Part 2 · The postulates

2.1 Measurement and probability

Source: Centini’s whiteboard of 9 Oct, p. 1; the dice, the expected value, the spread and the continuous density from the 2024/25 boards #1–#3

Objectives

Understand why a measurement at the atomic scale disturbs what it measures.
Recipe: the expected value and the spread of a distribution, discrete or continuous.
Picture: a histogram with its mean and a band one spread wide.
Vocabulary: probability distribution, normalisation, expected value, uncertainty (spread), probability density, confidence interval.

Quantum mechanics predicts probabilities, not outcomes. Before the postulates, why it has to, and the arithmetic of probability it uses.

Why measurement needs postulates

To see where the electron of a hydrogen atom is, light must resolve the size of the atom, $1$ Å $= 1 0^{- 10}$ m. So take $λ = 1$ Å: $ν = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{1 0 ^{- 10}} = 3 \times 1 0^{18} Hz, E = h ν = 1.2 \times 1 0^{4} eV .$ The electron is bound by its ionisation energy, 13.6 eV. The photon carries about a thousand times that: looking at the electron knocks it out of the atom. Visible light is gentle enough, a few eV, but its wavelength, $5 \times 1 0^{- 7}$ m for green, is far too long to resolve an atom (2024/25 board #3).

Key observation 2.1.1Measurement disturbs

Classically, position, velocity and acceleration are known at every time from the equation of motion. At the atomic scale ==a measurement is not a passive look: it changes the system==. “A small perturbation of the system is a big perturbation for small systems” (2024/25 board #3). Classical physics has no rule for this, so quantum mechanics states one, as postulates (§2.2 States and measurement).

Dice

The 2024/25 course starts here, and every idea returns: a list of possible results with their probabilities, their average, their spread, and a continuous density.

One die. Every face has probability $\frac{1}{6}$ . One result or another: add, so $P (2 to 5) = \frac{4}{6} = 0.66$ . Two independent throws both giving a set result: multiply, so two 3s have $\frac{1}{36}$ .

Two dice. The sum $n$ runs from 2 to 12, but the sums are not equally likely: each ordered pair has $\frac{1}{36}$ , and a sum can be made by several pairs. $p (4) = p (1) p (3) + p (2) p (2) + p (3) p (1) = \frac{3}{36}$ . In one line (beyond the boards), $p (n) = \frac{6 - ∣ n - 7∣}{36}$ .

Definition 2.1.2Probability distribution2024/25 board #2

A list of probabilities $p (n)$ of mutually exclusive results that adds up to 1, $n \sum p (n) = 1,$ is a probability distribution; the condition is normalisation. For the two dice, $\frac{1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1}{36} = 1$ .

Definition 2.1.3Expected value and spread2024/25 board #2

The expected value (average) and the spread (uncertainty, standard deviation) are $⟨ n ⟩ = n \sum n p (n), Δ n = ⟨ n^{2} ⟩ - ⟨ n ⟩^{2}, ⟨ n^{2} ⟩ = n \sum n^{2} p (n) .$

The board writes $Δ n = ∣ ⟨ n ⟩^{2} - ⟨ n^{2} ⟩ ∣$ ; the order above is never negative, so the absolute value gives the same number.

Why the spread is never imaginarybeyond the boards

The mean squared distance from the mean is $⟨(n - ⟨ n ⟩)^{2} ⟩ = ⟨ n^{2} ⟩ - 2 ⟨ n ⟩ ⟨ n ⟩ + ⟨ n ⟩^{2} = ⟨ n^{2} ⟩ - ⟨ n ⟩^{2}$ : an average of squares, so $\geq 0$ . $Δ n^{2}$ is that mean squared distance.

Example 2.1.4Two dice2024/25 board #2

$⟨ n ⟩ = \frac{1}{36} (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) = \frac{252}{36} = 7$ , as the symmetry $p (n) = p (14 - n)$ says. $⟨ n^{2} ⟩ = \frac{1974}{36}$ , so $Δ n^{2} = \frac{1974}{36} - 49 = \frac{35}{6}$ and $Δ n = 2.42$ . (Board 1 first reached $\frac{122}{18} = 6.8$ , a slip in combining fractions; board 2 redoes it and gets 7.)

Loading the demo…

Figure 2.1.1 The distribution of the sum of two dice, with the

mean

and the band

⟨ n ⟩ \pm Δ n

. “1000 copies” throws the pair a thousand times and outlines the frequencies: close to the bars, never exactly on them.

From counts. Throw $N$ times, with $N_{i}$ throws giving $i$ : the average is $\sum_{i} N_{i} i / N$ , and as $N$ grows $N_{i} / N \to p_{i}$ . ==An expected value is the average over many repetitions==, which is how the quantum expectation value is read.

Confidence intervals. $7 \pm 2.4$ , from 4.6 to 9.4, holds the sums 5 to 9: $\frac{24}{36} = 0.67$ . $7 \pm 4.8$ holds 3 to 11: $\frac{34}{36} = 0.94$ . The board’s round figures, 66% and 95%, are those of a Gaussian (68.3% and 95.4%), which the histogram resembles.

Continuous results

When the result can be any real number $x$ , a position say, a single value has zero probability. What is left is a density.

Definition 2.1.5Probability density2024/25 board #2

A probability density $f (x) \geq 0$ gives the probability $f (x) d x$ of a result in the narrow interval $d x$ at $x$ , and is normalised: $\int_{- \infty}^{+ \infty} f (x) d x = 1$ . Sums become integrals: $⟨ x ⟩ = \int x f (x) d x, ⟨ x^{2} ⟩ = \int x^{2} f (x) d x, Δ x = ⟨ x^{2} ⟩ - ⟨ x ⟩^{2} .$

In quantum mechanics the density is $∣ ψ (x) ∣^{2}$ (Definition 2.2.1) and the discrete $p (n)$ are the $∣ ⟨ u_{n} ∣ ψ ⟩ ∣^{2}$ (Theorem 2.2.4). The bell-shaped $f (x)$ comes back in §2.5 The uncertainty principle.

Exercises

Exercise 2.1.1beyond the boards

One die: find $⟨ n ⟩$ and $Δ n$ , and the probability of a throw within $⟨ n ⟩ \pm Δ n$ .

Solution

$⟨ n ⟩ = \frac{21}{6} = 3.5$ ; $⟨ n^{2} ⟩ = \frac{91}{6} = 15.17$ ; $Δ n = 15.17 - 12.25 = 1.71$ . The interval 1.79 to 5.21 holds the faces 2 to 5: probability $\frac{4}{6} = 0.67$ .

Exercise 2.1.2beyond the boards

A density is $f (x) = \frac{3}{4} (1 - x^{2})$ on $[- 1, 1]$ and zero elsewhere. Check the normalisation and find $⟨ x ⟩$ and $Δ x$ .

Solution

$\int_{- 1}^{1} \frac{3}{4} (1 - x^{2}) d x = \frac{3}{4} (2 - \frac{2}{3}) = 1$ . $⟨ x ⟩ = 0$ by symmetry. $⟨ x^{2} ⟩ = \frac{3}{4} \int_{- 1}^{1} (x^{2} - x^{4}) d x = \frac{3}{4} (\frac{2}{3} - \frac{2}{5}) = \frac{1}{5}$ , so $Δ x = 1/ 5 = 0.45$ .

Exercise 2.1.3beyond the boards

Shift every result of a distribution by a constant, $n \to n + c$ . What happens to $⟨ n ⟩$ and to $Δ n$ ?

Solution

$⟨ n + c ⟩ = ⟨ n ⟩ + c$ , but $Δ n$ is unchanged: it measures distances from the mean, and both move together. Formally, $⟨(n + c)^{2} ⟩ - ⟨ n + c ⟩^{2} = ⟨ n^{2} ⟩ + 2 c ⟨ n ⟩ + c^{2} - ⟨ n ⟩^{2} - 2 c ⟨ n ⟩ - c^{2}$ .