Every measurable quantity becomes the same integral with a different operator in the middle. The hat is not decoration — it is the instruction for what to do between the two Ψ’s.

Two sections, and they do very different jobs. §3.5 is the most practical page of the chapter: it turns the densities of §3.2 and §3.3 into numbers you can compute, and in doing so introduces the object that runs the rest of the book — the operator . §3.6 is the most philosophical: it steps back and says plainly what has just been assumed, and what remains unexplained.

3.5 Expectation Values

The outcome of a quantum measurement is a random variable. §3.1 already gave the machinery for averaging one: multiply each outcome by its probability and integrate. All §3.5 does at first is apply Eq. (3.6) to the densities we now have.

Position is a continuous random variable with density $|\Psi(x,t)|^2$ , so

\langle x\rangle = \int_{-\infty}^{+\infty} x\,|\Psi(x,t)|^2\,\mathrm dx\tag{3.23}

The book prints Eq. (3.23) with its limits reversed

On page 46 this equation appears as $\int_{+\infty}^{-\infty}$ — lower limit $+\infty$ , upper limit $-\infty$ . Swapping the limits of an integral flips its sign, so as printed it evaluates to $-\langle x\rangle$ .

It is a typesetting slip in that one equation. Eq. (3.24) directly beneath it has the limits the right way round, and every subsequent use of $\langle x\rangle$ in the book assumes the correct form, which is the one above.

This is the second of four misprints in this chapter. The others are the $2A^2$ that should be $4A^2$ in §3.2, the lost exponent flagged further down this page, and a sign error in the probability current of problem 8. None of them affects any conclusion; all four are worth catching because the habit of checking is the transferable skill.

Momentum is a continuous random variable too, with density $|\tilde\Psi(p,t)|^2$ , so the same reasoning gives

\langle p\rangle = \int_{-\infty}^{+\infty} p\,|\tilde\Psi(p,t)|^2\,\mathrm dp\tag{3.24}

Writing them as sandwiches

Using $|\Psi|^2 = \Psi^*\Psi$ and $|\tilde\Psi|^2 = \tilde\Psi^*\tilde\Psi$ , both integrals can be rewritten with the variable sitting between the conjugate and the function:

\langle x\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\,x\,\Psi(x,t)\,\mathrm dx\tag{3.25}

\langle p\rangle = \int_{-\infty}^{+\infty}\tilde\Psi^*(p,t)\,p\,\tilde\Psi(p,t)\,\mathrm dp\tag{3.26}

This looks like pointless rearrangement — $\Psi^*x\Psi$ is just $x|\Psi|^2$ — and for position it is. Its value shows up at the next step.

Equation (3.26) is symmetric and elegant, but inconvenient: it needs $\tilde\Psi$ , and computing a Fourier transform to get an average is a lot of work. The symmetry is therefore usually hidden, because $\langle p\rangle$ is almost always computed from $\Psi(x,t)$ directly:

\langle p\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi(x,t)\,\mathrm dx\tag{3.27}

Now the sandwich form earns its keep. In Eq. (3.25) the filling is the number $x$ ; in Eq. (3.27) it is the instruction differentiate and multiply by $-i\hbar$ . Written as $x|\Psi|^2$ there would be nowhere to put such a thing — it has to act on $\Psi$ , so it has to sit between the two factors.

step 1 of 4

The book offers this 'for the benefit of mathematically inclined readers'. It is four lines, and it explains where the operator comes from instead of leaving it a rule to memorise.

1Start from Eq. (3.19), which writes Ψ as a superposition of momentum waves, and apply the operator to it.
$(- i ℏ \frac{\partial}{\partial x}) Ψ (x, t) = (- i ℏ \frac{\partial}{\partial x}) \frac{1}{2 π ℏ} \int_{- \infty}^{+ \infty} \tilde{Ψ} (p, t) e^{+ i p x /ℏ} d p$
2Differentiate under the integral. Only the exponential depends on x, and ∂/∂x of e^{ipx/ħ} brings down ip/ħ — which the −iħ out front turns into exactly p. THIS is the whole trick: differentiation in x is multiplication by p in the transform.
$(- i ℏ \frac{\partial}{\partial x}) Ψ (x, t) = \frac{1}{2 π ℏ} \int_{- \infty}^{+ \infty} p \tilde{Ψ} (p, t) e^{+ i p x /ℏ} d p$
3Substitute that into Eq. (3.27) and swap the order of integration.
$⟨ p ⟩ = \int_{- \infty}^{+ \infty} (\frac{1}{2 π ℏ} \int_{- \infty}^{+ \infty} Ψ^{*} (x, t) e^{+ i p x /ℏ} d x) p \tilde{Ψ} (p, t) d p$
4The bracket is exactly the conjugate of Eq. (3.20) — that is, Ψ̃*(p,t).
$⟨ p ⟩ = \int_{- \infty}^{+ \infty} \tilde{Ψ}^{*} (p, t) p \tilde{Ψ} (p, t) d p$
Which is Eq. (3.26). The two expressions are the same thing in two bases — the operator is not a new postulate but a consequence of the transform.

Phillips also notes a result worth holding onto, left to problem 9:

\langle p\rangle = m\frac{\mathrm d\langle x\rangle}{\mathrm dt}\tag{3.28}

The averages obey the classical relation even though nothing else does. This is the first hint of Ehrenfest’s theorem — expectation values move classically — and it is why quantum mechanics reproduces Newtonian physics for large objects.

Operators

Phillips now states the idea that will dominate the rest of the book: observables in quantum mechanics are described by operators. At this stage it is presented as a recipe, and the plainness is deliberate:

step⇅	what you do⇅
1	Prepare a sandwich with $Ψ^{*}$ and $Ψ$ .ⓘ
2	To find $⟨ x ⟩$ insert $x$ ; to find $⟨ p ⟩$ insert $- i ℏ \partial / \partial x$ .ⓘ
3	Then integrate over $x$ .ⓘ

Both fillings act on the wave function: $x$ merely multiplies $\Psi$ by a factor, while $-i\hbar\,\partial/\partial x$ differentiates it and multiplies by $-i\hbar$ . Emphasising that, Eqs. (3.25) and (3.27) become

\langle x\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\,\hat x\,\Psi(x,t)\,\mathrm dx \quad\text{and}\quad \langle p\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\,\hat p\,\Psi(x,t)\,\mathrm dx\tag{3.29}

The circumflex denotes an operator, and

\hat x = x \quad\text{and}\quad \hat p = -i\hbar\frac{\partial}{\partial x}\tag{3.30}

symbol: $\overset{x}{^} = x$
is: the position operator is just multiplication by x. It looks like nothing is happening, and in the position basis nothing is — which is precisely why position is the "natural" variable for Ψ. In the momentum basis it would be the one that differentiates.
units: $m$
type: multiplication operator

In three dimensions nothing changes except the notation:

\hat{\mathbf r} = \mathbf r \quad\text{and}\quad \hat{\mathbf p} = -i\hbar\nabla\tag{3.31}

\langle\mathbf r\rangle = \int\Psi^*(\mathbf r,t)\,\hat{\mathbf r}\,\Psi(\mathbf r,t)\,\mathrm d^3\mathbf r \quad\text{and}\quad \langle\mathbf p\rangle = \int\Psi^*(\mathbf r,t)\,\hat{\mathbf p}\,\Psi(\mathbf r,t)\,\mathrm d^3\mathbf r\tag{3.32}

Order matters — and getting it wrong gives every particle zero momentum

Phillips emphasises that the order in which functions and operators are written down is important:

\Psi^*(\mathbf r,t)\,\hat{\mathbf p}\,\Psi(\mathbf r,t) \ne \hat{\mathbf p}\,\Psi^*(\mathbf r,t)\,\Psi(\mathbf r,t)

The right-hand side means “apply $\hat{\mathbf p}$ to the product $\Psi^*\Psi$ ”, which is $-i\hbar\,\nabla|\Psi|^2$ — the gradient of the probability density, a completely different object.

The consequence is worse than a wrong number. Integrating the right-hand side over all space gives $-i\hbar\big[|\Psi|^2\big]_{-\infty}^{+\infty} = 0$ for any normalizable state, because $\Psi$ must vanish at infinity. So the wrong order does not merely give an inaccurate momentum; it gives exactly zero for every particle that has ever existed, including one visibly moving to the right. The Compute block below runs both on a moving wave packet.

An operator is an instruction, not a factor, and instructions do not commute with the things around them.

Engineer's bridge — an operator is a matrix and ⟨A⟩ is a quadratic form

Discretize $x$ onto a grid of $N$ points and every object here becomes something you have used:

$\Psi$ becomes a complex vector $\psi \in \mathbb C^N$ .
$\hat x$ becomes $\mathrm{diag}(x_1,\dots,x_N)$ — multiplication by $x$ is a diagonal matrix.
$\hat p$ becomes $-i\hbar$ times a finite-difference matrix, tridiagonal with $[-1, 0, +1]/2\delta x$ on the bands.
The sandwich integral $\int\Psi^*\hat A\Psi\,\mathrm dx$ becomes $\psi^\dagger A\psi\,\delta x$ — a quadratic form, the same expression as the energy of a spring system or the cost in a least-squares problem.
Normalization $\int|\Psi|^2\mathrm dx = 1$ becomes $\psi^\dagger\psi\,\delta x = 1$ .

§0.3 already built $\hat H$ this way and diagonalized it; $\hat x$ and $\hat p$ are the same move applied to two simpler observables. “Order matters” becomes the utterly familiar $AB \ne BA$ , and $\langle A\rangle$ real for all $\psi$ becomes $A$ Hermitian — which is exactly the condition chapter 7 imposes on observables.

Where it breaks. $\hat p$ on the infinite line is unbounded — its spectrum runs over all of $\mathbb R$ — and it has no eigenvectors inside the space of normalizable functions at all. Its “eigenfunctions” $e^{ipx/\hbar}$ are the plane waves §3.2 already ruled out as physical states, because they cannot be normalized. A finite grid hides this completely: an $N\times N$ matrix always has $N$ eigenvectors and a bounded spectrum, so the numerics you write will look perfectly well behaved while the underlying operator is not. Chapter 7 deals with it properly, with delta-function normalization. Until then, treat the matrix picture as a reliable intuition and an unreliable proof.

Uncertainties

The uncertainties come from the same sandwiches. Equation (3.8) of §3.1, applied to each observable:

\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} \quad\text{and}\quad \Delta p = \sqrt{\langle p^2\rangle - \langle p\rangle^2}\tag{3.33}

with the second moments given by sandwiches with a squared operator inside:

\langle x^2\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\,\hat x^2\,\Psi(x,t)\,\mathrm dx,\qquad \langle p^2\rangle = \int_{-\infty}^{+\infty}\Psi^*(x,t)\,\hat p^2\,\Psi(x,t)\,\mathrm dx\tag{3.34}

where $\hat x^2$ means $\hat x \times \hat x$ and $\hat p^2$ means $\hat p\times\hat p$ — apply the operator twice. Using Eq. (3.30):

\hat x^2 = x^2 \quad\text{and}\quad \hat p^2 = -\hbar^2\frac{\partial^2}{\partial x^2}\tag{3.35}

Where the second derivative in the Schrödinger equation came from

$\hat p^2 = -\hbar^2\,\partial^2/\partial x^2$ is worth a pause. Applying $-i\hbar\,\partial/\partial x$ twice gives $(-i\hbar)^2\partial^2/\partial x^2 = -\hbar^2\,\partial^2/\partial x^2$ — the minus sign from $i^2$ .

Now recall that kinetic energy is $p^2/2m$ . So the kinetic energy operator is

\frac{\hat p^2}{2m} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}

which is exactly the first term of the Schrödinger equation. Chapter 2 arrived at it by demanding that plane waves satisfy $E = p^2/2m$ ; here it drops out of the operator definition. Two independent routes to the same term, and §4.1 makes the connection official by naming $\hat H$ the Hamiltonian operator.

measurements 0

running mean —

⟨x⟩ from Eq. (3.25) 0.5000

Δx: sampled / exact — / 0.2658

The first excited state, and the important case. |ψ|² is exactly ZERO at x = a/2, yet ⟨x⟩ = a/2 precisely, by symmetry. An expectation value is an average, not a prediction, and it need not be a likely outcome — or even a possible one. (Being pedantic: the density vanishes only AT the node, so outcomes very close to a/2 are rare rather than forbidden — about 1 in 19 000 lands within ±0.01a of it. In the histogram above, the bin straddling the node collects 175 times fewer counts than a peak bin.)

Put a number on it — the minimum-uncertainty Gaussian

Problem 4 takes the wave function

\Psi(x) = N e^{-x^2/2a^2}

and asks you to show that $\Delta x = a/\sqrt2$ and $\Delta p = \hbar/a\sqrt2$ , so that

\Delta x\,\Delta p = \frac{\hbar}{2}

exactly — in accordance with the Heisenberg uncertainty principle stated back in Eq. (1.15). For every other wave function the product is larger, as problems 5, 6 and 7 illustrate, and as the widget below lets you check by hand.

Compare the box states worked out in §3.3–3.4: the ground state manages $0.568\,\hbar$ , close to the floor but not on it; $n = 2$ gives $1.670\,\hbar$ and $n = 3$ gives $2.627\,\hbar$ . Only the Gaussian touches the bound.

A typo on this page of the book — though problem 4 itself is printed correctly

Where page 50 quotes problem 4’s wave function it gives $\Psi(x) = Ne^{-x/2a^2}$ — with $x$ , not $x^2$ , in the exponent. The problem statement itself, on page 53, has it right, so the slip is confined to this forward reference.

You can see which is intended two ways. Dimensionally: $x/a^2$ has units of 1/length, so that exponent is not dimensionless. From the answer: the stated results $\Delta x = a/\sqrt2$ and $\Delta p = \hbar/a\sqrt2$ follow only from $e^{-x^2/2a^2}$ , whose $|\Psi|^2 \propto e^{-x^2/a^2}$ is a Gaussian of standard deviation $a/\sqrt2$ . The Compute block confirms both to five decimals for the squared form, and problem 4(a) asks you to show $\langle x^2\rangle = a^2/2$ — which is $\sigma^2$ for exactly that Gaussian.

So a superscript was lost here and nothing downstream is affected. Note also that this $a$ is a length, one of the three meanings tabulated back in §3.1.

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.

Reproduce it in NumPy

Eq. (3.29) applied literally — psi*, then the operator acting on psi, then integrate — with the derivatives taken numerically rather than analytically, so the recipe is being tested rather than the algebra restated.

import numpy as np

hbar = 1.0

def sandwich(psi, x, op):
    """<A> = integral psi* (A psi) dx  -- Eq (3.29), done literally."""
    return np.trapezoid(np.conj(psi)*op(psi, x), x).real

d1 = lambda f, x: np.gradient(f, x, edge_order=2)
p_op  = lambda f, x: -1j*hbar*d1(f, x)
p2_op = lambda f, x: -hbar**2*np.gradient(d1(f, x), x, edge_order=2)

print("the sandwich recipe, Eq. (3.29), evaluated for three states")
print("=" * 66)
print(f"{'state':22} {'<x>':>9} {'<p>':>9} {'Dx':>9} {'Dp':>9} {'Dx*Dp/hbar':>11}")

# 1-3: particle in a box, a = 1
a = 1.0
x = np.linspace(0, a, 400001)
for n in (1, 2, 3):
    psi = np.sqrt(2/a)*np.sin(n*np.pi*x/a) + 0j
    mx  = sandwich(psi, x, lambda f, _: x*f)
    mx2 = sandwich(psi, x, lambda f, _: x*x*f)
    mp  = sandwich(psi, x, p_op)
    mp2 = sandwich(psi, x, p2_op)
    dx, dp = np.sqrt(mx2-mx**2), np.sqrt(mp2-mp**2)
    print(f"{'box n='+str(n):22} {mx:9.5f} {mp:9.5f} {dx:9.5f} {dp:9.5f} {dx*dp/hbar:11.5f}")

# 4: the Gaussian of problem 4 -- note the exponent
A = 1.0
x = np.linspace(-40*A, 40*A, 800001)
psi = np.exp(-x**2/(2*A**2)) + 0j              # Psi = N exp(-x^2/2a^2)
psi /= np.sqrt(np.trapezoid(np.abs(psi)**2, x))
mx  = sandwich(psi, x, lambda f, _: x*f)
mx2 = sandwich(psi, x, lambda f, _: x*x*f)
mp  = sandwich(psi, x, p_op)
mp2 = sandwich(psi, x, p2_op)
dx, dp = np.sqrt(mx2-mx**2), np.sqrt(mp2-mp**2)
print(f"{'gaussian (prob. 4)':22} {abs(mx):9.5f} {abs(mp):9.5f} {dx:9.5f} {dp:9.5f} {dx*dp/hbar:11.5f}")
print(f"\nproblem 4 claims Dx = a/sqrt(2) = {A/np.sqrt(2):.5f} and "
      f"Dp = hbar/(a*sqrt(2)) = {hbar/(A*np.sqrt(2)):.5f}")
print(f"  -> matches only for Psi = N exp(-x^2/2a^2); the book prints exp(-x/2a^2)")

print("\ndoes the ORDER in the sandwich matter?  (p.49)")
print("  a moving packet, Psi = exp(i*k0*x) exp(-x^2/2a^2) with k0 = 3:")
x = np.linspace(-40, 40, 800001)
k0 = 3.0
psi = np.exp(1j*k0*x)*np.exp(-x**2/2)
psi /= np.sqrt(np.trapezoid(np.abs(psi)**2, x))
right = np.trapezoid(np.conj(psi)*p_op(psi, x), x)          # Psi* (p_hat Psi)   -- Eq (3.29)
wrong = np.trapezoid(p_op(np.conj(psi)*psi, x), x)          # p_hat (Psi* Psi)   -- the wrong order
print(f"  integral Psi* (p_hat Psi) dx = {right.real:+.5f}   <- correct, = hbar*k0 = {k0:.1f}")
print(f"  integral p_hat (Psi* Psi) dx = {wrong.real:+.5f}   <- wrong order")
print("  the wrong order is -i*hbar*d|Psi|^2/dx, whose integral is ZERO for any")
print("  normalizable state -- it would give every particle zero momentum.")

print("\nis <x> a possible outcome?  (the n=2 box state)")
x = np.linspace(0, 1, 400001)
psi = np.sqrt(2)*np.sin(2*np.pi*x)
rho = psi**2
print(f"  <x>          = {np.trapezoid(x*rho, x):.6f}")
print(f"  |psi|^2 there= {2*np.sin(2*np.pi*0.5)**2:.6e}   <- the particle is NEVER found at <x>")

printsthe sandwich recipe, Eq. (3.29), evaluated for three states ================================================================== state <x> <p> Dx Dp Dx*Dp/hbar box n=1 0.50000 0.00000 0.18076 3.14159 0.56786 box n=2 0.50000 0.00000 0.26583 6.28319 1.67029 box n=3 0.50000 0.00000 0.27876 9.42478 2.62720 gaussian (prob. 4) 0.00000 0.00000 0.70711 0.70711 0.50000 problem 4 claims Dx = a/sqrt(2) = 0.70711 and Dp = hbar/(a*sqrt(2)) = 0.70711 -> matches only for Psi = N exp(-x^2/2a^2); the book prints exp(-x/2a^2) does the ORDER in the sandwich matter? (p.49) a moving packet, Psi = exp(i*k0*x) exp(-x^2/2a^2) with k0 = 3: integral Psi* (p_hat Psi) dx = +3.00000 <- correct, = hbar*k0 = 3.0 integral p_hat (Psi* Psi) dx = +0.00000 <- wrong order the wrong order is -i*hbar*d|Psi|^2/dx, whose integral is ZERO for any normalizable state -- it would give every particle zero momentum. is <x> a possible outcome? (the n=2 box state) <x> = 0.500000 |psi|^2 there= 2.999520e-32 <- the particle is NEVER found at <x>

3.6 Quantum States

The chapter opened with a puzzle: how can one wave function describe both wave-like and particle-like behaviour? The answer has been assembled — the wave function governs the potential outcomes of measurements — and §3.6 says outright what that cost.

Physics no longer tries to predict exactly what will happen; it is now content with predicting the probabilities of a range of possible outcomes.

The question the book leaves open

Is quantum probability like the probability in the kinetic theory of gases — covering up our ignorance of an underlying description that assigns the particle a definite position all along, which the measurement merely reveals? Or is it the complete and fundamental description?

Phillips does not answer, and is explicit that it has been debated since the inception of quantum mechanics. But he states the second horn vividly enough that it is worth quoting:

In the latter case, it is pointless or meaningless to speculate on where the particle is prior to the measurement. Its exact position is not revealed by the measurement, but brought into existence by the measurement; the particle, like the experimenter, is surprised by the outcome!

The book’s own position is pragmatic: these issues are “not our immediate concern”, and the aim is to appreciate quantum mechanics as a consistent theory of microscopic phenomena. A footnote points to F. Laloë, American Journal of Physics 69, 655 (2001) for readers who want the full debate.

Why the mathematics cannot settle it

Every equation in this chapter is compatible with both readings, which is exactly why the question survives.

$|\Psi|^2$ gives the probability of each outcome. That statement is unchanged whether the particle “really had” a position and you didn’t know it, or had none until you looked. No experiment described so far distinguishes the two, because they make identical predictions for everything computed here.

This is not a permanent stalemate — Bell’s theorem (1964) shows that a large class of “the particle really had a value” theories make different predictions from quantum mechanics for correlated pairs, and experiment has repeatedly favoured quantum mechanics. §1.4’s entangled photons are the setup those experiments use. But nothing in chapter 3 gets there, and it is worth being clear that the interpretation is currently an open question the formalism declines to answer rather than a subtlety you have failed to grasp.

Four properties of a quantum state

A wave function represents a quantum state — a state of motion bearing, in the book’s phrase, only a passing resemblance to the well-defined trajectories of classical physics. Phillips sets out four properties, and they are worth learning as a unit because the rest of the book uses all four constantly.

property⇅	what it says⇅	where you have already seen it⇅
Deterministic evolution	In the absence of measurements a quantum state evolves smoothly and deterministically, in accordance with the time-dependent Schrödinger equation, Eq. (2.17).	ch. 2 — the equation is first order in time.ⓘ
Potentialities	A quantum state describes potentialities which can become realities: it predicts the possible outcomes of any measurement and the probabilities of those outcomes.	§3.2 and §3.3 — $∣Ψ ∣^{2}$ and $∣ \tilde{Ψ} ∣^{2}$ .ⓘ
Linear superposition	A quantum state is a linear superposition of other quantum states, which means a particle in one quantum state is also simultaneously in other quantum states.	Eq. (3.11) — the two-slit wave function.ⓘ
Fragility	A quantum state is destroyed by measurement and replaced by a new state compatible with the random outcome — the collapse of the wave function.	§3.2 — which-path detection destroys the fringes.ⓘ

The fourth property is where the theory admits it is incomplete

Note how differently the four are stated. The first three are derived — they follow from the Schrödinger equation being linear and first order in time. The fourth is imposed, and Phillips does not disguise it:

To a considerable extent, the collapse of the wave function is an arbitrary rule that bridges the gap between the unobserved quantum system and the observed quantum system. The underlying mechanism for the collapse is not understood.

So quantum mechanics has two rules for how states change: smooth deterministic evolution when no one is looking, and the abrupt collapse of the wave function when someone is. Nothing in the theory says which applies when, or what counts as a measurement. This is the measurement problem , and it is still open — decoherence, many-worlds and objective-collapse models are all attempts at it.

It is worth registering that this is a genuine gap and not a gap in your understanding. The theory works spectacularly well; it just does not explain its own second rule.

A footnote worth reading twice

Attached to “quantum state”, Phillips adds:

In the most general formulation of quantum mechanics, quantum states are complex vectors. Some quantum states can also be represented by wave functions and others, like quantum states describing the spin properties of an electron, by matrices.

This is the whole of chapter 10 compressed into three lines, and it reframes everything you have done. The wave function is not fundamental — it is one representation of a state, the one that happens to be convenient when position is the observable of interest. Spin has no position dependence at all, so it has no wave function; its states are two-component vectors, and its observables are $2\times2$ matrices.

If you found the Bridge above natural — vector, matrix, quadratic form — that is because the linear algebra is the real structure, and the wave function is a coordinate system laid on top of it.

Where chapter 3 leaves us

Position and momentum are both described by one wave function; expectation values and uncertainties both come from sandwich integrals; and the observables themselves are operators. What is still missing is energy — and it is the one that matters most, because energy is what the Schrödinger equation is built around and what determines how a state evolves.

Chapter 4 takes $\hat H$ , applies exactly the machinery of this page to it, and finds something the position and momentum cases did not show: for a confined particle the possible outcomes are discrete. That is where quantization finally comes from.

First, though: the problems, which are where the four wave functions of this chapter get compared side by side and $\Delta x\,\Delta p$ is computed for each.

Check yourself

0 / 7 answered

1.Why must the operator sit *between* $Ψ^{*}$ and $Ψ$ rather than being pulled outside as a factor?
2.What happens if you compute $⟨ p ⟩$ as $\int \overset{p}{^} (Ψ^{*} Ψ) d x$ — the wrong order?
3.Why is $\overset{p}{^}^{2} = - ℏ^{2} \partial^{2} / \partial x^{2}$ worth noticing?
4.For the $n = 2$ box state, $⟨ x ⟩ = a /2$ while $∣Ψ (a /2) ∣^{2} = 0$ . What does that show?
5.The book says $Δ x Δ p = ℏ/2$ for the Gaussian and is larger for every other wave function. How do the box states compare?
6.Which of §3.6's four properties of a quantum state is *imposed* rather than derived?
7.Phillips' footnote says quantum states are, in general, complex vectors. Why does that reframe the chapter?