properties of the total HF energy. It means that whenever the interaction is invariant with respect to a symmetry group, the total HF energy is invariant with respect to transforming densities (or density matrices, in general) by the same symmetry group. However, the well-known phenomenon of the spontaneous symmetry breaking [32] may render densities and energy density themselves not invariant with respect to the symmetry group in question. We must, therefore, consider energy densities for symmetry-breaking densities.

This is best illustrated by the EDF derived for the Skyrme interaction [28,25]. The standard derivation treats the time-reversal and isospin symmetries in a different way than space symmetries (space inversion and other point symmetries or space rotation) [25,36,24]. Indeed, for the time reversal, the nonlocal densities are first split into the time-even and time-odd parts as

$$\rho (\vec {r},\vec {r}') = \rho _+(\vec {r},\vec {r}') + \rho _-(\vec {r},\vec {r}') ,$$ (63)

$$\vec {s}(\vec {r},\vec {r}') = \vec {s}_+(\vec {r},\vec {r}') + \vec {s}_-(\vec {r},\vec {r}') ,$$ (64)

where

$$\rho _\pm(\vec {r},\vec {r}') = \tfrac{1}{2}\left[\rho (\vec {r},\vec {r}') \pm \rho^T (\vec {r},\vec {r}')\right] ,$$ (65)

$$\vec {s}_\pm(\vec {r},\vec {r}') = \tfrac{1}{2}\left[\vec {s}(\vec {r},\vec {r}') \pm \vec {s}^T(\vec {r},\vec {r}')\right] ,$$ (66)

such that

$$\rho ^T_\pm(\vec {r},\vec {r}') = \pm\rho _\pm(\vec {r},\vec {r}') ,$$ (67)

$$\vec {s}^T_\pm(\vec {r},\vec {r}') = \pm\vec {s}_\pm(\vec {r},\vec {r}') .$$ (68)

The superscript $T$ here means that the nonlocal densities are calculated for the time-reversed many-body states.

Then, in the derivation of the HF energy density, only squares of the time-even and time-odd densities appear, because, due to the time-reversal symmetry of the interaction, the cross terms do not contribute. As a consequence, the energy density itself is time-even. Similarly, for the isospin symmetry, the densities are first split into the isoscalar and isovector parts, for which no cross terms contribute, and the obtained energy density is an isoscalar. In what follows, we call this kind of derivation 'derivation after separation of symmetries', which implies that the symmetry-breaking terms are absent in the energy density.

The derivation after separation of symmetries can be illustrated by considering the simplest term of the Skyrme interaction - just the contact force,

$$V_\delta(\vec {r}_1,\vec {r}_2)=t_0\delta(\vec {r}_1-\vec {r}_2),$$ (69)

for which the energy density reads (we neglected the isospin degree of freedom, so only one type of particles is considered),

$${\cal H}_\delta(\vec {r})=\tfrac{1}{2}t_0\rho^2(\vec {r}) -\tfrac{1}{2}t_0\vec {s}^2(\vec {r}).$$ (70)

This energy density is invariant with respect to the time reversal of local densities, $\rho^T(\vec {r})=\rho(\vec {r})$ and $\vec {s}^T(\vec {r})=-\vec {s}(\vec {r})$.

Here, the coupling constant multiplying the time-even density, $C^\rho=\tfrac{1}{2}t_0$, is not independent of the coupling constant multiplying the time-odd density, $C^s=-\tfrac{1}{2}t_0$. This fact is not related to the time-reversal symmetry but results from the vanishing range of the contact force. Proper treatment of the finite range corrections render these two coupling constants independent of one another. [15,16]. Irrespective of zero or finite range, the isovector and isoscalar coupling constants are also independent of one another [36,24].

For space symmetries, the standard derivation proceeds in another way, namely, the energy density is determined directly for the broken-symmetry HF state. For the space-inversion symmetry, for example, this means that both parity-even and parity-odd densities,

$$\rho_ {\rule{0ex}{1.5ex}P=\pm1}(\vec {r}) \!\! = \tfrac{1}{2}\left[\rho (\vec {r})\pm\rho ^P(\vec {r})\right] = \tfrac{1}{2}\left[\rho (\vec {r})\pm\rho (-\vec {r})\right] ,$$ (71)

$$\vec {s}_{\rule{0ex}{1.5ex}P=\pm1}(\vec {r}) \!\! = \tfrac{1}{2}\left[\vec {s}(\vec {r})\pm\vec {s}^P(\vec {r})\right] = \tfrac{1}{2}\left[\vec {s}(\vec {r})\pm\vec {s}(-\vec {r})\right] ,$$ (72)

appear in $\rho(\vec {r})$ and $\vec {s}(\vec {r})$ in the energy density of Eq. (70),

$$\rho (\vec {r}) = \rho _{\rule{0ex}{1.5ex}P=+1}(\vec {r})+\rho _{\rule{0ex}{1.5ex}P=-1}(\vec {r}) ,$$ (73)

$$\vec {s}(\vec {r}) = \vec {s}_{\rule{0ex}{1.5ex}P=+1}(\vec {r})+\vec {s}_{\rule{0ex}{1.5ex}P=-1}(\vec {r}) .$$ (74)

The superscript $P$ here means that the nonlocal and local densities are calculated for the space-inversed many-body states. We call this kind of derivation 'derivation before separation of symmetries', which implies that the symmetry-breaking terms are then explicitly present in the energy density. If the symmetry is broken, which in the case of space inversion corresponds to $\rho_{\rule{0ex}{1.5ex}P=-1}(\vec {r})\neq0$ or $\vec {s}_{\rule{0ex}{1.5ex}P=-1}(\vec {r})\neq0$, then the energy density is not invariant with respect to space inversion.

The total energy, i.e., the integral of the energy density (1) is, of course, invariant with respect to space inversion, because the integration then picks up only the space-inversion-invariant parts of the integrand. Therefore, the energy densities derived before and after separation of symmetries are not equal, but they are equivalent. In the case of the space-inversion symmetry, the energy density (70) derived after separation of symmetries reads

$${\cal H}'_\delta(\vec {r}) = \tfrac{1}{2}t_0\rho^2 _{\rule{0ex}{1.5ex}P=+1}(\vec {r}) +\tfrac{1}{2}t_0\rho^2 _{\rule{0ex}{1.5ex}P=-1}(\vec {r})$$

$$- \tfrac{1}{2}t_0\vec {s}^2_{\rule{0ex}{1.5ex}P=+1}(\vec {r}) -\tfrac{1}{2}t_0\vec {s}^2_{\rule{0ex}{1.5ex}P=-1}(\vec {r}).$$ (75)

This energy density is invariant with respect to space inversion and the coupling constants multiplying densities of opposite parities are not independent of one another (see also discussion in Ref. [30]).

The same principle applies to other broken spatial symmetries. For example, for the broken rotational symmetry, the density can be split into the sum of terms belonging to different irreducible representations of the rotational group, which in this case corresponds to the standard multipole series [44],

$$\rho(\vec {r})=\sum_{\lambda\mu}\rho_{\lambda\mu}(\vec {r}),$$ (76)

for

$$\rho_{\lambda\mu}(\vec {r})=\rho_{\lambda\mu}({r})Y^*_{\lambda\mu}(\theta,\phi),$$ (77)

where the multipole densities $\rho_{\lambda\mu}({r})$,

$$\rho_{\lambda\mu}({r})=\int {\rm d}\theta{\rm d}\phi\rho(\vec {r})Y_{\lambda\mu}(\theta,\phi),$$ (78)

depend only on the radial coordinate $r$. Then, the first term in the energy density (70), which is derived before separation of rotational symmetries, and which is not rotationally invariant, is equivalent to the following energy density derived after separation of rotational symmetries,

$${\cal H}_\rho(\vec {r}) = \tfrac{1}{2}t_0\sum_{\lambda} \sqrt{2\lambda+1}[\rho_{\lambda}(\vec {r})\rho_{\lambda}(\vec {r})]_0,$$ (79)

(see Ref. [45] for an example application of this series). This energy density is rotationally invariant and the coupling constants multiplying different multipole densities are again not independent of one another.

We have presented a detailed analysis of the problem to arm ourselves with proper tools for discussing construction of EDFs in situations where there is no underlying interaction. Then, the only consideration is the requirement of invariance of the total energy with respect to all symmetries usually conserved by nuclear interactions. We proceed with such a construction in two different ways as described below.