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Given that I have some coefficient(i.e. a number) which is to be determined from a radial integral:

$$b_{n00} = \frac{(2\pi)^{1/4}}{\sigma^{3/2}} \frac{1}{\sqrt{3}} [C(000|000)]^2 \int^{\infty}_{r = 0} \frac{1}{(a_0)^{3/2}} \frac{2}{n^2}\sqrt{\frac{1}{n}} L^{1}_{n-1}\left(\frac{2r}{na_0}\right)e^{-\frac{r}{n{a_0}}}e^{-\frac{r^2}{4\sigma^2}}r^2 dr$$

Where $\sigma$ is the spread of my original 3D Gaussian function, $C(000|000)$ is just a Clebsch-Gordan coefficient and $a_0$ is the Bohr radius.

Now I know for a fact that $b_{n00}$ is supposed to just be a number because it originated from the superposition of eigenfunctions(it's the expansion coefficients so it must be a number), shown below:

$$\psi(r) = \sum_{n=0}^{\infty}\sum_{l=0}^{n-1}\sum_{m=-l}^{l}b_{nlm}\varphi_{nlm}(r)$$

Throughout my steps on finding $b_{nlm}$, it was realised that for my specific situation, a condition of $l = m = 0$ was needed, leaving the $n$ index, and hence $b_{n00}$. All I'm trying to do here is a dimensionality check to make sure my RHS units are the same as LHS.

My question: So I know that both of the exponentials shouldn't have any units because they naturally cancels out and so does the units of Laguerre Polynomial and $n$ is just a number. Now looking at the integral it leaves me with units of $$\frac{1}{m^{3/2}}$$ that comes from $$\frac{1}{{a_0}^{3/2}}$$ and also another $$\frac{1}{m^{3/2}}$$ from $$\frac{1}{{\sigma}^{3/2}}$$ which leaves me with $$\frac{1}{m^{3}}.$$ So will I be right in assuming that the $\frac{1}{r^2}$ integral will integrate into a number with units of $m^3$, which completely cancels out the units?

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    $\begingroup$ Do you have some reason to doubt whether your assumption is correct? I'm assuming you do, otherwise why are you asking us? But I don't see that mentioned in the question. It would be good to edit it in. $\endgroup$ – David Z Oct 24 '18 at 20:40
  • $\begingroup$ Frankly it's hard to imagine why you can't work this out on your own. Anyone attempting calculus of this sort ought to be able to check unit dimensions as an almost trivial task. $\endgroup$ – StephenG Oct 24 '18 at 20:49
  • $\begingroup$ @DavidZ My only reason to doubt my assumption is that it took me pages and pages of working out before I reached my $b_{n00}$, which is usually unusual however not impossible to get it right at the first attempt. Therefore that somehow made me feel dubious about my own assumption so I guess all I wanted was for someone to give me a pat on my shoulder, for which in hindsight I realised my question is somehow pointless. $\endgroup$ – user3613025 Oct 24 '18 at 21:20
  • $\begingroup$ FYI: There is an urban legend to the effect that a PhD thesis was once judged unacceptable because the candidate couldn’t spell Clebsch-Gordan (with an “a”) properly. Paul Gordan (with an “a”, see en.m.wikipedia.org/wiki/Paul_Gordan) was, amongst others, the advisor of Emmy Noether. I took the liberty of correcting the typo. $\endgroup$ – ZeroTheHero Oct 24 '18 at 21:40
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The integral $\int r^2 dr$ has the dimensions of a volume, $[L]^3$. This will cancel out the $\frac{1}{[L]^3}$ that you have correctly pointed out above. Thus, your coefficients are indeed dimensionless!

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    $\begingroup$ General policy is not to answer "check my work" or "homework-type" questions. $\endgroup$ – StephenG Oct 24 '18 at 20:50
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Just a tip, regarding the dimension analysis of radial integrals: because there's a natural length scale $a_0$, it's often convenient to nondimensionalise with the substitution $u=\frac{r}{a_0}$, which should reduce any remaining dependence on $a_0,\,\sigma$ to one on the dimensionless ratio $\frac{\sigma}{a_0}$. This typically allows you to write quantities that ought to be dimensionless as products of obviously dimensionless factors. In this case, $$b_{n00}=\frac{2[C(000|000)]^2(2\pi)^{1/4}}{\sqrt{3}n^{5/2}}\cdot(\frac{\sigma}{a_0})^{-3/2}\cdot\int_0^\infty L_{n-1}^1\bigg(\frac{2u}{n}\bigg)e^{-\frac{u}{n}-\frac{u^2}{4(\sigma/a_0)^2}}u^2du.$$

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