What is the correct separable Schrödinger equation in spherical coordinates?

What is the correct separable Schrödinger equation in spherical coordinates?

Some articles use this formula: $$\Psi(r,\theta,\phi) = R(r)\cdot\Theta(\theta)\cdot \Phi(\phi),$$ and some of them use: $$\Psi(r,\theta,\phi) =\frac{R(r)\cdot \Theta(\theta)\cdot \Phi(\phi) }{r}.$$

So which is true?

Both formulas are correct. And it is essentially a matter of taste which one you prefer:

• If you use the separation approach $$\Psi(r,\theta,\phi) = R(r) \cdot \Theta(\theta) \cdot \Phi(\phi)$$ then you get the radial differential equation $${\frac {1}{r^{2}}}{\frac {d}{dr}}\left(r^{2}{\frac {dR}{dr}}\right)-{\frac {\ell (\ell +1)}{r^{2}}R+{\frac {2m}{\hbar ^{2}}}\left[E-V(r)\right]R=0}$$ (like in Wikipedia - Particle in a spherically symmetric potential - Derivation of the radial equation) with the normalization condition $$\int_0^\infty dr\ r^2 |R(r)|^2=1.$$
• If you use the separation approach $$\Psi(r,\theta,\phi) = \frac{u(r)}{r} \cdot \Theta(\theta) \cdot \Phi(\phi)$$ then you get the radial differential equation $$\frac {d^2u}{dr^2}-{\frac {\ell (\ell +1)}{r^{2}}u+{\frac {2m}{\hbar ^{2}}}\left[E-V(r)\right]u=0}$$ with the normalization condition $$\int_0^\infty dr\ |u(r)|^2=1.$$

At the end you get the same solutions $$\Psi(r,\theta,\phi)$$ in both cases because $$R(r)=\frac{u(r)}{r}$$. However, the approach with $$u(r)$$ leads to simpler math and has more similarity with a 1-dimensional Schrödinger equation.

If I am not wrong, then the second one is used for (I guess) when we solve the radial part. We have to put $$R(r)=u(r)/r$$, so maybe in second one they have written $$R(r)/r$$ instead of $$R(r)$$ is because they put directly $$R(r)/r$$. I am not sure though.

• Firstly I need to find radial, angular and azimuthal parts, then solve them. Jun 21 at 16:09
• Generally the angular and azimuthal parts are solved first and then radial but u can do either. Jun 21 at 16:23
• You will get a different equation because you have to derive $R/r$ twice. you can take a look at a laplacian equation. Jun 21 at 16:30
• U can solve radial eqn using the following method, Near r=0 the differential equation for radial part is d^2u/dx^2=l(l+1)/r^2, now let u=r^s,then putting the value of u in the diiferential eqn, we gets(s-1)=l(l+1), therefore s=l+1, and the solution is u(r)=r^(l+1) or R(r)=r^l Jun 21 at 17:30