Ground State Energy of Quantum Harmonic Oscillator Letting $u=rR(r)$, the radial part of the SE becomes:
$$-\frac{\hbar^2}{2m}u_{rr}+\frac{1}{2}m\omega^2r^2 u+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}u=Eu$$
I am interested in obtaining the energy of the ground state (which I know is $3\hbar \omega/2$). As such, I set $l=0$ to get
$$-\frac{\hbar^2}{2m}u_{rr}+\frac{1}{2}m\omega^2r^2 u=Eu$$
which is identical the 1D harmonic oscillator problem. The lowest energy of the 1D oscillator is $\hbar \omega/2$, which is not the right energy for the 3D case. Why does this method not give me the proper energy for the 3D case? How can I find the ground state energy using the spherical equations?
 A: It is true that you obtain an equation for $u$ that is exactly equal to the equation for the 1D harmonic oscillator; what changes are boundary conditions.
In fact, in solving these kind of equations, you require that the radial solution $R(r)$ goes like a certain power, that turns out to be
$$
R(r) \xrightarrow[r \to 0]{} r^l\, .
$$
Then necessarily we have $u(r)\xrightarrow[r \to 0]{} r^{l+1}$, that in particular for $l=0$ means
$$
u(0)=0\,, \qquad u(r)\xrightarrow[r \to 0]{} r \, \,.
$$
From here it can be noticed that this solution does not correspond to the ground state of the 1D harmonic oscillator, that being a Gaussian is not null at $r=0$. 
The first eigenfunction of the 1D harmonic oscillator that fulfills the boundary conditions is actually the one corresponding to $n=1$, with an energy given by
$$
E=\hbar \omega\left(1+\frac12\right)= \frac32 \hbar \omega \, ,
$$
which is the expected result.
A: The difference is in the boundary conditions: in the spherical problem, the radius $r$ is necessarily $\ge 0$ and it is not possible to have a wavefunction for $r<0$.  This forces the eigenfunction to have a node at $r=0$ so the solution does not “leak” into $r<0$, just like they must have a node at both ends of an infinite well since the wavefunctions cannot beyond the well.
As a result, the lowest eigenfunction allowed by the boundary condition is the lowest solution with a node at $r=0$, and this is the harmonic oscillator solution with energy $\frac{3}{2}\hbar\omega$.  Note also that, as a result, the next $\ell=0$ solution will occur at $\frac{7}{2}\hbar \omega$, since the energy $\frac{5}{2}\hbar \omega$ corresponds to a solution that does not satisfy the boundary condition at $r=0$.  Of course there are solutions with this energy $\frac{5}{2}\hbar \omega$: they just don’t have $\ell=0$.
