Position and momentum are not two properties of a particle. They are two representations of one function, related by a transform you have used before.

§3.2 established that $\Psi$ tells you where a particle might be found. This section makes the striking claim that the same function already tells you what momentum it might have — no new object, no extra postulate. You just have to look at it in a different basis.

3.3 Momentum Probabilities

If a wave function can represent a particle with a range of possible positions, it is reasonable to expect it can represent a range of possible momenta too. The simplest case makes the point: the two-term wave function

\Psi(x,t) = A_1 e^{+i(k_1x - \omega_1 t)} + A_2 e^{+i(k_2 x - \omega_2 t)}

describes a free particle with two possible momenta and energies, $p_1 = \hbar k_1,\ E_1 = \hbar\omega_1$ and $p_2 = \hbar k_2,\ E_2 = \hbar\omega_2$ .

The general free-particle solution, already derived as Eq. (2.16), does the same thing with a continuum of them:

\Psi(x,t) = \int_{-\infty}^{+\infty} A(k)\,e^{i(kx - \omega t)}\,\mathrm dk,\quad\text{with}\quad \hbar\omega = \frac{\hbar^2k^2}{2m}\tag{3.18}

In wave language this is a superposition of sinusoids, and $|A(k)|^2$ measures how much of each wave number $k$ is present. In particle language each $k$ is a momentum $p = \hbar k$ . Reading those two sentences together, in analogy with the Born interpretation, gives the assumption of the section: the most probable momenta are the ones for which $|A(k)|^2$ is large.

The transform pair

Rather than leave momentum tied to the particular form (3.18), Phillips treats position and momentum symmetrically. Any one-dimensional wave function can be written as a Fourier transform:

\Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty} \tilde\Psi(p,t)\,e^{+ipx/\hbar}\,\mathrm dp\tag{3.19}

with inverse

\tilde\Psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty} \Psi(x,t)\,e^{-ipx/\hbar}\,\mathrm dx\tag{3.20}

symbol: $\tilde{Ψ} (p, t)$
is: the momentum-space wave function — the SAME state, written in terms of momentum instead of position. The tilde is the book's marker for "Fourier transformed". Not a new physical object: it carries exactly the information Ψ does, no more and no less.
units: $(k g m s^{- 1})^{- 1/2}$
type: complex function of p and t

The pair has a property that makes the whole idea work: if $\Psi$ is normalized, so is $\tilde\Psi$ .

\int_{-\infty}^{+\infty}|\Psi(x,t)|^2\,\mathrm dx = 1 \quad\text{then}\quad \int_{-\infty}^{+\infty}|\tilde\Psi(p,t)|^2\,\mathrm dp = 1

That is exactly the licence needed to call $|\tilde\Psi|^2$ a probability density. So, by symmetry with position:

$|\tilde\Psi(p,t)|^2$ is the probability density for momentum — $|\tilde\Psi|^2\,\mathrm dp$ is the probability of a momentum outcome between $p$ and $p + \mathrm dp$ ;
$\tilde\Psi(p,t)$ is the probability amplitude for momentum , exactly as $\Psi$ is for position.

This is the Born interpretation generalised, and it extends to three dimensions without changing anything essential.

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.

Engineer's bridge — this is the time–bandwidth product, and you already believe it

Equations (3.19) and (3.20) are an ordinary Fourier transform pair with $k = p/\hbar$ . Every fact you know about transforms therefore transfers intact:

A narrow pulse has a broad spectrum. Halve the duration of a signal and you double its bandwidth. That is the widget above, with $x$ for time and $p$ for frequency.
A Gaussian is the minimum-uncertainty shape. In DSP this is why the Gaussian window uniquely minimises the time–bandwidth product; here it is why $\Delta x\,\Delta p = \hbar/2$ exactly, and nothing does better.
Sharp edges cost bandwidth. The square pulse in the widget has infinite $\Delta p$ for the same reason a rectangular window has sidelobes that decay only as $1/f$ .

So $\Delta x\,\Delta p \ge \hbar/2$ is not a statement about measurement disturbing things. It is a theorem about Fourier transforms, and it would hold for audio signals if you renamed the axes. What physics contributes is only the identification $p = \hbar k$ — which is why $\hbar$ , and nothing else, sets the scale.

Where the analogy breaks. In DSP the signal and its spectrum are two descriptions of one waveform, and you can measure the waveform directly — sample it, plot it, low-pass it. Here $\Psi$ is not observable at all. You never measure the “waveform”; you measure either a position or a momentum, one particle at a time, and get a single number drawn from one of the two densities. There is no experiment that returns $\Psi(x)$ the way an oscilloscope returns a voltage. The mathematics is identical; what the mathematics is about is not.

3.4 A Particle in a Box I

Now the first real calculation: a particle of mass $m$ confined to the region $0 < x < a$ . Section 4.4 will show that such a particle has an infinite number of states with discrete energies, labelled by a quantum number $n = 1, 2, 3, \dots$ . Here we take that result and use it.

A particle in the state $n$ has energy

E_n = \frac{\hbar^2k_n^2}{2m},\quad\text{where}\quad k_n = \frac{n\pi}{a}\tag{3.21}

and wave function

\Psi_n(x,t) = \begin{cases} N\sin k_nx\,e^{-iE_nt/\hbar} & \text{if } 0 < x < a\\ 0 & \text{elsewhere}\end{cases}\tag{3.22}

V(x)
energy level Eₙ
ψₙ(x)
classically forbidden (V > E)

selected level: E₀ = 4.9348
nodes in ψ: 0
bound states: infinitely many
E₁ − E₀: 14.8043

A flat box with impenetrable walls. Every state is bound, the levels go up as n², and ψ must vanish at both walls — which is the entire reason the energies are quantized.

Fixing the normalization constant

$N$ is a normalization constant , and Eq. (3.17) determines it.

step 1 of 4

The first normalization anyone does. Note how the time factor disappears before the integral is even set up.

1Form |Ψ|² = Ψ*Ψ. The exponentials are e^{+iEt/ħ} and e^{−iEt/ħ}, whose product is 1 — the state is stationary, exactly as §3.2 predicted for anything of the form ψ(x)e^{−iωt}.
$∣ Ψ_{n} (x, t) ∣^{2} = N^{*} sin k_{n} x e^{+ i E_{n} t /ℏ} N sin k_{n} x e^{- i E_{n} t /ℏ} = ∣ N ∣^{2} sin^{2} k_{n} x$
2Integrate over all space. Ψ vanishes outside the box, so the infinite range collapses to 0 → a.
$\int_{- \infty}^{+ \infty} ∣ Ψ_{n} ∣^{2} d x = ∣ N ∣^{2} \int_{0}^{a} sin^{2} k_{n} x d x$
3The integral of sin² over a whole number of half-periods is exactly half the interval — and k_n = nπ/a guarantees a whole number of half-periods fits. This is the one place the quantization condition does real work in the calculation.
$\int_{0}^{a} sin^{2} k_{n} x d x = \frac{a}{2}$
4Set the total probability to 1 and solve. Note the answer does not depend on n: every state of the box carries the same normalization constant.
$N = \frac{2}{a}$
Units: a is a length, so N has units of m^(−1/2) — which is what |Ψ|² needs to be a density in m^(−1).

The momentum side

For the momentum density, feed Eq. (3.22) into Eq. (3.20):

\tilde\Psi_n(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{+\infty}\Psi_n(x,t)\,e^{-ipx/\hbar}\,\mathrm dx

With $N = \sqrt{2/a}$ the prefactor tidies to $1/\sqrt{\pi\hbar a}$ , and the wave function’s vanishing outside the box again truncates the range:

\tilde\Psi_n(p,t) = \frac{1}{\sqrt{\pi\hbar a}}\,e^{-iE_nt/\hbar}\int_0^a e^{-ipx/\hbar}\sin k_nx\,\mathrm dx

The remaining integral is elementary once the sine is written as exponentials:

\sin k_nx = \frac{e^{+ik_nx} - e^{-ik_nx}}{2i}

which turns it into two ordinary exponential integrals — one peaked near $p = +\hbar k_n$ , the other near $p = -\hbar k_n$ .

Figure 3.3

position — |Ψ(x)|²

momentum — |Ψ̃(p)|²

Δx 0.1808

Δp 3.1368

Δx·Δp / ħ 0.5670 ≥ 0.5

Δp = nπħ/a = 3.142, peaks at p = ±nħπ/a = ±1, Δx = a√(1/12 − 1/2n²π²) = 0.1808

Eq. (3.22) with a = 1. For n = 1 the particle is most likely mid-box and its most likely momentum is zero. For n = 3 there are three position peaks — at a/6, a/2 and 5a/6 — and TWO momentum peaks, near p = ±3ħπ/a (the exact maxima sit at ±2.79; see the page). A high-n state is roughly a particle rattling between the walls with two possible momenta, ±nħπ/a. The Δp readout comes from a central-difference derivative on a 900-point grid, so it lands a few parts in a thousand below the exact nπħ/a printed above it — the gap is the discretization, not the physics.

Reading the two states off the figure:

$n = 1$ . One position peak, at $x = a/2$ : the most likely place to find the particle is the middle of the box. The momentum density is a single hump centred on zero, so the most likely momentum is zero — which is only sensible for something that is going nowhere on average, but note it does not mean the particle is at rest. $\Delta p$ is large; it is $\langle p\rangle$ that vanishes, by symmetry.

$n = 3$ . Three position peaks, at $x = a/6$ , $a/2$ and $5a/6$ — the maxima of $\sin^2(3\pi x/a)$ . But two momentum peaks, near $p = \pm3\hbar\pi/a$ . Three humps in position, two in momentum: there is no reason for the counts to match, because the two pictures are transforms of each other, not two views of one trajectory.

Phillips draws the moral: a state with a high $n$ can be roughly pictured as a particle trapped between the walls with two possible momenta, $p = -n\hbar\pi/a$ and $p = +n\hbar\pi/a$ . It is the closest thing to a classical picture the chapter permits — a ball bouncing between two walls, going left half the time and right half the time.

The n = 3 momentum peaks are not exactly at ±3ħπ/a

The book states the most likely momenta for $n = 3$ are $p = \pm3\hbar\pi/a$ . Read the figure carefully, or the widget above, and the peaks sit slightly inside that: the true maximum is at $p = \pm2.79\,\hbar\pi/a$ .

It is not a rounding of the plot. Evaluating the exact transform, at $p = 3\hbar\pi/a$ precisely the density is 89% below its peak value — the numerator $\cos^2(\pi v/2)$ happens to vanish there at the same rate as the denominator, leaving a modest finite value rather than a maximum. The <Compute> block below evaluates this at eight million points.

Why it happens. The naive expectation is that $\sin k_nx$ is an equal mixture of $e^{+ik_nx}$ and $e^{-ik_nx}$ , so the momentum density should be two delta spikes at exactly $\pm\hbar k_n$ . That would be true for a sine wave filling all space. But the particle is confined: the wave is chopped off at $x = 0$ and $x = a$ , and truncating a sinusoid smears each spike into a broad peak — which also shifts its maximum. Confinement is what gives the momentum a spread at all, and the same confinement moves the peak.

Phillips’ word “roughly” is carrying this. The picture of two definite momenta $\pm n\hbar\pi/a$ is a good one, and it gets better as $n$ grows — but it is never exact, and the widget shows why.

Engineer's bridge — that is spectral leakage

The effect in the Caution has a name you will recognise: windowing.

Multiplying an infinite sinusoid by a rectangular window of length $a$ convolves its spectrum — two delta functions at $\pm k_n$ — with the window’s transform, a sinc. The deltas become sinc-shaped lobes: broadened, with sidelobes, and with maxima pulled off the true frequency. Every spectrum analyser does this, and it is why you taper with a Hann or Kaiser window instead of just truncating.

Here the “window” is the box itself, and it is not a choice — the particle really is confined, so the leakage is physical rather than an artefact of analysis. The width of the momentum peak is not a measurement limitation; it is the actual spread of momenta the particle has.

Where it breaks: in DSP you can pick a better window. A physicist cannot. The box is the potential, and changing it changes the state.

Why three position peaks and only two momentum peaks

The temptation is to imagine the particle at $a/6$ , $a/2$ or $5a/6$ , moving left or right, and expect the counts to line up. They do not, and the reason is worth internalising because it recurs throughout the book.

$|\Psi|^2$ and $|\tilde\Psi|^2$ are not two views of one trajectory. They are two different questions you could ask, with two different answers, and the particle has no state of affairs in which both are settled at once. Asking “where is it and how fast is it going” presupposes a classical picture the formalism does not supply.

The counts have separate causes. Three position peaks because $\sin^2(3\pi x/a)$ has three maxima in $[0, a]$ — a statement about a standing wave. Two momentum peaks because a standing wave is a sum of two travelling waves — a statement about the decomposition of that same wave. Increase $n$ and the position peaks multiply while the momentum peaks stay at two, just moving further apart.

Put a number on it — an electron in a 1 nm box

Take $a = 1\ \mathrm{nm}$ , roughly a small molecule or a quantum dot.

E_1 = \frac{\hbar^2\pi^2}{2m_ea^2} \approx 0.376\ \mathrm{eV}

Comfortably in the range of infrared photon energies, so transitions between these levels are optically accessible — which is exactly how quantum dots work, and why their colour depends on their size. Shrink the box and $E_n \propto 1/a^2$ pushes the levels apart.

The ground state’s widths:

\Delta x = a\sqrt{\tfrac{1}{12} - \tfrac{1}{2\pi^2}} = 0.1808\ \mathrm{nm},\qquad \Delta p = \frac{\pi\hbar}{a} = 3.31\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}

giving $\Delta x\,\Delta p = 0.568\,\hbar$ — above the floor of $\hbar/2$ , but not by much. The box’s ground state is nearly a minimum-uncertainty state. Go to $n = 3$ and the product rises to $2.63\,\hbar$ : excited states are far from the limit, because the momentum spread grows as $n$ while the position spread cannot exceed the box.

For scale, $\Delta p$ corresponds to a speed spread of about $3.6\times10^5\ \mathrm{m\,s^{-1}}$ — a tenth of a percent of $c$ . Confining an electron to a nanometre makes it move fast, and that is not a statement about how roughly you measured it.

Reproduce it in NumPy

Eqs. (3.21)–(3.22) in real units, and the exact momentum transform evaluated at eight million points to locate the n = 3 peaks. The last block is why Fig. 3.3's momentum axis stops at ±6: essentially all of the probability is inside it.

import numpy as np

hbar, me, eV = 1.054571817e-34, 9.1093837015e-31, 1.602176634e-19
a = 1e-9                                     # a 1 nm box

print("an electron in a box of width a = 1 nm")
print("=" * 58)
for n in (1, 2, 3):
    k = n*np.pi/a
    E = hbar**2*k**2/(2*me)                  # Eq (3.21)
    print(f"  n={n}:  k_n = {k:.4e} 1/m   E_n = {E/eV:7.4f} eV   "
          f"p = hbar*k_n = {hbar*k:.4e} kg m/s")

print("\nnormalization of Eq (3.22):  integral |Psi|^2 dx = |N|^2 a/2 = 1")
x  = np.linspace(0, a, 200001)
for n in (1, 3):
    N = np.sqrt(2/a)
    rho = N**2*np.sin(n*np.pi*x/a)**2
    print(f"  n={n}:  N = sqrt(2/a) = {N:.4e} m^-1/2   "
          f"integral = {np.trapezoid(rho, x):.6f}")

print("\nwidths for the box eigenstate (exact formulae)")
for n in (1, 3):
    dx = a*np.sqrt(1/12 - 1/(2*n**2*np.pi**2))
    dp = n*np.pi*hbar/a
    print(f"  n={n}:  Dx = {dx/a:.4f} a = {dx*1e9:.4f} nm   "
          f"Dp = n*pi*hbar/a = {dp:.4e}   Dx*Dp/hbar = {dx*dp/hbar:.4f}")

print("\nwhere is the momentum density ACTUALLY largest?")
print("  |Psi~_n(p)|^2 in units v = p/(hbar*pi/a):")
def dens(v, n):
    num = n*np.pi*(1 - (-1)**n*np.exp(-1j*np.pi*v))
    den = np.pi**2*(n**2 - v**2)
    with np.errstate(divide='ignore', invalid='ignore'):
        out = np.abs(num/den)**2
    out[~np.isfinite(out)] = 1/(4*n**2)   # removable singularity at v = n
    return out
v = np.linspace(0, 8, 800001)
for n in (1, 3, 5):
    d = dens(v, n); i = int(np.argmax(d))
    at_n = dens(np.array([float(n)]), n)[0]
    print(f"  n={n}:  peak at v = {v[i]:.4f}  (density {d[i]:.4f});  "
          f"at v = {n} the density is {at_n:.4f}, {100*(1-at_n/d[i]):.1f}% lower")

print("\nfraction of the momentum probability inside |p| <= 6 hbar*pi/a")
v = np.linspace(-600, 600, 6000001)
for n in (1, 3):
    d = dens(v, n)
    print(f"  n={n}:  {np.trapezoid(d[abs(v)<=6], v[abs(v)<=6])/np.trapezoid(d, v):.5f}")

printsan electron in a box of width a = 1 nm ========================================================== n=1: k_n = 3.1416e+09 1/m E_n = 0.3760 eV p = hbar*k_n = 3.3130e-25 kg m/s n=2: k_n = 6.2832e+09 1/m E_n = 1.5041 eV p = hbar*k_n = 6.6261e-25 kg m/s n=3: k_n = 9.4248e+09 1/m E_n = 3.3843 eV p = hbar*k_n = 9.9391e-25 kg m/s normalization of Eq (3.22): integral |Psi|^2 dx = |N|^2 a/2 = 1 n=1: N = sqrt(2/a) = 4.4721e+04 m^-1/2 integral = 1.000000 n=3: N = sqrt(2/a) = 4.4721e+04 m^-1/2 integral = 1.000000 widths for the box eigenstate (exact formulae) n=1: Dx = 0.1808 a = 0.1808 nm Dp = n*pi*hbar/a = 3.3130e-25 Dx*Dp/hbar = 0.5679 n=3: Dx = 0.2788 a = 0.2788 nm Dp = n*pi*hbar/a = 9.9391e-25 Dx*Dp/hbar = 2.6272 where is the momentum density ACTUALLY largest? |Psi~_n(p)|^2 in units v = p/(hbar*pi/a): n=1: peak at v = 0.0000 (density 0.4053); at v = 1 the density is 0.2500, 38.3% lower n=3: peak at v = 2.7916 (density 0.2589); at v = 3 the density is 0.0278, 89.3% lower n=5: peak at v = 4.8772 (density 0.2531); at v = 5 the density is 0.0100, 96.0% lower fraction of the momentum probability inside |p| <= 6 hbar*pi/a n=1: 0.99933 n=3: 0.99168

Where this leaves us

One wave function, two densities, and no extra assumptions: $|\Psi|^2$ answers “where?” and $|\tilde\Psi|^2$ answers “how fast?”, and they are Fourier transforms of one another. That single fact is the source of the uncertainty principle — not measurement clumsiness, but the impossibility of a function and its transform both being narrow.

What is still missing is a way to get numbers out — an average position, an average momentum, a spread — without plotting a density and squinting at it. §3.5 supplies it, and in doing so introduces the object that runs the rest of the book: the operator.

Check yourself

0 / 7 answered

1.Where does the momentum information in a wave function come from?
2.Why is $Δ x Δ p \geq ℏ/2$ better described as a theorem than as a measurement limitation?
3.For the box eigenstate, what fixes $N = 2/ a$ ?
4.The $n = 3$ box state has three peaks in $∣Ψ ∣^{2}$ but only two in $∣ \tilde{Ψ} ∣^{2}$ . Why don't the counts match?
5.The book says the most likely momenta for $n = 3$ are $\pm 3ℏ π / a$ . What does the exact transform actually give?
6.The broadening of those momentum peaks has a familiar name in signal processing. Which, and where does the analogy stop?
7.For the box ground state, $Δ x Δ p \approx 0.568 ℏ$ , and for $n = 3$ it is $2.63 ℏ$ . What does that comparison show?