4.8 Exercises¶

Difficulty markers: $\star$ = warm-up, $\star\star$ = standard, $\star\star\star$ = harder. Full solutions are given in collapsed admonitions. Try each exercise before opening the solution.

What problem are we solving?

These exercises check that you can reproduce, not just recognise, the central results of Chapter 4: that the textbook wavefunctions really are normalised and orthogonal, that the harmonic-oscillator and particle-in-a-box energies come out where the chapter said, and that the "exponential wall" of many-electron quantum mechanics is a countable, concrete number rather than a slogan. Every formula you need is restated inside the problem, so you can work each one without flipping back through the chapter.

How these exercises are graded

The original difficulty stars ($\star$ to $\star\star\star$) are kept. On top of them, each problem is tagged with a level that says what kind of thinking it asks for, so you can build up gradually:

Level A — recall. Reproduce a definition or a one-line result.
Level B — single calculation. Turn one formula into one number or one function.
Level C — multi-step. Chain two or three steps (an integral, then a substitution, then a limit).
Level D — apply / code. Run or modify a calculation and read the result.
Level E — critique / estimate. Judge an approximation, estimate an order of magnitude, or explain why a method scales the way it does.

A rough order of attack for a first pass: do all the Level A and B problems first (4.1, 4.2, 4.3, the recall parts of 4.4), then the Level C ones (4.4, 4.5), then the Level D/E ones (4.6, 4.7, 4.8).

New vocabulary

Normalisation — scaling a wavefunction so that $\int|\psi|^2\,dx = 1$, i.e. the particle is somewhere with total probability one. See the beginner glossary entry on Wavefunction.
Orthogonality — two states with $\langle\phi_m|\phi_n\rangle = 0$; they have "no overlap" and represent independent measurement outcomes.
Eigenstate / eigenvalue — a state the operator merely rescales, $\hat A\psi = a\psi$; the number $a$ is the measurable value. See the glossary entries on Eigenvalue and Eigenvector.
Expectation value — the probability-weighted average of a quantity in a given state, $\langle x\rangle = \int \psi^* x\,\psi\,dx$.

The symbols below recur across several exercises; this table collects them once so none appears undefined.

Symbol	Meaning	Units (SI)
$L$	width of the 1-D box	m
$n$	quantum number (state label), $n = 1, 2, \dots$ (box) or $0, 1, 2, \dots$ (oscillator)	—
$m$, $m_e$	particle mass; electron mass $9.109\times10^{-31}$ kg	kg
$\hbar$	reduced Planck constant, $1.055\times10^{-34}$	J s
$e$	elementary charge, $1.602\times10^{-19}$	C (1 eV $= e$ joule)
$\omega$	angular frequency of the oscillator	rad s$^{-1}$
$\ell$	oscillator length $\sqrt{\hbar/(m\omega)}$	m
$\xi$	dimensionless oscillator coordinate $x/\ell$	—
$H_n(\xi)$	$n$th Hermite polynomial	—
$\hat A$, $\hat H$	a Hermitian operator; the Hamiltonian (energy operator)	varies; J
$N_g$, $N$	number of grid points; number of electrons	—

Exercise 4.1 — Normalisation of particle-in-a-box states ($\star$, Level B)¶

Verify directly that the analytical eigenfunctions

\[\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right), \qquad n = 1, 2, 3, \ldots,\]

are normalised on $[0, L]$. Self-contained statement: "normalised on $[0,L]$" means $\int_0^L |\psi_n(x)|^2\,dx = 1$, i.e. the probability of finding the particle somewhere in the box is exactly one. You will need the double-angle identity $\sin^2\theta = \tfrac12 - \tfrac12\cos 2\theta$.

Hint

Pull the constant $2/L$ outside the integral, replace $\sin^2$ by the double-angle identity, and integrate term by term. The cosine term integrates to a sine that vanishes at both ends because its argument is a whole number of half-periods. You are left with $\tfrac12 \times L$.

Solution

Compute $$\int_0^L |\psi_n|^2\, dx = \frac{2}{L}\int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right) dx.$$ Using the identity $\sin^2\theta = \tfrac12 - \tfrac12 \cos 2\theta$, $$\int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right) dx = \int_0^L \!\left[\tfrac12 - \tfrac12 \cos\!\left(\tfrac{2n\pi x}{L}\right)\right] dx = \tfrac{L}{2} - \tfrac12 \cdot \frac{L}{2n\pi}\sin\!\left(\tfrac{2n\pi x}{L}\right)\Big|_0^L = \tfrac{L}{2},$$ since $\sin(2n\pi) = \sin(0) = 0$ for any integer $n$. Therefore $\int_0^L |\psi_n|^2 dx = (2/L)(L/2) = 1$. $\blacksquare$

Exercise 4.2 — Orthogonality of non-degenerate eigenstates ($\star$, Level C)¶

Prove from first principles that two eigenfunctions of a Hermitian operator $\hat A$ belonging to different eigenvalues are orthogonal. (This was proved as a theorem in §4.2.6; rewrite the proof in your own words, without consulting the text.)

Self-contained statement: you may use two facts established earlier in the chapter. (i) Hermiticity means $\langle \phi_m | \hat A \phi_n\rangle = \langle \hat A \phi_m | \phi_n\rangle$ for all states — the operator can act "to the left" or "to the right" inside an inner product with the same result. (ii) The eigenvalues of a Hermitian operator are real, so $a_m^* = a_m$. "Orthogonal" means $\langle \phi_m | \phi_n\rangle = 0$.

Hint

Evaluate the same number $\langle \phi_m | \hat A \phi_n\rangle$ two ways: once letting $\hat A$ act on $\phi_n$ (giving $a_n$), once using Hermiticity to let it act on $\phi_m$ (giving $a_m$). Subtract. You get $(a_n - a_m)\langle \phi_m | \phi_n\rangle = 0$; now use $a_m \ne a_n$.

Solution

Let $\hat A\phi_m = a_m\phi_m$ and $\hat A\phi_n = a_n\phi_n$ with $a_m \neq a_n$. Compute $$\langle \phi_m | \hat A \phi_n\rangle = a_n \langle \phi_m | \phi_n\rangle.$$ By Hermiticity, $$\langle \phi_m | \hat A \phi_n\rangle = \langle \hat A \phi_m | \phi_n\rangle = a_m^* \langle \phi_m | \phi_n\rangle = a_m \langle \phi_m | \phi_n\rangle,$$ where we used the fact (also proved in §4.2.5) that Hermitian eigenvalues are real, so $a_m^* = a_m$. Equating the two expressions, $$(a_n - a_m)\langle \phi_m | \phi_n\rangle = 0,$$ and since $a_m \neq a_n$, the inner product must vanish: $\langle \phi_m | \phi_n\rangle = 0$. $\blacksquare$

Exercise 4.3 — Ground-state energy of an electron in a 1 Å box ($\star$, Level B)¶

Estimate the ground-state energy of an electron confined to a 1D box of width $L = 1\ \mathrm{\AA} = 10^{-10}$ m. Give the answer in eV.

Self-contained statement: the particle-in-a-box energy levels (derived in the chapter as equation (4.3.7)) are

\[E_n = \frac{n^2\pi^2\hbar^2}{2 m L^2}, \qquad n = 1, 2, 3, \ldots \tag{4.8.1}\]

so the ground state ($n=1$) is $E_1 = \pi^2\hbar^2/(2 m_e L^2)$. Use $\hbar = 1.055\times10^{-34}\ \mathrm{J\,s}$, $m_e = 9.109\times10^{-31}\ \mathrm{kg}$, and convert joules to electronvolts by dividing by $e = 1.602\times10^{-19}$ (since $1\ \mathrm{eV} = e$ joule).

Hint

Square $\hbar$ and $L$ first, keep track of the powers of ten, and only convert to eV at the very end. A useful sanity check: the answer should be tens of eV, comparable to atomic binding energies, not the milli-eV of thermal motion.

Solution

From (4.3.7), $E_1 = \pi^2 \hbar^2/(2 m_e L^2)$. Plug in: $$E_1 = \frac{\pi^2 \cdot (1.055\times 10^{-34})^2}{2 \cdot 9.109\times 10^{-31} \cdot (10^{-10})^2}\ \mathrm{J} = 6.025\times 10^{-18}\ \mathrm{J}.$$ Divide by $e = 1.602\times 10^{-19}$ C to convert to eV: $$E_1 \approx 37.6\ \mathrm{eV}.$$ For comparison, the ionisation energy of hydrogen is 13.6 eV; the particle-in-a-box ground state at $L = 1$ Å is in the same ballpark, which is why the box is a halfway-reasonable cartoon of an atom.

Exercise 4.4 — First three Hermite polynomials from the recursion ($\star\star$, Level C)¶

Starting from $H_0(\xi) = 1$ and the recursion

\[H_{n+1}(\xi) = 2\xi H_n(\xi) - 2n H_{n-1}(\xi), \tag{4.8.2}\]

derive $H_1(\xi)$, $H_2(\xi)$, and $H_3(\xi)$ explicitly. Verify that $H_2(\xi)$ has two real roots and identify their locations.

Self-contained statement: the Hermite polynomials $H_n(\xi)$ are the polynomial factors in the harmonic-oscillator eigenfunctions $\psi_n(x) \propto H_n(\xi)\,e^{-\xi^2/2}$ with $\xi = x/\ell$ (Section 4.4). The recursion (4.8.2) generates each one from the two below it. The roots of $H_n$ are the nodes (zero crossings) of the $n$th excited state, so $H_2$ having two roots says $\psi_2$ crosses zero twice.

Hint 1 — getting started

The recursion needs two lower polynomials. You are given $H_0 = 1$. To use (4.8.2) for $n = 0$ you also need $H_{-1}$: take the standard convention $H_{-1} \equiv 0$ (its coefficient $2n = 0$ at $n=0$ kills it anyway). Then the $n=0$ case gives $H_1$ directly.

Hint 2 — climbing the ladder

Substitute $n = 0$ into (4.8.2) to get $H_1$, then $n = 1$ to get $H_2$ (it needs $H_1$ and $H_0$), then $n = 2$ to get $H_3$ (it needs $H_2$ and $H_1$). Expand the brackets fully and collect like powers of $\xi$ at each step — do not leave $2\xi(4\xi^2 - 2)$ unmultiplied.

Hint 3 — the roots of $H_2$

Set your expression for $H_2$ equal to zero. It is a quadratic in $\xi$ with no linear term, so solve $4\xi^2 - 2 = 0$ for $\xi^2$, then take both square roots. Two real roots, symmetric about the origin, confirm the two nodes.

Solution

Apply the recursion with $H_{-1} \equiv 0$ and $H_0 = 1$. - $H_1 = 2\xi \cdot H_0 - 2\cdot 0 \cdot H_{-1} = 2\xi$. - $H_2 = 2\xi \cdot H_1 - 2\cdot 1 \cdot H_0 = 2\xi(2\xi) - 2 = 4\xi^2 - 2$. - $H_3 = 2\xi \cdot H_2 - 2\cdot 2 \cdot H_1 = 2\xi(4\xi^2 - 2) - 4(2\xi) = 8\xi^3 - 4\xi - 8\xi = 8\xi^3 - 12\xi$. Solving $H_2(\xi) = 4\xi^2 - 2 = 0$ gives $\xi^2 = 1/2$, hence $\xi = \pm 1/\sqrt{2}$. These are the two nodes of $\psi_2(x)$ — the wavefunction of the second excited state of the harmonic oscillator, which has two zero crossings, as expected.

Exercise 4.5 — Expectation value of position for an SHO ground state ($\star\star$, Level C)¶

Compute $\langle x\rangle$ and $\langle x^2\rangle$ in the ground state $\psi_0$ of the 1D harmonic oscillator, and use them to evaluate $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2}$. Express your answer in units of the oscillator length $\ell = \sqrt{\hbar/(m\omega)}$.

Self-contained statement: the harmonic-oscillator ground state (equation (4.4.9), written here in terms of $\ell$) is the normalised Gaussian

\[\psi_0(x) = \frac{1}{(\pi\ell^2)^{1/4}}\,\exp\!\left(-\frac{x^2}{2\ell^2}\right). \tag{4.8.3}\]

An expectation value is the probability-weighted average $\langle f\rangle = \int_{-\infty}^{\infty}\psi_0^*\,f\,\psi_0\,dx = \int_{-\infty}^{\infty} f(x)\,|\psi_0(x)|^2\,dx$, and $\Delta x$ is the root-mean-square spread about that average. You may use the standard Gaussian integral

\[\int_{-\infty}^{\infty} x^2\,e^{-x^2/\ell^2}\,dx = \frac{\ell^3\sqrt{\pi}}{2}. \tag{4.8.4}\]

Hint

Do not integrate $\langle x\rangle$ by brute force: $|\psi_0|^2$ is an even function and $x$ is odd, so the integrand is odd and the integral over the symmetric range $(-\infty,\infty)$ is zero by symmetry — hence $\langle x\rangle = 0$. For $\langle x^2\rangle$, note $|\psi_0|^2 = (\pi\ell^2)^{-1/2} e^{-x^2/\ell^2}$ and plug straight into (4.8.4). With $\langle x\rangle = 0$, $\Delta x = \sqrt{\langle x^2\rangle}$.

Solution

The ground state from (4.4.9) is $$\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\!\left(-\frac{m\omega x^2}{2\hbar}\right) = \frac{1}{(\pi\ell^2)^{1/4}}\exp\!\left(-\frac{x^2}{2\ell^2}\right).$$ Since $|\psi_0|^2$ is even, $\langle x\rangle = 0$ by symmetry. For $\langle x^2\rangle$ use the standard Gaussian integral $$\int_{-\infty}^\infty x^2 e^{-x^2/\ell^2}\, dx = \frac{\ell^3 \sqrt\pi}{2}.$$ Then $$\langle x^2\rangle = \frac{1}{\ell \sqrt\pi}\int_{-\infty}^\infty x^2 e^{-x^2/\ell^2}\, dx = \frac{1}{\ell\sqrt\pi}\cdot \frac{\ell^3 \sqrt\pi}{2} = \frac{\ell^2}{2}.$$ Hence $\Delta x = \ell/\sqrt{2}$. By an analogous calculation $\Delta p = \hbar/(\sqrt{2}\ell)$, and the product saturates the uncertainty bound: $\Delta x\, \Delta p = \hbar/2$. The harmonic-oscillator ground state is a minimum-uncertainty state.

Exercise 4.6 — Modifying the FD code for a double well ($\star\star$, Level D)¶

Take the finite-difference code of §4.3 and modify it to solve the double-well potential

\[V(x) = V_0 \left[\left(\frac{x}{a}\right)^2 - 1\right]^2, \tag{4.8.5}\]

with $V_0 = 1.0$ eV and $a = 0.5$ nm. Plot the lowest four eigenstates and their probability densities, and explain the pattern of (near-)degeneracies you see in terms of tunnelling between the two wells.

Self-contained statement: the finite-difference method (Section 4.3) discretises $-\tfrac{\hbar^2}{2m}\psi'' + V(x)\psi = E\psi$ on a grid of spacing $h$, turning the Hamiltonian into a tridiagonal matrix: $2\,\mathrm{pre} + V_i$ on the diagonal and $-\mathrm{pre}$ on the two neighbours, with $\mathrm{pre} = \hbar^2/(2 m h^2)$. Diagonalising it (np.linalg.eigh) returns the eigenvalues (energies) and eigenvectors (wavefunctions). The potential (4.8.5) is a "W" shape: two minima at $x = \pm a$ separated by a hump of height $V_0$ at $x = 0$. The runtime is plain NumPy/Matplotlib, so this one does run live in the browser.

Hint

You only need to change the potential, not the solver. Replace the box or oscillator $V$ with V = V0 * ((x/a)**2 - 1)**2 (with V0 and a in SI units), keep a grid wide enough to contain both wells with room to spare (a few nm), and diagonalise as before. After solving, look at the spacing of the first four eigenvalues: you should find them in two close pairs, not four evenly spaced levels.

Solution

A complete script:

import numpy as np
import matplotlib.pyplot as plt

HBAR = 1.054_571_817e-34
M_E  = 9.109_383_7e-31
EV   = 1.602_176_634e-19

def build_H(x, V, mass=M_E):
    h = x[1] - x[0]
    pre = HBAR**2 / (2 * mass * h**2)
    n = x.size
    H = (np.diag(2*pre*np.ones(n) + V)
         + np.diag(-pre*np.ones(n-1), k=1)
         + np.diag(-pre*np.ones(n-1), k=-1))
    return H

L = 3.0e-9
n_grid = 1200
x = np.linspace(-L/2, L/2, n_grid)

V0 = 1.0 * EV
a  = 0.5e-9
V  = V0 * ((x/a)**2 - 1)**2

H = build_H(x, V)
eigvals, eigvecs = np.linalg.eigh(H)
eigvecs /= np.sqrt(x[1] - x[0])

fig, ax = plt.subplots(figsize=(8, 5))
ax.plot(x*1e9, V/EV, "k", lw=1.5)
scale = 0.15
for n in range(4):
    ax.plot(x*1e9, eigvals[n]/EV + scale*eigvecs[:, n]/np.max(np.abs(eigvecs[:, n])),
            label=f"n={n}, E={eigvals[n]/EV:.4f} eV")
    ax.axhline(eigvals[n]/EV, color="gray", ls=":", lw=0.5)
ax.set_xlabel("x (nm)")
ax.set_ylabel("E (eV)")
ax.set_ylim(-0.05, 1.5)
ax.legend()
plt.tight_layout()
plt.show()

What you should see. The lowest two eigenvalues come in a closely-spaced pair (symmetric and antisymmetric combinations of "left-well" and "right-well" states), and similarly the third and fourth. The splitting within each pair is the tunnel splitting: even though the two wells are separated by a barrier of height $V_0$, the wavefunction has a small but non-zero amplitude in the classically forbidden region, allowing the particle to be in a coherent superposition of "in the left well" and "in the right well". The smaller the barrier or the lighter the particle, the larger the tunnel splitting. This is the prototype of every quantum-mechanical double well — ammonia inversion, hydrogen bonds, ferroelectric phase transitions.

Exercise 4.7 — Energies via Hellmann–Feynman ($\star\star$, Level C)¶

For a normalised energy eigenstate $|\psi_\lambda\rangle$ of a Hamiltonian $\hat{H}(\lambda)$ depending on a parameter $\lambda$, prove the Hellmann–Feynman theorem:

\[\frac{dE(\lambda)}{d\lambda} = \langle\psi_\lambda | \frac{\partial \hat{H}}{\partial \lambda} | \psi_\lambda\rangle.\]

Self-contained statement: "normalised" means $\langle\psi_\lambda|\psi_\lambda\rangle = 1$ for every $\lambda$, and the eigenvalue equation is $\hat H(\lambda)|\psi_\lambda\rangle = E(\lambda)|\psi_\lambda\rangle$. The notation $\partial_\lambda \equiv \partial/\partial\lambda$. The point of the theorem is that to get $dE/d\lambda$ you only differentiate the operator, never the (hard-to-find) wavefunction.

Hint

Start from $E(\lambda) = \langle\psi_\lambda|\hat H|\psi_\lambda\rangle$ and differentiate using the product rule — three terms appear, because $\langle\psi_\lambda|$, $\hat H$, and $|\psi_\lambda\rangle$ each depend on $\lambda$. In the term where $\hat H$ hits $|\partial_\lambda\psi_\lambda\rangle$, use the eigenvalue equation; in the term where it hits $\langle\partial_\lambda\psi_\lambda|$, use Hermiticity to move it onto $|\psi_\lambda\rangle$. The two wavefunction-derivative terms then combine into $E\,\partial_\lambda\langle\psi_\lambda|\psi_\lambda\rangle = E\,\partial_\lambda(1) = 0$.

Solution

By assumption, $\hat{H}(\lambda)|\psi_\lambda\rangle = E(\lambda)|\psi_\lambda\rangle$ with $\langle\psi_\lambda|\psi_\lambda\rangle = 1$. Take the inner product with $\langle\psi_\lambda|$: $$E(\lambda) = \langle\psi_\lambda | \hat{H} | \psi_\lambda\rangle.$$ Differentiate with respect to $\lambda$: $$\frac{dE}{d\lambda} = \langle \partial_\lambda \psi_\lambda | \hat{H} | \psi_\lambda\rangle + \langle\psi_\lambda | \partial_\lambda \hat{H} | \psi_\lambda\rangle + \langle\psi_\lambda | \hat{H} | \partial_\lambda \psi_\lambda\rangle.$$ Use the eigenvalue equation in the first and third terms, and the Hermiticity of $\hat{H}$: $$\langle \partial_\lambda \psi_\lambda | \hat{H} | \psi_\lambda\rangle + \langle\psi_\lambda | \hat{H} | \partial_\lambda \psi_\lambda\rangle = E(\lambda)\bigl[\langle \partial_\lambda \psi_\lambda | \psi_\lambda\rangle + \langle\psi_\lambda | \partial_\lambda \psi_\lambda\rangle\bigr] = E(\lambda)\, \partial_\lambda \langle\psi_\lambda|\psi_\lambda\rangle = E(\lambda)\cdot 0 = 0,$$ using normalisation. Hence $$\frac{dE}{d\lambda} = \langle\psi_\lambda | \partial_\lambda \hat{H} | \psi_\lambda\rangle. \quad\blacksquare$$ Application. Taking $\lambda \to \mathbf R_I$ (a nuclear position) gives the force formula (4.6.14) that we used in the BO section. The force on a nucleus is the expectation value of the gradient of the Hamiltonian — no need to differentiate the wavefunction, which is the gift that makes ab-initio MD feasible.

Exercise 4.8 — Estimating the size of a many-electron Hilbert space ($\star\star\star$, Level E)¶

Consider $N$ spinless electrons confined to a 1D box of width $L$, discretised on a grid of $N_g$ points. How many antisymmetric basis states are there? Evaluate this number for $N = 20$ electrons and $N_g = 100$ grid points, and compare with $N_g^N$ (the dimension if antisymmetry were not enforced).

Self-contained statement: a many-electron wavefunction must be antisymmetric under swapping any two electrons (the Pauli principle, Section 4.5); for spinless electrons this forbids two electrons sitting on the same grid point. The number of ways to choose $N$ distinct points out of $N_g$, order not mattering, is the binomial coefficient

\[\binom{N_g}{N} = \frac{N_g!}{N!\,(N_g - N)!}. \tag{4.8.6}\]

Without antisymmetry, each of the $N$ electrons could independently sit on any of the $N_g$ points, giving $N_g^N$ arrangements.

Hint 1 — why a binomial coefficient

Antisymmetry (a Slater determinant) is built from $N$ distinct single-particle states, and swapping two of them only flips the overall sign — it is not a new physical state. So you are choosing an unordered set of $N$ occupied grid points out of $N_g$. "Choose $N$ from $N_g$, order irrelevant" is exactly $\binom{N_g}{N}$.

Hint 2 — evaluating $\binom{100}{20}$ without overflow

Do not compute $100!$ directly — it overflows ordinary floating point. Instead build the product incrementally, $\binom{N_g}{N} = \prod_{i=0}^{N-1}\frac{N_g - i}{i + 1}$, multiplying and dividing as you go so the running value stays moderate. In Python, math.comb(100, 20) gives the exact integer ($\approx 5.36\times10^{20}$).

Hint 3 — the comparison

Compare $\binom{100}{20}\approx 5.4\times10^{20}$ with $N_g^N = 100^{20} = 10^{40}$. The ratio is about $1.9\times10^{19}$, which is of order $N! = 20! \approx 2.4\times10^{18}$ (a few times larger). So enforcing antisymmetry removes roughly a factor of $N!$ — a huge saving in relative terms, yet the surviving count is still astronomically large. That is the quantitative face of the "exponential wall".

Solution

Place a single electron on the grid in one of $N_g$ ways. With $N$ electrons and antisymmetry, two electrons cannot share the same grid point (Pauli, since we are taking them to be spinless or all of one spin). The number of antisymmetric basis states is therefore the number of ways to choose $N$ distinct grid points out of $N_g$ — independent of order, because the determinantal structure fixes the sign: $$\dim \mathcal H_{\mathrm{antisym}} = \binom{N_g}{N}.$$ For $N_g = 100$, $N = 20$: $$\binom{100}{20} = \frac{100!}{20!\,80!} \approx 5.36\times 10^{20}.$$ Without antisymmetry, the unrestricted Hilbert space would have $N_g^N = 100^{20} = 10^{40}$ basis states. Antisymmetry reduces the count by a factor of $N! \cdot (\text{double-occupancy combinatorics}) \sim N!$, but the resulting number is still astronomical — 20 electrons on a 100-point grid require half a sextillion basis states. This is a quantitative version of the exponential wall of §4.5, and explains why FCI is impossible for any system with more than ~20 electrons in modest basis sets. Take-away. Pauli antisymmetry helps — it removes a factor of $\sim N!$ from the basis count — but it does not save us from exponential scaling. Approximate methods (HF, DFT, MP2, CCSD, …) that scale polynomially in $N$ are the only way forward.

Check yourself¶

Before moving on, make sure you can answer these without looking back at the solutions. They test the threads running through the whole exercise set.

Common misunderstandings

Normalisation is about $|\psi|^2$, not $\psi$. In Exercise 4.1 the condition is $\int|\psi_n|^2\,dx = 1$. The wavefunction $\psi$ itself can be negative; only its squared modulus is a probability density.
"Orthogonal" needs different eigenvalues. The proof in Exercise 4.2 relies on $a_m \ne a_n$. For degenerate states (equal eigenvalues) the theorem says nothing — you have to orthogonalise them by hand.
A node count is not an energy. In Exercise 4.4, $H_2$ having two roots tells you $\psi_2$ has two nodes; it does not directly give the energy. More nodes does correlate with higher energy, but the energy comes from the eigenvalue, not from counting zeros.
Antisymmetry helps but does not rescue you. In Exercise 4.8 the Pauli principle shrinks the basis by a factor of order $N!$, yet $\binom{100}{20}$ is still $\sim10^{20}$. The wall is exponential; a factorial prefactor does not defeat it.

Check yourself

In Exercise 4.1, where does the factor $\sqrt{2/L}$ in $\psi_n = \sqrt{2/L}\,\sin(n\pi x/L)$ come from, and what would the integral $\int_0^L|\psi_n|^2\,dx$ equal if you forgot it?
State the two facts about a Hermitian operator that the orthogonality proof of Exercise 4.2 relies on.
The ground-state energy of an electron in a $1\,\mathrm{\AA}$ box (Exercise 4.3) came out near $38\,\mathrm{eV}$. If you doubled the box width to $2\,\mathrm{\AA}$, by what factor would $E_1$ change?
Using the recursion (4.8.2), how many real roots does $H_3(\xi)$ have, and what does that tell you about $\psi_3$?
For the oscillator ground state (Exercise 4.5), why is $\langle x\rangle$ zero without doing any integral?
In Exercise 4.8, why is the antisymmetric count a binomial coefficient $\binom{N_g}{N}$ rather than $N_g^{\,N}$?

Answer

It is the normalisation constant: $\int_0^L \sin^2(n\pi x/L)\,dx = L/2$, so to make $\int_0^L|\psi_n|^2\,dx = 1$ we need $(2/L)(L/2) = 1$. Without the factor, the integral would equal $L/2$, not $1$ — the state would not be normalised.
(i) Hermiticity, $\langle\phi_m|\hat A\phi_n\rangle = \langle\hat A\phi_m|\phi_n\rangle$; and (ii) the eigenvalues are real, $a_m^* = a_m$.
From $E_1 = \pi^2\hbar^2/(2 m_e L^2)\propto 1/L^2$, doubling $L$ divides $E_1$ by $2^2 = 4$, giving roughly $9.4\,\mathrm{eV}$. Wider box, lower confinement energy.
$H_3(\xi) = 8\xi^3 - 12\xi = 4\xi(2\xi^2 - 3)$ has three real roots, at $\xi = 0$ and $\xi = \pm\sqrt{3/2}$. So $\psi_3$ has three nodes — the standard "$n$ nodes in the $n$th excited state" pattern.
Because $|\psi_0|^2$ is even in $x$ while $x$ is odd, so the integrand $x\,|\psi_0|^2$ is odd and integrates to zero over the symmetric range $(-\infty,\infty)$. The state is symmetric about the origin, so its average position is the centre.
Antisymmetry (a Slater determinant) is built from $N$ distinct occupied points, and reordering them only flips a sign rather than giving a new state. So you choose an unordered set of $N$ points from $N_g$ — exactly $\binom{N_g}{N}$. The $N_g^{\,N}$ count would treat ordered, possibly repeated occupations as distinct, which antisymmetry forbids.

Going further¶

If you want more practice, try:

Solve numerically for an electron in a finite square well of depth $V_0 = 5$ eV and width $L = 1$ nm. Count the number of bound states and compare with the analytical transcendental-equation result.
Repeat the SHO calculation in atomic units ($\hbar = m_e = 1$). Confirm that $E_n = \omega(n + 1/2)$ with $\omega = 1$ gives eigenvalues $0.5, 1.5, 2.5, \ldots$ hartree.
For the double-well potential of Exercise 4.6, plot the tunnel splitting $\Delta E_{01} = E_1 - E_0$ as a function of barrier height $V_0$. Verify that it decays exponentially with $\sqrt{V_0}$ — the WKB prediction.
Compute the Hartree–Fock ground-state energy of the helium atom analytically using a single-zeta Slater orbital $\phi(\mathbf r) = (\zeta^3/\pi)^{1/2} e^{-\zeta r}$, minimising over $\zeta$. Compare with the experimental value (-79.0 eV) and identify the correlation energy.

In Chapter 5 we will leave wavefunctions behind and reformulate everything in terms of the electron density. The mathematical machinery developed in this chapter — Hermitian operators, eigenvalue problems, finite-difference discretisations, SCF cycles — will all reappear, but in a form that can finally be applied to real materials.

Symbol	Meaning	Units (SI)
\(L\)	width of the 1-D box	m
\(n\)	quantum number (state label), \(n = 1, 2, \dots\) (box) or \(0, 1, 2, \dots\) (oscillator)	—
\(m\), \(m_e\)	particle mass; electron mass \(9.109\times10^{-31}\) kg	kg
\(\hbar\)	reduced Planck constant, \(1.055\times10^{-34}\)	J s
\(e\)	elementary charge, \(1.602\times10^{-19}\)	C (1 eV \(= e\) joule)
\(\omega\)	angular frequency of the oscillator	rad s\(^{-1}\)
\(\ell\)	oscillator length \(\sqrt{\hbar/(m\omega)}\)	m
\(\xi\)	dimensionless oscillator coordinate \(x/\ell\)	—
\(H_n(\xi)\)	\(n\)th Hermite polynomial	—
\(\hat A\), \(\hat H\)	a Hermitian operator; the Hamiltonian (energy operator)	varies; J
\(N_g\), \(N\)	number of grid points; number of electrons	—

4.8 Exercises¶

Exercise 4.1 — Normalisation of particle-in-a-box states (\(\star\), Level B)¶

Exercise 4.2 — Orthogonality of non-degenerate eigenstates (\(\star\), Level C)¶

Exercise 4.3 — Ground-state energy of an electron in a 1 Å box (\(\star\), Level B)¶

Exercise 4.4 — First three Hermite polynomials from the recursion (\(\star\star\), Level C)¶

Exercise 4.5 — Expectation value of position for an SHO ground state (\(\star\star\), Level C)¶

Exercise 4.6 — Modifying the FD code for a double well (\(\star\star\), Level D)¶

Exercise 4.7 — Energies via Hellmann–Feynman (\(\star\star\), Level C)¶

Exercise 4.8 — Estimating the size of a many-electron Hilbert space (\(\star\star\star\), Level E)¶

Check yourself¶

Going further¶