The Rigid Rotor

A rotating molecule is a particle on a sphere¶

Lecture 4.4 reduced a rigid dumbbell to one particle of reduced mass $\mu$ on a circle. Let the dumbbell point in any direction and the particle moves on a sphere of radius $R_e$, the bond length. The potential energy does not change as the molecule turns, so all of its energy is the kinetic energy of rotation, $L^2/2I$:

Lecture 4.4 already found the eigenfunctions of $\hat{L}^2$, the spherical harmonics, so the problem is solved. For rotating molecules, spectroscopists write $J$ and $M_J$ in place of $l$ and $m$:

Source

# synced: rotor_levels
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

TEAL, CARDINAL, GRAY, PURPLE, ORANGE = "#107895", "#C8102E", "#6c757d", "#6a3d9a", "#e07b00"
fig, ax = plt.subplots(figsize=(5.6, 4.4))
w, g = 0.3, 0.1                                             # width of one state's bar, gap between bars
for J in range(6):
    E = J * (J + 1)
    n = 2 * J + 1
    x0 = -0.5 * (n * w + (n - 1) * g)
    for k in range(n):
        ax.plot([x0 + k * (w + g), x0 + k * (w + g) + w], [E, E], color=PURPLE, lw=3, solid_capstyle="butt")
    ax.text(-2.6, E, rf"$J = {J}$", ha="right", va="center", fontsize=19)
    ax.text(2.6, E, "0" if J == 0 else rf"${E}\tilde{{B}}$", ha="left", va="center", fontsize=19, color=PURPLE)
ax.set_xlim(-4.6, 4.3); ax.set_ylim(-1.5, 32); ax.axis("off")
fig.subplots_adjust(left=0.01, right=0.99, top=0.99, bottom=0.01)
plt.show()

Fig. The levels of a rigid rotor, in units of the rotational constant $\tilde{B}$ (defined below). Level $J$ holds $2J + 1$ states, one bar for each value of $M_J$.

• The gaps grow. $E_{J+1} - E_J = (\hbar^2/I)(J + 1)$: two, four, six, ... times $\hbar^2/2I$. Each step up in rotation costs more than the last.

• No zero-point energy. $J = 0$ has $E = 0$. $Y_0^0$ is the same in every direction: the bond axis points anywhere with equal probability, and nothing rotates. As on the ring of Lecture 4.4, no wall and no potential squeezes the wavefunction. A vibrating bond, by contrast, always keeps $\frac{1}{2}\hbar\omega$ (Lecture 4.2).

• What a state describes. $\lvert Y_J^{M_J}(\theta, \phi)\rvert^2$ is the probability density for the bond axis to point in the direction $(\theta, \phi)$.

Spectroscopists count in wavenumbers¶

As for vibrations in Lecture 4.3, divide each energy by $hc$:

Rotation is cheap¶

• Rotational steps are a hundred to ten thousand times smaller than vibrational ones. At room temperature $k_BT = 208$ cm$^{-1}$ is far above $\tilde{B}$, so many $J$ levels are occupied (below), while nearly every molecule sits in $v = 0$ (Lecture 4.1).

• The light hydrides have the largest $\tilde{B}$: the moment of inertia depends on the reduced mass, which a hydrogen atom keeps small.

Which transitions light can drive¶

A rotating dipole is an antenna¶

Light pushes on the dipole moment $M$ of a molecule (Lecture 4.3). A polar diatomic such as HCl or CO carries a permanent dipole $M_e$ along its axis. As the molecule turns, the dipole turns with it: seen along any fixed direction it oscillates at the rotation frequency, and it can exchange energy with light at that frequency. A nonpolar molecule (H$_2$, N$_2$, O$_2$, CO$_2$, CH$_4$) has no permanent dipole, so its rotation is invisible in absorption. (Raman spectroscopy still sees it, through the polarizability.)

One rung at a time: $\Delta J = \pm 1$¶

A field along $z$ pushes on the $z$ component of the dipole, $M_e\cos\theta$, where $\theta$ is the angle between the bond axis and $z$. The transition dipole of Lecture 4.3 becomes an integral over orientations:

• From $J = 0$: $\cos\theta\,Y_0^0 = Y_1^0/\sqrt{3}$ exactly. The function the field produces is a $J = 1$ state, so $J = 1$ is the only level it reaches.

• In general, $\cos\theta\,Y_J^{M_J}$ is a mixture of $Y_{J+1}^{M_J}$ and $Y_{J-1}^{M_J}$ (a recursion of the spherical harmonics), so $J$ changes by one and $M_J$ stays. Light polarized in the $xy$ plane couples through $\sin\theta\,e^{\pm i\phi}$ and changes $M_J$ by one as well.

• Physically, a photon carries one unit $\hbar$ of angular momentum. Absorbing it changes the molecule’s angular momentum by exactly one unit.

Source

# synced: rotor_selection
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

TEAL, CARDINAL, GRAY, PURPLE, ORANGE = "#107895", "#C8102E", "#6c757d", "#6a3d9a", "#e07b00"
plt.rcParams.update({"axes.spines.top": False, "axes.spines.right": False})
th = np.linspace(0, np.pi, 801)
dth = th[1] - th[0]
Y0 = np.full_like(th, 1 / np.sqrt(4 * np.pi))
Y1 = np.sqrt(3 / (4 * np.pi)) * np.cos(th)
Y2 = np.sqrt(5 / (16 * np.pi)) * (3 * np.cos(th) ** 2 - 1)
cases = [(r"$J = 0 \to 1$", Y1, 1, "allowed"), (r"$J = 0 \to 2$", Y2, 2, "lobes cancel, forbidden")]
fig, axes = plt.subplots(2, 2, figsize=(9.6, 3.7), gridspec_kw={"height_ratios": [0.85, 1], "hspace": 0.62, "wspace": 0.12})
for (ax, bx), (title, Yj, n, verdict) in zip(axes.T, cases):
    ax.axhline(0, color=GRAY, lw=0.8)
    ax.plot(th, Y0, color=GRAY, lw=1.8, ls="--")
    ax.plot(th, Yj, color=TEAL, lw=2.2)
    ax.plot(th, 0.3 * np.cos(th), color=CARDINAL, lw=2.6)
    for xk, lab, col in ((1.0, r"$Y_0^0$", GRAY), (1.55, r"$\cos\theta$", CARDINAL), (2.15, rf"$Y_{n}^0$", TEAL)):
        ax.text(xk, 0.5, lab, color=col, fontsize=15, ha="center", va="center")
    ax.set_title(title, loc="left", fontsize=16)
    f = 2 * np.pi * Yj * np.cos(th) * Y0 * np.sin(th)       # the phi integral gives 2 pi
    area = np.sum(f) * dth
    bx.fill_between(th, f, 0, where=f > 0, color=PURPLE, alpha=0.45, lw=0)
    bx.fill_between(th, f, 0, where=f < 0, color=ORANGE, alpha=0.55, lw=0)
    bx.plot(th, f, color="k", lw=1.6)
    bx.axhline(0, color=GRAY, lw=0.8)
    bx.set_title(rf"multiply: $Y_{n}^0\,\cos\theta\;Y_0^0\,\sin\theta$", loc="left", fontsize=15)
    bx.set_xlabel(("area = 0" if abs(area) < 1e-3 else f"area = {abs(area):.2f}") + f": {verdict}", fontsize=15,
                  color=PURPLE if abs(area) > 1e-3 else GRAY, weight="bold")
    for a in (ax, bx):
        a.set_xlim(0, np.pi); a.set_yticks([]); a.spines["left"].set_visible(False)
    ax.set_xticks([]); ax.spines["bottom"].set_visible(False)
    bx.set_xticks([0, np.pi / 2, np.pi]); bx.set_xticklabels(["0", r"$\pi/2$", r"$\pi$"]); bx.tick_params(labelsize=12)
    ax.set_ylim(-0.62, 0.62); bx.set_ylim(-0.75, 0.75)
fig.subplots_adjust(left=0.02, right=0.99, top=0.9, bottom=0.16)
plt.show()

Fig. Why $J$ changes by one. Top: the state $Y_0^0$ (gray dashed), the factor $\cos\theta$ from the dipole along the bond axis (red), and a final state (teal), against the angle $\theta$ between the axis and the field. Bottom: their product times $\sin\theta$ from the surface element; its area is the transition dipole over $M_e$. It is $1/\sqrt{3} = 0.58$ for $J = 0 \to 1$ and zero for $J = 0 \to 2$, whose lobes cancel.

The microwave spectrum¶

Absorption from $J$ to $J + 1$ takes a photon of wavenumber

Source

# synced: microwave_lines
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

TEAL, CARDINAL, GRAY, PURPLE, ORANGE = "#107895", "#C8102E", "#6c757d", "#6a3d9a", "#e07b00"
plt.rcParams.update({"axes.spines.top": False, "axes.spines.right": False})
fig, (ax, bx) = plt.subplots(1, 2, figsize=(9.6, 4.0), gridspec_kw={"width_ratios": [0.85, 1.3], "wspace": 0.22})
for J in range(6):
    E = J * (J + 1)
    ax.hlines(E, 0, 6.2, color=PURPLE, lw=2.6)
    ax.text(-0.15, E, rf"$J = {J}$", ha="right", va="center", fontsize=12.5)
for J in range(5):
    x = 0.9 + 1.05 * J
    ax.annotate("", xy=(x, (J + 1) * (J + 2)), xytext=(x, J * (J + 1)),
                arrowprops=dict(arrowstyle="-|>", color=ORANGE, lw=2.2, mutation_scale=15, shrinkA=0, shrinkB=0))
    ax.text(x + 0.14, 0.5 * (J * (J + 1) + (J + 1) * (J + 2)), rf"${2 * (J + 1)}\tilde{{B}}$", fontsize=12,
            color=ORANGE, va="center", bbox=dict(fc="white", ec="none", pad=0.5))
ax.set_xlim(-1.6, 6.4); ax.set_ylim(-1.5, 32); ax.axis("off")
for J in range(5):
    bx.vlines(2 * (J + 1), 0, 1, color=ORANGE, lw=3)
    bx.text(2 * (J + 1), 1.05, rf"${J}\to{J + 1}$", ha="center", va="bottom", fontsize=12, color=ORANGE)
bx.annotate("", xy=(6, 0.5), xytext=(4, 0.5), arrowprops=dict(arrowstyle="<->", color=GRAY, lw=1.4, shrinkA=2, shrinkB=2))
bx.text(5, 0.55, r"$2\tilde{B}$", ha="center", va="bottom", fontsize=14, color=GRAY)
bx.set_xlim(0, 11.5); bx.set_ylim(0, 1.35); bx.set_yticks([])
bx.spines["left"].set_visible(False)
bx.set_xticks([0, 2, 4, 6, 8, 10]); bx.set_xticklabels(["0", r"$2\tilde{B}$", r"$4\tilde{B}$", r"$6\tilde{B}$",
                                                        r"$8\tilde{B}$", r"$10\tilde{B}$"])
bx.tick_params(labelsize=12.5)
bx.set_xlabel(r"wavenumber $\tilde{\nu}$", fontsize=13)
fig.subplots_adjust(left=0.01, right=0.99, top=0.95, bottom=0.14)
plt.show()

Fig. Left: absorptions climb one rung at a time, with gaps $2\tilde{B}$, $4\tilde{B}$, $6\tilde{B}$, .... Right: the lines they make, evenly spaced by $2\tilde{B}$.

• The gaps grow by the same amount each step, so the lines are evenly spaced: the fingerprint of a rotating molecule. Measure the spacing and you have $\tilde{B}$, then $I$, then the bond length.

• Lines between 1 and 100 cm$^{-1}$ have frequencies $\nu = c\tilde{\nu}$ from 30 GHz to 3 THz: microwaves and the far infrared.

• Swapping $^{12}$C for $^{13}$C leaves the bond length alone but raises $\mu$, so every line moves (Problem 1). Radio astronomers map the cold gas between the stars through CO’s 115 GHz line.

Line intensities follow the populations¶

How strong a line is depends first on how many molecules start in level $J$. The Boltzmann factor with the degeneracy $2J + 1$ gives:

Source

# synced: rotor_populations
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

TEAL, CARDINAL, GRAY, PURPLE, ORANGE = "#107895", "#C8102E", "#6c757d", "#6a3d9a", "#e07b00"
plt.rcParams.update({"axes.spines.top": False, "axes.spines.right": False})
hck, T = 1.438777, 300.0                                   # hc/k_B in cm K
LIGHT = "#cdb9e3"
fig, axes = plt.subplots(1, 2, figsize=(9.6, 3.6), gridspec_kw={"width_ratios": [1, 1.6], "wspace": 0.12})
for ax, (name, B, Jn) in zip(axes, (("HCl", 10.44, 16), ("CO", 1.9225, 31))):
    J = np.arange(Jn)
    allJ = np.arange(400)
    q = np.sum((2 * allJ + 1) * np.exp(-hck * B * allJ * (allJ + 1) / T))
    f = (2 * J + 1) * np.exp(-hck * B * J * (J + 1) / T) / q
    top = int(np.argmax(f))
    ax.bar(J, f, width=0.75, color=[PURPLE if j == top else LIGHT for j in J])
    ax.text(top, f[top] + 0.004, rf"$J = {top}$", ha="center", va="bottom", fontsize=13, color=PURPLE)
    ax.set_title(name, fontsize=15, loc="left")
    ax.set_xlabel("$J$", fontsize=13)
    ax.set_ylim(0, f.max() * 1.22); ax.tick_params(labelsize=11.5)
    ax.set_xlim(-0.8, Jn - 0.2)
axes[0].set_ylabel("fraction of molecules", fontsize=12.5)
fig.subplots_adjust(left=0.08, right=0.99, top=0.9, bottom=0.15)
plt.show()

Fig. The fraction of molecules in each rotational level at 300 K, for HCl ($\tilde{B} = 10.44$ cm$^{-1}$) and CO ($\tilde{B} = 1.92$ cm$^{-1}$). The most populated level is darker.

• The degeneracy $2J + 1$ rises and the Boltzmann factor falls, so the populations peak: at $J = 3$ for HCl and $J = 7$ for CO at room temperature (Problem 3 finds the peak). The heavier molecule spreads over more levels.

• Each absorption line inherits the population of its starting level. (For microwave lines the photon energy adds a factor that pushes the strongest line a few $J$ higher.)

numpy checks the HCl populations:

import numpy as np

B, T = 10.44, 300.0                       # HCl: rotational constant (cm^-1) and temperature (K)
hc_k = 1.4388                             # hc/k_B in cm K
J = np.arange(0, 40)
N = (2 * J + 1) * np.exp(-hc_k * B * J * (J + 1) / T)
print("N_J / N_0 for J = 0 to 7:", np.round(N[:8] / N[0], 2))
print("most populated J:", N.argmax(), "  fraction of molecules there:", round(N.max() / N.sum(), 3))

N_J / N_0 for J = 0 to 7: [1.   2.71 3.7  3.84 3.31 2.45 1.59 0.91]
most populated J: 3   fraction of molecules there: 0.189

Rotation rides on vibration: infrared bands¶

An infrared photon changes $v$ (Lecture 4.3), but the molecule is also rotating, and a photon must change $J$ by one as well. The energies add (Lecture 4.1):

Starting from $(v = 0, J)$, a photon can reach $(1, J + 1)$ or $(1, J - 1)$:

Source

# synced: rovib_branches
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

TEAL, CARDINAL, GRAY, PURPLE, ORANGE = "#107895", "#C8102E", "#6c757d", "#6a3d9a", "#e07b00"
plt.rcParams.update({"axes.spines.top": False, "axes.spines.right": False})
B, nu0, hck, T = 10.59, 2886.0, 1.438777, 300.0             # HCl as a rigid rotor and harmonic oscillator
Jn = 5
s = 0.5                                                     # drawing units per cm^-1 of rotational energy
y1 = 185.0                                                  # where the v = 1 ladder starts (energy axis broken)
lines = [("R", J, nu0 + 2 * B * (J + 1)) for J in range(Jn - 1)] + [("P", J, nu0 - 2 * B * J) for J in range(1, Jn)]
fig, (ax, bx) = plt.subplots(2, 1, figsize=(9.6, 5.8), sharex=True, gridspec_kw={"height_ratios": [2.6, 1], "hspace": 0.04})
x0, x1 = 2780, 2990
for J in range(Jn):
    E = B * J * (J + 1) * s
    ax.hlines(E, x0, x1, color=PURPLE, lw=1.8)
    ax.hlines(y1 + E, x0, x1, color=PURPLE, lw=1.8)
    ax.text(x1 + 2, E, rf"$J = {J}$", va="center", fontsize=11)
    ax.text(x1 + 2, y1 + E, rf"$J = {J}$", va="center", fontsize=11)
ax.text(x0 - 3, 0.5 * B * 12 * s, "$v = 0$", ha="right", va="center", fontsize=14, color=TEAL)
ax.text(x0 - 3, y1 + 0.5 * B * 12 * s, "$v = 1$", ha="right", va="center", fontsize=14, color=TEAL)
for br, J, nu in lines:
    J2 = J + 1 if br == "R" else J - 1
    ax.annotate("", xy=(nu, y1 + B * J2 * (J2 + 1) * s), xytext=(nu, B * J * (J + 1) * s),
                arrowprops=dict(arrowstyle="-|>", color=ORANGE if br == "R" else TEAL, lw=1.8, mutation_scale=12,
                                shrinkA=0, shrinkB=0))
ax.axvline(nu0, color=GRAY, lw=1, ls=(0, (4, 3)))
ax.set_xlim(x0 - 40, x1 + 25); ax.set_ylim(-6, y1 + B * 20 * s + 6); ax.axis("off")
for br, J, nu in lines:
    S = J + 1 if br == "R" else J
    I = S * np.exp(-hck * B * J * (J + 1) / T)
    bx.vlines(nu, 0, I, color=ORANGE if br == "R" else TEAL, lw=3)
    bx.text(nu, I + 0.12, f"{br}({J})", ha="center", va="bottom", fontsize=11.5, color=ORANGE if br == "R" else TEAL)
bx.axvline(nu0, color=GRAY, lw=1, ls=(0, (4, 3)))
bx.text(nu0 + 3, 3.2, r"$\tilde{\nu}_0$", ha="left", va="bottom", fontsize=13, color=GRAY)
bx.set_ylim(0, 4.0); bx.set_yticks([]); bx.spines["left"].set_visible(False)
bx.tick_params(labelsize=11.5)
bx.set_xlabel(r"wavenumber (cm$^{-1}$)", fontsize=12.5)
fig.subplots_adjust(left=0.01, right=0.99, top=0.99, bottom=0.1)
plt.show()

Fig. Top: the rotational ladders of $v = 0$ and $v = 1$ (the gap between them is not to scale). R transitions (orange) end one rung higher, P transitions (teal) one rung lower. Bottom: the lines they make, lined up under the arrows, for HCl at 300 K. The two branches have lines $2\tilde{B}$ apart, with a gap at the band origin $\tilde{\nu}_0$.

• The gap. R(0) and P(1) are $4\tilde{B}$ apart, twice the usual spacing: the missing line at $\tilde{\nu}_0$ would need $\Delta J = 0$. Molecules with extra angular momentum about the axis, such as NO or the bend of CO$_2$, do show a central Q branch.

• The envelopes. The heights follow the populations of the starting levels, so each branch rises and falls like the curve of the last section.

Try it: pick a molecule and change the temperature. Heat spreads the molecules over higher $J$, so the branches widen; the “real molecule” setting adds the corrections of the next section.

Beyond the rigid rotor¶

The bond is longer in $v = 1$¶

A real bond stretches more easily than it compresses (the Morse well of Lecture 4.3), so the average bond length grows with $v$, and $\tilde{B}$ shrinks:

For HCl, $\tilde{B}_0 = 10.44$ and $\tilde{B}_1 = 10.14$ cm$^{-1}$. With $\tilde{B}_1$ in the upper level and $\tilde{B}_0$ in the lower one, the branches are no longer evenly spaced:

• Since $\tilde{B}_1 < \tilde{B}_0$, the R lines crowd together at high $J$ and the P lines spread apart.

• Pairs of lines that share an upper or a lower level give $\tilde{B}_1$ and $\tilde{B}_0$ separately (combination differences).

Fast rotation stretches the bond¶

A rapidly spinning molecule stretches, like a mass on a spring swung in a circle. Its moment of inertia grows and its levels fall slightly below $\tilde{B}J(J+1)$:

The centrifugal distortion constant $\tilde{D}$ is tiny, about $4\tilde{B}^3/\tilde{\nu}^2$: a stiff bond barely stretches (Problem 7).

Bigger molecules¶

• Linear molecules (OCS, HCN, CO$_2$) rotate like diatomics, with one moment of inertia and the same $J(J+1)$ ladder. CO$_2$ has no dipole, so it has no microwave spectrum.

• Nonlinear molecules turn about three axes (Lecture 4.1), with three moments of inertia. In symmetric tops such as NH$_3$ two of them are equal, and the levels need a second quantum number. Water, with three different moments, has a famously irregular spectrum. These are beyond this course.

Problems¶

Problem 1: Carbon-13¶

The $J = 0 \to 1$ line of $^{12}$C$^{16}$O is at 115.271 GHz. Predict the line of $^{13}$C$^{16}$O, assuming the same bond length, and compare with the measured 110.201 GHz. Why are such isotope lines useful to astronomers?

Problem 2: Which molecules have microwave spectra?¶

Which of these have a pure rotational spectrum: HBr, Cl$_2$, CO$_2$, OCS, H$_2$O, CH$_4$, N$_2$O, CH$_3$Cl? Explain each answer in one line.

Problem 3: The most populated level¶

Treating $J$ as continuous, show that $(2J+1)\,e^{-hc\tilde{B}J(J+1)/k_BT}$ is largest at

Evaluate it for HCl ($\tilde{B} = 10.44$ cm$^{-1}$) and CO ($\tilde{B} = 1.92$ cm$^{-1}$) at 300 K, and for HCl at 1000 K.

Problem 4: A thermometer in space¶

In a cold interstellar cloud, CO is at 10 K. Find $N_1/N_0$ and $N_2/N_0$. Why can the ratio of the $1 \to 2$ and $0 \to 1$ line strengths serve as a thermometer, and why would a hydride like HCl be a poor one at this temperature?

Problem 5: Deuterium chloride¶

From the HCl constants of Lecture 4.3 (band origin 2886 cm$^{-1}$, $\tilde{\nu}_e = 2990$ and $\tilde{\nu}_ex_e = 52$ cm$^{-1}$) and $\tilde{B}_e = 10.59$ cm$^{-1}$, predict $\tilde{B}_e$ and the band origin of D$^{35}$Cl. The measured values are 5.45 and 2091 cm$^{-1}$.

Problem 6: The microwave oven¶

A microwave oven runs at 2.45 GHz. Convert this to cm$^{-1}$ and compare with the lowest rotational absorption of water vapor, near 557 GHz. Does the oven drive a rotational transition of water? What do you think heats the food?

Problem 7: Centrifugal distortion in HCl¶

Estimate $\tilde{D} \approx 4\tilde{B}^3/\tilde{\nu}^2$ for HCl ($\tilde{B} = 10.59$, $\tilde{\nu}_e = 2990$ cm$^{-1}$). How much does it lower the level $J = 20$, and the line $J = 20 \to 21$, compared with the rigid rotor?

Problem 8: How fast does a molecule spin?¶

Treat HCl in the level $J = 10$ classically, with $L = \sqrt{J(J+1)}\,\hbar$ and $L = I\omega$. How many turns per second does it make? Compare with its vibrational frequency, $8.65\times10^{13}$ Hz. Does the bond have time to vibrate during one turn?