A Function expansion in Cartesian and spherical tensors

[Malamala]

Hello! My question stems from reading several physics papers (here are two relevant ones: https://journals.aps.org/prc/abstract/10.1103/PhysRevC.56.2820 and https://journals.aps.org/pra/abstract/10.1103/PhysRevA.65.032113). I will simplify the problem for the purpose of this question. Basically, I have an integral of the form:

$$V(\mathbf{R}) = \int_0^\infty f(\mathbf{r})/|\mathbf{R}-\mathbf{r}|d^3\mathbf{r}$$

In the first paper above (eq. A2) they expand the $1/|\mathbf{R}-\mathbf{r}|$ term in cartesian tensors and they claim that the $l=3$ term is an reducible tensor (in terms of spherical tensors), which can be written as the sum of a rank 1 irreducible tensor and a rank 3 tensor irreducible tensor.

In the second paper, they do the expansion in terms of Legendre polynomials (eq. 7), using:

$$1/|\mathbf{R}-\mathbf{r}| = \sum_l\frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta)$$

If I understand it correctly, in this case, the $l=3$ term is already an irreducible tensor of rank 3 (without any rank 1 tensor component). My questions are:

1. Why the terms for a give l are not the same between the 2 expansions? I assumed that they are only written in 2 different frames (i.e. they have a different form when written down), but it seems like the terms for the same l are not the same. Does this mean that the l=1 term in the Legendre expansion is equal to the l=1 term in the Cartesian expansion PLUS the extra rank 1 tensor coming from the l=1 term in the Cartesian expansion? Can there be other l>3 terms in the cartesian expansion that also contain rank 1 tensors?

2. In the Legendre expansion we have a clear separation between r>R and r<R. Where is this hidden in the Cartesian expansion?

 

[pasmith]

If we expand a function as \sum_{n=0}^\infty a_nx^n and also as \sum_{n=0}^\infty b_n P_n(x), do we expect a_n \equiv b_n for each n?

Because P_n is a polynomial of order n, expanding both expressions to n = N will get all terms of order x^N or lower, but the coefficients will in general be different. If we replace P_n by its definition in terms of powers of x, then the two expressions should simplify to the same result.

The same principle applies here.