Angular Momentum and Central Potentials

3.1Introduction

Chapter 2 solved the Schrödinger equation in one dimension. Atoms are three-dimensional, and the force binding them is directed toward a center. This chapter develops the machinery that this difference requires.

The resulting simplification is substantial, and is worth stating at the outset. A general three-dimensional problem is a partial differential equation in three variables. A problem with a central potential — one depending only on the distance $r$ from a fixed point — reduces to an ordinary differential equation in $r$ alone, with the angular part solved once and for all, for every central potential there will ever be. Chapter 4 then solves the radial part exactly for the Coulomb potential.

We proceed in two independent ways, which is instructive because each illuminates the other. Sections 3.2–3.4 obtain the entire spectrum of angular momentum from commutators alone, with no differential equations and no wave functions. Sections 3.5–3.6 then find the eigenfunctions explicitly in spherical coordinates. The two routes give the same answer, and the algebraic one gives slightly more — it admits half-integer values that the differential route excludes, and nature uses them for spin.

3.2The Angular Momentum Operators

Classically $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$, with components

Promote $\overrightarrow{r}$ and $\overrightarrow{p}$ to operators obeying the canonical commutation relations of Chapter 1,

The three components of $\widehat{\mathrm{𝑳}}$ do not commute. Let us compute $[{\hat{L}}_x,{\hat{L}}_y]$ in full, since this single calculation is the foundation of the chapter.

Take the four terms in turn. In the second, every operator involved ($\hat{y},{\hat{p}}_z,\hat{x}$) commutes with every other, so the term vanishes; the same is true of the third. The first and fourth survive:

Adding,

By cyclic permutation of $x\to y\to z\to x$,

Because no two components commute, Chapter 1's uncertainty relation forbids any state from having definite values of all three at once. But the total does commute with each.

Everything below uses only Eq. (3.8), so it applies equally to orbital angular momentum $\widehat{\mathrm{𝑳}}$, to spin $\widehat{\mathrm{𝑺}}$ (Sec. 3.7), and to any sum of them.

So ${\hat{J}}^2$ and one component — conventionally ${\hat{J}}_z$ — are compatible observables and possess simultaneous eigenkets. Write them $|j,m⟩$, with

where $\lambda$ and $m$ are dimensionless and to be determined.

3.3Ladder Operators

Define

which are not Hermitian; instead ${\hat{J}}_{+}^{†}={\hat{J}}_{-}$.

Their commutator with ${\hat{J}}_z$ is the engine of the whole construction:

that is

the second following from Theorem 3.2.1.

We also need the products ${\hat{J}}_{\mp }{\hat{J}}_{\pm }$. Expanding,

which rearranges to the identity used repeatedly below:

3.3.1Why ${\hat{J}}_{\pm }$ Steps $m$

Apply ${\hat{J}}_z$ to the state ${\hat{J}}_{\pm }|j,m⟩$ and use Eq. (3.14):

So ${\hat{J}}_{\pm }|j,m⟩$ is an eigenket of ${\hat{J}}_z$ with eigenvalue raised or lowered by one unit of $\hslash$. And because $[{\hat{J}}^2,{\hat{J}}_{\pm }]=0$, it still has the same ${\hat{J}}^2$ eigenvalue $\lambda {\hslash }^2$. The operators move states along a ladder of fixed $\lambda$.

3.3.2The Ladder Must Terminate

The ladder cannot run forever, because ${\hat{J}}_z^2$ cannot exceed ${\hat{J}}^2$. To make that precise, note that for any state

since a norm is non-negative. Evaluating the middle expression with Eq. (3.15),

For fixed $\lambda$ this bounds $m$ both above and below. Let $m_{max}$ be the largest value on the ladder. Since ${\hat{J}}_{+}|j,m_{max}⟩$ would have $m_{max}+1$, which does not exist, it must be the zero vector, and by Eq. (3.19)

Similarly for the smallest value $m_{min}$, with ${\hat{J}}_{-}|j,m_{min}⟩=0$,

Equating Eqs. (3.20) and (3.21) gives $m_{min}=-m_{max}$ (the other root, $m_{min}=m_{max}+1$, contradicts $m_{min}\le m_{max}$).

Finally, ${\hat{J}}_{+}$ climbs from $m_{min}$ to $m_{max}$ in integer steps, so $m_{max}-m_{min}=2m_{max}$ must be a non-negative integer. Writing $j\equiv m_{max}$, we have proved:

3.4Matrix Representations

For a given $j$ the states $|j,m⟩$ span a $(2j+1)$-dimensional space, and the operators become finite matrices. Two cases are worth writing out.

$j=\frac{1}{2}$. With basis $|\frac{1}{2},\frac{1}{2}⟩\equiv |\uparrow ⟩$, $|\frac{1}{2},-\frac{1}{2}⟩\equiv |\downarrow ⟩$, Eq. (3.24) gives ${\hat{J}}_{+}|\downarrow ⟩=\hslash |\uparrow ⟩$ and ${\hat{J}}_{+}|\uparrow ⟩=0$, so

Recovering ${\hat{J}}_x=\frac{1}{2}({\hat{J}}_{+}+{\hat{J}}_{-})$ and ${\hat{J}}_y=\frac{1}{2i}({\hat{J}}_{+}-{\hat{J}}_{-})$ gives $\widehat{\mathrm{𝑱}}=\frac{\hslash }{2}\overrightarrow{\sigma }$ with the Pauli matrices

$j=1$. With basis $|1,1⟩,|1,0⟩,|1,-1⟩$ and Eq. (3.24) giving ${\hat{J}}_{+}|1,0⟩=\hslash \sqrt{2}|1,1⟩$ and ${\hat{J}}_{+}|1,-1⟩=\hslash \sqrt{2}|1,0⟩$,

3.5Orbital Angular Momentum in Spherical Coordinates

For orbital angular momentum we can go further and find the eigenfunctions. Use

Consider ${\hat{L}}_z=-i\hslash (x{\partial }_y-y{\partial }_x)$ and apply the chain rule to $\partial /\partial \phi$ at fixed $r$ and $\theta$:

The right-hand side is exactly ${\hat{L}}_z/(-i\hslash )$, so

a simple and important result: ${\hat{L}}_z$ generates rotations about $z$, and $\phi$ is the angle of such a rotation. The same procedure applied to ${\hat{L}}^2$ (a longer calculation, left to Exercise E2) gives

Neither operator contains $r$. Angular momentum is a statement about angles only, which is exactly why it will separate from the radial problem in Sec. 3.8.

The eigenvalue equations

are solved in Chapter 4, Sec. 4.5, by separating $Y=\mathrm{\Theta }(\theta )\mathrm{\Phi }(\phi )$: the $\phi$ equation gives $\mathrm{\Phi }\propto e^{im\phi }$ with $m$ an integer by single-valuedness, and the $\theta$ equation is the associated Legendre equation, finite at the poles only for integer $l\ge |m|$. The normalized solutions are the spherical harmonics

orthonormal on the sphere,

The lowest few are listed in Eq. (4.31).

Note the agreement between the two routes: the differential equation gives $l=0,1,2,\dots$ with $|m|\le l$, which is Theorem 3.3.1 restricted to integers. The spectroscopic letters are

from the historical names sharp, principal, diffuse, fundamental; they carry no meaning beyond the value of $l$.

3.6Parity and Selection Rules

The parity operator $\hat{\mathrm{\Pi }}$ inverts space through the origin,

Since ${\hat{\mathrm{\Pi }}}^2=\hat{I}$, its eigenvalues are $+1$ (even) and $-1$ (odd) and nothing else. For a central potential $[\hat{H},\hat{\mathrm{\Pi }}]=0$, so energy eigenstates may be taken to have definite parity.

In spherical coordinates $\overrightarrow{r}\to -\overrightarrow{r}$ means $r\to r$, $\theta \to \pi -\theta$, $\phi \to \phi +\pi$. Under these, $\cos \theta \to -\cos \theta$, so $P_l^m(\cos \theta )\to (-1{)}^{l-|m|}{P}_l^m(\cos \theta )$, while $e^{im\phi }\to e^{im\pi }{e}^{im\phi }=(-1{)}^m{e}^{im\phi }$. The product of the two factors is $(-1{)}^{l-|m|}(-1{)}^m=(-1{)}^l$ for either sign of $m$, so

$s$ and $d$ states are even; $p$ and $f$ states are odd. Parity alternates with $l$.

This matters because light acts through the electric dipole operator, proportional to $\widehat{\mathrm{𝒓}}$, which is odd:

Evaluating the angular integral in full sharpens this into the rules quoted in every spectroscopy text:

The $\mathrm{\Delta }l=\pm 1$ rule is Laporte's rule made quantitative: the photon carries one unit of angular momentum, so $l$ must change by exactly one, which also flips the parity. The $\mathrm{\Delta }m$ rule records the polarization: $\mathrm{\Delta }m=0$ for light polarized along $z$ (whose operator is $z=r\cos \theta$, carrying no $\phi$ dependence), and $\mathrm{\Delta }m=\pm 1$ for $x\pm iy=r\sin \theta e^{\pm i\phi }$.

3.7Spin

Electrons carry an intrinsic angular momentum $\widehat{\mathrm{𝑺}}$ obeying Eq. (3.8) with $s=\frac{1}{2}$. It is not associated with motion through space and has no position-space wave function; it is an extra two-dimensional factor in the state space, attached by the tensor product of Chapter 1.

Theorem 3.3.1 with $j=\frac{1}{2}$ gives everything:

with the matrices of Eq. (3.26), $\widehat{\mathrm{𝑺}}=\frac{\hslash }{2}\overrightarrow{\sigma }$.

Two facts carry forward. Spin doubles the number of available states, which is what makes the Pauli principle of Chapter 5 restrictive. And the electric dipole operator does not act on spin, so spin is a spectator in every transition computed in this book — but a spectator that must still be counted.

3.8Central Potentials

A central potential satisfies $V(\overrightarrow{r})=V(r)$. Then

because ${\hat{L}}^2$ and ${\hat{L}}_z$ contain no $r$ (Sec. 3.5) and therefore commute with both $V(r)$ and the radial kinetic energy. So energy eigenstates can be labeled by $l$ and $m$ as well as by energy — the single most useful structural fact in atomic physics.

Because ${\hat{L}}^2$ collects all the angular derivatives, the Laplacian splits as

and the separable form

reduces the three-dimensional problem to one dimension. Chapter 4 carries this out in full, Eqs. (4.12)–(4.20); the result, after the substitution $U=rR$, is the following. Here $\mu$ is the mass of the orbiting particle — written $\mu$ rather than $m$ because in every application it is a reduced mass, the two-body separation being carried out in Sec. 4.2.

Three features are used throughout the rest of the book.

It is a one-dimensional problem. Every technique of Chapter 2 — matching across regions, counting nodes, reading off bound states — applies unchanged on the half-line.

The centrifugal barrier excludes high $l$ from the center. The term ${\hslash }^{2}l(l+1)/2\mu r^2$ is repulsive and diverges as $r\to 0$ for every $l>0$. An $s$ electron ($l=0$) has no barrier and reaches the origin; a $d$ electron is held out. This becomes the explanation of the quantum defect in Chapter 5 and of autoionization widths in Chapter 10.

Bound and continuum states. For $V\to 0$ at large $r$: if $E<0$, $U$ decays exponentially and solutions exist only at discrete energies, normalizable as ${\int }_0^{\mathrm{\infty }}|U{|}^{2}dr=1$. If $E>0$, $U$ oscillates to infinity, solutions exist at every energy, and normalization is replaced by

Equation (3.44) is unfamiliar now and becomes central in Chapter 10; for the moment only the qualitative picture is needed — a countable ladder below threshold, a continuum above it.

3.9Summary

• From $[{\hat{x}}_i,{\hat{p}}_j]=i\hslash {\delta }_{ij}$ follows $[{\hat{L}}_x,{\hat{L}}_y]=i\hslash {\hat{L}}_z$ and cyclic; hence $[{\hat{J}}^2,{\hat{J}}_i]=0$, so ${\hat{J}}^2$ and ${\hat{J}}_z$ have simultaneous eigenkets $|j,m⟩$.

• The ladder operators ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$ satisfy $[{\hat{J}}_z,{\hat{J}}_{\pm }]=\pm \hslash {\hat{J}}_{\pm }$ and step $m$ by one. Requiring the ladder to close at both ends gives ${\hat{J}}^2={\hslash }^{2}j(j+1)$, ${\hat{J}}_z=\hslash m$ with $|m|\le j$ and $2j+1$ values, and ${\hat{J}}_{\pm }|j,m⟩=\hslash \sqrt{j(j+1)-m(m\pm 1)}|j,m\pm 1⟩$.

• Half-integer $j$ is allowed by the algebra and realized by spin, $s=\frac{1}{2}$, with $\widehat{\mathrm{𝑺}}=\frac{\hslash }{2}\overrightarrow{\sigma }$.

• In spherical coordinates ${\hat{L}}_z=-i\hslash {\partial }_{\phi }$ and ${\hat{L}}^2$ is Eq. (3.31); neither contains $r$. The eigenfunctions are the spherical harmonics $Y_l^m$, with integer $l$ forced by single-valuedness in $\phi$ and finiteness at the poles.

• Parity: $\hat{\mathrm{\Pi }}{Y}_l^m=(-1{)}^l{Y}_l^m$. The dipole operator is odd, so one-photon transitions must change parity: $\mathrm{\Delta }l=\pm 1$, $\mathrm{\Delta }m=0,\pm 1$. A forbidden route makes a state metastable — hydrogen's $2s$ outlives its $2p$ by eight orders of magnitude.

• A central potential separates as $\psi =R_{nl}{Y}_l^m$, giving the radial equation (3.43) for $U=rR$ with the centrifugal barrier added to $V$. $E<0$ gives discrete states; $E>0$ a continuum normalized to $\delta (\epsilon -{\epsilon }^{\prime })$.

3.10Exercises

1. The commutators, by hand. Reproduce the calculation of Eqs. (3.3)–(3.6) for $[{\hat{L}}_y,{\hat{L}}_z]$, showing which of the four terms vanish and why. Then prove $[{\hat{L}}^2,{\hat{L}}_x]=0$ by the method of Theorem 3.2.1.

2. ${\hat{L}}^2$ in spherical coordinates. Starting from ${\hat{L}}^2={\hat{L}}_{-}{\hat{L}}_{+}+{\hat{L}}_z^2+\hslash {\hat{L}}_z$ and the spherical form ${\hat{L}}_{\pm }=\hslash e^{\pm i\phi }(\pm {\partial }_{\theta }+i\cot \theta {\partial }_{\phi })$, derive Eq. (3.31).

3. Ladder normalization. Derive Eq. (3.24) from Eqs. (3.18) and (3.19). Then verify it for $j=1$ by applying the matrix of Eq. (3.27) to each basis vector.

4. Parity and a selection rule.

   1. Verify Eq. (3.36) explicitly for $Y_0^0$, $Y_1^0$ and $Y_1^{\pm 1}$ using Eq. (4.31).

   2. Evaluate $\int Y_1^{0\ast }\cos \theta Y_2^{0}d\mathrm{\Omega }$ using $\cos \theta =\sqrt{4\pi /3}{Y}_1^0$ and the fact that the integral of three spherical harmonics is nonzero only when their $l$ values can form a triangle. Is $2p\to 3d$ allowed?

5. Spin matrices. Verify that ${\sigma }_i^2=\hat{I}$ for each $i$, that $\{{\sigma }_x,{\sigma }_y\}\equiv {\sigma }_x{\sigma }_y+{\sigma }_y{\sigma }_x=0$, and hence that ${\hat{S}}^2=\frac{3}{4}{\hslash }^2\hat{I}$. Find the eigenvectors of ${\sigma }_x$ and express them in the $\{|\uparrow ⟩,|\downarrow ⟩\}$ basis.

3.11Project: The Centrifugal Barrier

The problem. A particle moves in the attractive Coulomb potential $V(r)=-1/r$ in atomic units, so that the effective potential of Eq. (3.43) is

1. On paper. For $l>0$, find the position and depth of the minimum of $V_{\mathrm{eff}}$. Show that $r_{min}=l(l+1)$ and that the depth is $-1/2l(l+1)$.

2. On paper. For energy $E<0$ the classically allowed region is $V_{\mathrm{eff}}(r)<E$. Find the two turning points as functions of $E$ and $l$ by solving the resulting quadratic, and show that for $l>0$ the particle is excluded from a region around the origin whose size grows with $l$.

3. On the computer. Plot $V_{\mathrm{eff}}(r)$ for $l=0,1,2,3$ on one set of axes. Mark the energy $E=-1/2n^2$ for $n=3$ as a horizontal line and mark the turning points from (b).

4. On the computer. Discretize Eq. (3.43) on a grid and find the lowest bound-state energy for $l=0,1,2$. Compare with $-1/2n^2$ for the appropriate $n$ — Chapter 4 explains which.

5. On the computer. For each $l$, compute the fraction of the probability lying inside $r<2a_0$. Tabulate it against $l$ and comment.

6. What has to move. One animation: sweep $l$ continuously from $0$ to $3$ and show the barrier rising out of the well near the origin, with the lowest bound state moving with it. Caption it with what to watch.

The check. Your energies in (d) must come out equal to $-1/2n^2$ for integer $n$, and the lowest $l=1$ state must coincide with the second $l=0$ state. That coincidence is the accidental degeneracy of Chapter 4, and reproducing it is the sign your solver is right.

Be ready to answer. An $s$ electron and a $d$ electron in the same atom can have nearly the same energy, yet only one comes close to the nucleus. Which one, which term in $V_{\mathrm{eff}}$ decides it, and why will that matter as soon as the atom has more than one electron?