The book has used operators for six chapters without saying what qualifies one. Three requirements do it, and each exists to protect a prediction already being made.

Six chapters have used operators — $\hat r$ , $\hat p$ , $\hat H$ — without ever asking what an operator has to be. This chapter asks.

Phillips warns that it is “more abstract and mathematical than those encountered elsewhere” and “may be omitted without significant loss of continuity”. The second half of that is true of the results; it is not true of the method. Chapter 8 is built entirely on machinery this chapter supplies, and two of the book’s earlier assumptions get paid off here as consequences.

Physics you need first — what it means for an operator to 'be' an observable

Nothing new physically. What is new is the direction of the argument.

Until now the book has handed you an operator and shown what it does: $\hat p = -i\hbar\,\partial/\partial x$ gives momentum expectation values, $\hat H$ gives energies. This chapter runs the other way — it asks what properties an operator must have if its eigenvalues are to be measurement outcomes, and then checks that the familiar ones qualify.

Three facts from earlier chapters do all the work:

Eigenvalues are the possible outcomes of a measurement (§4.3). So they had better be real numbers.
Any state is a superposition of eigenfunctions (§4.5), with $|c_n|^2$ the probability of outcome $n$ . So the eigenfunctions had better span everything.
Expectation values are $\int\Psi^*\hat A\Psi$ (§3.5). So those had better be real too.

Each of those “had better” becomes a mathematical requirement below.

7.1 The three requirements

Take $\hat H$ as the prototype and ask what any operator $\hat A$ describing an observable must satisfy.

It must be linear

Every observable’s operator must be a linear operator :

\hat A(c_1\Psi_1 + c_2\Psi_2) = c_1\hat A\Psi_1 + c_2\hat A\Psi_2\tag{7.1}

It must be Hermitian

The second requirement is that $\hat A$ be a Hermitian operator :

\int\Psi_1^*\hat A\,\Psi_2\,\mathrm d^3\mathbf r = \int(\hat A\Psi_1)^*\,\Psi_2\,\mathrm d^3\mathbf r\tag{7.2}

Reproduce it in NumPy

Eq. (7.2) is an integral condition, but on a finite basis it is exactly A = Aᵀ*. Breaking it by a single entry is enough to produce complex eigenvalues — the requirement is sharp, not approximate.

import numpy as np

print("what Hermiticity actually buys: real eigenvalues, orthogonal eigenvectors")
print("=" * 72)
rng = np.random.default_rng(7)
M = rng.normal(size=(4, 4)) + 1j*rng.normal(size=(4, 4))
H = (M + M.conj().T)/2          # Hermitian by construction: H = H-dagger
print("  a Hermitian 4x4 (H equals its own conjugate transpose):")
print(f"    max |H - H-dagger|        = {np.abs(H - H.conj().T).max():.1e}")
w = np.linalg.eigvals(H)
print(f"    eigenvalues               = {np.round(np.sort(w.real), 4)}")
print(f"    max |imaginary part|      = {np.abs(w.imag).max():.1e}   <- REAL, as required")
_, V = np.linalg.eigh(H)
print(f"    eigenvectors orthonormal? max |V-dagger V - I| = "
      f"{np.abs(V.conj().T@V - np.eye(4)).max():.1e}")

print("\n  the same matrix with Hermiticity broken by ONE entry:")
B = H.copy(); B[0, 1] += 0.8j
wb = np.linalg.eigvals(B)
print(f"    max |B - B-dagger|        = {np.abs(B - B.conj().T).max():.2f}")
print(f"    eigenvalues               = {np.round(wb, 3)}")
print(f"    max |imaginary part|      = {np.abs(wb.imag).max():.3f}   <- COMPLEX")
print("  A complex eigenvalue is a measurement outcome that is not a number any")
print("  instrument can report.  That is the whole reason Eq. (7.2) is demanded.")

print("\nand the expectation value is real for ANY state, not just eigenstates")
print("=" * 72)
for k in range(3):
    psi = rng.normal(size=4) + 1j*rng.normal(size=4)
    psi /= np.linalg.norm(psi)
    a = psi.conj() @ H @ psi
    b = psi.conj() @ B @ psi
    print(f"  random state {k+1}:  <H> = {a.real:+.4f} {a.imag:+.1e}i     "
          f"<B> = {b.real:+.4f} {b.imag:+.4f}i")
print("  Hermitian: the imaginary part is zero to machine precision, every time.")

printswhat Hermiticity actually buys: real eigenvalues, orthogonal eigenvectors ======================================================================== a Hermitian 4x4 (H equals its own conjugate transpose): max |H - H-dagger| = 0.0e+00 eigenvalues = [-1.7149 -0.7352 0.7889 1.856 ] max |imaginary part| = 1.9e-16 <- REAL, as required eigenvectors orthonormal? max |V-dagger V - I| = 8.9e-16 the same matrix with Hermiticity broken by ONE entry: max |B - B-dagger| = 0.80 eigenvalues = [-1.971+0.098j 1.905+0.05j -0.622-0.11j 0.882-0.038j] max |imaginary part| = 0.110 <- COMPLEX A complex eigenvalue is a measurement outcome that is not a number any instrument can report. That is the whole reason Eq. (7.2) is demanded. and the expectation value is real for ANY state, not just eigenstates ======================================================================== random state 1: <H> = +0.6786 +2.8e-17i <B> = +0.8362 +0.1348i random state 2: <H> = +0.1415 +1.4e-17i <B> = +0.1731 -0.2573i random state 3: <H> = -0.7941 -8.3e-17i <B> = -0.7370 -0.1364i Hermitian: the imaginary part is zero to machine precision, every time.

Its eigenfunctions must be complete

\Psi(\mathbf r, t) = \sum_n c_{a_n}(t)\,\psi_{a_n}(\mathbf r) + \int c(a',t)\,\psi_{a'}(\mathbf r)\,\mathrm da'\tag{7.3}

with $|c_{a_n}|^2$ the probability of outcome $a_n$ and $|c(a',t)|^2\,\mathrm da'$ the probability of an outcome between $a'$ and $a' + \mathrm da'$ .

Engineer's bridge — this is the spectral theorem, and you already rely on it

Discretize and every statement above becomes standard numerical linear algebra.

Hermitian = self-adjoint = A == A.conj().T. The spectral theorem then guarantees exactly the two properties §7.1 demands: real eigenvalues, and an orthonormal eigenbasis that spans the space. That is completeness. NumPy splits eig from eigh for precisely this reason — eigh may assume the matrix is Hermitian, so it can return sorted real eigenvalues and orthonormal vectors, and it is both faster and better conditioned for it. Every eigh call in this site’s qm.ts is leaning on Eq. (7.2).

Linearity is what lets you use a matrix at all. A non-linear operator has no matrix representation, so there is nothing to diagonalize and the entire machinery is unavailable.

Change of basis is the physics. Expanding $\Psi$ in $\hat A$ ‘s eigenfunctions is projecting a vector onto an orthonormal basis, and $c_n = \langle\psi_n|\Psi\rangle$ is an inner product. Measuring $A$ instead of $H$ is a change of basis and nothing more — the state has not changed, only the axes you resolved it along.

Where it breaks, and §7.2 is about the break. Finite matrices have finite discrete spectra. Position and momentum have continuous spectra, and no finite-dimensional space can hold them — which is why Eq. (7.3) needs an integral, why the eigenfunctions turn out not to be normalizable, and why the next section has to introduce an object that is not a function.

7.2 Position and momentum

Now apply all three requirements to the two observables ch03 introduced. Both turn out to be awkward, and the awkwardness is instructive.

Position eigenfunctions

The eigenvalue equation is

\hat x\,\psi_{x'}(x) = x'\,\psi_{x'}(x)\tag{7.4}

and since $\hat x$ is simply multiplication by $x$ , this reads

x\,\psi_{x'}(x) = x'\,\psi_{x'}(x)\tag{7.5}

Read that literally: multiplying $\psi_{x'}$ by the variable $x$ must give the same thing as multiplying it by the constant $x'$ . For any $x \ne x'$ that forces $\psi_{x'}(x) = 0$ . So the function is zero everywhere except one point — and it cannot be zero there too, or it would be nothing at all. It must be infinite at $x'$ , in a way that keeps its integral finite.

\psi_{x'}(x) = \delta(x - x')\tag{7.6}

the Dirac delta function , defined not by its values but by what it does inside an integral:

\int_{-\infty}^{+\infty}f(x)\,\delta(x-x')\,\mathrm dx = f(x')\tag{7.7}

δ_ε(x − x′) — area exactly 1, always
the test function f(x)

ε0.3162

height 1.58

width 0.6325

area 0.999975

∫f·δ_ε 1.721982

Drag ε down and watch three numbers. The height runs away to infinity and the width collapses to zero — but the area stays at 1, to six decimal places, at every value of ε. That refusal to move is the entire design, and it is the reason the limit means anything at all.

The fourth readout is what a delta is for. The sampling integral ∫f(x)δ_ε(x−x′)dx is currently 1.721982, against the exact f(x′) = 1.644241 — an error of 7.77e-2. Narrow ε by ten and that error falls by a hundred: it is ε²f″(x′)/6, and nothing about the shape of the spike survives except its area.

Which is why the other two shapes work just as well. A Gaussian and a Lorentzian look nothing like Fig. 7.1's box, and all three converge to the same limit — because the limit is not a function being approached but a rule: sample f at x′. Notice also that nothing converges pointwise. At x = x′ the value diverges; everywhere else it was already zero. There is no function at the end of this process.

Put a number on it — how narrow is narrow enough?

The delta is a limit, and a limit is only useful if you know how fast it converges. Take $f(x) = e^{-x/2}\cos 3x + 2$ and sample it at $x' = 0.7$ , where $f(x') = 1.644241$ .

Using Fig. 7.1’s box of half-width $\epsilon$ , the integral $\int f\,\delta_\epsilon$ is just the average of $f$ over the window:

$\epsilon$	height $1/2\epsilon$	$\int f\,\delta_\epsilon$	error
$1$	$0.5$	$2.098567$	$+4.5\times10^{-1}$
$0.1$	$5$	$1.652424$	$+8.2\times10^{-3}$
$0.01$	$50$	$1.644323$	$+8.2\times10^{-5}$
$0.003$	$167$	$1.644248$	$+7.4\times10^{-6}$

Each tenfold narrowing buys a hundredfold accuracy. The error is $\epsilon^2f''(x')/6$ — second order, because the box is symmetric about $x'$ so the first-order term cancels.

So “infinitely narrow” is not a demand for the impossible: at $\epsilon = 0.01$ — a window one hundredth as wide as the feature it is sampling — the answer is already right to five significant figures. A delta function is a claim about a limit that converges quickly, which is why physicists use one without apology and why a numerical calculation can approximate it with a narrow spike and lose nothing that matters.

Reproduce it in NumPy

The last column is the check that matters: error/ε² settling on a constant is what proves the convergence is second order rather than merely fast. A first draft integrated on one shared grid and the quadrature error swamped that law near ε = 0.01 — so the box is integrated locally, which is what it is.

import numpy as np

# Fig. 7.1's delta_epsilon: height 1/2eps on |x - x'| < eps, zero outside.
# Integrating f against it is just the AVERAGE of f over that window, so it is
# computed locally and accurately rather than on one coarse global grid.
f = lambda x: np.exp(-0.5*x)*np.cos(3*x) + 2
x0 = 0.7

def sample(eps, n=200001):
    t = np.linspace(x0-eps, x0+eps, n)
    return np.trapezoid(f(t), t)/(2*eps)       # = integral of f * delta_eps

print("Eq. (7.7): a delta function is defined by what it DOES, not what it is")
print("=" * 72)
print(f"  test function f(x) = exp(-x/2) cos(3x) + 2,   sampling point x' = {x0}")
print(f"  exact f(x') = {f(x0):.8f}\n")
print(f"  {'epsilon':>9} {'height 1/2eps':>14} {'width 2eps':>11} "
      f"{'integral f d_eps':>17} {'error':>11} {'error/eps^2':>12}")
for eps in (1.0, 0.3, 0.1, 0.03, 0.01, 0.003):
    v = sample(eps)
    e = v - f(x0)
    print(f"  {eps:9.3f} {1/(2*eps):14.1f} {2*eps:11.3f} {v:17.8f} "
          f"{e:+11.2e} {e/eps**2:12.4f}")
print("\n  The last column is constant: the error is exactly (eps^2/6) f''(x'),")
print("  so ten times narrower is a hundred times more accurate.")
print("  The AREA, meanwhile, is 1 at every epsilon -- that is the design.  The")
print("  height runs to infinity and the width to zero while the area holds, and")
print("  the integral converges on f(x').")
print("  Nothing converges POINTWISE: at x = x' the value diverges and everywhere")
print("  else it was already 0.  That is why a delta is not a function at all but")
print("  a rule for evaluating integrals.")

printsEq. (7.7): a delta function is defined by what it DOES, not what it is ======================================================================== test function f(x) = exp(-x/2) cos(3x) + 2, sampling point x' = 0.7 exact f(x') = 1.64424096 epsilon height 1/2eps width 2eps integral f d_eps error error/eps^2 1.000 0.5 2.000 2.09856726 +4.54e-01 0.4543 0.300 1.7 0.600 1.71460915 +7.04e-02 0.7819 0.100 5.0 0.200 1.65242405 +8.18e-03 0.8183 0.030 16.7 0.060 1.64498125 +7.40e-04 0.8225 0.010 50.0 0.020 1.64432325 +8.23e-05 0.8229 0.003 166.7 0.006 1.64424837 +7.41e-06 0.8230 The last column is constant: the error is exactly (eps^2/6) f''(x'), so ten times narrower is a hundred times more accurate. The AREA, meanwhile, is 1 at every epsilon -- that is the design. The height runs to infinity and the width to zero while the area holds, and the integral converges on f(x'). Nothing converges POINTWISE: at x = x' the value diverges and everywhere else it was already 0. That is why a delta is not a function at all but a rule for evaluating integrals.

Expanding a state in position eigenfunctions, Eq. (7.8), and applying Eq. (7.7) gives $\Psi(x,t) = c(x,t)$ — the wave function simply is the position probability amplitude. That was assumed in §3.2; here it is a consequence.

Momentum eigenfunctions

The momentum eigenvalue equation is

\hat p\,\psi_{p'}(x) = p'\,\psi_{p'}(x)\tag{7.9}

and unlike Eq. (7.4) this one has real content, because $\hat p$ differentiates rather than multiplies. Substituting $\hat p = -i\hbar\,\partial/\partial x$ makes it a differential equation,

-i\hbar\frac{\partial\psi_{p'}(x)}{\partial x} = p'\,\psi_{p'}(x)\tag{7.10}

whose solutions are plane waves:

\psi_{p'}(x) = \frac{1}{\sqrt{2\pi\hbar}}e^{ip'x/\hbar}\tag{7.11}

with wave number $k' = p'/\hbar$ and wavelength $\lambda' = h/p'$ — de Broglie, arriving as an eigenvalue problem rather than an assumption.

Chapter 3's biggest assumption, now a consequence

Expanding a state in these, Eq. (7.12), and substituting Eq. (7.11) gives

\Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty}c(p',t)\,e^{ip'x/\hbar}\,\mathrm dp'

which is exactly the inverse Fourier transform ch03 wrote down at Eq. (3.19) — where $\tilde\Psi(p,t)$ was assumed to be the momentum probability amplitude on the strength of it being the natural thing to try.

Here it is derived. The momentum amplitude is the Fourier transform because the momentum eigenfunctions are plane waves, and they are plane waves because $\hat p$ is a derivative. Nothing was assumed; the assumption was a theorem waiting for chapter 7.

position — |Ψ(x)|²

momentum — |Ψ̃(p)|²

σ1.00

Δx 1.0000

Δp 0.5000

Δx·Δp / ħ 0.5000 ≥ 0.5

Δx = σ = 1.00, Δp = ħ/2σ = 0.500, product = ħ/2 exactly

The one shape that achieves the minimum. Its transform is another Gaussian, so squeezing one side widens the other by exactly the reciprocal factor and the product never moves off ½. §3.5 shows this is the minimum-uncertainty state; §7.4 proves no state can do better.

The awkwardness: neither basis is normalizable

\int_{-\infty}^{+\infty}\psi_{x'}^*(x)\,\psi_{x''}(x)\,\mathrm dx = \delta(x'-x'')\tag{7.13}

\int_{-\infty}^{+\infty}\psi_{p'}^*(x)\,\psi_{p''}(x)\,\mathrm dx = \delta(p'-p'')\tag{7.14}

Both are orthogonal — the overlap vanishes for different eigenvalues — but neither normalizes to 1. The overlap of an eigenfunction with itself is infinite. This condition, a delta on the right-hand side where an ordinary basis would have a 1, is called delta-function normalization .

So a state of definite position or definite momentum does not exist

This is not a technicality to be waved past. A wave function must satisfy $\int|\Psi|^2 = 1$ if it describes a particle that is somewhere. Neither $\delta(x-x')$ nor $e^{ip'x/\hbar}/\sqrt{2\pi\hbar}$ can be scaled to satisfy it.

Strictly, neither is a physically acceptable state. What is acceptable is a superposition of them — a wave packet — which can have very small uncertainty in position or momentum but never zero.

This settles a debt. §5.1b met unbound states that could not be normalized and its footnote deferred the problem to chapter 7; this is the resolution. The book also notes the practical dodge: put the particle in a very large box, and everything becomes discrete and normalizable again, with the box size dropping out of any physical answer.

Reproduce it in NumPy

The middle block is the honest version of 'orthogonal': at any finite window the overlap is a sinc, not zero. It becomes a delta only in the limit — and it does so by exactly the three properties the widget above shows for Fig. 7.1's box.

import numpy as np

hbar, me, h, eV = 1.054571e-34, 9.10938e-31, 6.626070e-34, 1.602177e-19

print("Eq. (7.11): the momentum eigenfunction is a plane wave")
print("=" * 72)
print("  psi_p'(x) = exp(i p' x/hbar)/sqrt(2 pi hbar), so |psi|^2 = 1/(2 pi hbar)")
print("  everywhere -- a state of definite momentum carries NO position")
print("  information at all.\n")
for E_eV in (1.0, 100.0, 1e4):
    p = np.sqrt(2*me*E_eV*eV)
    print(f"  an electron at {E_eV:>7.0f} eV:  p = {p:.4e} kg m/s,  "
          f"lambda = h/p = {h/p*1e9:8.4f} nm")
print("\n  A 1 eV momentum eigenstate oscillates every 1.23 nm, forever, in both")
print("  directions.  No particle is ever in that state; it is a basis vector.")

print("\nEq. (7.14): so it cannot be normalized -- and what replaces normalization")
print("=" * 72)
print("  Overlap of two momentum eigenfunctions over a window of width L")
print("  (hbar = 1):   I(dp) = (1/2 pi) int e^{i dp x} dx = (L/2 pi) sinc(dp L/2)")
print("  Watch the three properties of Fig. 7.1's delta appear as L grows:\n")
print(f"  {'L':>8} {'peak I(0)':>11} {'width to 1st zero':>19} {'total area':>12}")
for L in (20.0, 100.0, 500.0, 2500.0):
    dp = np.linspace(-40, 40, 4_000_001)
    I = np.empty_like(dp)
    nz = dp != 0                      # np.where would evaluate BOTH branches
    I[nz] = np.sin(dp[nz]*L/2)/(np.pi*dp[nz])
    I[~nz] = L/(2*np.pi)
    print(f"  {L:8.0f} {L/(2*np.pi):11.3f} {2*np.pi/L:19.4f} "
          f"{np.trapezoid(I, dp):12.4f}")
print("\n  Peak grows as L, width shrinks as 1/L, area stays 1.  Those are exactly")
print("  the three properties of delta_eps -- so the overlap IS becoming a delta,")
print("  in dp rather than in x.  At any FINITE L it is a sinc, not zero, which is")
print("  why the book says delta(p' - p'') and not 'zero unless equal'.")
print("\n  And why no constant can fix it: the norm itself diverges with the window.")
print(f"  {'L':>8} {'integral |psi|^2':>18}")
for L in (10.0, 100.0, 1000.0, 4000.0):
    print(f"  {L:8.0f} {L/(2*np.pi):18.4f}")
print("  Proportional to L, without bound.  Only superpositions -- wave packets --")
print("  are normalizable, which is the point section 7.2 closes on.")

printsEq. (7.11): the momentum eigenfunction is a plane wave ======================================================================== psi_p'(x) = exp(i p' x/hbar)/sqrt(2 pi hbar), so |psi|^2 = 1/(2 pi hbar) everywhere -- a state of definite momentum carries NO position information at all. an electron at 1 eV: p = 5.4027e-25 kg m/s, lambda = h/p = 1.2264 nm an electron at 100 eV: p = 5.4027e-24 kg m/s, lambda = h/p = 0.1226 nm an electron at 10000 eV: p = 5.4027e-23 kg m/s, lambda = h/p = 0.0123 nm A 1 eV momentum eigenstate oscillates every 1.23 nm, forever, in both directions. No particle is ever in that state; it is a basis vector. Eq. (7.14): so it cannot be normalized -- and what replaces normalization ======================================================================== Overlap of two momentum eigenfunctions over a window of width L (hbar = 1): I(dp) = (1/2 pi) int e^{i dp x} dx = (L/2 pi) sinc(dp L/2) Watch the three properties of Fig. 7.1's delta appear as L grows: L peak I(0) width to 1st zero total area 20 3.183 0.3142 1.0008 100 15.915 0.0628 1.0001 500 79.577 0.0126 1.0001 2500 397.887 0.0025 1.0000 Peak grows as L, width shrinks as 1/L, area stays 1. Those are exactly the three properties of delta_eps -- so the overlap IS becoming a delta, in dp rather than in x. At any FINITE L it is a sinc, not zero, which is why the book says delta(p' - p'') and not 'zero unless equal'. And why no constant can fix it: the norm itself diverges with the window. L integral |psi|^2 10 1.5915 100 15.9155 1000 159.1549 4000 636.6198 Proportional to L, without bound. Only superpositions -- wave packets -- are normalizable, which is the point section 7.2 closes on.

⇅	position⇅	momentum⇅
the operator	multiply by $x$ ⓘ	$- i ℏ \partial / \partial x$ ⓘ
eigenfunction	$δ (x - x^{'})$ ⓘ	$e^{i p^{'} x /ℏ} / 2 π ℏ$ ⓘ
expansion of a state	$Ψ (x, t)$ itselfⓘ	the Fourier transform of $Ψ$ ⓘ
normalization	$δ (x^{'} - x^{''})$ ⓘ	$δ (p^{'} - p^{''})$ ⓘ
spectrum	continuous, all of $R$ ⓘ	continuous, all of $R$ ⓘ

Where this goes next

§7.3–7.4 asks which observables can be sharp together, and answers with the commutator — turning the awkwardness above into the sharpest statement in the book: $[\hat x,\hat p] = i\hbar$ , from which the uncertainty principle follows as a theorem rather than an experimental fact.

Check yourself

0 / 6 answered

1.Why must an operator describing an observable be Hermitian?
In the delta widget, drag ε down and watch the four readouts.
2.Three numbers change and one refuses to. Which is the important one, and why?
3.Section 7.2 shows that expanding a state in momentum eigenfunctions gives its Fourier transform. Why does that matter, given that chapter 3 already used the Fourier transform this way?
4.Neither $δ (x - x^{'})$ nor $e^{i p^{'} x /ℏ} / 2 π ℏ$ can be normalized to 1. What follows?
5.A 1 eV electron's momentum eigenfunction has $λ = 1.23$ nm and $∣ ψ ∣^{2}$ constant everywhere. What does the second half of that tell you?
6.The bridge calls Eq. (7.3) — completeness — the spectral theorem. Where does that correspondence break down?