Eigenvalues of L-squared and Lz

Point a diatomic molecule at a microwave source. It absorbs energy only at sharply defined frequencies — never a continuum. That fact sits behind every rotational spectrum ever recorded. It is a direct fingerprint of a deceptively simple mathematical fact. The square of the angular momentum operator, $\hat L^2$ , and one of its components, $\hat L_z$ , can be diagonalized simultaneously. The eigenvalues that fall out are forced — by pure operator algebra — to be $\hbar^2 l(l+1)$ and $\hbar m$ . Here $l$ is a non-negative half-integer-or-integer, and $m$ runs in integer steps from $-l$ to $+l$ . No differential equation is needed to get this. No assumption about the shape of a wavefunction is needed either. No reference to spherical coordinates is needed. It is algebra alone — commutators and ladder operators — that manufactures the quantization. This matters because angular momentum is the organizing principle behind atomic and molecular structure. The quantum numbers $l$ and $m$ that appear here are exactly the ones chemists use. They write them as $s, p, d, f$ orbital labels. They also write them as $m_l = 0, \pm 1, \pm 2,\dots$ magnetic sublevels. Rotational spectroscopy, the Zeeman effect, and the vector model of angular momentum coupling all rely on these numbers. So do term symbols and eventually electron spin. All of them inherit their quantum numbers from this one derivation. The bridge from Ladder Operators and the Angular Momentum Algebra to a working set of quantum numbers is precisely the content here. Given only the commutation relations $[\hat L_x,\hat L_y]=i\hbar \hat L_z$ (cyclically), and the existence of raising/lowering operators $\hat L_\pm = \hat L_x \pm i\hat L_y$ , the entire ladder of allowed eigenvalues gets pinned down. This happens through a norm-cannot-be-negative argument. The payoff is enormous relative to the input. Starting from four lines of commutator algebra, one derives several key results. Angular momentum magnitude-squared eigenvalues are never continuous. For each allowed magnitude, there are exactly $2l+1$ orientations relative to any chosen axis. Half-integer values of $l$ are mathematically permitted by the algebra. Yet orbital angular momentum in three dimensions only realizes the integer branch. Spin fills in the half-integer branch later. This selection between integer and half-integer solutions matters greatly. So does the resulting $2l+1$ degeneracy. Together these show up as the number of orbitals in a given subshell. They also appear as the number of lines in a Zeeman-split spectrum. They further define the multiplicity structure of atomic term symbols. This article develops the derivation from the operator algebra alone. It does not invoke the explicit differential form of $\hat L^2$ or $\hat L_z$ in spherical coordinates. That explicit form and its eigenfunctions belong to Spherical Harmonics as Angular Wavefunctions. What is delivered here is the abstract, representation-independent skeleton. This includes the eigenvalue formulas. It includes the reason $m$ must lie between $-l$ and $+l$ . It includes the reason the ladder must terminate at both ends. It also includes the reason $l$ is quantized in half-integer steps starting from zero. The same skeleton, applied to intrinsic spin, produces the two-state structure. That structure is probed in Spin Angular Momentum and the Stern-Gerlach Experiment.

Key takeaways

$\hat L^2$ and $\hat L_z$ commute, so they share a complete set of eigenstates $|l,m\rangle$ .
Ladder operators $\hat L_\pm$ raise/lower $m$ by exactly $\hbar$ while leaving the $\hat L^2$ eigenvalue unchanged.
Positivity of $\langle\psi|\hat L_\pm^\dagger \hat L_\pm|\psi\rangle$ forces the ladder to terminate, giving $m_{\max}=l$ , $m_{\min}=-l$ .
Termination at both ends is only consistent if $2l$ is a non-negative integer: $l = 0, \tfrac12, 1, \tfrac32, \dots$ .
Eigenvalues: $\hat L^2|l,m\rangle = \hbar^2 l(l+1)|l,m\rangle$ and $\hat L_z|l,m\rangle = \hbar m|l,m\rangle$ , with $m = -l,-l+1,\dots,l$ .
Key idea: the entire discreteness of angular momentum — magnitudes, orientations, and the $2l+1$ degeneracy — is forced purely by the commutation relations and by demanding that no state have negative norm.

Where this lives today

Every rotational spectrum of a gas-phase molecule — the microwave-region absorption lines used to determine bond lengths of diatomics like CO or HCl to sub-picometer precision — is a direct readout of the $\hbar^2 l(l+1)$ eigenvalue spectrum (rewritten with quantum number $J$ instead of $l$ for molecular rotation, giving the rigid-rotor energy $E_J = BJ(J+1)$ with $B=\hbar^2/2I$ ). Radio astronomers detect interstellar CO and other molecules using exactly these transitions between adjacent rotational levels, mapping the cold gas of star-forming regions across the galaxy. The $2l+1$ degeneracy directly explains orbital counting in the periodic table: one $s$ -orbital, three $p$ -orbitals, five $d$ -orbitals, seven $f$ -orbitals, following $2l+1$ for $l=0,1,2,3$ — the entire architecture of electron configurations and the shape of the periodic table traces back to the counting argument in Step 7 of the derivation above. Electron spin resonance (EPR) and nuclear magnetic resonance (NMR) spectroscopy both split a single energy level into $2l+1$ (or, for spin, $2s+1$ ) sublevels under an applied magnetic field along $z$ , with transitions between adjacent $m$ -sublevels obeying the selection rule $\Delta m=\pm1$ inherited directly from the ladder operators' unit-step action — this selection rule is why NMR and EPR spectra show simple, evenly spaced multiplet patterns rather than arbitrary splittings. In materials science, the same eigenvalue structure underlies crystal-field and ligand-field theory: the five-fold degeneracy of $d$ -orbitals in a free ion, and how an octahedral or tetrahedral ligand field partially lifts that degeneracy, is a story that starts from the $l=2$ , $2l+1=5$ counting derived here before any splitting is introduced. In quantum computing and atomic physics, laser cooling and optical trapping schemes engineer transitions between specific $m$ sublevels of atomic hyperfine states, again governed by the same $\Delta m$ ladder selection rules. The abstract algebraic method itself — commutators, raising and lowering operators, norm-positivity arguments — is reused verbatim for the harmonic oscillator's number states, for coupling multiple angular momenta (Clebsch–Gordan theory), and for spin, making this derivation one of the most load-bearing pieces of algebraic machinery in the whole of quantum chemistry.

Ask the tutor: How does adding an external magnetic field turn the $2l+1$ -fold degenerate $m$ states into distinct, measurably split energy levels in the Zeeman effect?

Eigenvalues of L² and Lz

The physical picture

Angular momentum in quantum mechanics is not a vector that simply points somewhere, the way a spinning top's angular momentum does classically. Instead, $\hat L_x$ , $\hat L_y$ , $\hat L_z$ are three operators that fail to commute with one another — measuring one component precisely erases certainty about the other two. What does commute with all three components is the operator $\hat L^2 = \hat L_x^2+\hat L_y^2+\hat L_z^2$ , representing the squared magnitude of the angular momentum vector. This means a quantum state can have a sharp value of the total magnitude $|\vec L|$ and a sharp value of one chosen component, conventionally $\hat L_z$ , simultaneously — but not more than one component at a time. Picture this as a cone. The magnitude of the angular momentum vector is fixed (radius of the cone, related to $\sqrt{l(l+1)}\,\hbar$ ), and the $z$ -component is fixed (height of the cone's apex above the base plane, $m\hbar$ ), but the $x$ - and $y$ -components are maximally uncertain — the vector can be thought of as smeared uniformly around the cone's surface. This picture, the "vector model," is a mnemonic, not a literal trajectory; but it correctly encodes that only $L^2$ and $L_z$ are simultaneously sharp, and that the allowed cones come in a discrete family indexed by $l$ , each with $2l+1$ discrete allowed heights indexed by $m$ . The reason there are exactly $2l+1$ orientations, rather than a continuum of angles, is the ladder mechanism: $\hat L_+$ and $\hat L_-$ step $m$ up and down in units of exactly one $\hbar$ , and the process must terminate — a physical angular momentum component cannot exceed the total magnitude. Requiring the ladder to terminate cleanly on both ends, after an integer number of steps, is what forces $2l$ to be a whole number and pins the top and bottom rungs at $m=\pm l$ .

Figure

The angular momentum cone: fixed |L|, fixed Lz, undetermined Lx, Ly

The other essential picture is the ladder itself: a vertical stack of evenly spaced rungs, each rung a value of $m\hbar$ , with $\hat L_+$ moving a token up one rung and $\hat L_-$ moving it down one rung, for a fixed value of $l$ (the ladder operators never change the "cone," only where on the cone's rim the vector's projection sits).

Figure

The Lz ladder for a fixed l = 2 state: five rungs, m = -2 … +2

The core ideas

The starting point is the algebra alone — no coordinate representation is assumed:

[\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

where: $\hat L_x,\hat L_y,\hat L_z$ are Hermitian operators representing Cartesian components of angular momentum, and $\hbar$ is the reduced Planck constant. Reading: these three relations (cyclic in $x\to y\to z\to x$ ) are the entire definition of "angular momentum" in quantum mechanics — any operator triplet obeying them is an angular momentum, whether it comes from orbital motion, spin, or nuclear isospin. From these, $\hat L^2 \equiv \hat L_x^2+\hat L_y^2+\hat L_z^2$ commutes with every component:

[\hat L^2,\hat L_z]=0

where: this single fact is what licenses simultaneous eigenstates $|l,m\rangle$ of $\hat L^2$ and $\hat L_z$ — the Ladder Operators and the Angular Momentum Algebra machinery is built precisely to exploit this compatibility. Define the non-Hermitian ladder (raising/lowering) operators:

\hat L_\pm \equiv \hat L_x \pm i\hat L_y

where: $\hat L_+$ is the Hermitian adjoint of $\hat L_-$ and vice versa ( $\hat L_+^\dagger=\hat L_-$ ), and neither is individually a physical observable, but their products are useful combinations of observables. Two identities, derivable directly from the commutators above, do all the remaining work:

[\hat L_z,\hat L_\pm]=\pm\hbar \hat L_\pm ,\qquad [\hat L^2,\hat L_\pm]=0

where: the first says $\hat L_\pm$ shifts the $\hat L_z$ eigenvalue by $\pm\hbar$ ; the second says $\hat L_\pm$ leaves the $\hat L^2$ eigenvalue completely unchanged. Reading: $\hat L_\pm$ moves a state strictly along one ladder (fixed magnitude, changing orientation) — it never jumps between ladders. The final identity needed converts $\hat L^2$ into a form built from $\hat L_\pm$ and $\hat L_z$ :

\hat L_\mp \hat L_\pm = \hat L^2 - \hat L_z^2 \mp \hbar \hat L_z

where: this algebraic rearrangement (proved by expanding $\hat L_\pm\hat L_\mp = (\hat L_x\pm i\hat L_y)(\hat L_x\mp i\hat L_y)$ ) is the single most useful identity in the whole derivation — it converts a norm calculation into an eigenvalue calculation. The final results, derived in full below, are:

\hat L^2\,|l,m\rangle = \hbar^2\, l(l+1)\, |l,m\rangle ,\qquad l = 0,\tfrac12,1,\tfrac32,2,\dots

\hat L_z\,|l,m\rangle = \hbar\, m\, |l,m\rangle ,\qquad m = -l,-l+1,\dots,l-1,l

where: $l$ (the angular momentum quantum number) labels the magnitude, and $m$ (the magnetic quantum number) labels the projection onto the chosen $z$ -axis; for each allowed $l$ there are exactly $2l+1$ values of $m$ . Reading: this is the complete quantization law for any angular momentum operator satisfying the commutation algebra — orbital angular momentum realizes only the integer branch ( $l=0,1,2,\dots$ ), while intrinsic spin realizes both branches.

Derivation ASimultaneous eigenstates and the ladder shift rule

Goal. Show that $\hat L^2$ and $\hat L_z$ can be diagonalized together, label their common eigenstates $|l,m\rangle$ with eigenvalues to be pinned down, and show precisely how $\hat L_\pm$ moves between these states.

Step 1 — Compatibility. Because $[\hat L^2,\hat L_z]=0$ , the two operators possess a complete common set of eigenstates. Call such a state $|\lambda,m\rangle$ , defined for now only by

\hat L^2 |\lambda,m\rangle = \hbar^2\lambda\, |\lambda,m\rangle, \qquad \hat L_z |\lambda,m\rangle = \hbar m\, |\lambda,m\rangle

with $\lambda$ dimensionless and $m$ dimensionless (both eigenvalues are written with an explicit $\hbar^2$ or $\hbar$ pulled out because angular momentum has units of $\hbar$ ). At this stage $\lambda$ and $m$ are just unknown real numbers; nothing about their allowed range has been assumed yet. States are taken normalized, $\langle\lambda,m|\lambda,m\rangle=1$ .

Step 2 — Action of $\hat L_\pm$ on $\hat L_z$ -eigenvalue. Using $[\hat L_z,\hat L_\pm]=\hat L_z\hat L_\pm-\hat L_\pm\hat L_z=\pm\hbar\hat L_\pm$ , apply both sides to $|\lambda,m\rangle$ :

\hat L_z\big(\hat L_\pm|\lambda,m\rangle\big) = \hat L_\pm \hat L_z|\lambda,m\rangle \pm \hbar \hat L_\pm|\lambda,m\rangle = (\hbar m \pm \hbar)\big(\hat L_\pm|\lambda,m\rangle\big)

This says: $\hat L_\pm|\lambda,m\rangle$ is itself an eigenstate of $\hat L_z$ , with eigenvalue $\hbar(m\pm1)$ — one rung up or down from where we started. If the state $\hat L_\pm|\lambda,m\rangle$ is nonzero, it must therefore be proportional to $|\lambda,m\pm1\rangle$ .

Step 3 — $\hat L_\pm$ does not touch $\lambda$ . Since $[\hat L^2,\hat L_\pm]=0$ ,

\hat L^2\big(\hat L_\pm|\lambda,m\rangle\big) = \hat L_\pm \hat L^2 |\lambda,m\rangle = \hbar^2\lambda\big(\hat L_\pm|\lambda,m\rangle\big)

so the new state has the same $\hat L^2$ eigenvalue $\hbar^2\lambda$ . Combining Steps 2 and 3:

\hat L_\pm|\lambda,m\rangle = c_\pm(\lambda,m)\,|\lambda,m\pm1\rangle

for some constant $c_\pm$ to be found from normalization — $\hat L_\pm$ climbs the $m$ -ladder at fixed $\lambda$ , exactly the picture in the ladder figure above. This establishes the mechanism; the next derivation uses it to pin down the allowed range of $\lambda$ and $m$ .

Go deeper · Derivation B · 2 minTermination of the ladder and the quantization of l

Derivation BTermination of the ladder and the quantization of l

Goal. Use the requirement that no physical state can have negative norm to force the ladder to terminate at both ends, and show this termination is only self-consistent if $\lambda = l(l+1)$ for $l=0,\tfrac12,1,\tfrac32,\dots$ , with $m$ ranging over $-l$ to $+l$ in integer steps.

Step 1 — A norm that must be non-negative. Consider the norm of the state produced by $\hat L_+$ acting on $|\lambda,m\rangle$ :

\big\|\hat L_+|\lambda,m\rangle\big\|^2 = \langle \lambda,m|\hat L_+^\dagger \hat L_+|\lambda,m\rangle = \langle\lambda,m|\hat L_-\hat L_+|\lambda,m\rangle \ge 0

using $\hat L_+^\dagger=\hat L_-$ . This inequality — a norm in Hilbert space cannot be negative — is the single physical input beyond the pure algebra, and it is what forces quantization.

Step 2 — Evaluate using the operator identity. Insert $\hat L_-\hat L_+ = \hat L^2-\hat L_z^2-\hbar\hat L_z$ (the identity stated above, with the upper sign choice) and apply it to the eigenstate:

\hat L_-\hat L_+|\lambda,m\rangle = \big(\hbar^2\lambda - \hbar^2m^2 - \hbar^2 m\big)|\lambda,m\rangle = \hbar^2\big[\lambda - m(m+1)\big]|\lambda,m\rangle

so the norm condition of Step 1 becomes an ordinary algebraic inequality:

\lambda - m(m+1) \ge 0 \quad\Longrightarrow\quad \lambda \ge m(m+1)

Reading this physically: for fixed total-magnitude label $\lambda$ , the projection $m$ cannot grow without bound — eventually $m(m+1)$ would exceed $\lambda$ and violate the inequality. So raising must stop.

Step 3 — Same argument for $\hat L_-$ . Identically, using $\hat L_+\hat L_- = \hat L^2-\hat L_z^2+\hbar\hat L_z$ ,

\big\|\hat L_-|\lambda,m\rangle\big\|^2 = \hbar^2\big[\lambda - m(m-1)\big] \ge 0 \quad\Longrightarrow\quad \lambda \ge m(m-1)

so $m$ is also bounded below. There must exist a largest value $m_{\max}$ beyond which $\hat L_+$ produces zero (not merely a state of too-large norm, but literally the zero vector — otherwise Step 2's chain would generate an infinite ladder of ever-larger $m$ with $\lambda$ fixed, contradicting the bound). Likewise there must be a smallest value $m_{\min}$ where $\hat L_-$ annihilates the state.

Step 4 — Exact termination condition at the top. Setting $\hat L_+|\lambda,m_{\max}\rangle=0$ exactly (the norm hits zero, not just a bound) gives, from Step 2's equality,

\lambda = m_{\max}(m_{\max}+1)

Step 5 — Exact termination condition at the bottom. Setting $\hat L_-|\lambda,m_{\min}\rangle = 0$ gives, from Step 3's equality,

\lambda = m_{\min}(m_{\min}-1)

Step 6 — Equate the two expressions for $\lambda$ .

m_{\max}(m_{\max}+1) = m_{\min}(m_{\min}-1)

Expand both sides:

m_{\max}^2 + m_{\max} = m_{\min}^2 - m_{\min}

Rearrange into a difference of squares plus linear terms:

m_{\max}^2 - m_{\min}^2 = -m_{\min}-m_{\max}

(m_{\max}-m_{\min})(m_{\max}+m_{\min}) = -(m_{\max}+m_{\min})

This factors into two possibilities: either $m_{\max}+m_{\min}=0$ , or $m_{\max}-m_{\min}=-1$ . The second is impossible because $m_{\max}\ge m_{\min}$ (the top rung cannot lie below the bottom rung) — so the only physical solution is

m_{\min} = -m_{\max}

The ladder is symmetric about zero, exactly as the figure shows for $l=2$ ( $m=-2\ldots+2$ ).

Step 7 — Define $l\equiv m_{\max}$ and count rungs. Since $\hat L_+$ raises $m$ in steps of exactly $1$ (in units of $\hbar$ ), and the ladder runs from $m_{\min}=-l$ to $m_{\max}=l$ , the total number of steps taken is $2l$ , and this must be a non-negative integer (you cannot climb a fractional number of unit rungs and land exactly on the far end):

2l = 0,1,2,3,\dots \quad\Longrightarrow\quad l = 0,\tfrac12,1,\tfrac32,2,\dots

For each such $l$ , the number of $m$ -values from $-l$ to $l$ in integer steps is $2l+1$ — the degeneracy of a given $\hat L^2$ eigenvalue.

Step 8 — Final eigenvalue. Substituting $m_{\max}=l$ back into Step 4,

\lambda = l(l+1)

so relabeling $|\lambda,m\rangle \to |l,m\rangle$ throughout gives exactly the boxed results quoted:

\hat L^2|l,m\rangle=\hbar^2 l(l+1)|l,m\rangle, \qquad \hat L_z|l,m\rangle = \hbar m|l,m\rangle, \qquad m=-l,\dots,l

Nothing about spherical coordinates, differential operators, or associated Legendre functions was used anywhere in this argument — it is pure representation-independent algebra, exactly as promised.

Worked examples

Example 1 (foundational) — reading off $l$ , $m$ , and physical numbers for a $d$ -orbital-type state. Consider a state labeled $l=2$ (a $d$ -type orbital angular momentum) with magnetic quantum number $m=-1$ . Find the eigenvalues of $\hat L^2$ and $\hat L_z$ , the total degeneracy of this $l$ , and the angle the "vector model" cone makes with the $z$ -axis. The $\hat L^2$ eigenvalue is immediate from the formula:

\hat L^2\text{ eigenvalue} = \hbar^2 l(l+1) = \hbar^2(2)(3) = 6\hbar^2

so $|\vec L| = \hbar\sqrt{6} \approx 2.449\,\hbar$ . The $\hat L_z$ eigenvalue for $m=-1$ is simply

\hat L_z\text{ eigenvalue} = \hbar m = -\hbar

The degeneracy (number of allowed $m$ values, i.e. the number of orbitals in this subshell) is

2l+1 = 2(2)+1 = 5

matching the five real $d$ -orbitals ( $d_{z^2}, d_{x^2-y^2}, d_{xy}, d_{xz}, d_{yz}$ ) known from general chemistry. The cone angle $\theta$ between $\vec L$ and the $z$ -axis follows from the vector-model relation $\cos\theta = L_z/|\vec L|$ :

\cos\theta = \frac{m}{\sqrt{l(l+1)}} = \frac{-1}{\sqrt6} = -0.4082

\theta = \arccos(-0.4082) = 114.1^\circ

Limiting-case check: for $m=l$ (the maximum projection, "most aligned" state), $\cos\theta = l/\sqrt{l(l+1)} \to 1$ only as $l\to\infty$ — even the maximally aligned state is never perfectly parallel to the $z$ -axis at finite $l$ , a purely quantum feature (for $l=2$ , $m=2$ : $\cos\theta = 2/\sqrt6=0.8165$ , $\theta=35.3^\circ$ , not $0^\circ$ ). This reflects the uncertainty principle for $L_x, L_y$ : full alignment would require $L_x=L_y=0$ exactly, contradicting $[\hat L_x,\hat L_y]\ne0$ .

Example 2 (applied) — normalizing the ladder step and predicting a rotational-spectroscopy-style transition. Find the normalization constant $c_+$ in $\hat L_+|l,m\rangle = c_+|l,m+1\rangle$ for $l=3$ , $m=1$ (an $f$ -orbital-type state), and use it to state the relative "strength" (norm-squared) of the raising step compared to a step nearer the top of the ladder, $m=2\to m=3$ . From Step 2 of the second derivation, the norm-squared of $\hat L_+|l,m\rangle$ is $\hbar^2[l(l+1)-m(m+1)]$ , so

|c_+(l=3,m=1)|^2 = \hbar^2\big[3(4) - 1(2)\big] = \hbar^2(12-2) = 10\hbar^2

c_+(l=3,m=1) = \hbar\sqrt{10} \approx 3.162\,\hbar

taking the conventional (Condon–Shortley) choice of positive real phase. Compare with the step from $m=2$ to $m=3$ (the very top rung, since $l=3$ ):

|c_+(l=3,m=2)|^2 = \hbar^2\big[3(4)-2(3)\big] = \hbar^2(12-6) = 6\hbar^2

c_+(l=3,m=2) = \hbar\sqrt6 \approx 2.449\,\hbar

So the raising step gets weaker as the ladder approaches its top ( $3.162\hbar$ down to $2.449\hbar$ ), consistent with the physical requirement that the very last step, $m=3\to m=4$ , must vanish entirely:

|c_+(l=3,m=3)|^2 = \hbar^2\big[3(4)-3(4)\big] = 0

confirming the ladder terminates exactly at $m=l=3$ , with no sixth rung — the algebraic mechanism that produces the $2l+1=7$ -fold degeneracy of an $f$ -subshell. In the language of the Zeeman effect, these seven $m$ sublevels (for a one-electron atom in an external field along $z$ ) are what split a single spectral line into up to seven magnetic components, each separated in energy by $\Delta E = g\mu_B B\,\Delta m$ for a field $B$ along $z$ .

Example 3 (exam / competition level) — testing whether a candidate $\lambda$ is consistent, given only one endpoint. A problem states that some angular momentum operator (not necessarily orbital — it could represent a nuclear spin or a nonstandard nuclear multiplet) has a common eigenstate with $\hat L_z$ eigenvalue $m=\tfrac{5}{2}$ , and that this state is annihilated by $\hat L_+$ (i.e. it sits at the top of its ladder). A second measurement on a different member of the same $\hat L^2$ -multiplet gives $\hat L_z$ eigenvalue $-\hbar\left(\tfrac32\right)$ . Determine whether this second data point is consistent with the algebra, and if not, find the actual minimum $m$ value in this multiplet. Since the state is annihilated by $\hat L_+$ at $m=\tfrac52$ , this identifies $m_{\max}=l=\tfrac52$ directly (Step 4 of the second derivation: $\lambda = m_{\max}(m_{\max}+1)$ ):

\lambda = \frac52\left(\frac52+1\right) = \frac52\cdot\frac72 = \frac{35}{4}

By the symmetry result of Step 6 ( $m_{\min}=-m_{\max}$ ), the ladder must terminate at the bottom at

m_{\min} = -l = -\frac52

The proposed second value $m=-\tfrac32$ is not the minimum — it is consistent only as an intermediate rung of the same ladder, since the full allowed set for $l=\tfrac52$ is

m = -\frac52,\,-\frac32,\,-\frac12,\,\frac12,\,\frac32,\,\frac52

(six values total, since $2l+1 = 2(\tfrac52)+1=6$ , an even count — as expected for half-integer $l$ ). So $m=-\tfrac32$ is a legitimate rung (the second-from-bottom one), but it is not the terminal value; the true minimum is $m_{\min}=-\tfrac52$ , corresponding to

\hat L_z\text{ eigenvalue at bottom} = \hbar\left(-\frac52\right) = -2.5\,\hbar

This example also illustrates the half-integer branch of the general solution — a branch that, as emphasized throughout, is realized in nature only by intrinsic spin and never by orbital motion in three-dimensional space (a fact that the algebra alone cannot forbid, but that the explicit spherical-harmonic representation does forbid, since single-valuedness of $Y_l^m(\theta,\phi)$ under $\phi\to\phi+2\pi$ requires $m$ , and hence $l$ , to be an integer for orbital angular momentum).

Limiting cases, scope & common traps

Situation	What happens	Why
$l=0$	Only $m=0$ ; $\hat L^2=0$ , $\hat L_z=0$	No orientation possible — a single point on the "ladder," consistent with $s$ -orbitals having zero orbital angular momentum
$l\to\infty$	$\cos\theta_{\min}=l/\sqrt{l(l+1)}\to1$	Recovers the classical picture of a well-defined vector direction; quantum "cone smearing" becomes negligible
Half-integer $l$	Mathematically allowed by the algebra ( $2l+1$ even)	Realized physically only by spin, never by orbital motion (single-valuedness of $Y_l^m$ forbids it for orbital $\hat L$ )
$m=l$ (top rung)	$\hat L_+\|l,l\rangle=0$ exactly	Defines the termination condition that fixes $\lambda=l(l+1)$
Confusing $\|\vec L\|$ with $L_z^{\max}$	$L_z^{\max}=\hbar l < \|\vec L\|=\hbar\sqrt{l(l+1)}$ always	The vector can never point exactly along $z$ ; $L_x,L_y$ cannot both vanish because $[\hat L_x,\hat L_y]\ne0$
Two components at once	Cannot assign sharp values to, e.g., $L_x$ and $L_z$ simultaneously	$[\hat L_x,\hat L_z]=-i\hbar\hat L_y\ne0$ ; only $\hat L^2$ with one chosen component commute

The chemist behind it

The angular momentum eigenvalue problem, as an abstract operator-algebra argument independent of any particular representation, is usually credited to the joint development of matrix mechanics by Werner Heisenberg, Max Born, and Pascual Jordan (1925), with the specific ladder-operator technique for angular momentum crystallizing soon after in work associated with Wolfgang Pauli and Paul Dirac. Dirac's operator formalism — writing physical observables as abstract Hermitian operators subject only to commutation relations, independent of any wavefunction representation — is exactly the framework in which the derivation above lives natively; Dirac himself used essentially this ladder-operator method to derive the eigenvalues of angular momentum in his 1930 textbook The Principles of Quantum Mechanics, generalizing an approach Pauli had already applied to the hydrogen atom's angular momentum structure.

Origins · 1 minPaul Adrien Maurice Dirac and how this was discovered

Paul Adrien Maurice Dirac

1902–1984 · Cambridge, developed the abstract bra–ket / operator formalism of quantum mechanics and predicted the positron

Dirac trained first as an electrical engineer before turning to mathematics and then physics at Cambridge, and his engineering instinct for finding the most economical, structural description of a system shows clearly in how he treated angular momentum: rather than solving a differential equation, he treated $\hat L^2$ and $\hat L_z$ as abstract operators obeying commutation rules and extracted the eigenvalue spectrum from algebra alone — precisely the ladder-operator argument reproduced here. This representation-independent style became the standard modern treatment of angular momentum in quantum mechanics, later extended without modification to intrinsic spin, where the half-integer branch of the algebra (mathematically present but physically unrealized for orbital motion) turns out to be exactly what nature uses.

History

- 1925 — Heisenberg, Born, and Jordan formulate matrix mechanics, introducing non-commuting observables into physics.

- 1927 — Pauli develops spin matrices and a two-component formalism, exploring the half-integer branch of the angular momentum algebra.

- 1928 — Dirac derives the relativistic wave equation for the electron, from which spin-1/2 emerges naturally.

- 1930 — Dirac's Principles of Quantum Mechanics presents the abstract ladder-operator derivation of angular momentum eigenvalues in the form used throughout modern textbooks.

- 1950s — the ladder-operator technique becomes the standard pedagogical route in quantum chemistry texts (Pauling & Wilson, later McQuarrie), replacing purely differential-equation-based derivations for many audiences.

Why can't we know Lx, Ly, and Lz all at once, even though we can know L² and one component?

Because $\hat L_x$ , $\hat L_y$ , and $\hat L_z$ do not commute with each other — $[\hat L_x,\hat L_y]=i\hbar\hat L_z$ , and cyclic permutations — so no state can be a simultaneous eigenstate of any two Cartesian components unless that eigenvalue is trivially zero for both. In contrast, $\hat L^2$ commutes with every component individually (since $\hat L^2$ is built symmetrically from all three), so choosing any one axis (conventionally $z$ ) and specifying both $\hat L^2$ and $\hat L_z$ gives the maximal information consistent with the uncertainty relations. This is why the "vector model" pictures $\vec L$ as a cone of fixed length and fixed height but undetermined azimuthal angle.

Does the derivation assume the wavefunction is a spherical harmonic?

No — that is precisely the point of the algebraic approach. The derivation uses only the commutation relations $[\hat L_i,\hat L_j]=i\hbar\epsilon_{ijk}\hat L_k$ and the requirement that Hilbert-space norms cannot be negative. It never invokes a specific coordinate representation, a differential equation, or the explicit functional form $Y_l^m(\theta,\phi)$ . The explicit spherical-harmonic wavefunctions, and the extra constraint (single-valuedness under $\phi\to\phi+2\pi$ ) that restricts orbital angular momentum specifically to integer $l$ , belong to the companion treatment of Spherical Harmonics as Angular Wavefunctions.

Why does half-integer l show up in nature at all, if orbital motion forbids it?

Because the ladder-operator algebra by itself never distinguishes "orbital" from any other kind of angular momentum — it only requires the commutation relations to hold. Intrinsic spin angular momentum satisfies exactly the same commutation relations as orbital angular momentum but is not built from position and momentum operators acting on a spatial wavefunction, so the single-valuedness argument that excludes half-integer $l$ for orbital motion simply does not apply to spin. Nature realizes both branches: integer $l$ for orbital motion, and both integer and half-integer values for spin (electrons and protons have spin $s=\tfrac12$ ), as explored further in the treatment of Spin Angular Momentum and the Stern-Gerlach Experiment.

What physically stops the Lz ladder from continuing forever?

The stopping condition is a mathematical consequence of demanding that $\|\hat L_+|l,m\rangle\|^2 = \hbar^2[l(l+1)-m(m+1)]$ never be negative. As $m$ increases toward $l$ , this quantity shrinks toward zero; at $m=l$ it hits exactly zero, meaning $\hat L_+$ annihilates the state rather than producing a state of negative (impossible) norm. Physically, this reflects that the projection of angular momentum along any one axis can never exceed the total magnitude $\hbar\sqrt{l(l+1)}$ — an $m$ value beyond $l$ would require $L_z$ to exceed $|\vec L|$ , which is nonsensical.

How does this connect to the number of orbitals in an electron subshell?

The count $2l+1$ , derived purely from counting integer steps between $m=-l$ and $m=+l$ , is exactly the number of orbitals conventionally associated with each subshell label: one for $s$ ( $l=0$ ), three for $p$ ( $l=1$ ), five for $d$ ( $l=2$ ), seven for $f$ ( $l=3$ ). Each distinct $m$ value corresponds to a distinct spatial orbital with a different orientation (or, in the real combinations used in general chemistry, a different lobe pattern) but the same energy in a spherically symmetric potential, since energy in the hydrogen-like atom depends only on $l$ (and the principal quantum number $n$ ), not on $m$ .

Is Lz special, or could we have quantized Lx instead?

$\hat L_z$ is chosen purely by convention — nothing in the algebra privileges the $z$ -axis. Because $\hat L^2$ commutes with all three Cartesian components equally, one could just as validly build simultaneous eigenstates of $\hat L^2$ and $\hat L_x$ , or $\hat L^2$ and $\hat L_y$ , obtaining an identical spectrum of eigenvalues $\hbar^2 l(l+1)$ and $\hbar m$ . The physical choice of axis matters only when there is an external symmetry-breaking direction, such as an applied magnetic field in the Zeeman effect — in that case, $z$ is conventionally chosen along the field direction because that is the physically preferred axis.

Why is the raising/lowering operator Hermitian conjugate structure (L₊† = L₋) essential to the whole argument?

The entire quantization argument rests on interpreting $\langle\lambda,m|\hat L_-\hat L_+|\lambda,m\rangle$ as a norm, $\|\hat L_+|\lambda,m\rangle\|^2$ , which is only valid if $\hat L_+^\dagger=\hat L_-$ . This adjoint relationship follows directly from $\hat L_x$ and $\hat L_y$ being Hermitian observables: $\hat L_+^\dagger = (\hat L_x+i\hat L_y)^\dagger = \hat L_x^\dagger - i\hat L_y^\dagger = \hat L_x - i\hat L_y = \hat L_-$ . Without this, the key inequality $\lambda \ge m(m+1)$ would have no physical justification, and the entire termination argument — and hence the quantization of $l$ and $m$ — would collapse.