Two operators commute out of three, so a state can carry a total and one component and never more. Everything in this chapter is bookkeeping for that single restriction.

Planck’s constant has the units of angular momentum. That is not a coincidence worth shrugging at — it suggests $\hbar$ may be the fundamental unit of angular momentum, and that angular momentum may be a fundamental observable. Chapter 8 takes the suggestion seriously.

It also introduces something with no classical counterpart at all. Some point-like particles carry an intrinsic angular momentum called spin , which cannot be the orbital motion of constituent parts, because there are no parts. It is simply a property, like charge.

The one idea: a fuzzy vector

The most important property of angular momentum in quantum mechanics is that the outcome of a measurement is at best a fuzzy vector — an object with two defining properties:

a definite magnitude, and
a definite value for just one of its three Cartesian components.

The other two components are uncertain — not merely unknown, but without a value to be known — though still quantized when you choose to measure them. So a quantum angular momentum needs two quantum numbers to specify, never three.

A fuzzy vector: definite length, definite height, indeterminate direction

quantum number l2which cone (m_l)2

Natural units (ħ = 1). Every vector drawn on the cone is equally consistent with the two measured numbers — the fuzziness is not ignorance about which one is real, it is that L_x and L_y have no value to be ignorant of.

Why a cone, and why it never closes

Fix the length of a vector and fix its height. In three dimensions that leaves one degree of freedom: the vector is free to point anywhere on a circle, so its tip sweeps a cone about the $z$ axis.

Every vector on that cone is equally consistent with the two numbers you measured. That is the whole content of “fuzzy”: it is not that the vector really points somewhere and you failed to look, it is that $L_x$ and $L_y$ have no values for you to have missed.

And the cone never closes onto the axis. The largest $L_z$ is $l\hbar$ , but the magnitude is $\sqrt{l(l+1)}\,\hbar$ , which is strictly bigger. The vector is forbidden from lying along $z$ — because if it did, you would know all three components at once ( $L_x = L_y = 0$ ), and ch07 problem 3 says you cannot.

That gap is the single most quotable fact in the chapter, and the widget’s third tab follows it out to the classical limit.

The magnitude, and the ladder

The only possible precise values for the magnitude of orbital angular momentum are

L = \sqrt{l(l+1)}\,\hbar,\qquad\text{where}\quad l = 0, 1, 2, 3, \dots\tag{8.1}

With the magnitude fixed by $l$ , the component in any chosen direction has $2l+1$ possible values. Choosing $z$ :

L_z = m_l\hbar,\qquad\text{where}\quad m_l = +l,\ +(l-1),\ \dots,\ -(l-1),\ -l\tag{8.2}

and once you have done that, $L_x$ and $L_y$ are uncertain — measure either and you get a quantized value somewhere in the range $-l\hbar$ to $+l\hbar$ .

Reproduce it in NumPy

The gap creeps toward ½ħ and never closes; the angle it subtends collapses like 1/√l. Both facts matter, and they point in opposite directions — which is exactly what a correspondence limit looks like.

import numpy as np

# hbar = 1.  Eq. (8.1) magnitude vs Eq. (8.2) largest z component.
print("why the cone never closes:  L = sqrt(l(l+1)).hbar  but  max Lz = l.hbar")
print("=" * 68)
print(f"  {'l':>4} {'L/hbar':>10} {'max Lz/hbar':>12} {'L - Lz':>9} {'min angle':>11}")
for l in [0, 1, 2, 3, 5, 10, 50, 86, 1000]:
    L = np.sqrt(l*(l+1))
    ang = np.degrees(np.arccos(l/L)) if l else float('nan')
    print(f"  {l:>4} {L:>10.4f} {l:>12} {L-l:>9.4f} {ang:>10.2f} deg" if l
          else f"  {l:>4} {L:>10.4f} {l:>12} {L-l:>9.4f} {'--':>14}")

print("\n  L > max Lz for EVERY l >= 1, so the angular momentum vector can never")
print("  lie along z.  If it could, all three components would be known at once")
print("  (Lx = Ly = 0), and chapter 7 problem 3 forbids that.")
print("\n  The gap L - Lz never vanishes, but the ANGLE does:")
for l in [1, 10, 100, 10000, 10**6]:
    print(f"    l = {l:>7}:  min angle from z = {np.degrees(np.arccos(l/np.sqrt(l*(l+1)))):8.4f} deg")
print("  -> at l = 86 (chapter 8 problem 2's classical electron) the cone is")
print("     already within 6 degrees of the axis, which is why a classical")
print("     spinning object looks like it has a definite direction.")

printswhy the cone never closes: L = sqrt(l(l+1)).hbar but max Lz = l.hbar ==================================================================== l L/hbar max Lz/hbar L - Lz min angle 0 0.0000 0 0.0000 -- 1 1.4142 1 0.4142 45.00 deg 2 2.4495 2 0.4495 35.26 deg 3 3.4641 3 0.4641 30.00 deg 5 5.4772 5 0.4772 24.09 deg 10 10.4881 10 0.4881 17.55 deg 50 50.4975 50 0.4975 8.05 deg 86 86.4986 86 0.4986 6.15 deg 1000 1000.4999 1000 0.4999 1.81 deg L > max Lz for EVERY l >= 1, so the angular momentum vector can never lie along z. If it could, all three components would be known at once (Lx = Ly = 0), and chapter 7 problem 3 forbids that. The gap L - Lz never vanishes, but the ANGLE does: l = 1: min angle from z = 45.0000 deg l = 10: min angle from z = 17.5484 deg l = 100: min angle from z = 5.7106 deg l = 10000: min angle from z = 0.5729 deg l = 1000000: min angle from z = 0.0573 deg -> at l = 86 (chapter 8 problem 2's classical electron) the cone is already within 6 degrees of the axis, which is why a classical spinning object looks like it has a definite direction.

Engineer's bridge — two commuting matrices, and the labels they hand you

§7.3 established that commuting Hermitian operators share an eigenbasis, and problem 3 established which angular-momentum operators commute:

[\hat L^2,\hat L_z] = 0\qquad\text{but}\qquad [\hat L_x,\hat L_y] = i\hbar\hat L_z \ne 0

In linear-algebra terms: $\hat L^2$ and $\hat L_z$ are simultaneously diagonalizable, so one basis diagonalizes both and each vector in it carries two eigenvalues — two labels, $l$ and $m_l$ . But $\hat L_x$ and $\hat L_y$ do not commute with $\hat L_z$ , so in that basis they are not diagonal. They are off-diagonal matrices, and an off-diagonal matrix has no definite value on a basis vector; it maps it onto a spread of others.

That is the fuzzy vector, stated as a sparsity pattern. Two labels, because exactly two of the four operators can be diagonal at once. The cone picture and the matrix picture are the same fact.

Where the analogy needs care: $\hat L^2$ is not “the vector $\hat{\mathbf L}$ ” — it is $\hat L_x^2 + \hat L_y^2 + \hat L_z^2$ , a single scalar operator. There is no operator whose eigenvalue is “the direction of $\mathbf L$ ”, which is precisely why the question has no answer.

Spin, and the values integers cannot supply

The quantum numbers $s$ and $m_s$ are used when the angular momentum is due solely to spin. A particle has spin $s$ if the magnitude is $S = \sqrt{s(s+1)}\,\hbar$ and the $z$ components are

S_z = m_s\hbar,\qquad\text{where}\quad m_s = +s,\ +(s-1),\ \dots,\ -(s-1),\ -s\tag{8.3}

Structurally identical to Eqs. (8.1)–(8.2) — same magnitude rule, same ladder, same $2s+1$ values. With one difference that changes everything: $s$ may be a half-integer. The W boson has $s = 1$ with $m_s = +1, 0, -1$ ; the electron has $s = \tfrac12$ with $m_s = \pm\tfrac12$ , and therefore just two states.

Half-integer spin is the one result this book does not establish

Every other quantization in this book was derived: energy levels came from boundary conditions killing all but a discrete set of $k$ , the oscillator ladder came from the algebra of $[q \mp \mathrm d/\mathrm dq]$ , and the fuzzy-vector structure comes from ch07’s commutators. Half-integer spin does not.

The book’s footnote 2 says what would be needed: angular momentum operators generate rotations, and the geometry of rotations in three dimensions forces the commutation relations, which in turn require quantization in integer and half-integer multiples of $\hbar$ .

The honest summary: run the ladder argument on $[\hat L_x,\hat L_y] = i\hbar\hat L_z$ alone — the same argument §6.6 ran for the oscillator, with $\hat L_\pm = \hat L_x \pm i\hat L_y$ in place of the raising and lowering operators — and the ladder is forced to terminate at both ends. That termination allows $2j$ to be any non-negative integer, so $j$ itself may be a half-integer. Orbital angular momentum is then restricted to integers by a separate requirement — that $\psi(\phi + 2\pi) = \psi(\phi)$ , which problem 5 proves.

So: integers for orbital motion because a wave function must be single-valued; half-integers additionally allowed for spin because nothing about a spatial angle constrains it. Take that as the shape of the answer, not as a derivation — this book does not give one.

Adding them: total angular momentum

Orbital and spin angular momenta combine into a total angular momentum , with the same two-number structure:

J = \sqrt{j(j+1)}\,\hbar\qquad\text{and}\qquad J_z = m_j\hbar\tag{8.4}

where in general

j = 0,\ \tfrac12,\ 1,\ \tfrac32,\ 2,\ \dots\qquad\text{and}\qquad m_j = +j,\ +(j-1),\ \dots,\ -j\tag{8.5}

Which values of $j$ actually arise depends on what is being combined. For an orbital $l$ with a spin $s$ :

j = l+s,\ l+s-1,\ \dots,\ |l-s|\tag{8.6}

So $l = 1$ with $s = \tfrac12$ gives $j = \tfrac32$ and $\tfrac12$ ; and $l = 1$ with $s = 1$ gives $j = 2, 1$ and $0$ . More generally, two angular momenta $j_1$ and $j_2$ combine to give $j = j_1+j_2,\ j_1+j_2-1,\ \dots,\ |j_1-j_2|$ .

Eq. (8.6): combining two angular momenta, and checking the states add up

first (l, or j₁)1second (s, or j₂)1/2

j = 3/2, 1/2 (from 3/2 down to 1/2)

uncoupled: (2l+1)(2s+1) = 6 states

coupled: Σ(2j+1) = 6 states

✓ equal — as they must be, because the two labellings describe one set of states

Each row is one value of m_j = m_l + m_s. The stack of squares is how many uncoupled states share it; the colours are the multiplets peeled off from the top, largest j first.

3/2

1/2

-1/2

-3/2

j = 3/2 (4 states)j = 1/2 (2 states)

The staircase is why Eq. (8.6) stops where it does. Peel the widest multiplet off the top of the stack, then the next, and the columns run out exactly when j reaches |l − s|. Try l = 3, s = 1 — problem 1's case: 21 states either way.

Why the series stops exactly at |l − s|

Eq. (8.6) looks like a rule to memorize. It is a consequence of counting.

Line up all the uncoupled states $|m_l, m_s\rangle$ — there are $(2l+1)(2s+1)$ of them — and sort them by their total $m_j = m_l + m_s$ . You get a staircase: one state at the top ( $m_j = l+s$ , achievable only one way), two at the next level down, and so on until the staircase plateaus.

Now peel off multiplets. The single state at the top must belong to $j = l+s$ , which drags along its whole ladder of $2(l+s)+1$ states, one from each level. Remove them, and the new top has $m_j = l+s-1$ , so there is a $j = l+s-1$ multiplet. Keep going, and the columns run out exactly when $j$ reaches $|l-s|$ .

That is the widget’s colour tiling, and it is a proof rather than a decoration: the number of states in each $m_j$ row equals the number of multiplets that reach it, which is what makes the tiling possible at all.

Problem 1's case: l = 3 with s = 1

Uncoupled count. $m_l$ has $2l+1 = 7$ values and $m_s$ has $2s+1 = 3$ , so there are $7 \times 3 = \mathbf{21}$ distinct states labelled by $(m_l, m_s)$ .

Coupled values. Eq. (8.6) gives $j$ from $l+s = 4$ down to $|l-s| = 2$ :

j = 4,\ 3,\ 2

Coupled count. Each $j$ carries $2j+1$ states:

(2\cdot4+1) + (2\cdot3+1) + (2\cdot2+1) = 9 + 7 + 5 = \mathbf{21}

The two counts agree, which is what problem 1’s note asks you to notice. It is not a coincidence and not a check on arithmetic: the uncoupled labels $(m_l, m_s)$ and the coupled labels $(j, m_j)$ describe the same 21 states. Eq. (8.6) is a change of basis, and a change of basis cannot create or destroy dimensions.

Reproduce it in NumPy

Exact rational arithmetic (fractions.Fraction), so the half-integer cases are not at the mercy of floating point. Note the ordering: js() builds from |l−s| upward, so the printed lists read low-to-high while Eq. (8.6) is written high-to-low.

import numpy as np
from fractions import Fraction as F

# Eq. (8.6): j = l+s, l+s-1, ..., |l-s|.  Does the state count survive?
def js(a, b):
    lo, hi = abs(a-b), a+b
    n = int(hi-lo)
    return [lo + k for k in range(n+1)]

def fmt(x):
    f = F(x).limit_denominator(2)
    return str(f.numerator) if f.denominator == 1 else f"{f.numerator}/{f.denominator}"

print("Eq. (8.6) preserves the number of states")
print("=" * 70)
print(f"  {'l':>4} {'s':>4} | {'(2l+1)(2s+1)':>13} | {'j values':<22} {'sum(2j+1)':>10}  match")
ok = True
for l, s in [(1,F(1,2)), (1,1), (3,1), (0,F(1,2)), (2,F(3,2)), (3,3), (F(1,2),F(1,2)), (5,2)]:
    prod = int((2*l+1)*(2*s+1))
    J = js(l, s)
    tot = int(sum(2*j+1 for j in J))
    m = prod == tot
    ok &= m
    print(f"  {fmt(l):>4} {fmt(s):>4} | {prod:>13} | {', '.join(fmt(j) for j in J):<22} {tot:>10}  {'OK' if m else 'FAIL'}")
print(f"\n  all match: {ok}")
print("\n  This is the check problem 1's note asks for, and it is not a")
print("  coincidence: counting (ml, ms) pairs and counting (j, mj) pairs are")
print("  two ways of labelling the SAME set of states.  Eq. (8.6) is a change")
print("  of basis, and a change of basis cannot create or destroy dimensions.")

printsEq. (8.6) preserves the number of states ====================================================================== l s | (2l+1)(2s+1) | j values sum(2j+1) match 1 1/2 | 6 | 1/2, 3/2 6 OK 1 1 | 9 | 0, 1, 2 9 OK 3 1 | 21 | 2, 3, 4 21 OK 0 1/2 | 2 | 1/2 2 OK 2 3/2 | 20 | 1/2, 3/2, 5/2, 7/2 20 OK 3 3 | 49 | 0, 1, 2, 3, 4, 5, 6 49 OK 1/2 1/2 | 4 | 0, 1 4 OK 5 2 | 55 | 3, 4, 5, 6, 7 55 OK all match: True This is the check problem 1's note asks for, and it is not a coincidence: counting (ml, ms) pairs and counting (j, mj) pairs are two ways of labelling the SAME set of states. Eq. (8.6) is a change of basis, and a change of basis cannot create or destroy dimensions.

Spectroscopic notation

The classification of atomic spectra left the subject with a lettering scheme — spectroscopic notation — that this book uses from here to the end, so it is worth learning once:

l⇅	letter⇅	state is called⇅	2l+1 values of $m_l$⇅	where the letter came from⇅
0	s	s-state	1	“sharp” spectral series
1	p	p-state	3	“principal”
2	d	d-state	5	“diffuse”
3	f	f-state	7	“fundamental”
4	g	g-state	9	alphabetical from f onward

The letters are historical accidents from 19th-century spectroscopy, not mnemonics for anything physical — but they are universal, and chapter 9 will name hydrogen’s states 1s, 2p, 3d and so on without further comment.

Three notation collisions to keep straight

$\theta$ and $\phi$ . From §8.3 onward this book uses the physics convention: $\theta$ is the polar angle measured from $z$ , $\phi$ the azimuthal angle in the $xy$ plane. Most engineering and mathematics texts use them the other way round. If you have met spherical coordinates before, this is the one place your prior knowledge will actively mislead you.

Capital $L$ , $S$ , $J$ . In this chapter they are magnitudes — the numbers $\sqrt{l(l+1)}\,\hbar$ and so on. The book’s own footnote 1 warns that chapter 11 will reuse the same capitals as quantum numbers for multi-electron atoms.

$s$ twice over. The spin quantum number is $s$ , and the $l = 0$ spectroscopic letter is also s. An “s-state with $s = \tfrac12$ ” is not a tautology — it is an electron with zero orbital angular momentum.

Engineer's bridge — a spin-½ is a qubit, and this is its state count

An electron’s spin has $s = \tfrac12$ and therefore exactly $2s+1 = 2$ states. Two basis states, complex superpositions allowed, norm preserved under evolution — that is a qubit, and the correspondence is exact rather than an analogy.

The two states $m_s = \pm\tfrac12$ are $|0\rangle$ and $|1\rangle$ . A general spin state $\alpha|{\uparrow}\rangle + \beta|{\downarrow}\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ is a point on the Bloch sphere. And the reason you cannot know $S_x$ , $S_y$ and $S_z$ together is the reason you cannot measure a qubit in three bases at once.

Where it breaks, and it matters: the Bloch sphere is a picture of the state, not of the angular momentum vector. The fuzzy-vector cone above is drawn in real 3-D space; the Bloch sphere is drawn in the 2-D complex state space. A spin-½ pointing “up” sits at the north pole of the Bloch sphere, yet in the cone picture it still leans $54.7°$ off the axis, since $\cos\theta = \tfrac12\big/\tfrac{\sqrt3}{2} = 1/\sqrt3$ . §8.2 builds the Stern–Gerlach widget where both pictures appear side by side.

Check yourself

0 / 6 answered

1.Why can a quantum angular momentum never point along the $z$ axis?
In the widget, switch to the classical-limit tab.
2.The gap $∣ L ∣ - L_{z}^{m a x}$ approaches ½ and never reaches zero, yet the angle shrinks to nothing. How do both facts fit together?
3.An electron ( $s = \frac{1}{2}$ ) has orbital angular momentum $l = 2$ . Which total angular momenta arise, and how many states are there?
4.What does it mean that $\hat{L}^{2}$ and $\hat{L}_{z}$ commute but $\hat{L}_{x}$ and $\hat{L}_{y}$ do not?
5.Which of these is NOT established anywhere in this book?
6.Why do the uncoupled count $(2 l + 1) (2 s + 1)$ and the coupled count $\sum_{j} (2 j + 1)$ always agree?