Reducing the central-force problem

Since Chapter 1 we have been carrying a debt: Bohr used two unjustified postulates to get $E_n=-13.6\ \text{eV}/n^2$, matching the spectrum to four significant figures — yet his picture was wrong through and through. The circular orbits he drew do not exist. In this chapter we solve the hydrogen atom completely, with real quantum mechanics, and see which results Bohr “happened” to get right, why he could, and what the things he could never see (degeneracy, orbital shapes, fine structure) actually look like.

Chapter 5 has just finished polishing the last tool we need: angular momentum. The eigenfunctions of $\hat L^2$ and $\hat L_z$ are the spherical harmonics $Y_l^m(\theta,\phi)$, with eigenvalues $l(l+1)\hbar^2$ and $m\hbar$ — see section 5.2. In this section you will watch that entire angular machine get carried into the hydrogen atom without re-deriving a single line.

First, an experimental fact: the nucleus is not nailed down

Asked to write the hydrogen Hamiltonian, most people’s first instinct is “an electron moving in the Coulomb field of a fixed proton”:

$$\hat H \overset{?}{=} \frac{\hat p^2}{2m_e} - \frac{e^2}{4\pi\varepsilon_0 r}\tag{6.1.1}$$

The “proton doesn’t move” approximation looks unimpeachable — the proton is 1836 times heavier than the electron. But spectroscopy is too precise to tolerate it.

In 1932, Harold Urey slowly evaporated and concentrated liquid hydrogen, photographed its spectrum, and found that every Balmer line had a faint companion beside it: an extra line 0.18 nm away from $\mathrm H\alpha$ (656.28 nm). It was not a new element — it was deuterium, the heavy isotope of hydrogen, whose nucleus is a proton plus a neutron, doubling the mass. Why should a heavier nucleus shift the spectral lines? Because the nucleus moves too: electron and nucleus orbit their common centre of mass, and the heavier the nucleus, the closer the centre of mass sits to it, nudging the energy levels by just a whisker. That 0.18 nm shift is exactly the whisker — and it won Urey the 1934 Nobel Prize in Chemistry.

So the honest starting point must be the two-body problem:

$$\hat H=\frac{\hat p_e^2}{2m_e}+\frac{\hat p_p^2}{2m_p} -\frac{e^2}{4\pi\varepsilon_0\,\lvert\vec r_e-\vec r_p\rvert}\tag{6.1.2}$$

Six coordinates (three for the electron, three for the proton) — bigger than any problem we have solved. Fortunately there is a surgical move, already familiar from classical mechanics, that slices it cleanly in two.

Step one: two bodies become one

Centre-of-mass + relative coordinates: an exact separationbasic~7 minexpand

Step 1: change coordinates. Define the centre-of-mass and relative coordinates:

$$\vec R=\frac{m_e\vec r_e+m_p\vec r_p}{m_e+m_p},\qquad \vec r=\vec r_e-\vec r_p\tag{6.1.3}$$

The potential energy depends only on $\vec r$ — that is the root of the separability. Rewriting the kinetic energy in the new coordinates (differentiate through the coordinate change directly, or copy the classical-mechanics result — at the operator level it is identical):

$$\frac{\hat p_e^2}{2m_e}+\frac{\hat p_p^2}{2m_p} =\frac{\hat P^2}{2M}+\frac{\hat p^2}{2\mu}\tag{6.1.4}$$

where $M=m_e+m_p$ is the total mass, and

$$\mu\equiv\frac{m_em_p}{m_e+m_p}\tag{6.1.5}$$

is called the reduced mass — in plain words, “the mass you should assign to a single particle orbiting a fixed centre, if it is to stand in for two particles that are both moving”. It is always a little lighter than the lighter of the two: $1/\mu=1/m_e+1/m_p$.

Step 2: the Hamiltonian splits in half.

$$\hat H=\underbrace{\frac{\hat P^2}{2M}}_{\hat H_\text{cm}} +\underbrace{\frac{\hat p^2}{2\mu}-\frac{e^2}{4\pi\varepsilon_0 r}}_{\hat H_\text{rel}}\tag{6.1.6}$$

$\hat H_\text{cm}$ contains only centre-of-mass variables, $\hat H_\text{rel}$ only relative variables, and the two halves commute. So the solution can be written as a product $\Psi(\vec R,\vec r)=\chi(\vec R)\,\psi(\vec r)$, with the total energy the sum of the two.

Step 3: read off the physics. The centre-of-mass half is a free particle — the atom drifting through space as a whole, irrelevant to its internal structure, so from now on we simply drop it. What remains is the problem we actually need to solve:

$$\left[-\frac{\hbar^2}{2\mu}\nabla^2-\frac{e^2}{4\pi\varepsilon_0 r}\right]\psi(\vec r)=E\,\psi(\vec r)\tag{6.1.7}$$

A single particle of mass $\mu$ moving in a fixed Coulomb potential. The two-body problem is gone, and the entire price was swapping $m_e$ for $\mu$.

Step two: three dimensions become one

We now face a three-dimensional partial differential equation. The way in is the symmetry of the potential: $V=V(r)$ depends only on distance, not direction. Such a potential is called a central potential (in plain words: a field that looks the same from every direction).

Separation of variables: the radial equationbasic~5 minexpand

Substitute $\psi=R\,Y_l^m$ into

$$-\frac{\hbar^2}{2\mu}\left[\frac{1}{r^2}\frac{\dd}{\dd r}\!\left(r^2\frac{\dd R}{\dd r}\right)Y_l^m -\frac{R\,\hat L^2Y_l^m}{\hbar^2r^2}\right]+V(r)\,R\,Y_l^m=E\,R\,Y_l^m\tag{6.1.11}$$

Use $\hat L^2Y_l^m=l(l+1)\hbar^2Y_l^m$ and divide through by $Y_l^m$:

$$-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\dd}{\dd r}\!\left(r^2\frac{\dd R}{\dd r}\right) +\left[V(r)+\frac{l(l+1)\hbar^2}{2\mu r^2}\right]R=ER\tag{6.1.12}$$

Every angular variable is gone — an ordinary differential equation in the single variable $r$. Notice two things:

1. $m$ does not appear. For a given $l$, all $2l+1$ values of $m$ share the same radial equation and the same energy — a degeneracy forced by isotropy (with no external field, space has no preferred direction, and it cannot matter along which axis you measure $\hat L_z$).
2. $l$ appears in a new term, $l(l+1)\hbar^2/2\mu r^2$. It was carved out of the kinetic energy, yet it stands alongside the potential — in the next section it takes centre stage.

Taking stock

A two-body partial differential equation in six coordinates, after two purely structural “splits”:

• Two bodies → one body: the freely moving centre of mass is discarded, leaving one particle of mass $\mu$;
• Three dimensions → one: the angular part is solved universally by $Y_l^m$, leaving an ordinary differential equation for $R(r)$.

No approximation was made anywhere — both steps are exact. Every remaining difficulty has been compressed into that one radial equation.

∎Key formulas

• Reduced mass

  $$\mu=\frac{m_em_p}{m_e+m_p}=\frac{m_e}{1+m_e/m_p}\approx0.999456\,m_e$$

  The entire price of two-bodies-to-one; levels ∝ μ, whence the isotope shift

• Relative-motion equation

  $$\left[-\frac{\hbar^2}{2\mu}\nabla^2+V(r)\right]\psi=E\psi$$

  Centre-of-mass translation separated off and discarded

• Separated form

  $$\psi(r,\theta,\phi)=R(r)\,Y_l^m(\theta,\phi)$$

  Angular part universal for every central potential; m does not enter the energy

• Radial equation

  $$-\frac{\hbar^2}{2\mu r^2}\frac{\dd}{\dd r}\!\left(r^2\frac{\dd R}{\dd r}\right)+\left[V+\frac{l(l+1)\hbar^2}{2\mu r^2}\right]R=ER$$

  Every remaining difficulty lives here; the next section tidies it up

?Self-check4 questions

1. 1.

   The two-body problem separates exactly into centre-of-mass motion plus relative motion. What is the key prerequisite?

   • The potential energy depends only on the relative position of the two particles
   • One particle is much heavier than the other
   • The potential must be a Coulomb potential
   • The two particles have equal masses

   Submit
2. 2.

   Why is the angular part of the wavefunction the same family of spherical harmonics for every central potential?

   • Spherical symmetry makes H commute with L² and L_z, and the angular part of their common eigenstates is fixed by angular momentum alone
   • Because the Coulomb potential happens to have this special property
   • Because spherical harmonics are the only functions defined on a sphere
   • Because the angular kinetic energy is negligible

   Submit
3. 3.

   The radial equation contains no magnetic quantum number m. What does that imply?

   • States with different m under the same (n, l) share the same energy — a (2l+1)-fold degeneracy forced by isotropy
   • m has no effect on the wavefunction
   • m can only be zero
   • The energy is completely independent of angular momentum

   Submit
4. 4.

   Positronium consists of an electron and a positron (equal masses). Using a reduced-mass argument, what is its ground-state binding energy in eV? (Take 13.6 eV for hydrogen.)

   eV20% relative tolerance

   Submit

What comes next

A new face has appeared in the radial equation: $l(l+1)\hbar^2/2\mu r^2$. It plainly comes from the kinetic energy, yet it looks like a potential barrier — and it diverges as $r\to0$. Plot it together with the Coulomb attraction as a single “effective potential” curve, and everything becomes visible at once: where the bound states live, why larger $l$ pushes the electron farther from the nucleus, and even why an s-state electron can “touch” the nucleus.