Angular Momentum

Angular momentum¶

Learning objectives¶

After this chapter, you should be able to:

• use the angular-momentum commutation relations and ladder operators;

• identify allowed $j,m$ values and measurement probabilities;

• distinguish orbital and intrinsic angular momentum;

• connect orbital eigenstates with spherical harmonics;

• add two angular momenta using coupled and uncoupled bases;

• construct the singlet and triplet states of two spin-$\tfrac12$ systems;

• recover the Pauli matrices of Chapter 3 as the spin-$\tfrac12$ case of the general ladder construction; and

• build explicit Clebsch–Gordan coefficients for coupling orbital and spin angular momentum.

9.1 The angular-momentum algebra¶

Chapters 1 through 4 introduced spin operationally, through Stern–Gerlach outcomes, and represented a spin-$\tfrac12$ system’s observables by $\hat{\mathbf S}=\tfrac{\hbar}{2}\boldsymbol\sigma$. That construction was built directly from the two-outcome apparatus; nothing in it referred to orbital motion. This chapter asks a more general question: what mathematical structure makes something “an angular momentum” in quantum mechanics at all? The answer is not a formula like $\mathbf r\times\mathbf p$—that expression reappears in Section 9.3, but only as one particular example—but a set of commutation relations. Anything obeying them, whatever its physical origin, inherits the entire apparatus developed below: discrete eigenvalues, ladder operators, and the role of generating rotations. Spin qualifies; so does orbital motion; so, as Section 9.4 shows, does any sum of angular momenta.

An angular momentum $\hat{\mathbf J}$ is, by definition, any vector operator whose Cartesian components satisfy

and cyclic permutations. This is exactly the same kind of relation already proved directly from the Pauli matrices in Chapter 3, equation (29): $[\hat S_x,\hat S_y]=i\hbar\hat S_z$. What is new here is the direction of the logic. Chapter 3 started from a concrete $2\times2$ matrix representation and derived the commutator. This chapter runs the argument in reverse: it starts from the commutator alone, treated as the defining property of “an angular momentum,” and derives everything else—the allowed eigenvalues, the ladder construction, even the existence of representations other than spin-$\tfrac12$—from that one algebraic fact. Nothing below assumes any particular matrix size in advance.

Before deriving anything from this relation, it is worth checking that it is physically plausible. Take an ordinary object—a book, a phone—and rotate it $90^\circ$ about $x$ and then $90^\circ$ about $y$. Now reset it, and perform the same two rotations in the opposite order. The final orientations differ. Finite rotations about different axes simply do not commute, and equation (1) is the quantum statement of the corresponding infinitesimal fact: the generators of rotations about different axes cannot commute either. Section 9.3 makes this connection exact.

Its squared magnitude,

commutes with every component. This is not an additional assumption; it follows from equation (1) alone. For example,

and substituting $[\hat J_x,\hat J_z]=-i\hbar\hat J_y$ and $[\hat J_y,\hat J_z]=i\hbar\hat J_x$ makes every term cancel in pairs. The same pattern holds for $\hat J_x$ and $\hat J_y$ by the cyclic symmetry of the algebra. Because $\hat J^2$ commutes with each component, we may choose simultaneous eigenstates of $\hat J^2$ and any one component—conventionally $\hat J_z$:

The allowed values are

Section 9.2 derives this restriction rather than simply asserting it. For fixed $j$, the space has dimension $2j+1$. Measuring $J_z$ in an arbitrary state in that space returns one of the $2j+1$ values $\hbar m$.

Concept check 9.1¶

If $J^2=2\hbar^2$, what values can a measurement of $J_z$ return?

Concept check 9.2¶

Does $m=0$ imply zero angular momentum?

9.2 Ladder operators¶

Define

These are not Hermitian—$\hat J_+^\dagger=\hat J_-$—so they do not represent directly measurable quantities. Their role is purely constructive: acting with $\hat J_+$ or $\hat J_-$ on an eigenstate produces another eigenstate with $m$ shifted by exactly one unit, which is what lets us build an entire $j$-multiplet from a single state. The commutators

show that $\hat J_\pm|j,m\rangle$ has $m$ shifted by one: applying $\hat J_z$ to $\hat J_\pm|j,m\rangle$ and using the commutator gives eigenvalue $\hbar(m\pm1)$.

To fix the normalization—and, along the way, to see why the ladder must terminate—compute the squared norm of $\hat J_+|j,m\rangle$ directly. Start by multiplying out the product $\hat J_-\hat J_+$ term by term, exactly as you would expand $(a-ib)(a+ib)$ for ordinary numbers:

The only difference from ordinary numbers is that $\hat J_x\hat J_y$ and $\hat J_y\hat J_x$ do not cancel, since $\hat J_x$ and $\hat J_y$ do not commute—their difference is exactly the commutator $[\hat J_x,\hat J_y]$ that survives in the last term. Substituting $[\hat J_x,\hat J_y]=i\hbar\hat J_z$ from equation (1) and $\hat J_x^2+\hat J_y^2=\hat J^2-\hat J_z^2$ then gives

Therefore

A squared norm can never be negative, so $j(j+1)-m(m+1)\ge0$ for every physically realized $m$. Equality—a state that $\hat J_+$ annihilates—occurs exactly when $m=j$, not before and not after. This is the actual reason the ladder “must terminate”: it is forced by the non-negativity of a norm, the same kind of argument used throughout this book to rule out states that a formula might otherwise seem to allow. The analogous calculation with $\hat J_+\hat J_-=\hat J^2-\hat J_z^2+\hbar\hat J_z$ shows that $\hat J_-$ annihilates a state exactly at $m=-j$. Together these fix both ends of the ladder, and since each application of $\hat J_-$ changes $m$ by exactly one, they force the number of steps from $m=j$ down to $m=-j$—namely $2j$—to be a non-negative integer. That is precisely the statement that $j$ itself must be an integer or a half-integer. Normalizing the resulting states gives

The transverse components follow from

Example 9.1: a $j=1$ measurement¶

Let

The $J_z$ probabilities are $1/4,1/2,1/4$, so $\langle J_z\rangle=0$ and $\langle J_z^2\rangle=\hbar^2/2$. Hence $\Delta J_z=\hbar/\sqrt2$. Every term has the same $j$, so a measurement of $J^2$ gives $2\hbar^2$ with certainty.

Example 9.2: recovering the Pauli matrices¶

For $j=\tfrac12$, the ladder relations should reproduce the Pauli-matrix spin operators of Chapter 3—after all, spin-$\tfrac12$ is nothing but the $j=\tfrac12$ representation of the same algebra. Order the two basis states as $|\tfrac12,\tfrac12\rangle$ then $|\tfrac12,-\tfrac12\rangle$. Equation (11) gives

and, by the same formula with the lower sign, $\hat J_-|\tfrac12,\tfrac12\rangle=\hbar|\tfrac12,-\tfrac12\rangle$ and $\hat J_-|\tfrac12,-\tfrac12\rangle=0$. In matrix form,

Combining these through $\hat J_x=(\hat J_++\hat J_-)/2$, $\hat J_y=(\hat J_+-\hat J_-)/2i$, and $\hat J_z|\tfrac12,\pm\tfrac12\rangle=\pm(\hbar/2)|\tfrac12,\pm\tfrac12\rangle$ gives

which is exactly $\hat{\mathbf J}=(\hbar/2)\boldsymbol\sigma$ with the Pauli matrices defined in Chapter 3. Nothing about spin was assumed beyond $j=\tfrac12$: the entire matrix structure of spin-$\tfrac12$ observables, including the noncommutativity that drove Chapters 2 through 4, is a special case of the abstract ladder construction.

Concept check 9.3¶

A calculation of $\langle j,m|\hat J_-\hat J_+|j,m\rangle$ for some candidate $m$ in the range $-j\le m\le j$ comes out negative. What does this signal?

9.3 Orbital angular momentum¶

For motion in space,

This is the historical origin of the name “angular momentum,” and it is one concrete realization of the abstract algebra of Section 9.1: substituting the canonical commutator $[\hat x,\hat p]=i\hbar$ (Chapter 6) into the components of $\hat{\mathbf r}\times\hat{\mathbf p}$ shows directly that they satisfy equation (1). In spherical coordinates,

and the simultaneous eigenfunctions of $\hat L^2$ and $\hat L_z$ are the spherical harmonics:

For orbital motion, single-valued spatial wavefunctions require integer $\ell=0,1,2,\ldots$. The reason is direct: a wavefunction must return to the same value after $\phi\to\phi+2\pi$, and $e^{im\phi}$ does that only for integer $m$, hence integer $\ell$. The intrinsic spin label $s$ may also be half-integer, precisely because spin is never built from a spatial wavefunction’s dependence on $\phi$ in the first place—it is defined directly by the abstract algebra of Section 9.1, with no analog of “returning to the same value” to enforce. Both $\hat{\mathbf L}$ and $\hat{\mathbf S}$ obey the same operator algebra, but spin is not generated by a particle orbiting around an internal path.

A state with sharp $L$ and $L_z$ but indefinite $L_x,L_y$ is often drawn as a static cone, which makes that indefiniteness look like mere ignorance of a vector sitting still somewhere on it. The classical system that gets the geometry right is not static at all.

Rotation generators¶

A rotation through angle $\alpha$ about $\mathbf n$ is

This is the precise sense in which angular momentum “generates” rotations, foreshadowed by the noncommuting-rotations argument in Section 9.1: expanding the exponential to first order in $\alpha$ gives $\hat R_{\mathbf n}(\alpha)\approx\hat I-i\alpha\,\mathbf n\cdot\hat{\mathbf J}/\hbar$, so $\hat{\mathbf J}$ is literally the infinitesimal generator, and composing two small rotations about different axes reproduces $[\hat J_x,\hat J_y]=i\hbar\hat J_z$ to leading order. For spin $\tfrac12$, $\hat{\mathbf J}=\hbar\boldsymbol\sigma/2$ and this reduces to the rotation operator of Chapter 4. Angular momentum is the generator of rotations for every quantum system.

A striking consequence follows immediately from equation (20). A full $2\pi$ rotation about any axis $\mathbf n$ gives $\hat R_{\mathbf n}(2\pi)=e^{-2\pi i\hat J_z/\hbar}$ once evaluated on an eigenbasis, and acting on $|j,m\rangle$ this is simply the phase $e^{-2\pi im}$. For integer $j$ (hence integer $m$) this equals +1: a full turn returns the state unchanged, matching classical intuition. For half-integer $j$, $m$ is also half-integer and $e^{-2\pi im}=-1$: the state acquires an overall minus sign after one full turn and needs a second full turn to return to itself. This is not an experimental defect. It is a genuine prediction of treating spin as a $j=\tfrac12$ angular momentum, and it has been confirmed directly with neutron interferometry.

Concept check 9.4¶

Orbital angular momentum requires integer $\ell$. Does this mean every total angular momentum encountered in this chapter must be an integer?

9.4 Adding two angular momenta¶

For subsystems with $\hat{\mathbf J}_1$ and $\hat{\mathbf J}_2$, define

Each subsystem might be two different particles’ spins, or one particle’s orbital and spin angular momenta, or the total angular momenta of two larger subsystems already built from more elementary pieces; the algebra does not care which. The uncoupled basis diagonalizes $\hat J_1^2,\hat J_{1z},\hat J_2^2,\hat J_{2z}$ and is written $|j_1,m_1;j_2,m_2\rangle$. It is the natural basis for describing the two subsystems separately. The coupled basis diagonalizes $\hat J_1^2,\hat J_2^2,\hat J^2,\hat J_z$ and is written $|j_1,j_2;j,m\rangle$. It is the natural basis whenever a physical interaction or a measurement cares about the total angular momentum rather than the two parts individually—a spin–orbit coupling being the leading example, taken up in Section 9.6.

The possible totals obey the triangle rule:

In every term of a coupled-state expansion, $m=m_1+m_2$, since $\hat J_z=\hat J_{1z}+\hat J_{2z}$. The coefficients connecting the two bases are Clebsch–Gordan coefficients. They are not arbitrary: each coupled state is built, as in the example below, by starting from the unique state of maximum $m$ and applying $\hat J_-=\hat J_{1-}+\hat J_{2-}$ repeatedly, exactly as one builds any $j$-multiplet from its top state.

A useful check on any such construction is dimension counting: the uncoupled basis has $(2j_1+1)(2j_2+1)$ states, and the coupled basis, summing the multiplicities $(2j+1)$ over the triangle-rule range of $j$, must total the same number. This bookkeeping catches many errors before a single coefficient is computed.

Example 9.3: coupling orbital motion and spin¶

Couple $j_1=\ell=1$ (a $p$ electron’s orbital angular momentum) with $j_2=s=\tfrac12$ (its spin). The triangle rule allows $j=\tfrac32$ and $j=\tfrac12$, and the dimension check confirms $(2\cdot1+1)(2\cdot\tfrac12+1)=6=(2\cdot\tfrac32+1)+(2\cdot\tfrac12+1)=4+2$. Abbreviate the spin states $|\tfrac12,\pm\tfrac12\rangle$ as $|{\uparrow}\rangle,|{\downarrow}\rangle$.

The top state is unique, since only one uncoupled combination reaches the maximum $m=\tfrac32$:

Lowering with $\hat J_-=\hat L_-+\hat S_-$ and equation (11) gives $\hat J_-|\tfrac32,\tfrac32\rangle=\hbar\sqrt3\,|\tfrac32,\tfrac12\rangle$ on the left, while on the right $\hat L_-|1,1\rangle=\hbar\sqrt2\,|1,0\rangle$ and $\hat S_-|{\uparrow}\rangle=\hbar|{\downarrow}\rangle$, so

Dividing both sides by the common factor $\hbar\sqrt3$ isolates the ket on the left and turns the two right-hand coefficients into the Clebsch–Gordan numbers themselves:

giving

Repeating the same lowering step (Exercise 18 asks you to check this instead by lowering from the bottom of the ladder) produces

At $m=\tfrac12$, the uncoupled basis has two states, $|1,0\rangle|{\uparrow}\rangle$ and $|1,1\rangle|{\downarrow}\rangle$; equation (26) uses one combination, and the state orthogonal to it must be the $j=\tfrac12$ state at the same $m$, since $j=\tfrac32$ and $j=\tfrac12$ are the only totals the triangle rule allows. Choosing the conventional phase (the coefficient of the state with the larger $m_1$ taken positive),

Equations (26) and (28) give the Clebsch–Gordan coefficients for $\ell=1$ coupled with $s=\tfrac12$; the same six coefficients reappear throughout atomic fine structure (Section 9.6) and in any $p$-orbital spin–orbit problem.

Concept check 9.5¶

Why does coupling $\ell=1$ with $s=\tfrac12$ produce one quadruplet ($j=\tfrac32$) and one doublet ($j=\tfrac12$), rather than, say, two doublets with a state left over?

9.5 Two spin-one-half systems¶

The highest state is uniquely

Applying the total lowering operator $\hat J_-=\hat S_{1-}+\hat S_{2-}$ gives

A second lowering gives $|1,-1\rangle=|-z,-z\rangle$. These three symmetric states form the $j=1$ triplet. The remaining orthogonal state is

the antisymmetric $j=0$ singlet.

It is worth pausing to recognize these states: equation (31) is exactly the entangled state $|\Psi^-\rangle$ from Chapter 5, and $|1,0\rangle$ above is exactly $|\Psi^+\rangle$. Chapter 5 introduced them as the states with the strongest possible nonclassical correlations between two separated spins, without yet having a name for why nature singles out precisely this pair of combinations. This section supplies that reason: the singlet and the $m=0$ triplet state are not an arbitrary choice of entangled states—they are exactly the states of definite total angular momentum, $j=0$ and $j=1$ respectively, that a rotationally symmetric interaction between two spins naturally produces and conserves. Entanglement and the addition of angular momentum turn out to be two views of the same construction for a two-spin system.

Example 9.4: measuring total spin¶

The product state $|+z,-z\rangle$ is not an eigenstate of $\hat J^2$. Inverting the coupled-basis relations,

A total-spin measurement therefore gives $j=1$ or $j=0$, each with probability $1/2$, while $J_z=0$ is certain.

Concept check 9.6¶

Why is the singlet unchanged by applying the same spatial rotation to both spins?

9.6 Coupling in atoms¶

An electron has orbital angular momentum $\hat{\mathbf L}$ and spin $\hat{\mathbf S}$. Their total is $\hat{\mathbf J}=\hat{\mathbf L}+\hat{\mathbf S}$, with

except that $\ell=0$ permits only $j=1/2$—consistent with the triangle rule of Section 9.4 applied to $j_1=\ell$ and $j_2=\tfrac12$. Example 9.3 constructed the explicit coupled states for $\ell=1$; the same lowering construction works for any $\ell$. A spin–orbit interaction is proportional to $\hat{\mathbf L}\cdot\hat{\mathbf S}$. Since

its energy shifts are simplest in the coupled basis: on a state $|j,m\rangle$ built from fixed $\ell$ and $s=\tfrac12$,

This is a general lesson: choose a basis that diagonalizes the dominant terms of the Hamiltonian and the measurement being asked about.

Example 9.5: fine-structure splitting of a $p$ level¶

Apply equation (35) to $\ell=1$. For $j=\tfrac32$,

while for $j=\tfrac12$,

For a spin–orbit coupling of the usual (positive) sign, the energy shift is proportional to $\langle\hat{\mathbf L}\cdot\hat{\mathbf S}\rangle$, so the $j=\tfrac32$ level is pushed up and the $j=\tfrac12$ level is pushed down: a single unperturbed $p$ level splits into two, with $j=\tfrac32$ higher. This is the same splitting pattern responsible for the sodium D-line doublet, where the $3p$ level’s $j=\tfrac32$ and $j=\tfrac12$ components acquire slightly different energies and hence emit at two closely spaced wavelengths.

Summary¶

• Angular momentum is defined by its commutation algebra, not by any single formula; $\mathbf r\times\mathbf p$ and $\tfrac{\hbar}{2}\boldsymbol\sigma$ are two realizations of the same structure.

• Angular momentum has simultaneous eigenstates of $J^2$ and one component.

• A non-negative-norm argument, not a separate postulate, forces the ladder to terminate and fixes the allowed $j,m$ values.

• Ladder operators generate all $m$ states for a fixed $j$, and for $j=\tfrac12$ they exactly reproduce the Pauli matrices.

• Orbital and spin angular momentum share an algebra but have different physical origins; only orbital motion is restricted to integer values.

• Quantum rotations are generated by angular momentum, and half-integer $j$ states pick up a sign under a full $2\pi$ rotation.

• Coupled and uncoupled bases organize the addition of angular momenta, with Clebsch–Gordan coefficients connecting them.

• Two spin-one-half systems combine into a symmetric triplet and an antisymmetric singlet; orbital and spin angular momentum combine into fine-structure multiplets with a predictable splitting pattern.

Exercises¶

1. Use equation (11) to build the matrix of $\hat J_x$ for $j=1$.

2. Calculate $\langle J_x\rangle$ in the state of Example 9.1.

3. Show that $[\hat J^2,\hat J_z]=0$ from the component commutators.

4. Verify that $Y_\ell^m\propto e^{im\phi}$ follows from the $L_z$ equation.

5. List the allowed total $j$ values for $j_1=1$ and $j_2=3/2$ and verify the dimensions on both sides.

6. Derive all triplet states by lowering from $|1,1\rangle$.

7. Verify directly that the singlet has zero total $J_z$, $J_+$, and $J_-$.

8. Find probabilities for total $j$ and $m$ in $|-z,+z\rangle$.

9. For $\ell=2$ and $s=1/2$, find the eigenvalues of $\hat{\mathbf L}\cdot\hat{\mathbf S}$ for both allowed $j$ values.

10. Explain why measuring $S_{1z}$ on the singlet destroys certainty in a subsequent total-spin measurement.

11. Show that a $2\pi$ rotation acts differently on integer-$j$ and half-integer-$j$ kets.

12. Construct projectors onto the singlet and triplet subspaces using $\hat{\mathbf S}_1\cdot\hat{\mathbf S}_2$.

13. Derive $\hat J_+\hat J_-=\hat J^2-\hat J_z^2+\hbar\hat J_z$ and use it to find the norm of $\hat J_-|j,m\rangle$, confirming it matches the lowering case of equation (11).

14. Using the $2\times2$ matrices from Example 9.2, compute $\hat J_x^2+\hat J_y^2+\hat J_z^2$ directly and verify it equals $\tfrac34\hbar^2\hat I$, consistent with $j=\tfrac12$.

15. Using the $j=1$ matrix for $\hat J_x$ found in Exercise 1, construct $\hat J_y$ and $\hat J_z$ in the same basis and verify $[\hat J_x,\hat J_y]=i\hbar\hat J_z$ explicitly.

16. Verify that the four $j=\tfrac32$ states and two $j=\tfrac12$ states of Example 9.3 are normalized and mutually orthogonal.

17. A $p$ electron is prepared in $|j=\tfrac32,m=\tfrac12\rangle$. Find the probability that a measurement of $L_z$ returns 0.

18. Starting instead from $|j=\tfrac32,-\tfrac32\rangle=|1,-1\rangle|{\downarrow}\rangle$ in Example 9.3 and raising with $\hat J_+$, rederive $|j=\tfrac32,-\tfrac12\rangle$ and confirm it agrees with the result found by lowering from the top.

19. For an $f$ electron ($\ell=3$), list the allowed $j$ values and use equation (35) to find $\langle\hat{\mathbf L}\cdot\hat{\mathbf S}\rangle$ for each.

20. Show that the two values found in Example 9.5 both follow from equation (35) evaluated at $j=\ell\pm\tfrac12$ with $\ell=1$.

21. Explain why the non-negative-norm argument behind Concept check 9.3 constrains $j$ itself, and not merely the range of $m$ available once $j$ is fixed.

22. A $2\pi$ rotation about an arbitrary axis is applied to a $j=\tfrac32$ state. State the result and compare it with the $j=1$ and $j=\tfrac12$ cases.

23. Two electrons occupy a $p$ subshell, so $\ell_1=\ell_2=1$. Use the triangle rule to list the allowed values of their total orbital angular momentum $L$.

24. A beam of atoms with $j=2$ has all five $m$ outcomes equally likely. Find $\langle J_z\rangle$ and $\langle J_z^2\rangle$.