Products of Dirac delta distributions in multiple dimensions Following up on Integral over product of Dirac delta functions, I have the following generalization of the question (arising from a problem in quantum mechanics):
If I have the product of two Dirac delta distributions with non-identical arguments
$$\tag{1}\label{eq:1}
\langle r_{rel} r_{cm} \vert r_1 r_2 \rangle 
    = \delta\big( (r_1-r_2) - r_{rel} \big) \times \delta\big( \tfrac12 (r_1+r_2) - r_{cm} \big),
$$
can I somehow argue that it is "equivalent" to the alternate form
$$\tag{2}\label{eq:2}
\langle r_1 r_2 \vert r_{rel} r_{cm} \rangle 
    = \delta\big( (r_{cm} + \tfrac12 r_{rel}) - r_1 \big) \times \delta\big( (r_{cm} - \tfrac12 r_{rel}) - r_2 \big) ?$$
(Note here that we can consider these as representing the change of variables $(r_1,r_2)\rightarrow(r_{rel},r_{cm})$ and $(r_{rel},r_{cm})\rightarrow(r_1,r_2)$, respectively.)

If I try to show that these two forms are equivalent by, e.g., integrating against test functions, I either end up with an integration variable appearing in the argument to two of the delta distributions or with some reordering of integration which I'm not sure I can justify. (In that sense, this may also be a question about Fubini's theorem in the theory of distributions.)
For instance, integrating each of these against the test functions $f(r_1,r_2)$, $g(r_{rel},r_{cm})$, I get
$$\tag{3}\label{eq:1int}
\int dr_1\, dr_2 \int dr_{rel}\, dr_{cm} \, f(r_1,r_2) g(r_{rel},r_{cm}) \delta\big( (r_1-r_2) - r_{rel} \big) \delta\big( \tfrac12 (r_1+r_2) - r_{cm} \big) \\= \int dr_1 \, dr_2 \, f(r_1,r_2) g(r_1-r_2, \tfrac12(r_1+r_2))
$$
$$\tag{4}\label{eq:2int}
\int dr_{rel}\, dr_{cm} \int dr_1\, dr_2 \, f(r_1,r_2) g(r_{rel},r_{cm}) \delta\big( (r_{cm} + \tfrac12 r_{rel}) - r_1 \big) \delta\big( (r_{cm} - \tfrac12 r_{rel}) - r_2 \big) \\= \int dr_{rel} \, dr_{cm} \, f(r_{cm}+\tfrac12 r_{rel},r_{cm}-\tfrac12 r_{rel}) g(r_{rel},r_{cm})
$$


*

*Is it a problem that I'm doing the multiple integrals in a different order between $\eqref{eq:1int}$ and $\eqref{eq:2int}$ if I want to show the equivalence of $\eqref{eq:1}$ and $\eqref{eq:2}$?

*Is it sufficient to do the somewhat trivial change of variables $(r_1,r_2)\rightarrow (r_{rel},r_{cm})$ on $\eqref{eq:1int}$ to show that it is equal to $\eqref{eq:2int}$, and is that sufficient to establish the equality/equivalence of $\eqref{eq:1}$ and $\eqref{eq:2}$?

 A: *

*Let us for concreteness consider a 3D position space $\mathbb{R}^3$. (The generalization to other dimensions is straightforward.) Consider position operators $\hat{\bf r}_1$ and $\hat{\bf r}_2$ for 2 non-identical particles.


*In a rigged Hilbert space we define eigenkets
$$|{\bf r}_1,{\bf r}_2\rangle, \tag{1} $$
such that
$$ \hat{\bf r}_1 |{\bf r}_1,{\bf r}_2\rangle~=~{\bf r}_1|{\bf r}_1,{\bf r}_2\rangle, \qquad 
\hat{\bf r}_2 |{\bf r}_1,{\bf r}_2\rangle~=~{\bf r}_2|{\bf r}_1,{\bf r}_2\rangle, \tag{2}$$
and
$$ \langle {\bf r}_1,{\bf r}_2 | {\bf r}^{\prime}_1,{\bf r}^{\prime}_2\rangle
~=~\delta^3({\bf r}_1\!-\!{\bf r}^{\prime}_1)\delta^3({\bf r}_2\!-\!{\bf r}^{\prime}_2) .\tag{3}$$


*Define operators
$$ \hat{\bf r}_{\rm cm}~:=~\frac{\hat{\bf r}_1+\hat{\bf r}_2}{2},\qquad \hat{\bf r}_{\rm rel}~:=~\hat{\bf r}_1-\hat{\bf r}_2, \tag{4} $$
and corresponding kets
$$|{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle~:=~|{\bf r}_1,{\bf r}_2\rangle, \qquad 
{\bf r}_{\rm cm}~\equiv~\frac{{\bf r}_1+{\bf r}_2}{2},\qquad 
{\bf r}_{\rm rel}~\equiv~{\bf r}_1-{\bf r}_2. \tag{5} $$


*One may use eqs. (2), (3), (4) & (5) to show that the kets (5) are eigenkets
$$ \hat{\bf r}_{\rm cm} |{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle~=~{\bf r}_{\rm cm}|{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle, \qquad 
\hat{\bf r}_{\rm rel} |{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle~=~{\bf r}_{\rm rel}|{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle, \tag{6}$$
and normalized
$$ \langle\!\langle {\bf r}_{\rm cm},{\bf r}_{\rm rel} | {\bf r}^{\prime}_{\rm cm},{\bf r}^{\prime}_{\rm rel}\rangle\!\rangle
~=~\delta^3({\bf r}_{\rm cm}\!-\!{\bf r}^{\prime}_{\rm cm})\delta^3({\bf r}_{\rm rel}\!-\!{\bf r}^{\prime}_{\rm rel}). \tag{7}$$
In eq. (7) we used how higher-dimensional Dirac delta distributions transform under coordinate transformations, cf. e.g. my Math.SE answer here. It is important that the Jacobian of the transformation (4) is one.


*Similarly, one may show that the overlap is
$$ \langle {\bf r}_1,{\bf r}_2 | {\bf r}^{\prime}_{\rm cm},{\bf r}^{\prime}_{\rm rel}\rangle\!\rangle
~=~\delta^3(\frac{{\bf r}_1+{\bf r}_2}{2}\!-\!{\bf r}^{\prime}_{\rm cm})\delta^3({\bf r}_1-{\bf r}_2\!-\!{\bf r}^{\prime}_{\rm rel}), \tag{8}$$
etc.
