Cartesian representation of the SO and tensor pseudopotentials

In this Section, we present construction of the SO ($\tilde{S}$=1) and tensor ($\tilde{S}$=2) components of the regularized pseudopotential. Following the same methodology as that introduced for central terms in Sec. 5, we define the Cartesian forms of the (non-antisymmetrized) SO and tensor pseudopotential, respectively, as

$\displaystyle \mathcal{V}_{SO}$ $\textstyle =$ $\displaystyle \sum_{nj} p^{(n)}_j
\left( \hat{1}_\sigma\hat{1}_\tau
-w^{(n)}_j \hat{1}_\sigma\hat{P}^\tau\right)$  
    $\displaystyle ~~~~~~~~\times \hat{P}^{(n)}_j(\bm{k}',\bm{k})\delta(\bm{r}'_1-\bm{r}_1)\delta(\bm{r}'_2-\bm{r}_2)
g_a(\bm{r}_1-\bm{r}_2)$ (106)

and
$\displaystyle \mathcal{V}_T$ $\textstyle =$ $\displaystyle \sum_{nj} q^{(n)}_j
\left( \hat{1}_\sigma\hat{1}_\tau
-v^{(n)}_j \hat{1}_\sigma\hat{P}^\tau\right)$  
    $\displaystyle ~~~~~~~~\times \hat{Q}^{(n)}_j(\bm{k}',\bm{k})\delta(\bm{r}'_1-\bm{r}_1)\delta(\bm{r}'_2-\bm{r}_2)
g_a(\bm{r}_1-\bm{r}_2) .$ (107)

The spin-dependent differential operators $\hat{P}^{(n)}_j(\bm{k}',\bm{k})$ and $\hat{Q}^{(n)}_j(\bm{k}',\bm{k})$ are built in the following way. First of all, they must be scalar, hermitian, and time-even operators that are obtained by coupling the space part to standard spin-vector and spin-tensor operators,

$\displaystyle \hat{S}_m$ $\textstyle =$ $\displaystyle \sigma^{(1)}_m + \sigma^{(2)}_m ,$ (108)
$\displaystyle \hat{{\mathsf S}}_{mn}$ $\textstyle =$ $\displaystyle {\textstyle{\frac{3}{2}}}\left(\sigma^{(1)}_m\sigma^{(2)}_n + \si...
...}_n\sigma^{(2)}_m\right)
-\delta_{mn} \bm{\sigma}^{(1)}\cdot\bm{\sigma}^{(2)} ,$ (109)

respectively, where indices $m$ and $n$ denote Cartesian components of the Pauli spin matrices. At this point we note that $\hat{\bm{S}}\hat{P}^\sigma\equiv\hat{\bm{S}}$ and $\hat{{\mathsf S}}\hat{P}^\sigma\equiv\hat{{\mathsf S}}$; therefore, the spin-exchange operators $\hat{P}^\sigma$ do not give independent terms and thus do not appear in Eqs. (107) and (108).

Because spin-vector operator $\hat{S}_m$ is even-parity and time-odd, we must build from relative-momentum operators an elementary even-parity and time-odd vector, which is only one,

$\displaystyle i(\bm{k}'^*\times\bm{k})_m$ $\textstyle =$ $\displaystyle i\sum_{nl}\varepsilon_{mnl} k_n'^* k_l.$ (110)

Similarly, because spin-tensor operator $\hat{{\mathsf S}}_{mn}$ is even-parity and time-even, we must built from relative-momentum operators all elementary even-parity and time-even traceless tensors, which are three,
$\displaystyle (\bm{k} \otimes\bm{k} )_{mn}$ $\textstyle =$ $\displaystyle k _m k _n - {\textstyle{\frac{1}{3}}} \delta_{mn} \bm{k} \cdot\bm{k} ,$ (111)
$\displaystyle (\bm{k}'^*\otimes\bm{k} )_{mn}$ $\textstyle =$ $\displaystyle k'^*_m k _n - {\textstyle{\frac{1}{3}}} \delta_{mn} \bm{k}'^*\cdot\bm{k} ,$ (112)
$\displaystyle (\bm{k}'^*\otimes\bm{k}'^*)_{mn}$ $\textstyle =$ $\displaystyle k'^*_m k'^*_n - {\textstyle{\frac{1}{3}}} \delta_{mn} \bm{k}'^*\cdot\bm{k}'^* .$ (113)

Contracting spin-vector and space-vector, as well as spin-tensor and space-tensor operators, we now obtain all suitable elementary scalar operators. Selecting convenient combinations of tensor terms, cf. Eqs. (39)-(41), we define them as,

$\displaystyle \hat{T}_s$ $\textstyle =$ $\displaystyle i(\bm{k}'^*\times\bm{k})\cdot \left(\bm{\sigma}^{(1)} + \bm{\sigma}^{(2)}\right) ,$ (114)
$\displaystyle \hat{T}_e$ $\textstyle =$ $\displaystyle {\textstyle{\frac{1}{2}}}(\bm{k} \cdot\hat{{\mathsf S}}\cdot\bm{k}
+\bm{k}'^*\cdot\hat{{\mathsf S}}\cdot\bm{k}'^*),$ (115)
$\displaystyle \hat{T}_o$ $\textstyle =$ $\displaystyle \bm{k}'^*\cdot\hat{{\mathsf S}}\cdot\bm{k} ,$ (116)
$\displaystyle \hat{T}_a$ $\textstyle =$ $\displaystyle {\textstyle{\frac{1}{2}}}(\bm{k} \cdot\hat{{\mathsf S}}\cdot\bm{k}
-\bm{k}'^*\cdot\hat{{\mathsf S}}\cdot\bm{k}'^*).$ (117)

Of course, at second order we have the standard SO (115), tensor-even (116), and tensor-odd (117) elementary operators, see, e.g., Ref. [23], whereas the tensor-antihermitian operator (118) does not appear. Owing to the GCH theorem [24], we can now build all higher-order operators by multiplying elementary operators (115)-(118) and scalars (39)-(41). The hermiticity is then enforced by always using even numbers of antihermitian factors.

Finally, up to N$^3$LO, we obtain all possible SO terms,

$\displaystyle \hat{P}^{(2)}_1(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_s ,$ (118)
$\displaystyle \hat{P}^{(4)}_1(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_1 \hat{T}_s ,$ (119)
$\displaystyle \hat{P}^{(4)}_2(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_2 \hat{T}_s ,$ (120)
$\displaystyle \hat{P}^{(6)}_1(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2+ \hat{T}_2^2) \hat{T}_s ,$ (121)
$\displaystyle \hat{P}^{(6)}_2(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle 2\hat{T}_1 \hat{T}_2 \hat{T}_s ,$ (122)
$\displaystyle \hat{P}^{(6)}_3(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2- \hat{T}_2^2) \hat{T}_s ,$ (123)
$\displaystyle \hat{P}^{(6)}_4(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_3^2 \hat{T}_s ,$ (124)

and all possible tensor terms,
$\displaystyle \hat{Q}^{(2)}_1 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_e ,$ (125)
$\displaystyle \hat{Q}^{(2)}_2 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_o ,$ (126)
$\displaystyle \hat{Q}^{(4)}_1 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_1 (\hat{T}_e - \hat{T}_o) ,$ (127)
$\displaystyle \hat{Q}^{(4)}_2 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_2 (\hat{T}_e - \hat{T}_o) ,$ (128)
$\displaystyle \hat{Q}^{(4)}_3 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_1 (\hat{T}_e + \hat{T}_o) ,$ (129)
$\displaystyle \hat{Q}^{(4)}_4 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_2 (\hat{T}_e + \hat{T}_o) ,$ (130)
$\displaystyle \hat{Q}^{(4)}_5 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_3 \hat{T}_a ,$ (131)
$\displaystyle \hat{Q}^{(6)}_1 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2+ \hat{T}_2^2) (\hat{T}_e - \hat{T}_o) ,$ (132)
$\displaystyle \hat{Q}^{(6)}_2 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle 2\hat{T}_1 \hat{T}_2 (\hat{T}_e - \hat{T}_o) ,$ (133)
$\displaystyle \hat{Q}^{(6)}_3 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2- \hat{T}_2^2) (\hat{T}_e - \hat{T}_o) ,$ (134)
$\displaystyle \hat{Q}^{(6)}_4 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_3^2 (\hat{T}_e - \hat{T}_o) ,$ (135)
$\displaystyle \hat{Q}^{(6)}_5 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2+ \hat{T}_2^2) (\hat{T}_e + \hat{T}_o) ,$ (136)
$\displaystyle \hat{Q}^{(6)}_6 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle 2\hat{T}_1 \hat{T}_2 (\hat{T}_e + \hat{T}_o) ,$ (137)
$\displaystyle \hat{Q}^{(6)}_7 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle (\hat{T}_1^2- \hat{T}_2^2) (\hat{T}_e + \hat{T}_o) ,$ (138)
$\displaystyle \hat{Q}^{(6)}_8 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_3^2 (\hat{T}_e + \hat{T}_o) ,$ (139)
$\displaystyle \hat{Q}^{(6)}_9 (\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_1 \hat{T}_3 \hat{T}_a ,$ (140)
$\displaystyle \hat{Q}^{(6)}_{10}(\bm{k}',\bm{k})$ $\textstyle =$ $\displaystyle \hat{T}_2 \hat{T}_3 \hat{T}_a .$ (141)

We note that we here recover the numbers of SO and tensor terms given in Table 1. We also note that the difference of the tensor-even and tensor-odd elementary operators depends only on the sum of relative momenta, that is,
$\displaystyle \hat{T}_e - \hat{T}_o$ $\textstyle =$ $\displaystyle {\textstyle{\frac{1}{2}}}(\bm{k}'+\bm{k}) \cdot\hat{{\mathsf S}}\cdot(\bm{k}'+\bm{k}) .$ (142)

This has motivated our choice of the $j=1$ and $j=2$ tensor terms, for which at $n>0$ we have,
$\displaystyle \hspace*{-1cm}
\hat{Q}^{(n)}_1(\bm{k}',\bm{k}) - \hat{Q}^{(n)}_2(\bm{k}',\bm{k})$ $\textstyle \equiv$ $\displaystyle \frac{1}{2^n}(\bm{k}'+\bm{k})^{n-2}\left[(\bm{k}'+\bm{k}) \cdot\hat{{\mathsf S}}\cdot(\bm{k}'+\bm{k})\right] .$ (143)

Therefore, these particular combinations of terms commute with the locality deltas, see Sec. 5, and thus are equivalent to local tensor potentials. On the other hand, none of the SO terms has such a property.

The tensor interaction presented here may be compared with the one discussed in a recent work by Davesne et al. [27], which extends to N$^3$LO the zero-range Cartesian pseudopotential. First we note that the $\hat T_e$ and $\hat T_o$ operators defined in the aforementioned article differ from those in Eqs. (116) and (117) by factors of 2. When $a\rightarrow 0$, one has the following correspondence between the coupling constants $q_j^{(n)}$ appearing in Eq. (108) and those used in Ref. [27], denoted by $t_e^{(n)}$ and $t_o^{(n)}$. (The coupling constants $w_j^{(n)}$ can be disregarded, because at the zero-range limit, the action of operator $\hat P^\tau$ reduces to a phase.) At second order, one recovers the pseudopotential from Ref. [27] with

$\displaystyle q_1^{(2)}$ $\textstyle =$ $\displaystyle t_e^{(2)}\,,$ (144)
$\displaystyle q_2^{(2)}$ $\textstyle =$ $\displaystyle t_o^{(2)}\,,$ (145)

at fourth order with
$\displaystyle q_1^{(4)}$ $\textstyle =$ $\displaystyle 2\left(t_e^{(4)}-t_o^{(4)}\right)\,,$ (146)
$\displaystyle q_2^{(4)}$ $\textstyle =$ $\displaystyle 2\left(t_o^{(4)}-t_e^{(4)}\right)\,,$ (147)
$\displaystyle q_3^{(4)}$ $\textstyle =$ $\displaystyle 2\left(t_e^{(4)}+t_o^{(4)}\right)\,,$ (148)
$\displaystyle q_4^{(4)}$ $\textstyle =$ $\displaystyle 2\left(t_e^{(4)}+t_o^{(4)}\right)\,,$ (149)
$\displaystyle q_5^{(4)}$ $\textstyle =$ $\displaystyle 0\,,$ (150)

and at sixth order with
$\displaystyle q_1^{(6)}$ $\textstyle =$ $\displaystyle t_e^{(6)}-t_o^{(6)}\,,$ (151)
$\displaystyle q_2^{(6)}$ $\textstyle =$ $\displaystyle t_o^{(6)}-t_e^{(6)}\,,$ (152)
$\displaystyle q_5^{(6)}$ $\textstyle =$ $\displaystyle t_e^{(6)}+t_o^{(6)}\,,$ (153)
$\displaystyle q_6^{(6)}$ $\textstyle =$ $\displaystyle t_e^{(6)}+t_o^{(6)}$ (154)

and $q_3^{(6)}=q_4^{(6)}=q_7^{(6)}=q_8^{(6)}=q_9^{(6)}=q_{10}^{(6)}=0$.

Jacek Dobaczewski 2014-12-07