Part III — The physics

Angular momentum

Radio telescopes pointed at the cold clouds between the stars pick up carbon monoxide by a sharp line at 115.27 GHz, with further lines at almost exactly twice and three times that frequency, and nothing in between. The molecules are tumbling end over end. Why can a molecule only rotate at certain rates?

Rotation is measured by angular momentum. This short essay finds what values it can take in quantum mechanics, and how much of it can be known at once. The answer has two parts: the component along any one axis comes in whole units of $\hbar$, and the total has its own set of allowed values that never quite match the largest component. Both reappear in every atom, starting with hydrogen in the next essay.

One component: whole units of $\hbar$

Momentum and position form a pair: momentum is the operator $-i\hbar\,d/dx$, and a state of definite momentum is a wave $e^{ipx/\hbar}$. Angular momentum about the $z$ axis and the angle $\phi$ around that axis form the same kind of pair. Its operator is $\hat L_z = -i\hbar\,\partial/\partial\phi$, and a state of definite $L_z$ has the angular dependence

$$\Phi(\phi) = e^{i L_z\phi/\hbar}.$$

There is one difference from a straight line, and it decides everything. Going once round the axis brings you back to where you started, so the wavefunction must come back to the same value: $\Phi(\phi + 2\pi) = \Phi(\phi)$. That requires $e^{2\pi i L_z/\hbar} = 1$, so $L_z/\hbar$ must be a whole number:

$$L_z = m\hbar,\qquad m = 0, \pm1, \pm2, \dots$$

It is the particle in a box again, with the walls replaced by the requirement that the wave join up with itself around a circle. A whole number of wavelengths must fit round the loop, which is the de Broglie picture of Bohr's rule, now as an exact statement about one component of angular momentum.

The total, and why it never lines up

The three components $\hat L_x$, $\hat L_y$, $\hat L_z$ do not commute with one another; for example $[\hat L_x, \hat L_y] = i\hbar\hat L_z$. By the formalism essay, they cannot share a complete set of sharp states. But each of them commutes with the squared total, $\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2$. So the best we can do is to make the total and one component sharp together, conventionally $L_z$. Solving for the allowed values, which the optional section sketches, gives

$$L^2 = l(l+1)\,\hbar^2,\qquad l = 0, 1, 2, \dots$$

and for each $l$, $m$ runs over $-l, -l+1, \dots, l$. The commutators on their own would also allow half-whole values, $l = \tfrac12, \tfrac32, \dots$; it is the requirement that an orbital wavefunction join up with itself around the axis that rules those out for orbital motion. That loophole will matter in the essay on spin.

So each value of $l$ comes with $2l + 1$ possible values of $m$. The surprise is the $l(l+1)$: the total angular momentum has size $\sqrt{l(l+1)}\,\hbar$, which is always larger than the largest possible component, $l\hbar$. The angular momentum can never point exactly along the $z$ axis. If it did, $L_x$ and $L_y$ would both be exactly zero, three sharp components at once, which the commutators forbid.

There is one exception, and it illustrates a point from the formalism essay. For $l = 0$, every component is zero and all of them are sharp together. Operators that do not commute can still share some eigenstates. What they cannot do is share all of them.

The "vector model" for $l = 2$. The angular momentum has length $\sqrt6\,\hbar$ and a sharp $z$-component $m\hbar$ with $m$ from $-2$ to $2$. The picture is a memory aid, not a photograph: in a state with sharp $L_z$ the $x$ and $y$ components are not sharp, and the arrow is best imagined smeared around a cone about the $z$ axis.

Optional: where $l(l+1)$ comes from

Define $\hat L_\pm = \hat L_x \pm i\hat L_y$. From the commutators, applying $\hat L_+$ to a state with $L_z = m\hbar$ gives a state with $L_z = (m+1)\hbar$ and the same $L^2$: it climbs a ladder in steps of $\hbar$, as the oscillator's raising operator did in steps of $\hbar\omega$. Because $L_z^2$ cannot exceed $L^2$, the ladder has a top rung, $m = l$, where $\hat L_+$ gives zero. Using $\hat L_-\hat L_+ = \hat L^2 - \hat L_z^2 - \hbar\hat L_z$ on that top state gives $0 = L^2 - l^2\hbar^2 - l\hbar^2$, so $L^2 = l(l+1)\hbar^2$. The bottom rung is $m = -l$ by the same argument, and since the steps are whole units the ladder has $2l+1$ rungs. For orbital motion, the single-valuedness around the axis then forces $l$ to be a whole number.

A rotating molecule

A diatomic molecule like carbon monoxide rotating about its centre has kinetic energy $L^2/2I$. With $L^2 = l(l+1)\hbar^2$, its rotational energies are

$$E_l = l(l+1)\,\frac{\hbar^2}{2I}.$$

The lines seen by the radio telescopes are electric-dipole transitions: the molecule's electric dipole, the slight separation of charge between carbon and oxygen, couples to the radiation's electric field as it rotates. For transitions of this kind between rotational levels of a diatomic molecule, $l$ changes by exactly one. (This is a selection rule for this particular kind of transition, not a law for every photon; weaker processes obey other rules.) A drop from $l+1$ to $l$ releases $E_{l+1} - E_l = 2(l+1)\,\hbar^2/2I$. The lines are evenly spaced, in the ratio $1 : 2 : 3$: exactly the pattern of the opening.

Worked example: sizing a molecule from its radio line

From frequency to moment of inertia. The lowest line, $l = 1 \to 0$, has energy $2\hbar^2/2I = h\nu$ with $\nu = 115.27$ GHz. So $\hbar^2/2I = h\nu/2 = 2.38\times10^{-4}$ eV, and

$$\begin{aligned} I &= \frac{\hbar^2}{h\nu}\\[4pt] &= \frac{(1.055\times10^{-34})^2}{(6.626\times10^{-34})(1.153\times10^{11})}\\[4pt] &= 1.456\times10^{-46}\ \text{kg m}^2 . \end{aligned}$$

From moment of inertia to bond length. For two atoms a distance $r$ apart, $I = \mu r^2$ with the reduced mass $\mu = 6.86$ u $= 1.139\times10^{-26}$ kg. Then $r = \sqrt{I/\mu} = 113$ pm. That is the bond length used in the oscillator essay, measured here from a radio telescope's line alone.

Check: the next lines. The formula predicts $2\times115.27 = 230.54$ GHz and $3\times115.27 = 345.81$ GHz. Laboratory measurements give 230.54 and 345.80 GHz. The tiny shortfall in the third comes from the bond stretching slightly as the molecule spins faster, which the rigid-rotor model ignores.

Check: everyday rotation. A spinning top with angular momentum $0.01$ J s has $l \approx L/\hbar \approx 10^{32}$. Adjacent allowed values differ by one part in $10^{32}$, and the rotation looks continuous.

Two questions

1. A heavier isotope. A small fraction of carbon monoxide contains carbon-13 instead of carbon-12. Its reduced mass is 7.17 u instead of 6.86 u, with the same bond length. At what frequency should its lowest rotational line appear?

Worked answer

The line frequency is $\nu = \hbar^2/(hI)$ and $I = \mu r^2$, so with $r$ unchanged, $\nu \propto 1/\mu$: $115.27 \times 6.86/7.17 = 110.2$ GHz. The measured line is at 110.20 GHz, and astronomers use exactly this shift to count carbon isotopes in distant clouds.

The slip to avoid is scaling by the carbon mass alone, $12/13$, which gives 106.4 GHz. The molecule rotates about its centre of mass, and both atoms move; only the reduced mass captures that.

2. Pointing along $z$? An electron is in a state with $l = 1$ and $m = 1$, so $L_z = \hbar$, the largest value allowed. Is its angular momentum pointing along the $z$ axis?

Worked answer

No. The total has size $\sqrt{l(l+1)}\,\hbar = \sqrt2\,\hbar \approx 1.41\hbar$, larger than $L_z = \hbar$. The remaining $L_x^2 + L_y^2 = L^2 - L_z^2 = \hbar^2$ is not zero, but neither $L_x$ nor $L_y$ is sharp: measured, each gives scattered results averaging to zero.

The tempting wrong turn is to picture a definite arrow that happens to be tilted, with some hidden $x$ and $y$ components. Sharp values of $L_x$, $L_y$ and $L_z$ at once are forbidden by the commutators for any $l > 0$. The tilted arrow in the figure is a way of remembering the numbers, not a description of the state.

Where this leaves us

A molecule can only rotate at certain rates because its angular momentum is quantised: the component along an axis in whole units of $\hbar$, because the wavefunction must join up with itself around a full turn, and the total at $\sqrt{l(l+1)}\,\hbar$. Its rotational energies follow, and the radio lines of carbon monoxide are their fingerprint, precise enough to measure a bond length from across the galaxy.

An electron moving around a nucleus has angular momentum too, described by the same $l$ and $m$. But it also has a distance from the nucleus to settle, and the Coulomb attraction to balance against. Putting all of that into Schrödinger's equation gives the hydrogen atom, the one real atom that can be solved exactly, and the answer to the question the Bohr model raised and could not close. That is the next essay.

Quantum Mechanics, Explained