# 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?

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.
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 $$\ell=0$$.