# Radial Schrödinger equation: from $R_l(r)$ to $u_l(r)$

I am in the 3-dimensional radial Schrödinger equation, in the spherical coordinates, where we try to find the separable solutions

$$\psi(r) = R_l(r) Y_l^m(\theta, \varphi) \equiv \frac{u_l(r)}{r}Y_l^m(\theta, \varphi).$$ ($$\theta$$ is the colatitude and $$\varphi$$ the azimuthal angle.)

I've been trying for an hour to get from this line to the next one:

$$\left[-\frac{\hbar}{2m} \left( \frac{1}{r} \frac{d^2}{dr^2} r - \frac{l(l+1)}{r^2} \right) + V(r)\right] R_l(r) Y_l^m(\theta, \varphi) = ER_l(r) Y_l^m(\theta, \varphi)$$

$$\Longleftrightarrow \left[-\frac{\hbar}{2m} \left( \frac{d^2}{dr^2} - \frac{l(l+1)}{r^2} \right) + V(r)\right] u_l(r) = Eu_l(r)$$ How did the $$\frac{1}{r}$$ and $$r$$ dissapear with $$u_l(r)$$?

• Hint: Check the definition in the first equation. Apr 23 at 7:56
• Are you sure that the line you are starting from is right? According to Griffiths 2nd edition Eq. (4.35) the second derivative term is different. It should be $$\frac{1}{r^{2}}\frac{d}{dr}\left( r^{2}\frac{dR}{dr}\right)$$ Apr 23 at 9:16
• I just checked in my textbook, the equation is the one I wrote. Maybe they are equivalent. Apr 23 at 10:42

## 2 Answers

While it seems like you've resolved your particular issue, this might be a good opportunity to think about how operations work more generally, as you will come to see a lot of operator algebra in quantum mechanics.

Taking your first equation and substituting $$R(r) \rightarrow u(r)/r$$ and dividing by $$Y(\theta,\varphi)$$, we get $$$$\left[ -\frac{\hbar}{2m} \left( \frac{1}{r} \frac{d^2}{dr^2} r - \frac{l(l+1)}{r^2} \right) + V(r) \right]\frac{u(r)}{r} = E \frac{u(r)}{r}.$$$$ Someone who is used to regular (scalar) algebra might be tempted to simply cancel the $$1/r$$ factors multiplying $$u(r)$$ on both sides. This would be fine if not for a certain term on the left-hand side; $$$$\tag{1} \label{term} -\frac{\hbar}{2m} \left( \frac{1}{r} \frac{d^2}{dr^2} r \right)\frac{u(r)}{r}.$$$$ There are two important reasons why we can't just multiply this term by $$r$$ to get rid of the $$r$$ dividing $$u(r)$$.

1. Operators are applied from the left, by convention.
2. The differential operator $$d/dr$$ does not commute with multiplication by a factor which involves $$r$$;

$$$$\frac{d}{dr}f(r) \neq f(r)\frac{d}{dr}.$$$$ With this in mind, if we multiply (\ref{term}) by $$r$$ we should do so from the left (you can think of it as applying a multiplication operator), which cancels the $$1/r$$ factor there (scalar multiplications do, of course, commute). The remaining part then tells us to first multiply $$u(r)/r$$ by $$r$$ and then differentiate the result twice with respect to $$r$$, which is why we get $$$$-\frac{\hbar}{2m}\frac{d^2u}{dr^2}.$$$$

A friend just explained me that actually, turning $$R_l(r)$$ into $$u_l(r)/r$$ made a $$\frac{1}{r}$$ appear everywhere, except in the $$\frac{d^2}{dr^2}$$ term where the $$r R_l(r)$$ turns into $$u_l(r)$$. As there was already a $$\frac{1}{r}$$ in front of $$\frac{d^2}{dr^2}$$, all the $$\frac{1}{r}$$ and $$Y_l^m(\theta, \varphi)$$ just dissapear.