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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 $u_l(r)$?

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  • $\begingroup$ Hint: Check the definition in the first equation. $\endgroup$
    – Qmechanic
    Apr 23 at 7:56
  • $\begingroup$ 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 $$ \frac{1}{r^{2}}\frac{d}{dr}\left( r^{2}\frac{dR}{dr}\right) $$ $\endgroup$
    – erikasan
    Apr 23 at 9:16
  • $\begingroup$ I just checked in my textbook, the equation is the one I wrote. Maybe they are equivalent. $\endgroup$ Apr 23 at 10:42

2 Answers 2

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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 \begin{equation} \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}. \end{equation} 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; \begin{equation} \tag{1} \label{term} -\frac{\hbar}{2m} \left( \frac{1}{r} \frac{d^2}{dr^2} r \right)\frac{u(r)}{r}. \end{equation} 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 $r$;

\begin{equation} \frac{d}{dr}f(r) \neq f(r)\frac{d}{dr}. \end{equation} 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 \begin{equation} -\frac{\hbar}{2m}\frac{d^2u}{dr^2}. \end{equation}

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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.

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