The Radial Equation

Section 16.1 The Radial Equation

So far we have obtained solutions to the angular (\(\theta\) and \(\phi\)) parts of Schrödinger’s equation. The \(Y_\ell^m(\theta,\phi)\) describe the spatial variation of the wave function on the surface of a sphere. Now we turn to the radial part, (6.5.15):

\begin{equation} \frac{d}{dr}\left(r^2 \frac{dR}{dt}\right) + \frac{2\mu}{\hbar^2} \bigl(E-U(r)\bigr) r^2R + AR = 0\tag{16.1.1} \end{equation}

Here \(A\) is the separation constant we obtained from the solution to the \(\theta\) equation: \(A=-\ell(\ell+1)\text{.}\) The solution to (16.1.1) will depend on the potential energy \(U(r)\text{;}\) for central forces, the potential energy \(U\) depends only on \(r\) and is independent of \(\theta\) and \(\phi\text{.}\) As a result, the are solutions to the angular part for any central force.

We begin by using the product rule to rewrite (16.1.1) as:

\begin{equation} \frac{d^2R}{dr^2} + \frac{2}{r}\frac{dR}{dr} + \frac{2\mu}{\hbar^2} \left( E-U(r)-\frac{\hbar^2\ell(\ell+1)}{2\mu r^2} \right) R = 0\tag{16.1.2} \end{equation}

Equation (16.1.2) resembles the one-dimensional Schrödinger equation corresponding to an effective potential energy \(U_\text{eff}\text{:}\)

\begin{equation} U_\text{eff}(r) = U(r) + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2}\tag{16.1.3} \end{equation}

The term \(\hbar^2\ell(\ell+1)/2\mu r^2\) is the centrifugal contribution to the effective potential energy. It behaves like a repulsive force, and it increases with \(\ell\) in exact analogy with classical mechanics, see (11.3.4).

We will solve the radial equation for a potential energy characteristic of an atom consisting of a single electron interacting with a nucleus of charge \(+Ze\) (i.e. \(\text{H}\text{,}\) \(\text{He}^+\text{,}\) \(\text{Li}^{++}\text{,}\) etc.):

\begin{equation} U(r) = -\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r}\tag{16.1.4} \end{equation}

Figure 14 shows \(U(r)\) and \(U_\text{eff}(r)\) for this case, exactly as it does in the classical case. Note that \(U_\text{eff}\) forms a potential energy minimum that permits negative-energy bound states to exist.

Reflection.

As \(\ell\) increases, do you think the expectation value of \(r\) will increase or decrease? Later we will use the radial wave functions to calculate the expectation value of \(r\text{,}\) and we will see if the calculated value is consistent with this expectation.

With this choice of \(U(r)\text{,}\) we can write the radial equation as:

\begin{equation} \frac{d^2R}{dr^2} + \frac{2}{r}\frac{dR}{dr} + \frac{2\mu}{\hbar^2} \left(E+\frac{k}{r}-\frac{\hbar^2\ell(\ell+1)}{2\mu r^2} \right) R = 0\tag{16.1.5} \end{equation}

where \(k=\frac{1}{4\pi\epsilon_0}\, Ze^2\text{.}\)

Prev Top Next