The effective potential

Intuition

The radial equation is a one-dimensional problem in an effective potential: the real one plus the centrifugal barrier $\hbar^{2}\ell(\ell+1)/2mr^{2}$ , which grows without limit at the centre. For an attractive Coulomb potential the sum has a minimum, at the radius where a classical particle with that angular momentum would circle. The higher $\ell$ , the further out the minimum and the shallower it is — which is why high angular momentum states sit far from the nucleus and are loosely bound.

A skater pulling in her arms spins faster; to come closer to the centre with the same angular momentum costs rotational energy. The centrifugal barrier is that cost, felt as a wall that keeps the particle away from the centre.

The effective potential $-\frac{1}{r}+\frac{\ell(\ell+1)}{2r^{2}}$ of an attractive Coulomb field in units with and unit strength. For it is the bare ; for it has a minimum of at ; for a minimum of at .

For the radial motion the particle feels the true potential plus a centrifugal barrier, repulsive and growing as $1/r^{2}$ towards the centre.

V_{\text{eff}}(r)=V(r)+\frac{\hbar^{2}\ell(\ell+1)}{2mr^{2}}

Properties

For $\ell\ge1$ the barrier dominates at small $r$ , so the particle is pushed away from the centre.
For $V=-k/r$ the minimum is at $r_{0} = \frac{ℏ^{2} ℓ (}{}$ with depth : the classical circular orbit with .

The minimum of the Coulomb effective potential

\begin{aligned}&V_{\text{eff}}=-\frac{k}{r}+\frac{\hbar^{2}\ell(\ell+1)}{2mr^{2}}\\ &\Longrightarrow\ r_{0}=\frac{\hbar^{2}\ell(\ell+1)}{mk},\quad V_{\text{eff}}(r_{0})=-\frac{k}{2r_{0}}\end{aligned}

Differentiate and set the derivative to zero: the attraction $k/r^{2}$ balances the centrifugal push, which fixes $r_{0}$ . At that radius the barrier term is half the attraction in size, so the sum is minus half of $k/r_{0}$ .

Proof steps

V_{\text{eff}}'(r)=\frac{k}{r^{2}}-\frac{\hbar^{2}\ell(\ell+1)}{mr^{3}}

Differentiate each term.

V_{\text{eff}}'(r_{0})=0\ \Rightarrow\ r_{0}=\frac{\hbar^{2}\ell(\ell+1)}{mk}

Attraction balances the centrifugal term.

\frac{\hbar^{2}\ell(\ell+1)}{2mr_{0}^{2}}=\frac{k}{2r_{0}}

Use $\hbar^{2}\ell(\ell+1)=mkr_{0}$ .

V_{\text{eff}}(r_{0})=-\frac{k}{r_{0}}+\frac{k}{2r_{0}}=-\frac{k}{2r_{0}}

Add the two terms.

Applications

Practice

The Effective Potential

For radial motion the particle feels its potential plus the centrifugal barrier. With a Coulomb attraction the sum has a minimum at the radius of a classical circular orbit.

V_{\text{eff}}=V(r)+\frac{\hbar^{2}\ell(\ell+1)}{2mr^{2}}

Try it

With $\hbar=m=k=1$ , at what radius is $V_{\text{eff}}=-\frac1r+\frac{\ell(\ell+1)}{2r^{2}}$ smallest for $\ell=1$ ?