Products of Dirac delta distributions in multiple dimensions

Question

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.)

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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 $$\tag{3}\label{eq:1int} \int dr_1\, dr_2 \int dr_{rel}\, dr_{cm} \, f(r_1,r_2) g(r_{rel},r_{cm}) \delta\big( (r_1-r_2) - r_{rel} \big) \delta\big( \tfrac12 (r_1+r_2) - r_{cm} \big) \\= \int dr_1 \, dr_2 \, f(r_1,r_2) g(r_1-r_2, \tfrac12(r_1+r_2)) $$ $$\tag{4}\label{eq:2int} \int dr_{rel}\, dr_{cm} \int dr_1\, dr_2 \, f(r_1,r_2) g(r_{rel},r_{cm}) \delta\big( (r_{cm} + \tfrac12 r_{rel}) - r_1 \big) \delta\big( (r_{cm} - \tfrac12 r_{rel}) - r_2 \big) \\= \int dr_{rel} \, dr_{cm} \, f(r_{cm}+\tfrac12 r_{rel},r_{cm}-\tfrac12 r_{rel}) g(r_{rel},r_{cm}) $$

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1. 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 $\eqref{eq:2}$?
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 $\eqref{eq:2}$?

Answer 1

1. 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.

2. 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 $$ \langle {\bf r}_1,{\bf r}_2 | {\bf r}^{\prime}_1,{\bf r}^{\prime}_2\rangle ~=~\delta^3({\bf r}_1\!-\!{\bf r}^{\prime}_1)\delta^3({\bf r}_2\!-\!{\bf r}^{\prime}_2) .\tag{3}$$

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 $$|{\bf r}_{\rm cm},{\bf r}_{\rm rel}\rangle\!\rangle~:=~|{\bf r}_1,{\bf r}_2\rangle, \qquad {\bf r}_{\rm cm}~\equiv~\frac{{\bf r}_1+{\bf r}_2}{2},\qquad {\bf r}_{\rm rel}~\equiv~{\bf r}_1-{\bf r}_2. \tag{5} $$

4. 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.

5. 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.