---
title: "Quantum Angular Momentum"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 10
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/10-quantum-angular-momentum
---

# Chapter 10 — Quantum Angular Momentum

Point a radio telescope at a dark patch of sky and tune it to $115\,\mathrm{GHz}$: a bright, needle-sharp line appears — carbon monoxide molecules, ten kelvin above absolute zero, stepping down one rung of a ladder of *rotational* states. Their tumbling, like everything that turns in quantum mechanics, is quantised twice over: the magnitude of the angular momentum can only be $\sqrt{l(l+1)}\,\hbar$, and its component along any chosen axis only $m\hbar$. This chapter derives that double quantisation — once algebraically, from the [commutators](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-commutator) inherited from the [Poisson brackets](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#def-b3-hamiltonian-mechanics-poisson) of [Chapter 2](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#ch-b3-hamiltonian-mechanics), and once concretely, as the [spherical harmonics](#prop-b3-quantum-angular-momentum-harmonics) that shape every atomic orbital. The algebra will hand us more than we ask: it permits half-integer values that no orbital motion can realise, a vacancy nature fills two chapters from now with spin. On the way we meet the molecules’ rotational ladders — the millimetre-wave lines by which astronomers weigh the galaxies’ cold gas — and the magnetic moments by which angular momentum first showed itself split.

## 10.1 The algebra of rotation

**Definition 10.1 (Angular momentum operators).**

Orbital angular momentum is the operator $\hat{\vect L} =
\hat{\vect r}\wedge\hat{\vect p}$, componentwise $\hat L_z = \hat
x\hat p_y - \hat y\hat p_x$ and cyclic. From $[\hat x, \hat p_x] =
\iu\hbar$:

$$
[\hat L_x, \hat L_y] = \iu\hbar\,\hat L_z
\quad\text{(and cyclic)} , \qquad
[\hat L^2, \hat L_z] = 0 :
$$

the components are mutually incompatible — no state has two of them sharp — but the total square is compatible with any one of them: the pair $(\hat L^2, \hat L_z)$ is the standard choice of labels. These [commutators](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-commutator) are exactly $\iu\hbar$ times the [Poisson brackets](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#def-b3-hamiltonian-mechanics-poisson) computed in [Example 2.12](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#ex-b3-hamiltonian-mechanics-brackets): Dirac’s dictionary at work.

**Theorem 10.2 (The spectrum, from the algebra alone).**

Let $\hat J_x, \hat J_y, \hat J_z$ be *any* three [Hermitian operators](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) obeying the commutation relations above. Then the joint [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) of $(\hat J^2, \hat J_z)$ are

$$
\hat J^2:\ j(j+1)\hbar^2 , \qquad
\hat J_z:\ m\hbar , \quad m = -j, -j+1, \dots, +j ,
$$

where $j$ is a non-negative *integer or half-integer*: $2j + 1$ values of $m$ for each $j$. The [ladder operators](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#def-b3-harmonic-oscillator-ladder) $\hat J_\pm = \hat
J_x \pm \iu\hat J_y$ step $m$ by $\pm1$ at fixed $j$:

$$
\hat J_\pm\ket{j,m} = \hbar\sqrt{j(j+1) - m(m\pm1)}\,\ket{j,m\pm1} .
$$

**Partial proof.** $[\hat J_z, \hat J_\pm] = \pm\hbar\hat J_\pm$: acting with $\hat
J_\pm$ shifts the $\hat J_z$ [eigenvalue](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) by $\pm\hbar$, at fixed $\hat J^2$ (which commutes with everything built from the $\hat
J_i$). The norm $\|\hat J_\pm\ket{j,m}\|^2 = \hbar^2[j(j+1) -
m(m\pm1)]$ (computed from $\hat J_\mp\hat J_\pm = \hat J^2 - \hat
J_z^2 \mp \hbar\hat J_z$) must stay non-negative: the ladder must terminate above at some $m_{\max}$ with $m_{\max}(m_{\max}+1) =
j(j+1)$, i.e. $m_{\max} = j$, and below at $m_{\min} = -j$. Climbing from $-j$ to $+j$ in unit steps forces $2j$ to be a non-negative integer. The [eigenvalue](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) notation $j(j+1)\hbar^2$ is thereby justified after the fact. ∎

**Remark 10.3 (A vacancy in the catalogue).**

The algebra allows $j = \tfrac12, \tfrac32, \dots$ — ladders with an even number of rungs. Orbital motion, we show next, uses only integers. Nature, however, wastes nothing: the electron itself carries $j = \tfrac12$, with no orbit behind it ([Chapter 12](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ch-b3-spin-two-level)); the algebra derived here, unchanged, will run atomic magnetism, nuclear spins and the qubit.

## 10.2 Orbital angular momentum: spherical harmonics

**Proposition 10.4 (Spherical harmonics).**

In spherical coordinates $\hat L_z = -\iu\hbar\,\partial_\varphi$, and the joint eigenfunctions of $(\hat L^2, \hat L_z)$ on the sphere are the *[spherical harmonics](#prop-b3-quantum-angular-momentum-harmonics)* $Y_l^m(\theta, \varphi)$:

$$
\hat L^2\,Y_l^m = l(l+1)\hbar^2\,Y_l^m , \qquad
\hat L_z\,Y_l^m = m\hbar\,Y_l^m ,
$$

with $l = 0, 1, 2, \dots$ *integer* — single-valuedness of $\eu^{\iu m\varphi}$ forces integer $m$, hence integer $l$. The first few, up to normalisation: $Y_0^0 = \text{const}$ (the $s$ shape, a sphere); $Y_1^0 \propto \cos\theta$ and $Y_1^{\pm1}
\propto \sin\theta\,\eu^{\pm\iu\varphi}$ (the $p$ shapes, two lobes); $Y_2^0 \propto 3\cos^2\theta - 1$ and its partners (the $d$ family). Parity: $Y_l^m(-\vect r$ direction$) = (-1)^l\,Y_l^m$.

**Proof.** *Admitted at this level.* ∎

![Polar diagrams of |Y_lm( )| (section in a plane containing the z axis; the full shape is the figure of revolution). These angular skeletons, independent of any potential, will dress every atom in .](https://one-course.com/images/onecourse/chapters/physics-5/b3-quantum-angular-momentum/fig-c23cfe51657f.svg)

*Polar diagrams of $|Y_l^m(\theta)|$ (section in a plane containing the $z$ axis; the full shape is the figure of revolution). These angular skeletons, independent of any potential, will dress every atom in [Chapter 11](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#ch-b3-hydrogen-atom).*

**Example 10.5 (The rigid rotor and its ladder).**

A diatomic molecule tumbling with moment of inertia $I$ has $\hat H
= \hat L^2/2I$: energies

$$
E_J = \frac{\hbar^2}{2I}\,J(J+1) = B\,J(J+1) ,
\qquad J = 0, 1, 2, \dots
$$

each $(2J+1)$-fold degenerate. Photon absorption obeys $\Delta J =
\pm1$, so the absorption frequencies are $\nu_{J+1\leftarrow J} =
2B(J+1)/h$: a comb of *equally spaced* lines — the signature by which a rotational spectrum is recognised at a glance. For carbon monoxide, $B/h = 57.6\,\mathrm{GHz}$: the fundamental line falls at $115\,\mathrm{GHz}$ ($\lambda = 2.6\,\mathrm{mm}$), the workhorse line of millimetre radio astronomy ([Problem 10.1](#pb-b3-quantum-angular-momentum-1)).

## 10.3 Central potentials, completed

**Theorem 10.6 (Separation in any central potential).**

For $V(r)$, the stationary states can be taken as

$$
\psi(\vect r) = \frac{u(r)}{r}\,Y_l^m(\theta, \varphi) ,
$$

where $u$ solves the one-dimensional [radial equation](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-swave)

$$
-\frac{\hbar^2}{2m}\,u'' + \Big[V(r) +
\frac{\hbar^2\,l(l+1)}{2mr^2}\Big]u = E\,u , \qquad u(0) = 0 .
$$

The angular problem is solved once and for all by the $Y_l^m$; each $l$ adds the repulsive *[centrifugal barrier](#thm-b3-quantum-angular-momentum-radial)* $\hbar^2l(l+1)/2mr^2$ — the quantum version of the effective potential of [Example 1.12](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#ex-b3-lagrangian-mechanics-central) — and every level of given $l$ is $(2l+1)$-fold degenerate, because no central force can care about the orientation of the $z$ axis. The $s$-wave trick of [Proposition 7.8](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-swave) was the $l = 0$ row of this theorem.

**Partial proof.** The Laplacian splits as $\Delta = \tfrac1r\partial_r^2(r\,\cdot) -
\hat L^2/\hbar^2r^2$ (admitted; it is the statement that $\hat L^2$ is the angular part of $-\hbar^2\Delta$). Insert $\psi = (u/r)Y_l^m$ and use the [eigenvalue](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) of $\hat L^2$: the stated equation. The degeneracy follows because $\hat H$ commutes with all three $\hat
L_i$, whose [ladder operators](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#def-b3-harmonic-oscillator-ladder) move $m$ without changing the energy. ∎

![Effective radial potentials of the Coulomb problem: the centrifugal barrier 2l(l+1)/2mr2 walls off the origin for l 1. Only s states touch the nucleus — with consequences from atomic spectra to radioactive electron capture.](https://one-course.com/images/onecourse/chapters/physics-5/b3-quantum-angular-momentum/fig-0987f533f451.svg)

*Effective radial potentials of the Coulomb problem: the [centrifugal barrier](#thm-b3-quantum-angular-momentum-radial) $\hbar^2l(l+1)/2mr^2$ walls off the origin for $l \ge 1$. Only $s$ states touch the nucleus — with consequences from atomic spectra to radioactive electron capture.*

## 10.4 Magnetic moments: angular momentum made visible

**Proposition 10.7 (Orbital magnetic moment and the Zeeman effect).**

A particle of charge $q$ and mass $m$ with orbital angular momentum $\hat{\vect L}$ carries the magnetic moment

$$
\hat{\vect\mu} = \frac{q}{2m}\,\hat{\vect L} ;
$$

for the electron, the natural unit is the *[Bohr magneton](#prop-b3-quantum-angular-momentum-magnetic)* $\mu_{\text{B}} = e\hbar/2m_{\text{e}} = 5.79 \times 10^{-5}\,\mathrm{eV}/\mathrm{T}$ (compare [Exercise 7.5](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#exo-b3-schrodinger-three-dimensions-5)). In a field $B\vect e_z$ the energy $-\hat{\vect\mu}\cdot\vect B$ shifts each level by $+m\,\mu_{\text{B}}B$ (electron charge negative): a level of given $l$ splits into its $2l + 1$ components — the *[Zeeman effect](#prop-b3-quantum-angular-momentum-magnetic)*, the first direct display of the quantisation of $m$, and the reason $m$ is called the [magnetic quantum number](#thm-b3-quantum-angular-momentum-spectrum). At $B = 1\,\mathrm{T}$ the splitting is $58\,\text{µ}\mathrm{eV}$: small beside optical energies, easily resolved as a shift of spectral lines — and, read in reverse, a magnetometer: the Zeeman splitting of sunlight’s lines maps the magnetic fields of sunspots.

**Partial proof.** Classically a charge on an orbit is a current loop: $\mu = IA =
(qv/2\pi r)(\pi r^2) = qL/2m$; the operator statement inherits it (and follows from the $\hat{\vect A}$ coupling of [Proposition 1.16](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#prop-b3-lagrangian-mechanics-charge)). The energy shift is first-order perturbation theory, anticipated here and justified in [Chapter 13](https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory#ch-b3-perturbation-theory). ∎

![“Space quantisation” for l = 2: the angular momentum vector has length √l(l+1)\, = √6\, but can offer the z axis only the five projections m — never its full length: since the components are incompatible, L can never lie exactly along any axis.](https://one-course.com/images/onecourse/chapters/physics-5/b3-quantum-angular-momentum/fig-dd0eb4cbdad3.svg)

*“Space quantisation” for $l = 2$: the angular momentum vector has length $\sqrt{l(l+1)}\,\hbar = \sqrt6\,\hbar$ but can offer the $z$ axis only the five projections $m\hbar$ — never its full length: since the components are incompatible, $\vect L$ can never lie exactly along any axis.*

**Method 10.8 (Angular momentum in practice).**

(1) Label states by $(l, m)$ — or $(j, m)$ for the abstract algebra — and never ask for two components at once. (2) Matrix elements: use the ladder formulas; $\hat L_x = (\hat L_+ + \hat
L_-)/2$. (3) Central potential: quote the $Y_l^m$, solve only the [radial equation](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-swave) with the [centrifugal barrier](#thm-b3-quantum-angular-momentum-radial). (4) Rotational energies: $BJ(J+1)$, lines at $2B(J+1)$ — extract $B$, hence bond lengths, from equal spacings. (5) Magnetic questions: convert angular momenta to moments at $\mu_{\text{B}}$ per $\hbar$, energies at $\mu_{\text{B}}B$ per unit $m$.

## 10.5 Exercises

**Exercise 10.1 ★.**

(a) Derive $[\hat L_x, \hat L_y] = \iu\hbar\hat L_z$ from the canonical [commutators](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-commutator). (b) Show $[\hat L^2, \hat L_z] = 0$. (c) Show $[\hat L_z, \hat x] = \iu\hbar\hat y$: what does $\hat L_z$ generate? (d) Why do the three components admit no common eigenbasis — except for one particular state (which)?

**Solution of Exercise 10.1.**

(a) $[\hat y\hat p_z - \hat z\hat p_y,\ \hat z\hat p_x - \hat x\hat
p_z]$: only the terms sharing a $\hat z, \hat p_z$ pair survive, giving $\iu\hbar(\hat x\hat p_y - \hat y\hat p_x)$. (b) $[\hat
L^2, \hat L_z] = \sum_i[\hat L_i^2, \hat L_z] = \hat L_x[\hat L_x,
\hat L_z] + [\hat L_x, \hat L_z]\hat L_x + (x \to y)$: the four terms cancel pairwise. (c) $[\hat L_z, \hat x] = \iu\hbar\hat y$: $\hat L_z$ generates rotations about $z$ — of operators as of states. (d) Common sharpness of two components forces (by the [commutator](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-commutator)) sharpness of the third with value zero for all three: only $l = 0$ achieves it.

**Exercise 10.2 ★.**

For $l = 1$, in the basis $\{\ket{1,1}, \ket{1,0}, \ket{1,-1}\}$: (a) write the matrix of $\hat L_z$; (b) use the ladder formula to write $\hat L_+$ and $\hat L_-$; (c) assemble $\hat L_x$ and check its [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) are $\hbar, 0, -\hbar$; (d) a state with $L_x =
+\hbar$ is measured along $z$: give the probabilities of the three outcomes.

**Solution of Exercise 10.2.**

(a) $\hat L_z = \hbar\operatorname{diag}(1, 0, -1)$. (b) $\hat L_+ =
\hbar\sqrt2\,\begin{pmatrix}0&1&0\\0&0&1\\0&0&0
\end{pmatrix}$, $\hat L_-$ its transpose. (c) $\hat L_x =
\dfrac{\hbar}{\sqrt2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0
\end{pmatrix}$: characteristic polynomial $\lambda(\lambda^2 -
\hbar^2)$. (d) The $L_x = +\hbar$ eigenvector is $\tfrac12(1,
\sqrt2, 1)$: probabilities $\tfrac14, \tfrac12, \tfrac14$ for $m =
+1, 0, -1$.

**Exercise 10.3 ★.**

Carbon monoxide: bond length $0.113\,\mathrm{nm}$, reduced mass $1.14 \times 10^{-26}\,\mathrm{kg}$. (a) Compute $I$ and $B = \hbar^2/2I$ in meV. (b) The frequency and wavelength of the $J = 1 \leftarrow 0$ line. (c) The next two lines. (d) An astronomer measures the comb spacing to five digits: what molecular quantity does she obtain, and to what precision?

**Solution of Exercise 10.3.**

(a) $I = \mu r_0^2 = 1.46 \times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2}$; $B = \hbar^2/2I =
3.8 \times 10^{-23}\,\mathrm{J} = 0.24\,\mathrm{meV}$. (b) $2B/h = 115\,\mathrm{GHz}$, $\lambda = 2.6\,\mathrm{mm}$. (c) $231\,\mathrm{GHz}$ and $346\,\mathrm{GHz}$. (d) The spacing gives $B$, hence $I = \mu r_0^2$: the bond length of a molecule light-years away, to roughly half the spacing’s relative precision.

**Exercise 10.4 ★.**

A particle is in a state of $l = 2$. (a) List the possible outcomes of measuring $L_z$. (b) The smallest angle between $\vect L$ and the $z$ axis, using $\cos\theta = m/\sqrt{l(l+1)}$: evaluate it. (c) Why can the angle never be zero? (d) Show the minimal angle tends to zero as $l \to \infty$: classical vectors recovered.

**Solution of Exercise 10.4.**

(a) $m\hbar$ with $m = -2 \dots 2$. (b) $\cos\theta = 2/\sqrt6$: $\theta = 35.3^\circ$. (c) Perfect alignment would make $L_x = L_y
= 0$ sharp together with $L_z$ — forbidden by the [commutators](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-commutator) except in the trivial $l = 0$ case. (d) $l/\sqrt{l(l+1)} \to 1$: for large $l$ the cone closes onto the axis and the classical arrow returns.

**Exercise 10.5 ★★.**

The angular part of $-\hbar^2\Delta$ is $\hat L^2$, with

$$
\hat L^2 = -\hbar^2\Big[\frac{1}{\sin\theta}\,\partial_\theta
(\sin\theta\,\partial_\theta) +
\frac{1}{\sin^2\theta}\,\partial_\varphi^2\Big] .
$$

(a) Verify that $Y_1^0 \propto \cos\theta$ has $\hat L^2$-eigenvalue $2\hbar^2$. (b) Verify $Y_1^{\pm1} \propto \sin\theta\,
\eu^{\pm\iu\varphi}$ likewise, and their $\hat L_z$ [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable). (c) Check the parities. (d) Why must $\braket{Y_1^0}{Y_1^1} = 0$ without computing any integral?

**Solution of Exercise 10.5.**

(a) With no $\varphi$ dependence, the operator gives $-\hbar^2(\sin\theta)^{-1}\partial_\theta(\sin\theta\,(-\sin\theta))
= 2\hbar^2\cos\theta$. (b) The same computation with the $\eu^{\pm\iu\varphi}$ factor: [eigenvalue](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable) $2\hbar^2$; $\hat L_z$ gives $\pm\hbar$. (c) $\cos\theta$ and $\sin\theta\,\eu^{\pm\iu
\varphi}$ change sign under $\vect r \to -\vect r$ ($\theta \to \pi
- \theta$, $\varphi \to \varphi + \pi$): parity $-1 = (-1)^1$. (d) They are eigenvectors of the Hermitian $\hat L_z$ with different [eigenvalues](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#def-b3-quantum-formalism-observable).

**Exercise 10.6 ★★.**

Which rotational line is brightest? The population of level $J$ is $\propto (2J+1)\,\eu^{-BJ(J+1)/k_{\text{B}}T}$. (a) Explain the two factors. (b) Show the maximum sits near $J_{\max} \approx
\sqrt{k_{\text{B}}T/2B} - \tfrac12$. (c) For CO at $10\,\mathrm{K}$ (a dark cloud) and at $300\,\mathrm{K}$: which lines dominate? (d) Inverting: an observed CO ladder peaking at $J = 7$ betrays what temperature?

**Solution of Exercise 10.6.**

(a) $2J + 1$ states share the level (degeneracy); the Boltzmann factor taxes its energy. (b) Maximise the product: $\dd/\dd J = 0$ gives $2 = (2J+1)^2B/k_{\text{B}}T$, the stated $J_{\max}$. (c) At $10\,\mathrm{K}$: $J_{\max} \approx 0.8$, so $J = 1$ dominates and the $115\,\mathrm{GHz}$ and $230\,\mathrm{GHz}$ lines shine; at $300\,\mathrm{K}$: $J_{\max} \approx 7$. (d) $T \approx 2B(J_{\max} + \tfrac12)^2/
k_{\text{B}} \approx 310\,\mathrm{K}$ — a rotational thermometer.

**Exercise 10.7 ★★.**

Normal [Zeeman effect](#prop-b3-quantum-angular-momentum-magnetic). A spectral line at $500\,\mathrm{nm}$ comes from a transition $l = 2 \to l = 1$ in a field $B = 2\,\mathrm{T}$; ignore spin (valid for special “singlet” states). (a) Sketch the split levels. (b) With the selection rule $\Delta m = 0, \pm1$, show only *three* line positions appear, at $0, \pm\mu_{\text{B}}B/h$. (c) Compute the splitting in GHz and in picometres of wavelength. (d) Most real lines split into more than three components (“anomalous” Zeeman): what missing ingredient, carried by the electron itself, was historical evidence for?

**Solution of Exercise 10.7.**

(a) The upper level splits into five, the lower into three, all with the same spacing $\mu_{\text{B}}B$. (b) With equal spacings, $h\nu = h\nu_0 + (\Delta m)\mu_{\text{B}}B$ and $\Delta m \in \{0,
\pm1\}$: three positions only. (c) $\mu_{\text{B}}/h = 14\,\mathrm{GHz}/\mathrm{T}$, so at $B = 2\,\mathrm{T}$ the splitting is $28\,\mathrm{GHz}$, i.e. $\Delta\lambda =
\lambda^2\Delta\nu/c \approx 23\,\mathrm{pm}$ — resolvable since Zeeman’s 1896 gratings. (d) The electron’s own spin, with its anomalous factor $g \approx
2$: the “anomalous” patterns were spin’s fingerprints before spin was named ([Chapter 12](https://one-course.com/books/physics/5/en/chapter/12-spin-and-two-level-systems#ch-b3-spin-two-level)).

**Exercise 10.8 ★★.**

The centrifugal wall. For hydrogen ($V = -e^2/4\pi\varepsilon_0r$), write $U_{\text{eff}}$ for $l = 1$ in units of $E_{\text{I}}$ and $a_0$. (a) Locate its minimum and value. (b) Show $U_{\text{eff}} >
0$ for $r < \dots$ — find the radius inside which the barrier dominates. (c) Compare $|\psi|^2$ near $r = 0$ for $s$ and $p$ states ($u \sim r^{l+1}$, admitted): which states “touch” the nucleus? (d) Electron capture — a nucleus swallowing one of its atom’s electrons — proceeds overwhelmingly from $s$ shells: explain in one sentence.

**Solution of Exercise 10.8.**

In the stated units $U_{\text{eff}} = -2/x + l(l+1)/x^2$. (a) For $l = 1$: minimum at $x = 2$ ($r = 2a_0$), depth $-E_{\text{I}}/2$. (b) $U_{\text{eff}} > 0$ for $x < 1$: inside one Bohr radius the wall wins. (c) $\psi \sim r^l$ near the origin: only $l = 0$ states have nonzero density at $r = 0$. (d) Capturing an electron requires electron density *at* the nucleus: the $s$ electrons, alone unbarred by the centrifugal wall, are the ones swallowed.

**Exercise 10.9 ★★.**

On $\ket{l, m}$: (a) show $\langle\hat L_x\rangle = \langle\hat
L_y\rangle = 0$; (b) show $\langle\hat L_x^2\rangle = \langle\hat
L_y^2\rangle = \tfrac12[l(l+1) - m^2]\hbar^2$; (c) verify the [uncertainty relation](https://one-course.com/books/physics/5/en/chapter/8-the-formalism-of-quantum-mechanics#thm-b3-quantum-formalism-uncertainty) $\Delta L_x\,\Delta L_y \ge \tfrac\hbar2
|\langle\hat L_z\rangle|$ on the stretched state $m = l$; (d) interpret the “cone” picture of the figure in the light of (a) and (b).

**Solution of Exercise 10.9.**

(a) $\hat L_x = (\hat L_+ + \hat L_-)/2$ changes $m$: diagonal elements vanish. (b) By symmetry the two transverse squares are equal, and their sum is $\langle\hat L^2 - \hat L_z^2\rangle$. (c) At $m = l$: $\Delta L_x\Delta L_y = l\hbar^2/2 = \tfrac\hbar2
\,l\hbar$: equality — the stretched state is as aligned as the algebra permits. (d) The cone: sharp $L_z$, vanishing transverse means, equal transverse spreads — a vector of definite length and projection, democratically smeared in azimuth.

**Exercise 10.10 ★★★.**

Vibration–rotation bands. An infrared vibrational transition ($\hbar\omega_0$) of a diatomic changes $J$ by $\pm1$ simultaneously. (a) Show the absorption lines fall at $h\nu =
\hbar\omega_0 + 2B(J+1)$ (the $R$ branch, from $J \to J+1$) and $h\nu = \hbar\omega_0 - 2BJ$ (the $P$ branch, $J \to J-1$), $J \ge
1$. (b) Why is there a *gap* at $\hbar\omega_0$ itself? (c) Sketch the band: two combs flanking a missing central line, each line’s strength following [Exercise 10.6](#exo-b3-quantum-angular-momentum-6). (d) The $15\,\text{µ}\mathrm{m}$ band of [Problem 9.1](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#pb-b3-harmonic-oscillator-1) has exactly this anatomy: what sets its overall width, and hence the “wings” that make added carbon dioxide effective?

**Solution of Exercise 10.10.**

(a) Energy bookkeeping with $E_J = BJ(J+1)$: $J \to J + 1$ adds $2B(J+1)$; $J \to J - 1$ subtracts $2BJ$. (b) $\Delta J = 0$ is forbidden, and the two branches start at $\pm 2B$: nothing falls at the pure vibrational frequency. (c) Two combs of spacing $2B$ flanking a gap, intensities rising to $J_{\max}$ then falling. (d) The band’s width is the populated extent of the combs, growing as $\sqrt{T}$: those thermally populated wings are precisely where a saturated greenhouse band keeps absorbing.

**Exercise 10.11 ★★★.**

Finish the algebra. (a) From $\hat J_\mp\hat J_\pm = \hat J^2 -
\hat J_z^2 \mp \hbar\hat J_z$, compute $\|\hat J_\pm\ket{j,m}\|^2$ and the ladder coefficients. (b) Show that if the ladder failed to terminate, some norm would go negative: exhibit the offending state. (c) Conclude that $m_{\max} - m_{\min} = 2j$ must be a non-negative integer and enumerate the allowed $j$. (d) Where exactly does the argument *fail* to exclude half-integers — and which additional requirement (single-valuedness on the sphere) excludes them for orbital motion only?

**Solution of Exercise 10.11.**

(a) $\|\hat J_\pm\ket{j,m}\|^2 = \bra{j,m}\hat J_\mp\hat J_\pm
\ket{j,m} = \hbar^2[j(j+1) - m(m\pm1)]$. (b) Climbing past $m = j$ would give $j(j+1) - j(j+1) = 0$ then negative norms one step further: the chain must contain the annihilated top state. (c) The top and bottom differ by an integer number of unit steps: $2j \in
\mathbb N$: $j = 0, \tfrac12, 1, \tfrac32, \dots$ (d) The algebra never uses wave functions; only the demand that $\eu^{\iu m\varphi}$ be single-valued on the circle forces integer $m$ — a condition binding orbital motion and leaving intrinsic (spin) angular momenta free to be half-integer.

**Exercise 10.12 ★★★.**

Einstein–de Haas: angular momentum is mechanical. An iron cylinder (radius $a = 1.0\,\mathrm{cm}$, density $\rho = 7.9 \times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}$, $n = 8.5 \times 10^{28}\,\mathrm{atoms}/\mathrm{m}^{3}$) hangs from a torsion fibre; reversing its magnetisation flips about one $\hbar$ of angular momentum per atom. (a) By conservation, the rod must recoil: compute the angular momentum per unit volume transferred. (b) With the rod’s moment of inertia $\tfrac12 M
a^2$, show the rod acquires $\omega = 2n\hbar/\rho a^2$ and evaluate it. (c) The 1915 experiment drove the reversal at the fibre’s resonance to accumulate the tiny kicks: estimate the oscillation amplitude after $Q \sim 100$ resonant reversals, taking your $\omega$ per kick. (d) The measured ratio moment/angular-momentum came out twice the orbital prediction $e/2m_{\text{e}}$: what was that factor of two announcing?

**Solution of Exercise 10.12.**

(a) $n\hbar$ per unit volume. (b) $\omega = n\hbar V/(\tfrac12 Ma^2)
= 2n\hbar/\rho a^2 = 2 \times 8.5 \times 10^{28} \times 1.055 \times 10^{-34}/
(7900 \times 10^{-4}) \approx 2.3 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$ — invisible in one shot. (c) Reversing in step with the fibre’s resonance accumulates $\sim Q$ kicks: $\omega \sim 2 \times 10^{-3}\,\mathrm{rad}/\mathrm{s}$, milliradian swings on a period of seconds — visible with a mirror and a light beam, as Einstein and de Haas saw. (d) The measured gyromagnetic ratio was $e/m_{\text{e}}$, twice the orbital $e/2m_{\text{e}}$: iron’s magnetism is not orbital currents but electron *spin*, whose $g \approx 2$ the next chapters explain.

## 10.6 Problem: Weighing galaxies at 2.6 millimetres

**Problem 10.1.**

Weekend problem — carbon monoxide, radio astronomy and the cold universe

Most of a galaxy’s star-making matter is cold molecular hydrogen — and it is invisible: H$_2$, symmetric, has no dipole and no rotational lines at cold-cloud temperatures. Astronomy’s solution is its faithful companion. One CO molecule per $10^{4}$ H$_2$, with its $115\,\mathrm{GHz}$ ladder, lights up every molecular cloud in the sky. Data: for CO, $B/h = 57.6\,\mathrm{GHz}$ ($B =
0.238\,\mathrm{meV}$); bond length $r_0 = 0.113\,\mathrm{nm}$; reduced mass $\mu = 1.14 \times 10^{-26}\,\mathrm{kg}$; $k_{\text{B}}T$ at $10\,\mathrm{K}$ is $0.862\,\mathrm{meV}$.

**Part I — The quantum rotor.**

1. From $\hat H = \hat L^2/2I$ , give the energies $E_J$ and their degeneracies.
2. Verify $B = \hbar^2/2I$ for CO from $r_0$ and $\mu$ .
3. With the selection rule $\Delta J = \pm1$ , derive the absorption frequencies $\nu_{J+1\leftarrow J} = 2B(J+1)/h$ and list the first three.
4. Why are the lines *equally spaced* — and what does a measured spacing hand the astronomer (through $I$ )?
5. Compute the wavelength of the fundamental line, and explain why observing it needs high, dry sites (what absorbs millimetre waves in our own atmosphere — recall the molecule of [Problem 9.1](https://one-course.com/books/physics/5/en/chapter/9-the-quantum-harmonic-oscillator#pb-b3-harmonic-oscillator-1) ’s neighbour, water)?
6. The photon of the $1 \to 0$ line carries $\hbar$ of angular momentum: why is $\Delta J = \pm1$ not merely a rule of thumb but a conservation law?

**Part II — Why CO and not H$_2$.**

7. H $_2$ is homonuclear: state why it emits no rotational dipole radiation at all.
8. H $_2$ is also *light* : compute the rotational constant of H $_2$ ( $\mu = m_{\text{p}}/2$ , $r_0 = 0.074\,\mathrm{nm}$ ) and its first accessible transition energy; compare with $k_{\text{B}}T$ at $10\,\mathrm{K}$ : even by quadrupole back-doors, could cold H $_2$ radiate?
9. CO’s first rung costs $2B = 0.48\,\mathrm{meV}$ against $k_{\text{B}}T = 0.86\,\mathrm{meV}$ : is the ladder alive at $10\,\mathrm{K}$ ?
10. Compute the population ratio $n_1/n_0$ at $10\,\mathrm{K}$ (degeneracy included).
11. Near which $J$ does the population peak at $10\,\mathrm{K}$ (use $J_{\max} \approx \sqrt{k_{\text{B}}T/2B} - \tfrac12$ )?
12. Summarise in one sentence the pact of the trade: what CO provides, what must be assumed about the CO-to-H $_2$ ratio.

**Part III — Reading the sky.**

13. A cloud’s line arrives at $115.156\,\mathrm{GHz}$ instead of the laboratory $115.271\,\mathrm{GHz}$ : compute the cloud’s velocity along the line of sight, and its direction.
14. The line is Doppler- *broadened* by internal motions of $\pm2\,\mathrm{km}/\mathrm{s}$ : compute the linewidth in MHz, and compare with the thermal width expected at $10\,\mathrm{K}$ ( $v_{\text{th}} \sim \sqrt{k_{\text{B}}T/m_{\text{CO}}}  \approx 55\,\mathrm{m}/\mathrm{s}$ ): what dominates the broadening?
15. Mapping the Doppler shift of CO across a spiral galaxy’s disc yields its rotation curve $v(r)$ . The curves stay *flat* far beyond the visible disc: recall (Year 1 volume, gravitation) what $v(r)$ should do around a centrally concentrated mass, and state what the flatness implies.
16. The ratio of the $2\to1$ line’s brightness to the $1\to0$ ’s measures level populations: which single physical quantity of the cloud does it deliver?
17. ALMA resolves CO in galaxies whose light left when the universe was young; the observed frequency of the $1 \to 0$ line from a galaxy at “redshift $z = 1$ ” arrives at half its rest value — at what frequency, and (recalling [Example 4.13](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#ex-b3-relativistic-kinematics-redshift) ) what stretched it?
18. From a measured CO line luminosity astronomers quote cloud masses in solar masses: list the chain of conversions (line photons $\to$ CO count $\to$ H $_2$ mass) and the weakest link.

**Part IV — The cold nurseries.**

19. Star-forming cores sit at $10\,\mathrm{K}$ : from Wien’s law their thermal glow peaks near $290\,\text{µ}\mathrm{m}$ — invisible to every optical telescope. In one sentence: why is the rotational ladder the *only* thermometer and scale such a core offers?
20. The $J = 1$ level lies $5.5\,\mathrm{K}$ (in temperature units) above the ground state: why does that make the $1\to0$ line an exquisite thermometer precisely in the $10\,\mathrm{K}$ regime (rather than, say, an optical line at $2\,\mathrm{eV}$ $\sim 23\,000\,\mathrm{K}$ )?
21. Why does a *collapsing* core eventually stop being visible in CO ( $1\to0$ ) — consider what high density and dust do to the line and its escape.
22. Molecular clouds also show lines of NH $_3$ , HCN, and dozens more, each with its own $B$ and dipole: what does the richness of this “chemical radio dial” let astronomers disentangle?
23. The whole sky glows at $2.7\,\mathrm{K}$ (the cosmic microwave background): what happens to the contrast of a cloud’s CO lines as the cloud’s own temperature approaches $2.7\,\mathrm{K}$ , and why is that a hard floor for this thermometer?
24. A millimetre dish must hold its shape to about $\lambda/20$ : compute that tolerance at $115\,\mathrm{GHz}$ , and compare with the tolerance an optical mirror ( $500\,\mathrm{nm}$ ) must meet — why could ALMA’s twelve-metre dishes be built in the open air?
25. Summarise the named result: a ladder $BJ(J+1)$ with $B/h =  57.6\,\mathrm{GHz}$ , populated at ten kelvin and read at $2.6\,\mathrm{mm}$ , is how the mass, temperature, velocity and rotation of the cold universe — and the evidence for dark matter around spiral galaxies — are measured.

**Solution of Problem 10.1.**

**1.** $E_J = BJ(J+1)$, degeneracy $2J + 1$. **2.** $I = \mu r_0^2 = 1.46 \times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2}$: $B =
\hbar^2/2I = 3.8 \times 10^{-23}\,\mathrm{J}$, i.e. $B/h = 57.7\,\mathrm{GHz}$ — the measured constant. **3.** $115$, $231$, $346\,\mathrm{GHz}$. **4.** $E_{J+1} - E_J = 2B(J+1)$ grows linearly: consecutive lines differ by the constant $2B$. The spacing gives $I$, hence the bond length — structural chemistry by radio. **5.** $\lambda = c/\nu = 2.6\,\mathrm{mm}$. Atmospheric water vapour absorbs millimetre waves greedily; hence Atacama, Mauna Kea, the South Pole — high, cold, dry. **6.** The photon is a spin-$1$ particle carrying $\hbar$: total angular momentum conservation obliges the molecule’s $J$ to change by one unit. **7.** Its charge distribution is symmetric at every rotation angle: no oscillating dipole, no dipole line — perfect camouflage. **8.** $B_{\text{H}_2} = \hbar^2/2\mu r_0^2 \approx
7.6\,\mathrm{meV}$; the symmetric molecule’s weak quadrupole transitions require $\Delta J = 2$, costing $6B \approx
46\,\mathrm{meV}$ — fifty times $k_{\text{B}}T$ at $10\,\mathrm{K}$: cold hydrogen cannot radiate at all. **9.** $2B = 0.48\,\mathrm{meV} < k_{\text{B}}T = 0.86\,\mathrm{meV}$: collisions keep the first rungs populated — the ladder works exactly where the clouds live. **10.** $n_1/n_0 = 3\,\eu^{-0.478/0.862} = 1.7$: the emitting level is well stocked. **11.** $J_{\max} \approx \sqrt{0.86/0.48} - 0.5 \approx 0.8$: the population peaks at $J = 1$. **12.** CO delivers the photons; converting them to total gas mass rests on an assumed CO-to-H$_2$ abundance — luminous messenger, calibrated ransom. **13.** $\Delta\nu = 115\,\mathrm{MHz}$: $v = c\,\Delta\nu/\nu
\approx 300\,\mathrm{km}/\mathrm{s}$, receding (frequency lowered) — galactic orbital speeds. **14.** $\pm2\,\mathrm{km}/\mathrm{s}$ spans $1.5\,\mathrm{MHz}$; thermal motion at $10\,\mathrm{K}$ gives only $\sim40\,\mathrm{kHz}$: the width is turbulence, not temperature — linewidths map a cloud’s internal weather. **15.** Around a central mass $v \propto 1/\sqrt r$ should fall (Kepler); the measured flatness means the enclosed mass keeps growing with radius — invisible matter enveloping the luminous disc: dark matter. **16.** The ratio of populations, i.e. the excitation temperature of the gas. **17.** At $57.6\,\mathrm{GHz}$: cosmic expansion has doubled every wavelength in flight — the cosmological redshift the last chapter of this book returns to. **18.** Photon flux $\to$ CO luminosity (distance needed) $\to$ CO count (excitation model) $\to$ H$_2$ mass (the CO-to-H$_2$ “X-factor”: the weakest link, calibrated locally and exported cautiously). **19.** A $10\,\mathrm{K}$ core emits nothing an optical telescope can see; its rotational lines are simultaneously its only thermometer, speedometer and scale. **20.** A thermometer is sensitive where level spacing $\sim
k_{\text{B}}T$: $5.5\,\mathrm{K}$ spacing responds strongly across $5$–$50$ K, while a $2\,\mathrm{eV}$ optical level would be populated by nothing at all. **21.** The line saturates (optically thick) and, worse, CO freezes out onto dust grains in the densest, coldest gas: astronomers switch to rarer isotopologues ($^{13}$CO, C$^{18}$O) and other molecules. **22.** Each species needs its own density and temperature to shine: together they tomograph the cloud — outer envelopes in CO, dense cores in NH$_3$ and HCN. **23.** A line is seen in contrast against the cosmic background; as the gas temperature approaches $2.7\,\mathrm{K}$ its excitation equilibrates with the background and the contrast vanishes: no colder cloud can be read this way. **24.** $\lambda/20 = 130\,\text{µ}\mathrm{m}$ against $25\,\mathrm{nm}$ for optical work: five thousand times more forgiving — which is why twelve-metre millimetre dishes stand in the open desert while optical mirrors live in domes. **25.** A ladder $BJ(J+1)$ with $B/h = 57.6\,\mathrm{GHz}$, alive at ten kelvin and read at $2.6\,\mathrm{mm}$, measures the mass, temperature, turbulence and rotation of the cold universe — and its flat rotation curves are a standing exhibit for dark matter.