Higher-order derivative operators

We begin by constructing all possible higher-order and higher-rank tensor operators from powers of the derivative $\nabla_{1\mu}$, where $\mu=-1,0,+1$ are the spin-projection components of the vector (rank-1) operator $\nabla$. It is obvious that all possible $n$th-order powers of the derivative can be written as sums of terms $\nabla_{1\mu_1}\ldots\nabla_{1\mu_n}$. Therefore, any $(n+1)$th-order power is simply obtained by multiplying some $n$th-order power by a sum of $\nabla_{1\mu}$ operators. Then, powers of a given rank can be obtained iteratively by vector coupling.

In the second order, the two nabla operators can be coupled to angular momenta 0 and 2. The coupling to angular momentum 0, $[\nabla\nabla]_0=\Delta/\sqrt{3}$, corresponds to the Laplacian operator. Furthermore, the coupling to angular momentum 2, $[\nabla\nabla]_2$, gives the second-order, rank-2 derivative operator. The rank-1 coupling, $[\nabla\nabla]_1=0$, vanishes because the derivatives commute. Similarly, in each one higher order, a rank-$L$ symmetric operator can be coupled with $\nabla$ only to $L-1$ and $L+1$. Hence, all the $n$th-order powers have the form of $\Delta^{(n-L)/2}$ multiplied by the $L$th-order rank-$L$ (stretched) coupled operators for $L=n,n-2,\ldots,(1)0$. Then, up to N$^3$LO, one obtains 16 different operators $D_{nL}$ listed in Table 1. Any arbitrary tensor formed by coupled operators $\nabla$ can always be rewritten as a sum of operators $D_{nL}$ through the repeated use of the $6j$ symbols.

Exactly in the same way, we define 16 different operators $K_{nL}$, which are spherical tensors built of the relative momentum operators $k$ coupled up to N$^3$LO, i.e., for $n\leq6$ and $L\leq6$. In the remainder of this section, we only discuss operators $D_{nL}$, while all the results mutatis mutandis also pertain to operators $K_{nL}$.

Table 1: Derivative operators $D_{nL}$ up to N$^3$LO as expressed through spherical tensor representation of the operator $\nabla$.

No.	tensor $D_{nL}$	order $n$	rank $L$
1	$1$	0	0
2	$\nabla$	1	1
3	${[}\nabla\nabla{]}_{0}$	2	0
4	${[}\nabla\nabla{]}_{2}$	2	2
5	${[}\nabla\nabla{]}_{0} \nabla$	3	1
6	${[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}$	3	3
7	${[}\nabla\nabla{]}_{0} ^{2}$	4	0
8	${[}\nabla\nabla{]}_{0} {[}\nabla\nabla{]}_{2}$	4	2
9	${[}\nabla{[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}{]}_{4}$	4	4
10	${[}\nabla\nabla{]}_{0}^2 \nabla$	5	1
11	${[}\nabla\nabla{]}_{0} {[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}$	5	3
12	${[}\nabla{[}\nabla{[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}{]}_{4}{]}_{5}$	5	5
13	${[}\nabla\nabla{]}_{0} ^3$	6	0
14	${[}\nabla\nabla{]}_{0}^2 {[}\nabla\nabla{]}_{2}$	6	2
15	${[}\nabla\nabla{]}_{0} {[}\nabla{[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}{]}_{4}$	6	4
16	${[}\nabla{[}\nabla{[}\nabla{[}\nabla{[}\nabla\nabla{]}_{2}{]}_{3}{]}_{4}{]}_{5}{]}_{6}$	6	6

The stretched coupled operators $D_{LL}$ for $n=L$,

$$D_{LL} = [\nabla\ldots[\nabla[\nabla\nabla]_2]_3\ldots]_L,$$ (11)

play a central role in our derivations below. They correspond to irreducible symmetric traceless Cartesian tensors built of the derivative $\nabla$. They have $2L+1$ tensor components $D_{LLM}$ numbered by the quantum number $M=-L,\ldots,L$ that we most often do not show below explicitly. Moreover, since terms in the EDF up to N$^3$LO depend only on operators $D_{LL}$ and $K_{LL}$ up to fourth order, $L\leq4$, see Sec. 3.1, below we do not discuss stretched coupled operators of fifth or sixth orders.

Equivalently, derivative operators $D_{LL}$ can be written in the Cartesian representation, in which their components are numbered by $L$ Cartesian indices, $D_{LL,a_1\ldots a_L}$, $a_i=x,y,z$. The order of these indices does not matter (totally symmetric tensors) and all traces vanish,

$$\sum_a D_{LL,aaa_3\ldots a_L} = 0.$$ (12)

The Cartesian components $D_{LL,a_1\ldots a_L}$ can be calculated by using the detracer operator defined in Sec. 5 of Ref. [27]. Up to fourth order they read:

$$D_{00} = 1 ,$$ (13)

$$D_{11,a_1} = \nabla_{a_1} ,$$ (14)

$$D_{22,a_1a_2} = \nabla_{a_1}\nabla_{a_2}-\tfrac{1}{3}\delta_{a_1a_2}\Delta ,$$ (15)

$$D_{33,a_1a_2a_3} = \nabla_{a_1}\nabla_{a_2}\nabla_{a_3} - \tfrac{1}{5}\Delta\Big( \n... ...ta_{a_2a_3} + \nabla_{a_2}\delta_{a_1a_3} + \nabla_{a_3}\delta_{a_1a_2} \Big) ,$$ (16)

$$D_{44,a_1a_2a_3a_4} = \nabla_{a_1}\nabla_{a_2}\nabla_{a_3} \nabla_{a_4} - \tfrac{1}{7}\... ...abla_{a_1}\nabla_{a_4}\delta_{a_2a_3} + \nabla_{a_2}\nabla_{a_3}\delta_{a_1a_4}$$

$$+ \nabla_{a_2}\nabla_{a_4}\delta_{a_1a_3} + \nabla_{a_3}\nabla_{a... ..._3a_4} + \delta_{a_1a_3}\delta_{a_2a_4} + \delta_{a_1a_4}\delta_{a_2a_3}\Big) .$$ (17)

We note here in passing that we could have equally well used the Cartesian derivative operators with traces not subtracted out, i.e.,

$${\cal D}_{00} = 1 ,$$ (18)

$${\cal D}_{11,a_1} = \nabla_{a_1} ,$$ (19)

$${\cal D}_{22,a_1a_2} = \nabla_{a_1}\nabla_{a_2} ,$$ (20)

$${\cal D}_{33,a_1a_2a_3} = \nabla_{a_1}\nabla_{a_2}\nabla_{a_3} ,$$ (21)

$${\cal D}_{44,a_1a_2a_3a_4} = \nabla_{a_1}\nabla_{a_2}\nabla_{a_3}\nabla_{a_4} .$$ (22)

Representations (13)-(17) and (18)-(22) are equivalent in the sense that each operator ${D}_{LL,a_1\ldots{}a_L}$ is evidently a linear combination of operators $\Delta^{(L-L')/2}{\cal D}_{L'L',a_1\ldots a_{L'}}$ for $L'=L$, $L-2,\ldots,(1)0$.

In principle, below one could replace the spherical representations of derivative operators shown in Table 1 by their Cartesian counterparts (13)-(17) or (18)-(22), and work entirely in the Cartesian representation. However, in our opinion, the use of the spherical representation is superior and more economical. Moreover, whenever calculation of the Cartesian derivatives is more suitable, we may express spherical components of the derivative operators through the Cartesian derivatives, as shown in Table 2. An example of using the Cartesian representation (18)-(22) is given in Sec. 4.

Table 2: Spherical components of the derivative operators $D_{nLM}$ expressed through the Cartesian derivatives. Expressions for negative components can be obtained as $D_{nL,-M}=\left(-1\right)^{L-M} D^*_{nLM}$, see Eqs. (107) and (109).

$D_{nLM}$		Cartesian derivatives
$D_{110}$	=	$-i\partial_{z}$
$D_{111}$	=	$i\frac{1}{\sqrt{2}}\left(\partial_{x}+i \partial_{y}\right)$
$D_{200}$	=	$\frac{1}{\sqrt{3}}\Delta$
$D_{220}$	=	$\frac{1}{\sqrt{6}}\left(\partial_{x}^2+\partial_{y}^2-2 \partial_{z}^2\right)$
$D_{221}$	=	$\left(\partial_{x}+i \partial_{y}\right) \partial_{z}$
$D_{222}$	=	$- \frac{1}{2} \left(\partial_{x}+i \partial_{y}\right)^2$
$D_{330}$	=	$i\frac{1}{\sqrt{10}}\partial_{z} \left(-3 \partial_{x}^2-3 \partial_{y}^2+2 \partial_{z}^2\right)$
$D_{331}$	=	$i\frac{1}{2} \sqrt{\frac{3}{10}} \left(\partial_{x}+i \partial_{y}\right) \left(\partial_{x}^2+\partial_{y}^2-4 \partial_{z}^2\right)$
$D_{332}$	=	$i\frac{1}{2} \sqrt{3} \left(\partial_{x}+i \partial_{y}\right)^2\partial_{z}$
$D_{333}$	=	$-i\frac{1}{2 \sqrt{2}}\left(\partial_{x}+i \partial_{y}\right)^3$
$D_{440}$	=	$\frac{1}{2\sqrt{70}}\left(3 \partial_{x}^4+6 \left(\partial_{y}^2-4\partial_{z... ..._{x}^2+3 \partial_{y}^4+8\partial_{z}^4-24 \partial_{y}^2 \partial_{z}^2\right)$
$D_{441}$	=	$\frac{1}{\sqrt{14}}\left(\partial_{x}+i \partial_{y}\right)\partial_{z}\left(3 \partial_{x}^2+3 \partial_{y}^2-4\partial_{z}^2\right)$
$D_{442}$	=	$- \frac{1}{2 \sqrt{7}}\left(\partial_{x}+i \partial_{y}\right)^2\left(\partial_{x}^2+\partial_{y}^2-6 \partial_{z}^2\right)$
$D_{443}$	=	$- \frac{1}{\sqrt{2}}\left(\partial_{x}+i \partial_{y}\right)^3\partial_{z}$
$D_{444}$	=	$\frac{1}{4} \left(\partial_{x}+i \partial_{y}\right)^4$