Energy is the one observable whose operator also runs the clock. Everything odd about this chapter follows from Ĥ doing two jobs that no other operator does at once.

Chapter 3 handled position and momentum. Chapter 4 does energy — and energy turns out to be different in kind, because its operator is not merely another observable. It is also what drives time forward.

The chapter states its destination in its first paragraph, and it is worth holding onto:

the observable properties of a quantum state with a sharply defined energy never change.

4.1 The Hamiltonian Operator

§3.5 established that observables are described by operators and gave two of them:

\hat{\mathbf r} = \mathbf r \quad\text{and}\quad \hat{\mathbf p} = -i\hbar\nabla\tag{4.1}

The energy observable gets an operator too — the Hamiltonian operator , written $\hat H$ . Phillips builds it by assumption, and says so: the relation between the operators for energy, momentum and position is taken to mirror the relation between the classical quantities. Classically $E = p^2/2m + V(\mathbf r)$ , so

\hat H = \frac{\hat{\mathbf p}^2}{2m} + V(\hat{\mathbf r})\tag{4.2}

and substituting Eq. (4.1) turns that into something you can actually apply:

\hat H = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)\tag{4.3}

You have already met this operator twice

Equation (4.3) should look familiar from two directions.

From chapter 2: it is the entire right-hand side of the Schrödinger equation. Chapter 2 assembled that expression by demanding plane waves satisfy $E = p^2/2m$ ; here it drops out of one assumption about operators.

From §3.5: applying $\hat p = -i\hbar\,\partial/\partial x$ twice gives $\hat p^2 = -\hbar^2\,\partial^2/\partial x^2$ , so $\hat p^2/2m$ is the kinetic-energy term. That was noticed there as a curiosity; here it is the definition.

And from §0.3, where the site built $\hat H$ as an explicit matrix — a tridiagonal $[1, -2, 1]$ stencil for the $\nabla^2$ plus $V$ on the diagonal — and diagonalized it numerically. That was this operator all along. What this chapter adds is the analytic theory of what that diagonalization means.

The dual role

Here is the observation the whole chapter turns on. $\hat H$ does two different jobs.

First, it is the energy observable. The recipe of §3.5 — sandwich the operator between $\Psi^*$ and $\Psi$ , integrate — gives the energy expectation value directly:

\langle E\rangle = \int \Psi^*(\mathbf r, t)\,\hat H\,\Psi(\mathbf r, t)\,\mathrm d^3\mathbf r\tag{4.4}

Second, it governs time evolution, because the Schrödinger equation, Eq. (2.17), is nothing more than

i\hbar\frac{\partial\Psi}{\partial t} = \hat H\Psi\tag{4.5}

That is the fundamental connection between energy and time, and everything in this chapter is a consequence of it.

How we will solve it

Equation (4.5) is a partial differential equation in four variables. Phillips announces the strategy plainly, and notes it is the same one used for the classical wave equation or the diffusion equation: seek a separable solution, then solve an eigenvalue problem.

Rather than introduce that abstractly, he does something better — he solves a problem you can already picture.

4.2 Normal Modes of a String

Let $\Psi(x,t)$ be the transverse displacement of a stretched string. It obeys the classical wave equation

\frac{\partial^2\Psi}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2\Psi}{\partial t^2} = 0\tag{4.6}

with $c$ the wave speed, and if the ends are fixed at $x = 0$ and $x = a$ ,

\Psi(x,t) = 0 \text{ at } x = 0 \text{ and at } x = a, \text{ for all time } t\tag{4.7}

Two symbols to keep straight — c and ω

$c$ here is the wave speed on the string, in m s⁻¹. Chapter 4 uses $c$ for three unrelated things within eight pages: this one, the third side of a box in §4.4, and $c_n$ the energy probability amplitude in §4.5. None of them is the speed of light, which does not appear in this chapter at all.

$\omega$ is worse, because all three of its meanings are angular frequencies — so the units never warn you:

Where	$\omega$ means	Set by
§4.2, here	how fast a string mode vibrates, $\omega_n = ck_n$	the string’s tension and length
§4.3	how fast a wave function’s phase rotates, $\omega = E/\hbar$	the state’s energy
§4.6	how fast an observable oscillates, $\omega = \lvert E_1-E_2\rvert/\hbar$	the difference of two energies

The first is a real motion you could photograph. The second is a rotation in the complex plane that nothing can detect — §4.3 makes that its central point. The third is the only one of the three that both involves $\hbar$ and shows up in an experiment, and it is a difference rather than a value.

Watch for it especially at the string/particle analogy: $\omega_n = ck_n$ for a string but $E_n = \hbar^2k_n^2/2m$ for a particle, so the same $k_n$ gives frequencies going as $n$ in one case and energies going as $n^2$ in the other.

Separating the variables

step 1 of 5

This is the whole technique. Watch what forces each step — nothing here is a guess.

1ASSUME a solution in which every point of the string moves with the same time dependence, differing only in how much. That is what a normal mode is, and it is the only assumption in the derivation.
$Ψ (x, t) = ψ (x) T (t)$
Equation (4.8). ψ(x) is the shape, T(t) is the common time factor.
2Substitute into the wave equation and divide through, collecting everything that depends on t on the left and everything that depends on x on the right.
$\frac{1}{T} \frac{d ^{2} T}{d t ^{2}} = \frac{c ^{2}}{ψ} \frac{d ^{2} ψ}{d x ^{2}}$
3Now the trick. A function of t alone equals a function of x alone, for EVERY x and t. Vary x with t held fixed: the left side cannot move, so neither can the right. Both sides must therefore be the same constant. Call it −ω², and set ω = ck.
$\frac{1}{T} \frac{d ^{2} T}{d t ^{2}} = - ω^{2} = \frac{c ^{2}}{ψ} \frac{d ^{2} ψ}{d x ^{2}}$
The separation constant is not chosen for convenience — its existence is forced.
4The time half is simple harmonic motion.
$\frac{d ^{2} T}{d t ^{2}} = - ω^{2} T \Rightarrow T (t) = A cos ω t + B sin ω t$
5The space half is the same equation with k in place of ω/c, and it inherits the boundary conditions from Eq. (4.7).
$\frac{d ^{2} ψ}{d x ^{2}} = - k^{2} ψ, k = \frac{ω}{c}, ψ (0) = ψ (a) = 0$

The separated equation itself is Eq. (4.9); equating each side to $-\omega^2$ splits it into Eq. (4.10) for the time factor, whose solution $T(t) = A\cos\omega t + B\sin\omega t$ is Eq. (4.11), and an equation for the spatial shape:

\frac{\mathrm d^2\psi}{\mathrm dx^2} = -k^2\psi,\quad\text{where}\quad k = \frac{\omega}{c}\tag{4.12}

\psi(0) = \psi(a) = 0\tag{4.13}

Where quantization comes from

The general solution of Eq. (4.12) is $\psi(x) = M\cos kx + N\sin kx$ . Now apply the boundary conditions, and watch what happens:

$\psi(0) = 0$ kills the cosine, so $M = 0$ .
$\psi(a) = 0$ then requires $\sin ka = 0$ , so $ka$ must be a whole multiple of $\pi$ : $k = \pi/a,\ 2\pi/a,\ 3\pi/a,\dots$

$k$ can no longer take just any value. There are infinitely many solutions, one per integer:

\psi_n(x) = N\sin k_nx,\quad\text{with}\quad k_n = \frac{n\pi}{a}\tag{4.14}

for $n = 1, 2, 3, \dots$ , and combining with Eq. (4.11) at $\omega_n = ck_n$ gives the complete normal mode:

\Psi_n(x,t) = \left[A_n\cos\omega_nt + B_n\sin\omega_nt\right]\sin k_nx\tag{4.15}

This is the only place quantization ever comes from

Nothing was quantized by decree. A differential equation admits a continuum of solutions; a boundary condition then discards all but a discrete set of them. That is the entire mechanism, and it is worth being clear about now because it is exactly what will produce quantized energies two sections from now.

A guitar string does not sound a continuum of pitches. It sounds a fundamental and its harmonics, because those are the only shapes that can fit between two fixed ends. No physicist finds that mysterious. §4.4 will show that a particle in a box is the same equation with the same boundary conditions, and its energies are discrete for the same unmysterious reason.

The strange part of quantum mechanics is not that some quantities are discrete. It is what the wave function means — and that argument was already had, in §3.2.

A₁1.00A₂0.00A₃0.00A₄0.00

One mode. The curve only scales between the dashed envelope and its mirror image — every point shares the single factor cos ωₙt, and the nodes never move. This is what makes Ψ(x, t) = ψ(x)·T(t) possible, and it is the classical shadow of a stationary state.

Figure 4.1 — the four shapes

The general motion, and a change of vocabulary

Because Eq. (4.6) is a homogeneous linear partial differential equation, any superposition of normal modes is also a solution — and in fact the general motion of the string is exactly that:

\Psi(x,t) = \sum_{n=1,2,3\dots}\left[A_n\cos\omega_nt + B_n\sin\omega_nt\right]\sin k_nx\tag{4.16}

Given the initial displacement and velocity of every point, ordinary Fourier series techniques find every $A_n$ and $B_n$ .

Phillips then does something quietly clever: he renames what he has just done. No new mathematics, only quantum vocabulary attached to steps you have already taken.

What we did	What it is called
Solved Eq. (4.12) subject to Eq. (4.13)	solved an eigenvalue problem
Found solutions exist only for $k = n\pi/a$	found the eigenvalues $k_n$
$\psi_n(x) = N\sin k_nx$	the eigenfunction belonging to $k_n$
Wrote the general motion as Eq. (4.16)	expanded in a superposition of eigenfunctions

Engineer's bridge — separation of variables is diagonalization

Discretize the string onto $N$ points and the wave operator $\partial^2/\partial x^2$ becomes the familiar tridiagonal $[1, -2, 1]/\delta x^2$ matrix — the same one §0.3 built. Then:

Eigenvalue problem $\psi'' = -k^2\psi$ with $\psi(0)=\psi(a)=0$ becomes $D\psi = -k^2\psi$ : an ordinary matrix eigenvalue problem.
Normal modes are its eigenvectors. There are $N$ of them, and the first few are discrete samples of $\sin(n\pi x/a)$ .
Eq. (4.16) — writing the general motion as a superposition — is a change of basis into the eigenvector basis, where the coupled system decouples into $N$ independent oscillators.
Separation of variables is diagonalization. It works for the same reason: in the right basis the operator is diagonal, so each coordinate evolves on its own, at its own $\omega_n$ .

This is why modal analysis works in structural engineering, why you solve coupled linear ODEs by diagonalizing the system matrix, and why the DFT diagonalizes any circular convolution.

Where the analogy breaks — and it matters for what follows. A string’s $\Psi$ is a real, measurable displacement. You can photograph it. Each mode’s $T(t) = \cos\omega_nt$ is real and passes through zero, so the string genuinely goes flat twice a cycle. The quantum analogue, coming in §4.3, has $T(t) = e^{-iEt/\hbar}$ instead — modulus one, never zero. Nothing about the state’s magnitude ever changes. That single difference between $\cos\omega t$ and $e^{-i\omega t}$ is what makes a stationary state stationary, and it is the payoff the chapter is building to.

Put a number on it — a guitar string, and why the analogy is only an analogy

Take the A string of a guitar: $a = 0.65$ m, fundamental 110 Hz. Then $\omega_1 = 2\pi(110) = 691\ \mathrm{s^{-1}}$ , and $c = \omega_1 a/\pi = 143\ \mathrm{m\,s^{-1}}$ . The harmonics sit at 110, 220, 330, 440 Hz — $\omega_n = n\omega_1$ , exactly Eq. (4.15).

Now the quantum case, and the contrast is the point. For a particle in a box §4.4 will find $E_n \propto n^2$ , not $n$ . The string’s frequencies are equally spaced; the box’s energies are not, because $\omega = ck$ for a string but $E = \hbar^2k^2/2m$ for a particle. Same eigenfunctions, different dispersion relation — the shapes $\sin k_nx$ carry over exactly, and the spectrum does not.

Which is a good reminder of what an analogy is for. §4.2 borrows the method, not the physics.

Reproduce it in NumPy

Everything §4.2 asserts, checked numerically — including the claim that separation of variables and matrix diagonalization are the same operation.

import numpy as np

a = 1.0
x = np.linspace(0, a, 200001)

print("normal modes: the string's boundary conditions quantize k")
print("=" * 60)
for n in (1, 2, 3, 4):
    k = n*np.pi/a
    psi = np.sqrt(2/a)*np.sin(k*x)
    interior = psi[1:-1]
    nodes = np.sum(np.sign(interior[:-1]) * np.sign(interior[1:]) < 0)
    lbl = f"{n}pi/a" if n > 1 else "pi/a"
    print(f"  n={n}  k_n = {lbl} = {k:<7.4f} nodes(interior) = {nodes}   "
          f"peak |psi| = {np.abs(psi).max():.4f}")
print("  (peak = sqrt(2/a) with a = 1, matching Fig. 4.1's 1.41)")

print("\northonormality of the eigenfunctions")
psi = {n: np.sqrt(2/a)*np.sin(n*np.pi*x/a) for n in (1, 2, 3)}
ip = lambda i, j: np.trapezoid(psi[i]*psi[j], x)
for n in (1, 2, 3):
    off = {1: (1, 2), 2: (1, 3), 3: (2, 3)}[n]
    print(f"  <psi_{n}|psi_{n}> = {ip(n, n):.6f}     "
          f"<psi_{off[0]}|psi_{off[1]}> = {abs(ip(*off)):.6f}")

print("\nseparation of variables IS diagonalization")
print("  discretize d2/dx2 on 400 interior points, solve the matrix eigenproblem:")
N = 400
h = a/(N+1)
D = (np.diag(-2*np.ones(N)) + np.diag(np.ones(N-1), 1) + np.diag(np.ones(N-1), -1))/h**2
w, v = np.linalg.eigh(-D)                       # -d2/dx2 has eigenvalues k^2
print(f"    {'mode':>4} {'k from matrix':>15} {'k = n*pi/a':>12} {'rel. error':>13}")
xi = np.linspace(h, a-h, N)
for n in (1, 2, 3, 4):
    k_num = np.sqrt(w[n-1])
    k_ex = n*np.pi/a
    vec = v[:, n-1]; vec = vec/np.linalg.norm(vec)
    ex = np.sin(k_ex*xi); ex = ex/np.linalg.norm(ex)
    dot = abs(vec @ ex)
    print(f"    {n:>4} {k_num:>15.6f} {k_ex:>12.6f} {abs(k_num-k_ex)/k_ex:>13.2e}")
print(f"  the eigenvectors are the sines, to a dot product of {dot:.6f}")

printsnormal modes: the string's boundary conditions quantize k ============================================================ n=1 k_n = pi/a = 3.1416 nodes(interior) = 0 peak |psi| = 1.4142 n=2 k_n = 2pi/a = 6.2832 nodes(interior) = 1 peak |psi| = 1.4142 n=3 k_n = 3pi/a = 9.4248 nodes(interior) = 2 peak |psi| = 1.4142 n=4 k_n = 4pi/a = 12.5664 nodes(interior) = 3 peak |psi| = 1.4142 (peak = sqrt(2/a) with a = 1, matching Fig. 4.1's 1.41) orthonormality of the eigenfunctions <psi_1|psi_1> = 1.000000 <psi_1|psi_2> = 0.000000 <psi_2|psi_2> = 1.000000 <psi_1|psi_3> = 0.000000 <psi_3|psi_3> = 1.000000 <psi_2|psi_3> = 0.000000 separation of variables IS diagonalization discretize d2/dx2 on 400 interior points, solve the matrix eigenproblem: mode k from matrix k = n*pi/a rel. error 1 3.141585 3.141593 2.56e-06 2 6.283121 6.283185 1.02e-05 3 9.424561 9.424778 2.30e-05 4 12.565856 12.566371 4.09e-05 the eigenvectors are the sines, to a dot product of 1.000000

What comes next

Everything on this page was classical physics. Not one line of it mentioned $\hbar$ , and a vibrating string is as concrete an object as physics has.

§4.3 now runs this exact procedure on the Schrödinger equation — same separation, same eigenvalue problem, same superposition at the end. Only one thing changes: because Eq. (4.5) is first order in time rather than second, the time factor comes out as $e^{-iEt/\hbar}$ instead of $\cos\omega t$ .

That one difference is the whole chapter.

Check yourself

0 / 7 answered

1.What is meant by the Hamiltonian's "dual role"?
2.In the separation of variables, why must both sides of $\frac{1}{T} \frac{d ^{2} T}{d t ^{2}} = \frac{c ^{2}}{ψ} \frac{d ^{2} ψ}{d x ^{2}}$ equal a constant?
3.What actually quantizes $k$ for the string?
4.In Fig. 4.1 the eigenfunctions peak at about 1.41 rather than 1. Why?
5.A single normal mode keeps its shape as it vibrates. What does that correspond to mathematically?
6.Discretized, separation of variables turns out to be which linear-algebra operation?
7.The string's frequencies go as $ω_{n} \propto n$ , but the box's energies will go as $E_{n} \propto n^{2}$ . Where does the difference come from?