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Learning Objectives

By the end of this chapter, you should be able to:

Introduction

Chapter 8 solved the Schrödinger equation for several one-dimensional potentials, but every atom, molecule, and nucleus is a three-dimensional object, and the most important potential in atomic physics — the Coulomb attraction between an electron and a nucleus — depends only on the distance rr from a fixed center, not on a single Cartesian coordinate. This chapter extends the Schrödinger equation to three dimensions and develops the machinery — separation of variables in spherical coordinates, the centrifugal barrier in the radial equation, and the angular momentum quantum numbers that emerge from it — needed to solve any central-potential problem, of which the hydrogen atom (Chapter 10) is the most important example. The results obtained here, especially the quantization of angular momentum, apply unchanged to every central potential, not just the Coulomb potential, and reappear throughout atomic, molecular, and nuclear physics.

9.1Three Dimensions and Central Potentials

The Schrödinger Equation in Three Dimensions

Generalizing Chapter 8 to three spatial dimensions, the time-independent Schrödinger equation for a particle of mass mm in a potential V(x,y,z)V(x,y,z) is

22m(2ψx2+2ψy2+2ψz2)+V(x,y,z)ψ=Eψ,-\frac{\hbar^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) + V(x,y,z)\,\psi = E\psi,

or, compactly, 22m2ψ+Vψ=Eψ-\dfrac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi, where 2\nabla^2 is the Laplacian operator. As in one dimension, ψ(x,y,z)2dV|\psi(x,y,z)|^2\,dV gives the probability of finding the particle in the volume element dVdV about (x,y,z)(x,y,z), and ψ\psi must be normalizable: ψ2dV=1\int |\psi|^2\, dV = 1.

When the potential is separable in Cartesian coordinates, V(x,y,z)=V1(x)+V2(y)+V3(z)V(x,y,z) = V_1(x) + V_2(y) + V_3(z), the equation can be solved by seeking product solutions ψ(x,y,z)=X(x)Y(y)Z(z)\psi(x,y,z) = X(x)Y(y)Z(z), which reduces the problem to three independent one-dimensional Schrödinger equations, one per coordinate, each solved exactly as in Chapter 8. For a 3D infinite box of side lengths Lx,Ly,LzL_x, L_y, L_z (a direct generalization of the infinite square well), this gives

Enx,ny,nz=h28m(nx2Lx2+ny2Ly2+nz2Lz2),nx,ny,nz=1,2,3,,E_{n_x,n_y,n_z} = \frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right), \qquad n_x, n_y, n_z = 1, 2, 3, \ldots,

with three independent quantum numbers, one per dimension. A notable feature not seen in one dimension: for a cubic box (Lx=Ly=Lz=LL_x=L_y=L_z=L), distinct combinations of (nx,ny,nz)(n_x,n_y,n_z) — e.g., (2,1,1)(2,1,1), (1,2,1)(1,2,1), (1,1,2)(1,1,2) — can give the same total energy. This is called degeneracy, and it is a recurring feature of higher-dimensional quantum systems, generally traceable to an underlying symmetry of the potential (here, the equivalence of the three directions in a cube).

Worked Example: A Rectangular Quantum Dot

A semiconductor quantum dot confines an electron in a box with Lx=Ly=5.0 nmL_x = L_y = 5.0\ \text{nm} but Lz=10.0 nmL_z = 10.0\ \text{nm} (a “square” cross-section, but elongated along zz) — a shape with less symmetry than a cube, but more than a fully generic box. Using En(eV)=0.376 eVnm2×n2/L2(nm)E_n(\text{eV}) = 0.376\ \text{eV}\cdot\text{nm}^2 \times n^2/L^2(\text{nm}) (a convenient rewriting of Equation (9.2) for an electron), the ground state has

E1,1,1=(0.376 eVnm2)[1(5.0 nm)2+1(5.0 nm)2+1(10.0 nm)2]=(0.376 eVnm2)(0.090 nm2)34 meV,E_{1,1,1} = (0.376\ \text{eV}\cdot\text{nm}^2)\left[\frac{1}{(5.0\ \text{nm})^2}+\frac{1}{(5.0\ \text{nm})^2}+\frac{1}{(10.0\ \text{nm})^2}\right] = (0.376\ \text{eV}\cdot\text{nm}^2)(0.090\ \text{nm}^{-2}) \approx 34\ \text{meV},

a confinement energy of a few tens of millielectronvolts, comparable to kBTk_BT at room temperature (26 meV\approx 26\ \text{meV}) — which is why quantum-dot devices are often cooled to enhance the visibility of their discrete energy levels. Because Lx=LyLzL_x=L_y \ne L_z, the box retains a partial symmetry: states (nx,ny,nz)(n_x,n_y,n_z) and (ny,nx,nz)(n_y,n_x,n_z) are degenerate for any choice of nzn_z (swapping the two equal dimensions changes nothing), but states that instead permute the unequal dimension, such as (1,1,2)(1,1,2) versus (2,1,1)(2,1,1), are generally not degenerate with (1,1,1)(1,1,1) or with each other. A box with Lx=Ly=LzL_x=L_y=L_z would restore full degeneracy among all permutations of a given (nx,ny,nz)(n_x,n_y,n_z) triple, while a fully generic box (LxLyLzL_x \ne L_y \ne L_z) would show no degeneracy at all — the amount of degeneracy is a direct fingerprint of how much geometric symmetry the confining potential actually has.

Central Potentials and Spherical Coordinates

The Coulomb potential, and most potentials of physical interest in atomic and nuclear physics, depend only on the distance from a fixed point: V(x,y,z)=V(r)V(x,y,z) = V(r), where r=x2+y2+z2r = \sqrt{x^2+y^2+z^2}. Such a potential is generally not separable in Cartesian coordinates (the isotropic oscillator discussed later is an important exception), but it is separable in spherical coordinates (r,θ,ϕ)(r,\theta,\phi), precisely because its symmetry matches that coordinate system. Written in spherical coordinates, the Schrödinger equation for a central potential admits solutions of the separable form

ψ(r,θ,ϕ)=R(r)Y(θ,ϕ),\psi(r,\theta,\phi) = R(r)\, Y(\theta,\phi),

where R(r)R(r), the radial wave function, depends on the specific form of V(r)V(r) and carries the information about the particle’s radial probability distribution, while Y(θ,ϕ)Y(\theta,\phi), the angular wave function, turns out to be completely independent of the specific form of V(r)V(r) — it is determined entirely by the requirement that ψ\psi be single-valued and well-behaved on the sphere, and is therefore the same set of functions for the hydrogen atom, a 3D harmonic oscillator, or any other central potential.

Spherical Harmonics: The Lowest Few Explicitly

The angular functions Y(θ,ϕ)Y(\theta,\phi) that solve the angular equation are called spherical harmonics, conventionally written Ym(θ,ϕ)Y_\ell^{m_\ell}(\theta,\phi) and labeled by exactly the two angular quantum numbers introduced below. The lowest few, up to overall normalization constants, are

Y00=constant,Y10cosθ,Y1±1sinθe±iϕ,Y_0^0 = \text{constant}, \qquad Y_1^0 \propto \cos\theta, \qquad Y_1^{\pm1} \propto \sin\theta\, e^{\pm i\phi},

and already illustrate the general pattern. The =0\ell=0 harmonic, Y00Y_0^0, is completely independent of θ\theta and ϕ\phi — an ss-state wave function is spherically symmetric, with a probability density depending only on rr, the same in every direction from the center. The =1\ell=1 harmonics depend on angle: Y10Y_1^0, proportional to cosθ\cos\theta, is largest along the ±z\pm z-axis and vanishes in the xyxy-plane (a pzp_z-type angular distribution, in the language used for atomic orbitals in Chapter 10), while Y1±1Y_1^{\pm1}, proportional to sinθe±iϕ\sin\theta\,e^{\pm i\phi}, is largest in the xyxy-plane and vanishes along the zz-axis. Each higher \ell introduces additional angular structure — more lobes, more angular nodes — but, crucially, the same three functions Y10,Y1+1,Y11Y_1^0, Y_1^{+1}, Y_1^{-1} describe the angular dependence of a pp-state electron in hydrogen, a pp-state neutron in a nuclear shell-model potential, or a pp-state particle in the isotropic oscillator discussed later in this chapter, since (as already emphasized) the angular equation and its solutions never reference V(r)V(r) at all.

The Radial Equation and the Centrifugal Barrier

Substituting ψ=R(r)Y(θ,ϕ)\psi = R(r)Y(\theta,\phi) into the full Schrödinger equation and separating variables reduces the radial part to an ordinary differential equation for R(r)R(r). It is standard, and illuminating, to write this in terms of u(r)rR(r)u(r) \equiv rR(r), in which case the radial equation takes a form directly analogous to a one-dimensional Schrödinger equation for u(r)u(r) on the half-line r>0r>0:

22md2udr2+[V(r)+2(+1)2mr2]u=Eu,-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V(r) + \frac{\hbar^2\,\ell(\ell+1)}{2mr^2}\right] u = Eu,

where \ell is the orbital angular momentum quantum number introduced below. The extra term, 2(+1)/2mr2\hbar^2\ell(\ell+1)/2mr^2, is called the centrifugal barrier: it behaves as an additional repulsive potential that grows without bound as r0r\to 0 (for any >0\ell>0), pushing the particle’s radial probability density away from the origin. This is the quantum-mechanical counterpart of the classical fact that a particle with nonzero angular momentum cannot reach the force center while its angular momentum remains nonzero. For =0\ell=0 (an s-state, in the spectroscopic notation below), the centrifugal barrier vanishes entirely, and for the usual regular central potentials only =0\ell=0 states can have a nonvanishing probability density exactly at r=0r=0 — a fact used directly in Chapter 10 to explain, for instance, which hydrogen atomic states can undergo processes that require the electron to overlap with the nucleus.

Deriving the Radial Equation for u(r) = rR(r)

Separating ψ=R(r)Y(θ,ϕ)\psi=R(r)Y(\theta,\phi) in the full Schrödinger equation leaves R(r)R(r) obeying

22m1r2ddr ⁣(r2dRdr)+[V(r)+2(+1)2mr2]R=ER,-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right) + \left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]R = ER,

with the (+1)\ell(\ell+1) eigenvalue supplied by the angular equation. The first term mixes a first and second derivative of RR, which is exactly what the substitution u(r)rR(r)u(r)\equiv rR(r), i.e. R=u/rR=u/r, is designed to clean up:

ddr ⁣(r2dRdr)=ddr ⁣[r2ddr(ur)]=ddr ⁣(rdudru)=rd2udr2.\frac{d}{dr}\!\left(r^2 \frac{dR}{dr}\right) = \frac{d}{dr}\!\left[r^2\frac{d}{dr}\left(\frac{u}{r}\right)\right] = \frac{d}{dr}\!\left(r\frac{du}{dr}-u\right) = r\frac{d^2u}{dr^2}.

Multiplying the radial equation by rr and substituting this result gives

22md2udr2+[V(r)+2(+1)2mr2]u=Eu,-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V(r)+\frac{\hbar^2\ell(\ell+1)}{2mr^2}\right]u = Eu,

exactly the one-dimensional-looking form quoted in the main text. The simplification comes with a boundary condition: for a usual regular central potential, physical regularity requires R(r)=u(r)/rR(r)=u(r)/r to stay finite as r0r\to0, and hence u(0)=0u(0)=0 — playing the same role here that ψ(0)=0\psi(0)=0 plays at the rigid wall of a one-dimensional infinite well.

9.2Orbital Angular Momentum

The separation above is not a mathematical accident: it reflects the fact that a central potential exerts no torque about the force center (the force is always radial), so orbital angular momentum, L=r×p\vec L = \vec r\times\vec p, is conserved, exactly as in classical central-force motion (e.g., Kepler orbits). Solving the angular equation subject to the single-valuedness of Y(θ,ϕ)Y(\theta,\phi) shows that the magnitude and one Cartesian component (conventionally the zz-component) of L\vec L are simultaneously quantized:

L=(+1),=0,1,2,,L = \sqrt{\ell(\ell+1)}\,\hbar, \qquad \ell = 0, 1, 2, \ldots,
Lz=m,m=,+1,,0,,1,,L_z = m_\ell\hbar, \qquad m_\ell = -\ell, -\ell+1, \ldots, 0, \ldots, \ell-1, \ell,

where \ell is the orbital angular momentum quantum number and mm_\ell is the magnetic quantum number, so named because LzL_z determines how the system’s energy shifts in an external magnetic field (Chapter 11). For a given \ell, there are 2+12\ell+1 allowed values of mm_\ell, corresponding to 2+12\ell+1 distinct orientations of the angular momentum vector relative to the chosen zz-axis — a specific, testable manifestation of space quantization: the orbital angular momentum vector does not merely have a quantized length, it can only point in a discrete set of directions relative to an external axis, rather than any direction whatsoever as classical mechanics would allow.

Two features are worth emphasizing, since both run against classical intuition. First, L=(+1)L = \sqrt{\ell(\ell+1)}\hbar, not \ell\hbar — the “extra” factor means that for any >0\ell>0 the angular momentum vector’s length is slightly larger than its maximum possible zz-component, m,max=m_{\ell,\max}\hbar = \ell\hbar; the vector can never point exactly along the zz-axis. Second, because LxL_x and LyL_y are not simultaneously measurable with LzL_z, only the magnitude LL and a single component LzL_z can be assigned definite values at once — the other two components remain fundamentally indeterminate, consistent with the vector never lying exactly along any single axis.

Historically, states of a given \ell are labeled by spectroscopic letters inherited from early atomic spectroscopy: =0,1,2,3,4,\ell = 0,1,2,3,4,\ldots are denoted s,p,d,f,g,s, p, d, f, g,\ldots respectively — a labeling convention used throughout atomic physics (Chapters 1011) and retained today purely by tradition.

Why mm_\ell Is Called the Magnetic Quantum Number

The name “magnetic quantum number” is not arbitrary bookkeeping. A charged particle with orbital angular momentum L\vec L circulating about a center behaves, classically and quantum mechanically alike, as a tiny current loop with an associated orbital magnetic dipole moment,

μL=e2mL\vec\mu_L = -\frac{e}{2m}\vec L

for an electron of charge e-e, directed opposite to L\vec L because the electron’s charge is negative. Placed in an external magnetic field B\vec B along the zz-axis, this moment contributes an additional term to the electron’s energy, μLBm-\vec\mu_L\cdot\vec B \propto m_\ell, proportional directly to the magnetic quantum number: each of the 2+12\ell+1 otherwise energy-degenerate orientations of L\vec L acquires a distinct energy shift once a magnetic field is applied, splitting a single spectral line into 2+12\ell+1 closely spaced components. This splitting — the (normal) Zeeman effect — was observed spectroscopically well before quantum mechanics existed and is precisely why mm_\ell earned its name: it is the quantum number that controls how atomic energy levels respond to an external magnetic field, a connection developed quantitatively in Chapter 10 and Chapter 11, alongside the further complication (electron spin) required to explain the full richness of observed atomic spectra in a magnetic field.

That a magnetic field sorts a beam into a discrete number of orientations — rather than the continuum a classical dipole would give — is the content of Figure 9.1. Firing spin-12\tfrac12 atoms through a single analyzer splits the beam in two; chaining analyzers at different angles shows that the measurement does not simply read a pre-existing orientation.

Screenshot of the Stern Gerlach simulation

Figure 9.1:The Stern–Gerlach experiment, assembled from analyzers, magnets, and counters. Monte Carlo counts accumulate alongside the analytic quantum prediction, so the 2+12\ell+1 (or, for spin, 2s+12s+1) discrete outcomes can be checked against the statistics.

Interactive simulation: Stern Gerlach

Worked Example: The Vector Model for =2\ell=2

An electron is in a state with orbital angular momentum quantum number =2\ell=2 (a dd-state). Its angular momentum magnitude is

L=(+1)=62.449,L = \sqrt{\ell(\ell+1)}\,\hbar = \sqrt{6}\,\hbar \approx 2.449\,\hbar,

while the maximum possible zz-component is Lz,max=m,max=2L_{z,\max} = m_{\ell,\max}\hbar = 2\hbar. The smallest possible angle between L\vec L and the zz-axis, achieved when mm_\ell takes its largest value, is

cosθ=LzL=22.449=0.816θ35.3°,\cos\theta = \frac{L_z}{L} = \frac{2\hbar}{2.449\,\hbar} = 0.816 \quad \Longrightarrow \quad \theta \approx 35.3°,

confirming explicitly that L\vec L can never point exactly along zz (which would require θ=0°\theta=0°) no matter how large mm_\ell is chosen for a given \ell; the discrepancy \ell\hbar versus (+1)\sqrt{\ell(\ell+1)}\hbar shrinks only in a relative sense as \ell grows large, another instance of the correspondence principle encountered in Chapter 8.

The cone that this worked example describes is usually drawn as a static picture, which makes the indeterminacy of LxL_x and LyL_y look like an admission of ignorance about a vector that is really sitting still somewhere on the cone. The classical system that gets the geometry right is a gyroscope, Figure 9.2: gravity applies a torque perpendicular to the spin angular momentum, and the response is not to tip the axis over but to walk it around a cone at fixed polar angle, with L|\vec L| and LzL_z both constant while LxL_x and LyL_y oscillate. Put a magnetic moment in a field instead of a top in gravity and the equation of motion is the same one; the quantum content is only that the cone’s opening angle may take just 2+12\ell+1 values.

Screenshot of the Precession simulation

Figure 9.2:A spinning top under gravity. In steady precession the torque stays perpendicular to L\vec L, so it changes the direction of the angular momentum without changing either its magnitude or its vertical component — the classical motion behind the vector model’s cone, and behind the Larmor precession of a magnetic moment in a field (Chapter 10).

Interactive simulation: Precession

Angular Momentum Commutation Relations

The impossibility of simultaneously sharp values for more than one Cartesian component of L\vec L is not merely observed; it follows from the algebra of the angular momentum operators themselves, which satisfy the commutation relations

[L^x,L^y]=iL^z,[L^y,L^z]=iL^x,[L^z,L^x]=iL^y[\hat L_x, \hat L_y] = i\hbar \hat L_z, \qquad [\hat L_y,\hat L_z] = i\hbar \hat L_x, \qquad [\hat L_z,\hat L_x] = i\hbar \hat L_y

(a cyclic pattern in xyzxx\to y\to z\to x), together with [L^2,L^z]=0[\hat L^2, \hat L_z] = 0. Two operators that do not commute cannot, in general, have simultaneous eigenstates — precisely the mathematical statement of the uncertainty principle applied to angular momentum, directly analogous to the position–momentum commutator underlying the Heisenberg relation of Chapter 7. Because L^2\hat L^2 commutes with L^z\hat L_z (and likewise with L^x\hat L_x and L^y\hat L_y), a quantum state can be prepared with simultaneously sharp values of LL and one chosen component — exactly the states labeled by \ell and mm_\ell above — but, for >0\ell>0, never with sharp values of LL, LzL_z, and LxL_x all at once.

9.3Quantum Numbers and the Isotropic Oscillator

The Three Quantum Numbers of a Central-Potential Bound State

Solving the full three-dimensional problem for a bound state in a central potential V(r)V(r) produces three quantum numbers, each arising from a separate boundary condition in the separation of variables:

This trio of labels, with the same angular ranges and physical meanings, appears in every central-potential problem in this book — it is a consequence of three-dimensional rotational symmetry, not a special feature of any one potential. Only the Coulomb problem promotes the combination nr++1n_r+\ell+1 to a principal quantum number whose energy is independent of \ell; Chapter 10 specializes the general machinery to that case.

A Second Exactly Solvable Central Potential: The Isotropic Harmonic Oscillator

The Coulomb potential of Chapter 10 is the most important central potential in atomic physics, but it is not the only one that can be solved exactly. The isotropic three-dimensional harmonic oscillator, V(r)=12mω2r2V(r) = \tfrac12 m\omega^2 r^2, is a second example, and it is central (in the technical sense of depending only on rr) even though it is also separable in Cartesian coordinates, since r2=x2+y2+z2r^2 = x^2+y^2+z^2 splits into three independent one-dimensional oscillators. Solved in Cartesian form, using the one-dimensional harmonic-oscillator result of Chapter 8 three times over, the energy levels are

Enx,ny,nz=(nx+ny+nz+32)ω(N+32)ω,Nnx+ny+nz=0,1,2,,E_{n_x,n_y,n_z} = \left(n_x+n_y+n_z+\tfrac32\right)\hbar\omega \equiv \left(N + \tfrac32\right)\hbar\omega, \qquad N \equiv n_x+n_y+n_z = 0,1,2,\ldots,

depending only on the sum NN, not on how it is distributed among nxn_x, nyn_y, nzn_z individually. The isotropic oscillator has continuous rotational symmetry, which guarantees the 2+12\ell+1 degeneracy of each angular-momentum multiplet; its larger degeneracy across different \ell values at fixed NN reflects an additional oscillator symmetry. The same energy levels can equally well be labeled, via the spherical-coordinate separation of this chapter, by a radial quantum number and \ell, with N=2nr+N = 2n_r+\ell; the two labeling schemes describe the same physical states, related by a change of basis, and the total degeneracy of a given level NN works out to (N+1)(N+2)/2(N+1)(N+2)/2.

This may look like a mathematical curiosity, but the isotropic harmonic oscillator is directly useful: it is the starting point for the nuclear shell model (Chapter 13), in which each nucleon moves, to a first approximation, in an average central potential produced by all the other nucleons — a potential that resembles a finite well but is often approximated, for the purpose of a first, analytically tractable calculation, by an isotropic harmonic oscillator. The oscillator’s degenerate energy levels, once a spin-orbit correction (introduced in Chapter 13) is added, reproduce the empirically observed nuclear magic numbers — proton or neutron counts (2, 8, 20, 28, 50, 82, 126) at which nuclei are unusually stable — in much the same way that filled electron shells explain the chemical stability of the noble gases (Chapter 11).

Comparing Degeneracy Across Three Central Potentials

It is worth pausing to compare the degeneracy patterns of the three central (or Cartesian-separable) potentials encountered so far, since the comparison sharpens exactly what “degeneracy reflects underlying symmetry” means in practice. The cubic infinite box has a discrete symmetry (the box is invariant only under swapping xx, yy, zz and reflections, not under arbitrary rotations), and correspondingly modest degeneracies, arising only from specific numerical coincidences among sums of squares of integers. The isotropic harmonic oscillator has full continuous rotational symmetry plus an additional symmetry responsible for its larger degeneracy across different \ell values at fixed N=nx+ny+nzN=n_x+n_y+n_z. The hydrogen atom’s Coulomb potential (Chapter 10) has an additional degeneracy as well: not only is the energy independent of mm_\ell (true for any central potential, since no central potential singles out a preferred direction in space), but for the Coulomb potential specifically, the energy is also independent of \ell for fixed principal nn — an extra, “accidental” degeneracy not explained by rotational symmetry alone, and one of the most distinctive mathematical features of the 1/r1/r potential, examined further in Chapter 10.

9.4Summary

9.5Problems

Solution to Exercise 9.1 #

The box energy in the stated unit is nx2+ny2+nz2n_x^2+n_y^2+n_z^2. The three lowest distinct values are 3 from (1,1,1)(1,1,1) (degeneracy 1), 6 from permutations of (1,1,2)(1,1,2) (degeneracy 3), and 9 from permutations of (1,2,2)(1,2,2) (degeneracy 3). Therefore, the first three energy levels are 3, 6, and 9 times h2/(8mL2)h^2/(8mL^2), with degeneracies 1, 3, and 3.

Solution to Exercise 9.2 #

Using E=0.376 eV nm2[nx2/Lx2+ny2/Ly2+nz2/Lz2]E=0.376\ \text{eV nm}^2[n_x^2/L_x^2+n_y^2/L_y^2+n_z^2/L_z^2],

E=0.376(125+125+4100)eV=0.0451 eV=45.1 meV.E=0.376\left(\frac1{25}+\frac1{25}+\frac4{100}\right)\text{eV}=0.0451\ \text{eV}=45.1\ \text{meV}.

Interchanging xx and yy changes nothing, but this state already has nx=ny=1n_x=n_y=1; permutations involving zz change the energy because LzLxL_z\ne L_x. Therefore, (1,1,2)(1,1,2) has energy 45.1 meV45.1\ \text{meV} and no distinct symmetry partner at that energy.

Solution to Exercise 9.3 #

For =2\ell=2, m=2,1,0,1,2m_\ell=-2,-1,0,1,2. The magnitude is L=2(3)=6L=\sqrt{2(3)}\hbar=\sqrt6\hbar, while the largest component is Lz=2L_z=2\hbar. Since 2<62\hbar<\sqrt6\hbar, the vector retains an unavoidable transverse component, as drawn in Figure 9.3.

Left: five allowed vector orientations of the orbital angular momentum for ell equals 2, drawn from the origin to a circle of radius square root of 6 hbar, at heights corresponding to each allowed m sub ell. Right: comparison of the minimum angle to the z axis for ell equals 2 and ell equals 3, showing the angle shrinking as ell increases.

Figure 9.3:Left: the five allowed orientations of L\vec L for =2\ell=2; even the steepest one, m=2m_\ell=2, sits 35.335.3^\circ off the zz-axis. Right: Problem 9’s comparison — the minimum angle shrinks to 30.030.0^\circ for =3\ell=3, as LL grows faster than its maximum projection \ell\hbar.

Therefore, the allowed mm_\ell values are -2 through +2, and even the largest LzL_z is strictly less than LL.

Solution to Exercise 9.4 #

For a Coulomb bound state with n=3n=3, =0,1,2\ell=0,1,2. The combinations are s:(0,0)s:(0,0); p:(1,1),(1,0),(1,1)p:(1,-1),(1,0),(1,1); and d:(2,2),(2,1),(2,0),(2,1),(2,2)d:(2,-2),(2,-1),(2,0),(2,1),(2,2). Their total is 1+3+5=91+3+5=9. Therefore, there are nine allowed (,m)(\ell,m_\ell) states in the n=3n=3 shell.

Solution to Exercise 9.5 #

Because [Lx,Lz]=iLy[L_x,L_z]= -i\hbar L_y is generally nonzero, exact values of LxL_x and LzL_z cannot be simultaneous observables. A state with definite LL and LzL_z therefore has uncertain transverse components, so its angular-momentum vector cannot lie exactly along the zz axis unless L=0L=0. Therefore, component uncertainty is precisely why orbital angular momentum is represented by a cone rather than a fixed vector direction.

Solution to Exercise 9.6 #

For a Coulomb state with n=4n=4, \ell can be 0,1,2,30,1,2,3, so the maximum is =3\ell=3. The total number is =03(2+1)=1+3+5+7=16=n2\sum_{\ell=0}^{3}(2\ell+1)=1+3+5+7=16=n^2. Therefore, the n=4n=4 shell permits maximum orbital quantum number 3 and contains sixteen (,m)(\ell,m_\ell) states.

Solution to Exercise 9.7 #

When the Schrödinger equation separates, every angular derivative occurs in the angular equation and fixes the spherical harmonics Ym(θ,ϕ)Y_\ell^m(\theta,\phi). The potential V(r)V(r) appears only in the radial equation. Therefore, all central potentials share the same angular functions, while their radial functions and energies differ.

Solution to Exercise 9.8 #

For =0\ell=0, the centrifugal term 2(+1)/(2mr2)\hbar^2\ell(\ell+1)/(2mr^2) is zero, so there is no angular-momentum barrier at r=0r=0. For =1\ell=1, it diverges positively as 1/r21/r^2, suppressing the wave function near the nucleus.

The centrifugal term plotted against radius, flat at zero for ell equals 0 and diverging as one over r squared for ell equals 1 as r approaches zero.

Figure 9.4:For =0\ell=0 the centrifugal term is identically zero, so nothing prevents ψ(0)0\psi(0)\neq0; for =1\ell=1 it diverges as r0r\to0, an infinitely steep repulsive wall that forces ψ(0)=0\psi(0)=0.

Therefore, ss electrons can penetrate to the nucleus whereas pp electrons cannot.

Solution to Exercise 9.9 #

The smallest angle occurs for m==3m_\ell=\ell=3:

cosθ=LzL=33(4)=32,θ=30.0.\cos\theta=\frac{L_z}{L}=\frac{3\hbar}{\sqrt{3(4)}\hbar}=\frac{\sqrt3}{2},\qquad\theta=30.0^\circ.

For =2\ell=2, cosθ=2/6\cos\theta=2/\sqrt6 and θ=35.3\theta=35.3^\circ, both shown together in Figure 9.3. Therefore, an ff state reaches 30.030.0^\circ, closer to the zz axis than a dd state; the cone narrows as \ell increases.

Solution to Exercise 9.10 #

Cycling xyzxx\to y\to z\to x gives [Ly,Lz]=iLx[L_y,L_z]=i\hbar L_x and then [Lz,Lx]=iLy[L_z,L_x]=i\hbar L_y, consistent with the stated relation. In contrast, [L2,Li]=0[L^2,L_i]=0 for each component i=x,y,zi=x,y,z. Therefore, commuting of L2L^2 with one Cartesian component guarantees that LL and one chosen component can be simultaneously sharp.

Solution to Exercise 9.11 #

The nonnegative triples summing to 2 are (2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1). There are six, and the formula gives (2+1)(2+2)/2=6(2+1)(2+2)/2=6. Figure 9.5 carries this same degeneracy count into Problem 13’s nuclear-shell application. Therefore, the N=2N=2 oscillator shell has exactly six spatial states.

Solution to Exercise 9.12 #

The cubic box has only discrete rotational/permutation symmetry, and its energy depends on the separate squares nx2+ny2+nz2n_x^2+n_y^2+n_z^2. The isotropic oscillator has continuous rotational symmetry and energy depending only on the sum N=nx+ny+nzN=n_x+n_y+n_z, so many more partitions share one energy. Therefore, the oscillator’s larger symmetry produces its larger degeneracy.

Solution to Exercise 9.13 #

The spatial degeneracies are 1, 3, and 6 for N=0,1,2N=0,1,2. Including spin gives shell capacities 2, 6, and 12, so cumulative capacities are 2, 8, and 20.

Three oscillator shells N equals 0, 1, and 2 with their spatial degeneracies 1, 3, and 6, spin capacities 2, 6, and 12, and cumulative totals 2, 8, and 20 marked beside each level.

Figure 9.5:Filling the oscillator shells in order, the running total after each shell — 2, 8, 20 — lands exactly on the first three nuclear magic numbers.

Therefore, the simple oscillator model reproduces the first three nuclear magic numbers 2, 8, and 20 before spin--orbit corrections are needed.

Solution to Exercise 9.14 #

For the Coulomb n=2n=2 shell, the states are (0,0)(0,0) and (1,1),(1,0),(1,1)(1,-1),(1,0),(1,1). The maximum LzL_z is therefore obtained for =1\ell=1, m=1m_\ell=1. Then L=2L=\sqrt2\hbar and cosθ=/(2)=1/2\cos\theta=\hbar/(\sqrt2\hbar)=1/\sqrt2, so θ=45\theta=45^\circ. Therefore, the maximum-projection state has (,m)=(1,1)(\ell,m_\ell)=(1,1), L=2L=\sqrt2\hbar, and a 4545^\circ angle to zz.

Solution to Exercise 9.15 #

Because Y102cos2θ|Y_1^0|^2\propto\cos^2\theta, it vanishes where cosθ=0\cos\theta=0, namely at θ=90\theta=90^\circ. This nodal xyxy plane leaves two lobes on the positive and negative zz axes, hence the name pzp_z.

Polar plots comparing the angular probability density of an s-state, a filled circle uniform in every direction, with a p_z state, a dumbbell shape with two lobes along the vertical axis and a node at the equator.

Figure 9.6:The ss-state angular density is a circle — the same in every direction. The pzp_z-state density is a dumbbell with a node at θ=90\theta=90^\circ, concentrating probability along ±z\pm z.

Therefore, pzp_z is directional with an equatorial node, unlike an ss state whose angular probability is uniform in every direction.

Solution to Exercise 9.16 #

For =1\ell=1, m=1,0,+1m_\ell=-1,0,+1, so a shift proportional to mm_\ell produces three distinct levels. For =0\ell=0, only m=0m_\ell=0 exists and the orbital shift is zero.

Left: a single ell equals 1 level splitting into three levels labeled m sub ell equals minus 1, 0, and 1 when a magnetic field is turned on. Right: a single ell equals 0 level staying single with no splitting when the same field is turned on.

Figure 9.7:Turning on the field splits the =1\ell=1 level into three, one for each mm_\ell; the =0\ell=0 level has only m=0m_\ell=0 to begin with, so there is nothing to split.

Therefore, a pp level splits into three orbital Zeeman components while an ss level does not split orbitally.

Solution to Exercise 9.17 #

Substitution gives

E=(nx+ny+nz+32)ω=(0+0+0+32)ω=32ω.E=(n_x+n_y+n_z+\tfrac32)\hbar\omega=(0+0+0+\tfrac32)\hbar\omega=\tfrac32\hbar\omega.

Each independent one-dimensional coordinate contributes 12ω\tfrac12\hbar\omega. Therefore, the three-dimensional ground state has 3(12ω)=32ω3(\tfrac12\hbar\omega)=\tfrac32\hbar\omega.

Solution to Exercise 9.18 #

For a general central potential, the radial equation can yield energies that depend separately on the radial and angular quantum numbers, E=E(nr,)E=E(n_r,\ell); then the extra degeneracy among different \ell values found in a Coulomb shell is absent. Coulomb’s 1/r1/r potential is special in giving energy depending only on n=nr++1n=n_r+\ell+1. The isotropic oscillator instead depends on N=2nr+N=2n_r+\ell, so it has degeneracy among different (nr,)(n_r,\ell) combinations but not the Coulomb all-\ell degeneracy at fixed nn. Therefore, its degeneracy is special but not the same Coulomb accidental degeneracy.

Solution to Exercise 9.19 #

For an ss electron, =0\ell=0 and hence L=(+1)=0L=\sqrt{\ell(\ell+1)}\hbar=0. The orbital magnetic moment is then μL=(e/2m)L=0\vec\mu_L=-(e/2m)\vec L=0, so there is no orbital force μzBz/z\mu_z\,\partial B_z/\partial z to split the beam. Therefore, silver’s Stern--Gerlach splitting requires electron spin, not orbital angular momentum.