Probability · Quantum Mechanics (Phillips)

Two numbers summarise any distribution, and both are the same integral with a different power in it. Almost nothing else is used for the rest of the book.

Chapter 2 left a hole and said so. We have an equation, its solutions are necessarily complex, and Phillips ended the chapter admitting that when a measurement happens to a wave packet, “no one knows how this happens.”

Chapter 3 fills the hole. It opens by stating the puzzle as sharply as it can be put: a wave function is spread out and non-localized, so it describes wave-like behaviour easily —

but how can it describe an electron that arrives as a lump on the screen?

The answer is the one that changed physics: measurements have random outcomes, and the wave function supplies the probabilities. Before that can be made precise, §3.1 spends four pages on probability itself.

If you have done any statistics this section holds nothing new, and you can skim to §3.2. It is worth at least checking the notation, because §3.5 will apply every formula here to operators, and the resemblance is the whole point.

Discrete outcomes

Suppose a process has possible outcomes $x_0, x_1, x_2, \dots$ occurring with probabilities $p_0, p_1, p_2, \dots$ . The set $p_n$ is a probability distribution , and it must satisfy the normalization condition

\sum_{\text{all }n} p_n = 1\tag{3.1}

because something happens. Phillips’ example is disarming: if $p_n$ is the probability that a reader of this book has $n$ living grandparents, then $p_0 + p_1 + p_2 + p_3 + p_4 = 1$ .

The expectation value is the average outcome over infinitely many repetitions:

\langle x\rangle = \sum_{\text{all }n} x_n p_n\tag{3.2}

and the spread about it is measured by the variance , whose square root is the standard deviation, written $\Delta x$ :

(\Delta x)^2 = \sum_{\text{all }n}(x_n - \langle x\rangle)^2 p_n\tag{3.3}

Expanding the square in Eq. (3.3) gives a form that is easier to compute with, and it is the one every later calculation uses:

step 1 of 4

Three lines, and the result is used on nearly every page from here on.

1Expand the square. Nothing has happened yet.
$(x_{n} - ⟨ x ⟩)^{2} = x_{n}^{2} - 2 ⟨ x ⟩ x_{n} + ⟨ x ⟩^{2}$
2Sum term by term, and pull ⟨x⟩ out of the sums — it is a fixed number, not a function of n.
$(Δ x)^{2} = \sum x_{n}^{2} p_{n} - 2 ⟨ x ⟩ \sum x_{n} p_{n} + ⟨ x ⟩^{2} \sum p_{n}$
3Recognise each sum: the first is ⟨x²⟩ by definition, the second is ⟨x⟩, and the third is 1 by normalization.
$(Δ x)^{2} = ⟨ x^{2} ⟩ - 2 ⟨ x ⟩^{2} + ⟨ x ⟩^{2}$
4Collect. The result is worth memorising in words as well as symbols.
$(Δ x)^{2} = ⟨ x^{2} ⟩ - ⟨ x ⟩^{2}$
Equation (3.4): the variance is the average of the square minus the square of the average.

Continuous outcomes

When the outcome is a continuous variable, individual values have zero probability and you need a density instead. The probability of an outcome between $x$ and $x + dx$ is $\rho(x)\,dx$ , where $\rho$ is the probability density . Every sum becomes an integral:

\int_{\text{all }x}\rho(x)\,dx = 1\tag{3.5}

\langle x\rangle = \int_{\text{all }x} x\,\rho(x)\,dx\tag{3.6}

\langle x^2\rangle = \int_{\text{all }x} x^2\rho(x)\,dx\tag{3.7}

\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2}\tag{3.8}

σ1.00

∫ ρ dx (3.5): 1.000000
⟨x⟩ (3.6): 0.0000
⟨x²⟩ (3.7): 1.0000
(Δx)² by Eq. (3.3): 1.000000
(Δx)² by Eq. (3.4): 1.000000
Δx: 1.0000

what the algebra says

⟨x⟩ = 0 by symmetry

Δx = σ = 1.00

Problem 2, and the shape that returns in problem 4 as a Gaussian wave function. ⟨x⟩ = 0 follows from symmetry alone: x·ρ(x) is odd, so its integral vanishes without any calculation.

Three things worth doing with that widget

Check that the two variance formulas agree. The readout evaluates Eq. (3.3) and Eq. (3.4) separately, on the same distribution, and they match to six decimals for every choice. That agreement is the identity the step-through proves.

Look at “one die”. The expectation value is 3.5 — not a possible outcome. An expectation value is an average, not a prediction, and nothing requires it to be something you could ever observe. This matters more in quantum mechanics than in dice: $\langle x\rangle$ for a particle can sit at a point where $|\Psi|^2$ is exactly zero.

Watch two dice. Adding two flat distributions produces a peak. Keep adding and it tends to a Gaussian — the central limit theorem, and the reason problem 2 describes the Gaussian as the distribution of “a multitude of small random contributions, such as the net distance travelled by a tottering drunk with very small legs.”

Notation warning — chapter 3 uses a for three unrelated things

Two of them are already on this page, so it is worth naming all three now:

Where	$a$ means	Units
§3.1, problem 1	the mean of a Poisson distribution	none — it is a count
§3.4, Fig. 3.3, problems 5 and 6	the width of the box a particle is confined to	m
problem 4	the length scale of a Gaussian wave function	m

Nothing connects them. This is ordinary textbook symbol reuse — Phillips is not being careless, there are only so many letters — but it means you cannot carry $a$ from one page to the next. Check its units before using any formula that contains it: the Poisson $a$ is dimensionless, while the box and Gaussian $a$ are lengths.

Watch also for the near-misses: problem 7’s decay constant is $\alpha$ , not $a$ , and problem 2’s Gaussian width is $\sigma$ . All three look alike in running text.

Each page here states which $a$ it is using.

Engineer's bridge — Eq. (3.4) is the one-pass variance, and it has a famous failure mode

Equation (3.4) is exactly the shortcut every numerics library once used: accumulate $\sum x$ and $\sum x^2$ in a single pass, then subtract at the end. It is fast, it needs no second pass over the data, and it is numerically treacherous.

The problem is catastrophic cancellation. When $\langle x\rangle \gg \Delta x$ , the two terms $\langle x^2\rangle$ and $\langle x\rangle^2$ are nearly equal, and subtracting nearly-equal floating-point numbers destroys precision. Take $x = 10^8 \pm 1$ : in float64, $\langle x^2\rangle \approx 10^{16}$ has an absolute precision of about 2, so a variance of 1 comes out as noise — sometimes even negative.

Production code uses Welford’s algorithm, a one-pass update that never forms the difference. numpy.var uses a two-pass method for the same reason.

Where it does not bite you here. In quantum mechanics you almost always choose the origin so that $\langle x \rangle \approx 0$ — at the centre of a box, the bottom of a well, the position of the nucleus — and then the cancellation never happens. The formula is safe because the physics is conveniently centred, not because the formula is safe.

Watch for it in the numbers on the next few pages: whenever the book writes $\langle x\rangle = 0$ “by symmetry”, it has quietly removed the failure mode too.

Put a number on it — the Poisson distribution and why counting is slow

Select Poisson in the widget. Its distribution is

p_n = \frac{e^{-a}a^n}{n!}, \qquad n = 0,1,2,\dots

and problem 1 asks you to show $\langle n\rangle = a$ and $\Delta n = \sqrt a$ . Slide $a$ and watch both readouts track those formulas exactly.

The consequence is worth having. The fractional uncertainty is

\frac{\Delta n}{\langle n\rangle} = \frac{\sqrt a}{a} = \frac{1}{\sqrt a}

Count 100 events and you know the rate to 10%. To reach 1% you need 10 000 — a hundred times the counting time for a tenfold improvement. Every photon counting experiment, every particle-physics event rate, every radioactive decay measurement is governed by this, and it is why experiments take as long as they do.

Reproduce it in NumPy

Every formula in §3.1, evaluated rather than asserted — the two variance routes agreeing, and the counting-statistics consequence.

import numpy as np
from math import factorial, exp, sqrt

a = 4.0
n = np.arange(0, 40)
p = np.array([exp(-a)*a**k/factorial(k) for k in n])

mean = np.sum(n*p)
m2   = np.sum(n**2*p)
sd   = sqrt(m2 - mean**2)                      # Eq (3.4)
print(f"Poisson with a = {a:.0f}:")
print(f"  sum p_n        = {p.sum():.6f}   (Eq 3.1)")
print(f"  <n>            = {mean:.6f}   (Eq 3.2)   analytic a     = {a:.0f}")
print(f"  Delta_n        = {sd:.6f}   (Eq 3.4)   analytic sqrt(a) = {sqrt(a):.1f}")
print(f"  fractional     = {sd/mean:.6f}")

print("\nfractional uncertainty 1/sqrt(a):")
for A in (10, 100, 10_000, 1_000_000):
    print(f"  a = {A:7d}   ->  {100/sqrt(A):5.1f}%")

s = 1.0
x = np.linspace(-8*s, 8*s, 20001)
rho = np.exp(-x**2/(2*s**2)) / np.sqrt(2*np.pi*s**2)
dx = x[1] - x[0]
mean_x = np.sum(x*rho)*dx
m2_x   = np.sum(x**2*rho)*dx
print(f"\nGaussian sigma = {s:.1f}, integrated numerically:")
print(f"  integral rho dx = {np.sum(rho)*dx:.6f}   <x> = {abs(mean_x):.6f}   "
      f"Delta_x = {sqrt(m2_x - mean_x**2):.6f}")

printsPoisson with a = 4: sum p_n = 1.000000 (Eq 3.1) <n> = 4.000000 (Eq 3.2) analytic a = 4 Delta_n = 2.000000 (Eq 3.4) analytic sqrt(a) = 2.0 fractional = 0.500000 fractional uncertainty 1/sqrt(a): a = 10 -> 31.6% a = 100 -> 10.0% a = 10000 -> 1.0% a = 1000000 -> 0.1% Gaussian sigma = 1.0, integrated numerically: integral rho dx = 1.000000 <x> = 0.000000 Delta_x = 1.000000

Why this section is here

Nothing above is quantum mechanics. It is the ordinary theory of random variables, and Phillips introduces it because the next four sections apply it to something unprecedented.

In the kinetic theory of gases, probability covers up ignorance: each molecule has a definite position and velocity and we simply do not know them. The question §3.6 will leave open is whether quantum probability is like that, or whether it is the complete story — whether the particle has a position we do not know, or no position at all until it is measured.

The mathematics on this page cannot tell the difference. That is worth remembering when the formulas start looking familiar: $\langle x\rangle$ and $\Delta x$ will mean exactly what they mean here, and the interpretation is a separate question the equations do not settle.

Next: §3.2 identifies $\rho(x)$ with $|\Psi(x,t)|^2$ — one line that turns the Schrödinger equation into a theory of measurable things.

Check yourself

0 / 6 answered

1.Why is the variance defined with the *square* of the deviation, rather than the deviation itself?
2.Rolling one die, $⟨ x ⟩ = 3.5$ . What does that tell you about the outcome of a single roll?
3. $ρ (x)$ for a narrow Gaussian reaches a peak value of 1.3. Is something wrong?
4.A Poisson process is counted for a fixed time and gives about 100 events. Roughly how much longer must you count to know the rate to 1% instead of 10%?
5.Equation (3.4), $(Δ x)^{2} = ⟨ x^{2} ⟩ - ⟨ x ⟩^{2}$ , is the classic one-pass variance formula. When does using it numerically get you into trouble?
6.What has §3.1 established about *quantum* mechanics?