2.3.12 · D5Modern Physics
Question bank — Hydrogen atom — solving in spherical coordinates
Before we start, a one-line reminder of the cast of symbols, so nothing appears unexplained:
A third picture, Figure 3, sketches the radial function for a couple of states so you can count its nodes — the zero-crossings between and . This is the visual behind the node-counting trap in the Edge cases section.
True or false — justify
The wavefunction being single-valued is what forces to be an integer.
True. Requiring on forces , which only holds for integer — no physics of the force needed. Look at Figure 2: only whole-number windings return to the start.
Quantization of energy is caused by the Coulomb force.
False. Quantization of and comes from boundary conditions and appears for any central potential; only the specific formula uses the shape of Coulomb.
For a state with , the electron has no angular kinetic energy.
True. With the centrifugal term vanishes, so all kinetic energy is radial. In Figure 1 this is the flat-bottom Coulomb curve with no barrier; it is why -states can have nonzero density at .
Two states with the same but different always have different energy in hydrogen.
False. In pure hydrogen they have the same energy — the potential has a hidden symmetry making energy depend only on . (This "accidental" degeneracy breaks in multi-electron atoms.)
The separation constant had to be written as ; any other name would give wrong physics.
False. The constant is just a number; we could call it . Writing it as is a choice of label that makes the Legendre solutions terminate neatly, but the physics is identical.
The number counts the allowed orientations of the angular momentum for a given .
True. runs from to in integer steps, giving values — each a distinct projection of onto the -axis.
Because factorizes, the electron's radial and angular behaviour are physically independent motions.
Partly false. Factorization is a mathematical convenience valid because depends on alone; the radial equation still contains through the centrifugal barrier (Figure 1), so the angular quantum number feeds back into the radial problem.
The energy levels are negative because we chose an odd sign convention.
False. They are negative because bound states sit below the free-electron energy (which is set to zero at ). Negative means "it costs energy to rip the electron away."
The boundary conditions on the radial function are and .
True. must vanish at the origin (else blows up there) and must decay at infinity so the state is normalizable. These two conditions are exactly what force the energy to take discrete values.
Spot the error
"For , the allowed values are ."
Error: is capped at , so for only . The radial (Laguerre) solution fails to terminate — it blows up as — for .
"The centrifugal barrier is potential energy stored in the field."
Error: it is kinetic energy of angular motion, algebraically repackaged into an effective 1D potential (the rising red/green curves in Figure 1). The real field energy is only the Coulomb term .
"We replace by the reduced mass because the electron is heavier than we thought."
Error: accounts for the proton also moving about the common centre of mass. It converts a two-body problem into a one-body one; slightly, it isn't a correction to the electron's own mass.
"Both sides of the separated equation equal a constant because we set them equal by hand."
Error: they equal a constant necessarily: a function of only equals a function of only angles for all values of the variables, which is impossible unless both are the same fixed number.
" ranges over — that's values."
Error: includes negative values too: , giving values, not .
"The Laplacian's angular terms are physics unique to the hydrogen atom."
Error: the spherical-coordinate Laplacian is a pure geometry fact from the chain rule; it is identical for any problem in 3D, hydrogen or not.
"Since means no angular momentum, the solution must depend on and ."
Error: is a constant (spherically symmetric) — no angular dependence at all. That is exactly what "zero angular momentum" looks like.
"The number of radial nodes of a state is just ."
Error: the radial part has exactly nodes (zero-crossings strictly between and ), not . See Figure 3.
Why questions
Why do we switch from to in the radial equation?
Because it removes the first-derivative term and turns the radial equation into the familiar 1D Schrödinger form (the potential drawn in Figure 1), with the clean boundary conditions and — which we already know how to analyse.
Why must return to its own value after ?
Because and label the same physical point in space; the wavefunction must assign it one definite value, so single-valuedness is a physical requirement, not a math nicety. Figure 2 shows the wrap: only integer windings close up.
Why is the degeneracy of level exactly (ignoring spin)?
Summing over to is the sum of the first odd numbers, which equals . Every state shares because energy ignores and in hydrogen.
Why does larger push the electron farther from the nucleus?
A larger makes the centrifugal barrier stronger near the origin, raising there. In Figure 1, compare the curve (dives to ) with the curves (a wall near the origin): the wall repels the wavefunction outward, like a faster-spinning skater resisting being pulled to the centre.
Why did we choose spherical coordinates rather than solving the PDE in ?
Because the potential depends only on ; matching the coordinate system to the symmetry lets the single 3D PDE separate into three independent one-variable ODEs, turning an intractable problem into three solvable ones.
Why is nicknamed s, p, d, f rather than ?
Historical spectroscopy labels (sharp, principal, diffuse, fundamental) stuck; they carry the same information as the integer and connect the solution to periodic-table structure.
Why does the angular momentum magnitude come out as and not ?
Because (the operator for the squared magnitude) has eigenvalue ; taking the square root gives , always at least as large as — a genuinely quantum feature.
Edge cases
What are all the allowed states when ?
Only one: (capped at ) and . This single state is the ground state at eV.
What happens to the energy as ?
: levels crowd together toward zero from below. At exactly zero the electron becomes free (unbound) — this is the ionization threshold.
Is possible, and what does its angular part look like?
Yes; it is the -state whose spherical harmonic is a constant, so the probability cloud is perfectly spherical — no preferred direction.
Can ever exceed in magnitude, say with ?
No. The associated Legendre solutions are finite only when ; beyond that they blow up at the poles , so such a state does not exist.
How many radial nodes does the state have, and the state?
has nodes; has nodes. Same energy, very different radial shapes — see Figure 3.
What breaks the -degeneracy of a hydrogen level in a real experiment?
An external magnetic field (the Zeeman effect) splits states by , and departures from pure (in multi-electron atoms, or relativistic corrections) split by — the "accidental" symmetry is fragile.
What is the smallest possible total angular momentum, and which state has it?
gives : the -states have exactly zero orbital angular momentum, the true minimum.
If the proton were infinitely heavy, how would change?
as ; the reduced mass becomes the plain electron mass, since an immovable proton means only the electron moves.
What are the boundary conditions that actually pick out the discrete energies?
(regular at the origin) and (normalizable). Only special energies let a solution satisfy both at once — that discreteness is energy quantization.
Recall One-line self-test
Cover every answer above. For each, first say which rule (single-valuedness, termination of Legendre/Laguerre, centrifugal barrier, Coulomb-specific energy, boundary conditions ) is doing the work — that is the real test of understanding.