Nothing new is derived here. Both halves spend the previous section’s result — once on a real molecule, and once on a third dimension that costs no extra mathematics.

§6.3 worked out what the oscillator’s states are. These two short sections spend that capital: first on a real system — two nuclei vibrating in a chemical bond, where $\hbar\omega$ becomes an infrared photon and Eq. (6.12) becomes an instrument — and then on the same oscillator in three dimensions, where the evenly spaced ladder acquires degeneracy.

Physics you need first — what holds a molecule together, and what 'reduced mass' buys you

A chemical bond has a potential energy curve. Two atoms far apart barely interact; pushed close, their nuclei repel fiercely; in between there is a separation $r_0$ where the energy is lowest. That curve, $V_e(r)$ , is the net result of the electrons’ quantum behaviour and all the Coulomb interactions — a calculation this book does not attempt. Chapter 6 needs only its shape near the minimum, and §6.1 established what any smooth minimum looks like: a parabola.

The reduced mass turns two bodies into one. Two particles vibrating about each other is a two-body problem, but only their separation matters. Writing $x = r - r_0$ and using the reduced mass $\mu = m_1m_2/(m_1+m_2)$ turns it into a single fictitious particle of mass $\mu$ on a spring — the same trick §5.2b used for two protons. For two equal masses $\mu = m/2$ ; for very unequal ones, like HCl, $\mu$ is close to the lighter mass, which is why hydrogen dominates the vibration.

An infrared photon carries $10^{-1}$ eV or so. Visible light is 1.5–3 eV; infrared is below that, wavelengths of microns rather than hundreds of nanometres. Molecular vibrations land there, which is why infrared spectroscopy is the tool for identifying chemical bonds.

6.4 Diatomic molecules

Near its minimum, the internuclear potential is quadratic:

V_e(r) \simeq \tfrac12kx^2,\qquad x = r - r_0

so $k$ is an effective elastic constant characterizing the strength of the bond. Classically the two nuclei have energy $p_1^2/2m_1 + p_2^2/2m_2 + \frac12kx^2$ , and in the centre-of-mass frame $p_1 = p_2 = p$ , which collapses to

E_{\text{classical}} = \frac{p^2}{2\mu} + \tfrac12kx^2,\qquad \mu = \frac{m_1m_2}{m_1+m_2}

— the energy of one particle of mass $\mu$ on a spring. So the quantum problem is the one already solved, with $m \to \mu$ :

i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2\mu}\frac{\partial^2}{\partial x^2} + \frac12kx^2\right]\Psi\tag{6.18}

Equation (6.18) is printed with k², and it should be k

Page 120 gives the potential term as $\frac12k^2x^2$ . The superscript does not belong there, and three independent checks agree:

Dimensions. $k$ is in $\mathrm{N\,m^{-1}}$ , so $\frac12k^2x^2$ carries $\mathrm{N^2}$ — not an energy. $\frac12kx^2$ gives joules, as Eq. (6.1) requires.
The book’s own next line. Eq. (6.19) states $\omega = \sqrt{k/\mu}$ . A potential $\frac12k^2x^2$ would give $\omega = k/\sqrt\mu$ instead.
Eq. (6.1), which defined the potential as $\frac12kx^2$ eight pages earlier and never changed it.

Verified from page-137.png, so it is the book’s slip and not the text extraction. This is the second error found in chapter 6’s printed text, after the “Section 4.8” cross-reference that §6.3 inherits from page 114.

Everything from §6.3 now applies. In particular the vibrational energy levels are

E_n = \left(n + \tfrac12\right)\hbar\omega,\qquad \omega = \sqrt{\frac{k}{\mu}}\tag{6.19}

Weighing a chemical bond

A transition between adjacent levels emits or absorbs a photon of energy $E = \hbar\sqrt{k/\mu}$ — adjacent because $x_{m,n}$ vanishes unless $|m-n| = 1$ , the selection rule of §6.3 appearing here as the book’s footnote 4. The corresponding wavelength is

\lambda = \frac{hc}{E} = 2\pi c\sqrt{\frac{\mu}{k}}\tag{6.20}

Put a number on it — the carbon monoxide bond

Carbon monoxide has reduced mass $\mu = 6.85\ \mathrm{amu}$ , and transitions between adjacent vibrational levels absorb at $\lambda = 4.6\ \mu\mathrm{m}$ .

Step 1 — invert Eq. (6.20).

k = \mu\left(\frac{2\pi c}{\lambda}\right)^2

Step 2 — put the numbers in. $\mu = 6.85 \times 1.6605\times10^{-27} = 1.137\times10^{-26}\ \mathrm{kg}$ , and $2\pi c/\lambda = 2\pi(2.998\times10^8)/(4.6\times10^{-6}) = 4.095\times10^{14}\ \mathrm{s^{-1}}$ :

k = (1.137\times10^{-26})(4.095\times10^{14})^2 = 1907\ \mathrm{N\,m^{-1}}

matching the book’s $1908\ \mathrm{N\,m^{-1}}$ .

Step 3 — sanity-check it. A stiff laboratory spring is a few hundred $\mathrm{N\,m^{-1}}$ ; a car suspension spring perhaps $10^4$ . The CO bond sits between them — which is worth pausing on, because it is a chemical bond behaving like an ordinary macroscopic spring, and it came out of a wavelength.

Step 4 — the quantum, and why the molecule is in its ground state. $\hbar\omega = 0.270\ \mathrm{eV}$ , while $k_BT$ at room temperature is $0.026\ \mathrm{eV}$ — a factor of ten smaller. So essentially every CO molecule sits in $n = 0$ , which is why the transition shows up in absorption.

Reproduce it in NumPy

Two molecules through Eq. (6.20), H₂'s real anharmonic spectrum against the harmonic one, and the 3-D degeneracies obtained by literally enumerating (nₓ, n_y, n_z) rather than by quoting a formula.

import numpy as np
from itertools import product

hbar, c, eV, u = 1.054571e-34, 2.997925e8, 1.602177e-19, 1.660539e-27
kB = 1.380649e-23

print("SS 6.4 -- weighing a chemical bond with a wavelength")
print("=" * 70)
print("  Eq. (6.20): lambda = 2 pi c sqrt(mu/k)   =>   k = mu (2 pi c/lambda)^2")
print(f"  {'molecule':<10} {'mu/amu':>7} {'lambda/um':>10} {'k / N m^-1':>11} {'hbar w / eV':>12} {'kT(300K)':>10}")
for name, mu_amu, lam_um, book in (("CO", 6.85, 4.6, 1908), ("NO", 7.46, 5.33, 1548)):
    mu = mu_amu*u
    k = mu*(2*np.pi*c/(lam_um*1e-6))**2
    w = np.sqrt(k/mu)
    print(f"  {name:<10} {mu_amu:7.2f} {lam_um:10.2f} {k:11.1f} {hbar*w/eV:12.4f} "
          f"{kB*300/eV:10.4f}")
    print(f"  {'':<10} book says k = {book} N/m")
print("  Vibrational quanta are ~10x kT at room temperature, so these molecules")
print("  sit in n = 0.  That is why the transition is seen in ABSORPTION.")

print("\nSS 6.4 -- where the harmonic model dies: H2's real (Morse) spectrum")
print("=" * 70)
De, beta, mu = 4.7446*eV, 1.9426e10, 0.50391*u
k = 2*De*beta**2
w = np.sqrt(k/mu)
hw = hbar*w
print(f"  hbar omega = {hw/eV:.4f} eV      dissociation De = {De/eV:.3f} eV")
print(f"  anharmonicity (hbar omega)^2/4De = {hw**2/(4*De)/eV:.4f} eV\n")
print(f"  {'n':>3} {'harmonic/eV':>12} {'Morse/eV':>10} {'gap to n-1':>11} {'gap harmonic':>13}")
def morse_E(n): return hw*(n+0.5) - hw**2*(n+0.5)**2/(4*De)
prev = None
for n in (0, 1, 2, 5, 10, 14):
    Eh, Em = hw*(n+0.5), morse_E(n)
    gap = f"{(Em - morse_E(n-1))/eV:11.4f}" if n and n-1 >= 0 else f"{'-':>11}"
    print(f"  {n:3d} {Eh/eV:12.4f} {Em/eV:10.4f} {gap} {hw/eV:13.4f}")
nmax = int(2*De/hw - 0.5)
print(f"\n  The gaps SHRINK -- that is anharmonicity, and it is what Fig. 6.4 shows.")
print(f"  The n=0 -> n=1 gap is {(morse_E(1)-morse_E(0))/eV:.4f} eV against a measured 0.516 eV,")
print(f"  where the pure harmonic model predicts {hw/eV:.4f} eV.  Keeping one")
print(f"  anharmonic term moves the answer the right way and most of the distance.")
print(f"  Levels stop at n = {nmax}: above that the molecule is dissociated and")
print(f"  the spectrum becomes a continuum.  A parabola would have offered")
print(f"  infinitely many.")

print("\nSS 6.5 -- the 3-D oscillator, by counting")
print("=" * 70)
print("  E = (nx + ny + nz + 3/2) hbar omega, so only the SUM N matters.")
print(f"  {'N':>3} {'E/hbar w':>9} {'degeneracy':>11} {'(N+1)(N+2)/2':>14} {'states':>28}")
for N in range(5):
    combos = [t for t in product(range(N+1), repeat=3) if sum(t) == N]
    shown = ' '.join(''.join(map(str, t)) for t in combos[:6])
    if len(combos) > 6: shown += ' ...'
    print(f"  {N:3d} {N+1.5:9.1f} {len(combos):11d} {(N+1)*(N+2)//2:14d} {shown:>28}")
print("  1, 3, 6, 10, 15 -- the triangular numbers, and Fig. 6.5's right-hand")
print("  column.  Degeneracy here is pure combinatorics: the number of ways to")
print("  write N as an ordered sum of three non-negative integers.")

printsSS 6.4 -- weighing a chemical bond with a wavelength ====================================================================== Eq. (6.20): lambda = 2 pi c sqrt(mu/k) => k = mu (2 pi c/lambda)^2 molecule mu/amu lambda/um k / N m^-1 hbar w / eV kT(300K) CO 6.85 4.60 1907.3 0.2695 0.0259 book says k = 1908 N/m NO 7.46 5.33 1547.2 0.2326 0.0259 book says k = 1548 N/m Vibrational quanta are ~10x kT at room temperature, so these molecules sit in n = 0. That is why the transition is seen in ABSORPTION. SS 6.4 -- where the harmonic model dies: H2's real (Morse) spectrum ====================================================================== hbar omega = 0.5450 eV dissociation De = 4.745 eV anharmonicity (hbar omega)^2/4De = 0.0157 eV n harmonic/eV Morse/eV gap to n-1 gap harmonic 0 0.2725 0.2686 - 0.5450 1 0.8175 0.7823 0.5137 0.5450 2 1.3626 1.2647 0.4824 0.5450 5 2.9976 2.5242 0.3885 0.5450 10 5.7228 3.9971 0.2320 0.5450 14 7.9029 4.6120 0.1068 0.5450 The gaps SHRINK -- that is anharmonicity, and it is what Fig. 6.4 shows. The n=0 -> n=1 gap is 0.5137 eV against a measured 0.516 eV, where the pure harmonic model predicts 0.5450 eV. Keeping one anharmonic term moves the answer the right way and most of the distance. Levels stop at n = 16: above that the molecule is dissociated and the spectrum becomes a continuum. A parabola would have offered infinitely many. SS 6.5 -- the 3-D oscillator, by counting ====================================================================== E = (nx + ny + nz + 3/2) hbar omega, so only the SUM N matters. N E/hbar w degeneracy (N+1)(N+2)/2 states 0 1.5 1 1 000 1 2.5 3 3 001 010 100 2 3.5 6 6 002 011 020 101 110 200 3 4.5 10 10 003 012 021 030 102 111 ... 4 5.5 15 15 004 013 022 031 040 103 ... 1, 3, 6, 10, 15 -- the triangular numbers, and Fig. 6.5's right-hand column. Degeneracy here is pure combinatorics: the number of ways to write N as an ordered sum of three non-negative integers.

Where the model breaks — Figure 6.4

A real bond is not a parabola, and the difference shows up in the spectrum. The harmonic ladder climbs forever; a real molecule has a dissociation energy $D_e$ , above which the two nuclei simply part company and the levels give way to a continuum. Both ladders are drawn on the same well below.

the real H₂ bond, and its true levels
harmonic ladder, ħω(n+½) — evenly spaced forever
dissociation energy Dₑ

ħω 0.545 eV

n = 0 → 1 gap 0.5137 eV

bound levels 17

Dₑ 4.74 eV

Two ladders on one well. The dashed one is Eq. (6.19), evenly spaced at ħω = 0.545 eV and continuing upward forever — straight through the dissociation energy and out of the picture. The solid one is what the real bond does: the levels converge as they climb, and they stop, at 17 of them. Above Dₑ the molecule is in pieces and the spectrum is a continuum.

Near the bottom the two agree, which is why §6.4 can measure a bond with Eq. (6.20) at all: the n = 0 → 1 gap is 0.5137 eV against the harmonic model’s 0.5450 eV, and the measured value for H₂ is 0.516 eV. The correction is a few percent at the bottom and total at the top — which is exactly the shape of every good approximation.

Real vibrational spectra are not one line, and the book says so

Two complications the book raises and does not pursue, both worth knowing before you look at a real spectrum:

The levels are not exactly evenly spaced. The widget above shows why — a bond is not a parabola, and the gaps shrink as you climb. So the $n = 0 \to 1$ , $1 \to 2$ , $2 \to 3$ transitions have slightly different wavelengths, and the single line of Eq. (6.20) is really the first of a converging series.

Each vibrational level is a band. A molecule also rotates, and rotational energies are far smaller than vibrational ones — so every vibrational level is split into a ladder of closely spaced rotational sublevels, and every vibrational transition becomes a band of lines rather than one. The rotational structure needs angular momentum, which is chapter 8.

Neither complication touches the argument: the first transition still measures $k$ , to a few percent.

6.5 Three-dimensional oscillators

Now the same potential in three dimensions:

V(\mathbf r) = \tfrac12kr^2 = \tfrac12k(x^2+y^2+z^2)\tag{6.21}

The Hamiltonian is then a sum of three one-dimensional Hamiltonians:

\hat H = \hat H_x + \hat H_y + \hat H_z\tag{6.22}

with $\hat H_x = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac12m\omega^2x^2$ and likewise for $y$ and $z$ . Stationary states have the usual form

\Psi(x,y,z,t) = \psi(x,y,z)\,e^{-iEt/\hbar}\tag{6.23}

with $\hat H\psi = E\psi$ , Eq. (6.24).

step 1 of 5

This is the whole of §6.5, and it is four lines. The technique is worth more than the result: it is how every separable problem in the rest of the book is solved.

1Each one-dimensional Hamiltonian has its own eigenvalue equation, already solved in §6.3. Note that Ĥ_x involves only x — it treats y and z as constants.
$\hat{H}_{x} ψ_{n_{x}} (x) = (n_{x} + \frac{1}{2}) ℏ ω ψ_{n_{x}} (x)$
And identically for y and z, with their own quantum numbers.
2GUESS a product. There is no guarantee this works — it works only because Ĥ splits into a sum with each term acting on one variable, which is exactly what Eq. (6.22) says.
$ψ_{n_{x}, n_{y}, n_{z}} (x, y, z) = ψ_{n_{x}} (x) ψ_{n_{y}} (y) ψ_{n_{z}} (z)$
Equation (6.25).
3Apply Ĥ to the product. Ĥ_x differentiates only x, so ψ_{n_y}ψ_{n_z} passes through it as a constant factor — and the same for the other two terms.
$\hat{H} ψ = [\hat{H}_{x} ψ_{n_{x}}] ψ_{n_{y}} ψ_{n_{z}} + ψ_{n_{x}} [\hat{H}_{y} ψ_{n_{y}}] ψ_{n_{z}} + ψ_{n_{x}} ψ_{n_{y}} [\hat{H}_{z} ψ_{n_{z}}]$
4Substitute the three one-dimensional results. Every term now carries the SAME product, so it factors out — and what is left multiplying it is a number, which makes the product an eigenfunction.
$\hat{H} ψ = [(n_{x} + \frac{1}{2}) + (n_{y} + \frac{1}{2}) + (n_{z} + \frac{1}{2})] ℏ ω ψ$
5Collect the three halves into a 3/2. The energy depends on the three quantum numbers ONLY through their sum — which is the entire origin of the degeneracy below.
$E_{n_{x}, n_{y}, n_{z}} = (n_{x} + n_{y} + n_{z} + \frac{3}{2}) ℏ ω$
Written out as Eqs. (6.26) and (6.27) below. The ground state is 3/2 ħω — three independent zero-point contributions of ½ħω each.

So the product of Eq. (6.25) does satisfy the three-dimensional eigenvalue equation,

\hat H\psi_{n_x,n_y,n_z}(x,y,z) = E_{n_x,n_y,n_z}\,\psi_{n_x,n_y,n_z}(x,y,z)\tag{6.26}

provided that

E_{n_x,n_y,n_z} = \left(n_x + n_y + n_z + \tfrac32\right)\hbar\omega\tag{6.27}

Degeneracy is combinatorics, not physics

$E$ depends on $n_x$ , $n_y$ , $n_z$ only through $N = n_x + n_y + n_z$ . So every distinct triple with the same sum gives a different state at the same energy — which is exactly what degeneracy means.

Counting them is a problem with no quantum mechanics in it at all: how many ways can $N$ be written as an ordered sum of three non-negative integers? The answer is $\binom{N+2}{2} = \frac{(N+1)(N+2)}{2}$ — the triangular numbers 1, 3, 6, 10, 15, … , which is Fig. 6.5’s right-hand column and which the code above obtains by literal enumeration.

The book notes that this degeneracy “arises because the Hamiltonian has rotational and other symmetries”, and that is the deeper statement: $V$ depends only on $r$ , so nothing distinguishes the $x$ , $y$ and $z$ directions, and states that differ only by how excitation is shared between them must have equal energy. Chapter 9 makes this precise, and chapter 11 turns it into the periodic table.

⇅	1-D oscillator⇅	3-D oscillator⇅	3-D box (§4.4)⇅
energy	$(n + \frac{1}{2}) ℏ ω$ ⓘ	$(N + \frac{3}{2}) ℏ ω$ ⓘ	$\propto n_{x}^{2} + n_{y}^{2} + n_{z}^{2}$ ⓘ
ground state	$\frac{1}{2} ℏ ω$ ⓘ	$\frac{3}{2} ℏ ω$ ⓘ	$3 \times \frac{ℏ ^{2} π ^{2}}{2 m a ^{2}}$ ⓘ
degeneracy pattern	noneⓘ	$1, 3, 6, 10, 15 \dots$ ⓘ	irregularⓘ
what breaks it	—	different $ω_{x}, ω_{y}, ω_{z}$ ⓘ	different side lengthsⓘ

Engineer's bridge — separable means tensor product, and degeneracy means repeated eigenvalues

Eq. (6.25) is a tensor product. When $\hat H = \hat H_x \otimes I \otimes I + I \otimes \hat H_y \otimes I + I \otimes I \otimes \hat H_z$ , the eigenvectors of the whole are the Kronecker products of the eigenvectors of the parts, and the eigenvalues are the sums. That is a linear-algebra fact with no physics in it, and it is why separation of variables works whenever it works — and fails the moment a cross term like $\lambda xy$ appears, because then $\hat H$ no longer splits.

Numerically it is the difference between tractable and hopeless. Discretize each axis on $N$ points and the 3-D Hamiltonian is $N^3 \times N^3$ — for $N = 600$ , a matrix with $4\times10^{16}$ entries. Separability says you never build it: solve three $N\times N$ problems and take sums and products. Every solver that handles a separable PDE is exploiting exactly this.

Degeneracy is a repeated eigenvalue, and repeated eigenvalues are where numerical linear algebra gets interesting. The eigenvectors are then not unique — any basis of the degenerate subspace will do, and eigh returns an arbitrary one. That is not a defect of the algorithm; it is the physics. The three $N = 1$ states can equally well be described as $(1,0,0), (0,1,0), (0,0,1)$ or as combinations like $\psi_{1,0,0} \pm i\psi_{0,1,0}$ — and problem 8(c) asks for exactly that combination, because those are the states of definite angular momentum.

Where it breaks. A tensor-product state is separable in the linear-algebra sense, which in quantum information is precisely the definition of not entangled. But these three factors are three coordinates of one particle, not three particles, so no entanglement is on offer here either way. The identical mathematics describes an entangled two-particle state, and telling the two situations apart is a matter of what the factors are — which is chapter 10’s subject.

Where this goes next

§6.6 finally derives Eq. (6.12) — the result §6.3 took on credit and this section has now spent twice. It does so with raising and lowering operators, which build the whole spectrum from the ground state by algebra alone, never solving a differential equation. The book marks the section optional. It is the most reusable thing in the chapter: chapter 8 constructs angular momentum by the identical argument.

Check yourself

0 / 6 answered

1.Equation (6.20) lets a wavelength measure a chemical bond. Which quantity does it deliver, and what has to be known already?
Open the "Fig. 6.4 — where it dies" view.
2.The true levels converge as they climb while the harmonic ladder stays evenly spaced. What does that convergence tell you?
3.Why is the three-dimensional oscillator's ground state $\frac{3}{2} ℏ ω$ rather than $\frac{1}{2} ℏ ω$ ?
4.The degeneracies of the 3-D oscillator go 1, 3, 6, 10, 15. Where does that sequence come from?
5.Page 120 prints Eq. (6.18) with a potential term $\frac{1}{2} k^{2} x^{2}$ . How can you tell it is a misprint without leaving the page?
6.Separation of variables turns one 3-D eigenproblem into three 1-D ones. When does that trick fail?