Take the 2-minute tour ×  When trying to solve the Schrödinger
equation for hydrogen, one usually splits up the wave function
into two parts:  $\psi(r,\phi,\theta)= R(r)Y_{l,m}(\phi,\theta)$
I understand that the radial part usually has a singularity for
the 1s state at $r=0$ and this is why you remove it by writing:
$R(r) = \frac{U(r)}{r}$  But what is the physical meaning of
$R(r=0) = \infty$. Wouldn't this mean that the electron cloud is
only at the centre of the atomic nucleolus?  Thanks in advance!
share|improve this question  5 Answers 5  The infinitesimal
probability for the electron to be in the volume $dV$ around a
point $(r,\theta,\phi)\leftrightarrow (x,y,z)$ is given by $$ dP
= dV\cdot |\psi(x,y,z)|^2 = dV\cdot |R(r)|^2\cdot
|Y_{lm}(\theta,\phi)|^2 =\dots$$ as you can see if you
substitute your Ansatz for the wave function. However, the
infinitesimal volume $dV=dx\cdot dy\cdot dz$ may be rewritten in
terms of differentials of the spherical coordinates as $$ dV =
dr\cdot r^2 \cdot d\Omega = dr\c