Part 2 · The postulates

2.2 States and measurement

Source: Centini’s whiteboards of 6 Oct, p. 5–6, and 9 Oct, p. 1–6; the amplitude of probability, the vector picture and the many copies from the 2024/25 boards #3, #4 and #20

Objectives

Understand a state as a normalised vector, and a measurement as a projection on the eigenstates of an observable.
Recipe: from $∣ ψ ⟩ = \sum_{n} b_{n} ∣ u_{n} ⟩$ , the possible results, their probabilities, $⟨ \hat{A} ⟩$ and $Δ A$ .
Picture: a measurement collapsing a distribution onto one bar.
Theorem: $⟨ ψ ∣ \hat{A} ∣ ψ ⟩ = \sum_{n} P (a_{n}) a_{n}$ .
Vocabulary: wavefunction, normalisation, amplitude of probability, collapse.
Essential vocabulary: state, observable, expectation value.

The state

Definition 2.2.1StateEssential

The state of a system is its wavefunction $ψ (\overset{r}{ˉ}, t)$ , or ket $∣ ψ ⟩$ , normalised: $⟨ ψ ∣ ψ ⟩ = \int ψ^{*} ψ d^{3} r = 1.$ $∣ ψ ∣^{2} = ψ^{*} ψ$ is the probability density of finding the particle at $\overset{r}{ˉ}$ .

The area under $∣ ψ ∣^{2}$ is 1 because the particle is certainly somewhere. In one dimension, the probability of finding it between $x_{1}$ and $x_{2}$ is the area over that stretch (2024/25 board #20): $P (x_{1} < x < x_{2}) = \int_{x_{1}}^{x_{2}} ∣ ψ ∣^{2} d x .$ $∣ ψ ∣^{2}$ is the density $f (x)$ of Definition 2.1.5.

Recipe 2.2.2Normalising a state

If $∣ ψ ⟩ = \sum_{n} b_{n} ∣ u_{n} ⟩$ in an orthonormal basis, $⟨ ψ ∣ ψ ⟩ = \sum_{n} ∣ b_{n} ∣^{2}$ : the bra has the conjugate coefficients, and $⟨ u_{n} ∣ u_{m} ⟩ = δ_{nm}$ keeps only the diagonal products.
If $⟨ φ ∣ φ ⟩ \neq = 1$ , divide by the length: $∣ ψ ⟩ = ∣ φ ⟩ / ⟨ φ ∣ φ ⟩$ .

Example 2.2.3The board’s state

$∣ ψ ⟩ = \frac{2}{15} ∣ ϕ_{1} ⟩ + \frac{3}{15} ∣ ϕ_{2} ⟩ + \frac{6}{15} ∣ ϕ_{3} ⟩ + \frac{2}{15} ∣ ϕ_{4} ⟩$ has $⟨ ψ ∣ ψ ⟩ = \frac{2 + 3 + 6 + 4}{15} = 1$ .

Postulate 1: observables and measurement

Theorem 2.2.4Postulate 1

Each physical quantity, an observable, is described by a Hermitian operator $\hat{A}$ .
The result of a single measurement is one of the eigenvalues $a_{n}$ of $\hat{A}$ .
The system then collapses to the eigenstate $∣ u_{n} ⟩$ of the measured eigenvalue.
The probability of measuring $a_{n}$ is $P (a_{n}) = ∣ ⟨ u_{n} ∣ ψ ⟩ ∣^{2} .$

Why Hermitian: its eigenvalues are real, as a measured number is, and its eigenstates are a basis, so every state can be split along them (Theorem 1.4.2).

$∣ ψ ⟩ \hat{A} a_{3}, ∣ u_{3} ⟩ \hat{A} a_{3}, ∣ u_{3} ⟩$

One run gives one eigenvalue, say $a_{3}$ ; afterwards the system is in $∣ u_{3} ⟩$ , and measuring $\hat{A}$ again straight away gives $a_{3}$ with certainty.

With $∣ ψ ⟩ = \sum_{n} b_{n} ∣ u_{n} ⟩$ , $⟨ u_{n} ∣ ψ ⟩ = b_{n}$ (Recipe 1.4.5), so $P (a_{n}) = ∣ b_{n} ∣^{2}$ , and these add up to $⟨ ψ ∣ ψ ⟩ = 1$ . The number $⟨ u_{n} ∣ ψ ⟩$ itself is the amplitude of probability; the probability is its squared modulus (2024/25 board #3). As an ordinary vector is rebuilt from its projections on the axes, $\overset{v}{ˉ} = \sum_{i} (\overset{u}{^}_{i} \cdot \overset{v}{ˉ}) \overset{u}{^}_{i}$ , a state is rebuilt from its projections on the eigenstates (Figure 1.4.1).

Loading the demo…

Figure 2.2.1 The board’s state, energies

n^{2} E_{0}

with

E_{0} = 1

eV. The bars are

P (E_{n}) = ∣ ⟨ ϕ_{n} ∣ ψ ⟩ ∣^{2}

, the

orange line

⟨ \hat{H} ⟩

and the band

\pm Δ H

. Press Measure: one energy comes out and the state collapses onto its eigenstate; measure again and it repeats. Drag the bars to make other states; drag one to the top to make an eigenstate.

Expectation value and spread

Definition 2.2.5Expectation valueEssential

The expectation value of $\hat{A}$ in the state $∣ ψ ⟩$ is $⟨ \hat{A} ⟩ = ⟨ ψ ∣ \hat{A} ∣ ψ ⟩ = \int ψ^{*} \hat{A} ψ d^{3} r .$

Theorem 2.2.6The expectation value is the average result

$⟨ \hat{A} ⟩ = n \sum P (a_{n}) a_{n} .$

Proofbeyond the boards

With $∣ ψ ⟩ = \sum_{n} b_{n} ∣ u_{n} ⟩$ , $\hat{A} ∣ ψ ⟩ = \sum_{n} b_{n} a_{n} ∣ u_{n} ⟩$ , so $⟨ ψ ∣ \hat{A} ∣ ψ ⟩ = \sum_{m, n} b_{m}^{*} b_{n} a_{n} ⟨ u_{m} ∣ u_{n} ⟩ = \sum_{n} ∣ b_{n} ∣^{2} a_{n}$ .

It is the ordinary weighted average, the dice’s $\sum n p (n)$ . Many copies (2024/25 board #4): prepare the same $∣ ψ ⟩$ again and again and measure the energy each time, getting $E_{1}$ , $E_{2}$ , $E_{1}$ , $E_{4}$ , … The fraction of runs giving $E_{n}$ tends to $P (E_{n})$ , and their average to $⟨ \hat{H} ⟩$ . One copy measured once says nothing about the average.

Warning

==A single measurement gives an eigenvalue; the expectation value is an average and need not be any eigenvalue.== In the figure the energies are 1, 4, 9 and 16 eV and the average is 8.8 eV.

A result is reported as $⟨ \hat{A} ⟩ \pm Δ A$ , with the spread $Δ A = ⟨ \hat{A}^{2} ⟩ - ⟨ \hat{A} ⟩^{2}, ⟨ \hat{A}^{2} ⟩ = n \sum P (a_{n}) a_{n}^{2},$ since applying $\hat{A}$ twice squares each eigenvalue. The board notes that about 66% of the results fall within $\pm Δ A$ , as for the dice. $Δ A = 0$ exactly when $∣ ψ ⟩$ is an eigenstate of $\hat{A}$ .

Example 2.2.7Momentum in a box

In a box, $\overset{p}{^}_{x}$ has eigenstates $∣ u_{n} ⟩$ with $p_{n} = 2 π ℏ n / L$ (Theorem 1.5.1). In $∣ ψ ⟩ = \sum_{n} a_{n} ∣ u_{n} ⟩$ , a measurement of $\overset{p}{^}_{x}$ gives $p_{n}$ with probability $∣ a_{n} ∣^{2}$ , and leaves the state in $∣ u_{n} ⟩$ . The coefficients may carry time phases, $a_{n} e^{iω t}$ ; they change no probability, since $∣ e^{iω t} ∣^{2} = 1$ (Proposition 1.1.10).

A preview: spin ½

The board uses an electron’s spin (§4.1 Spin) to show how measurement works. An electron has an intrinsic angular momentum $\overset{ˉ}{S}$ with $\hat{S}^{2} = \frac{3}{4} ℏ^{2}$ , so $∣ \overset{ˉ}{S} ∣ = \frac{3}{2} ℏ \approx 0.87ℏ$ ; its component along $z$ can only be measured as $S_{z} = \pm \frac{ℏ}{2}$ .

Key observation 2.2.8The cone

$∣ \overset{ˉ}{S} ∣ = 0.87ℏ$ is longer than $S_{z} = 0.5ℏ$ , so $\overset{ˉ}{S}$ never points along $z$ . In the state with $S_{z} = + \frac{ℏ}{2}$ it lies somewhere on a cone around $z$ : $S_{z} = \frac{ℏ}{2} \pm 0, S_{x} = 0 \pm \frac{ℏ}{2}, S_{y} = 0 \pm \frac{ℏ}{2} .$

Classically all three components would have definite values. Here they cannot all be sharp: $\hat{S}_{x}$ , $\hat{S}_{y}$ , $\hat{S}_{z}$ are incompatible (§2.3 Commutators and compatible observables).

Example 2.2.9Measurements in sequence

Stern–Gerlach magnets, each splitting a beam by one spin component:

Measure $S_{z}$ and keep $+ \frac{ℏ}{2}$ : the state is $∣ S_{z}, + ⟩$ .
Measure $S_{x}$ on it: the beam splits into $\pm \frac{ℏ}{2}$ . Keep $+ \frac{ℏ}{2}$ : the state is $∣ S_{x}, + ⟩$ .
Measure $S_{z}$ again. Classically it would still be $+ \frac{ℏ}{2}$ . Instead the beam splits again into both values.

==Measuring $S_{x}$ collapsed the state onto an $S_{x}$ eigenstate and erased what was known about $S_{z}$ .== The splits are 50/50 (Example 4.1.4).

Exercises

Exercise 2.2.1Board exercise, 9 Oct

A system is in $∣ ψ ⟩ = \frac{1}{15} [2 ∣ ϕ_{1} ⟩ + 3 ∣ ϕ_{2} ⟩ + 6 ∣ ϕ_{3} ⟩ + 2∣ ϕ_{4} ⟩]$ , with $\hat{H} ∣ ϕ_{n} ⟩ = n^{2} E_{0} ∣ ϕ_{n} ⟩$ , $E_{0} = 1$ eV. (a) If energy is measured, what values and probabilities? Find $⟨ \hat{H} ⟩$ and $Δ H$ . (b) $\hat{A} ∣ ϕ_{n} ⟩ = 2 n a_{0} ∣ ϕ_{n} ⟩$ . Values and probabilities of $\hat{A}$ , and $⟨ \hat{A} ⟩$ ? © A measurement of energy gives $9 E_{0}$ . Immediately after, $\hat{A}$ is measured: what value?

Solution

(a) $1, 4, 9, 16$ eV. $P (E_{0}) = ∣ ⟨ ϕ_{1} ∣ ψ ⟩ ∣^{2}$ ; expanding, only $⟨ ϕ_{1} ∣ ϕ_{1} ⟩ = 1$ survives, so $P = \frac{2}{15}$ ; likewise $\frac{3}{15}$ , $\frac{6}{15}$ , $\frac{4}{15}$ , adding to 1. $∣ ϕ_{5} ⟩$ is not in $ψ$ , so $25 E_{0}$ never comes out. $⟨ \hat{H} ⟩ = E_{0} \frac{2 + 12 + 54 + 64}{15} = 8.8 E_{0}, ⟨ \hat{H}^{2} ⟩ = E_{0}^{2} \frac{2 + 48 + 486 + 1024}{15} = 104 E_{0}^{2},$ $Δ H = 104 - 77.44 E_{0} = 5.15 E_{0}$ : $E = (8.8 \pm 5.2)$ eV. (The board lists the eigenvalues $n^{4} E_{0}^{2}$ of $\hat{H}^{2}$ ; the rest of $Δ H$ is beyond the boards.) (b) Same eigenstates, so same probabilities: $2 a_{0}, 4 a_{0}, 6 a_{0}, 8 a_{0}$ with $\frac{2}{15}, \frac{3}{15}, \frac{6}{15}, \frac{4}{15}$ . $⟨ \hat{A} ⟩ = 2 a_{0} \cdot \frac{42}{15} = 5.6 a_{0}$ . © $9 E_{0}$ collapses the state to $∣ ϕ_{3} ⟩$ , which is an eigenstate of $\hat{A}$ too: $6 a_{0}$ , with certainty. $\hat{H}$ and $\hat{A}$ are compatible.

Exercise 2.2.2Board exercise, 2024/25

$∣ ψ ⟩ = \frac{5}{19} ∣ ϕ_{1} ⟩ + \frac{3}{19} ∣ ϕ_{2} ⟩ + \frac{7}{19} ∣ ϕ_{3} ⟩ + \frac{4}{19} ∣ ϕ_{4} ⟩$ , with $\hat{H} ∣ ϕ_{n} ⟩ = (n + 1)^{2} E_{0} ∣ ϕ_{n} ⟩$ , $E_{0} = 1$ eV, and $\hat{A} ∣ ϕ_{n} ⟩ = 3 n a_{0} ∣ ϕ_{n} ⟩$ . Energies and probabilities, $⟨ E ⟩$ , $⟨ A ⟩$ ; and what happens after a measurement gives $E_{2}$ ?

Solution

$4, 9, 16, 25$ eV with $\frac{5}{19}, \frac{3}{19}, \frac{7}{19}, \frac{4}{19}$ ( $n$ starts at 1, so $E_{1} = 4 E_{0}$ ). $⟨ E ⟩ = \frac{20 + 27 + 112 + 100}{19}$ eV $= 13.6$ eV; beyond the boards, $⟨ \hat{H}^{2} ⟩ = \frac{4615}{19}$ eV², $Δ E = 7.6$ eV. $⟨ A ⟩ = \frac{a _{0}}{19} (15 + 18 + 63 + 48) = 7.6 a_{0}$ . After $E_{2}$ the state is $∣ ϕ_{2} ⟩$ : measuring the energy again gives $E_{2}$ , and $\hat{A}$ would give $6 a_{0}$ , both with certainty.

Exercise 2.2.3Exercise 1, sample3

$∣ ψ ⟩ = \frac{1}{15} [3 ∣ ϕ_{1} ⟩ + 7 ∣ ϕ_{2} ⟩ + 2 ∣ ϕ_{3} ⟩ + 3 ∣ ϕ_{4} ⟩]$ , $\hat{H} ∣ ϕ_{n} ⟩ = n^{2} E_{0} ∣ ϕ_{n} ⟩$ , $\hat{A} ∣ ϕ_{n} ⟩ = 3 n a_{0} ∣ ϕ_{n} ⟩$ . (a) Energy values and probabilities. (b) $\hat{A}$ values and probabilities. © The energy gives $9 E_{0}$ ; then $\hat{A}$ is measured. Value?

Solution

(a) Normalised: $\frac{3 + 7 + 2 + 3}{15} = 1$ . $E_{0}, 4 E_{0}, 9 E_{0}, 16 E_{0}$ with $\frac{3}{15}, \frac{7}{15}, \frac{2}{15}, \frac{3}{15}$ ; $⟨ \hat{H} ⟩ = \frac{97}{15} E_{0} = 6.47 E_{0}$ , $⟨ \hat{H}^{2} ⟩ = 69.67 E_{0}^{2}$ , $Δ H = 5.28 E_{0}$ . (b) $3 a_{0}, 6 a_{0}, 9 a_{0}, 12 a_{0}$ with the same probabilities; $⟨ \hat{A} ⟩ = 7 a_{0}$ . © Collapse to $∣ ϕ_{3} ⟩$ : $9 a_{0}$ with certainty.

Exercise 2.2.4Sample questions, Q2 (iii)

What is the difference between the result of a single measurement and an expectation value? Write the expectation value for a 1D wavefunction and in Dirac notation.

Solution

A single measurement returns one eigenvalue $a_{n}$ and collapses the state; the expectation value is the average over many identically prepared systems, $\sum_{n} P (a_{n}) a_{n}$ , and need not be an eigenvalue. $⟨ \hat{A} ⟩ = \int ψ^{*} (x) \hat{A} ψ (x) d x = ⟨ ψ ∣ \hat{A} ∣ ψ ⟩$ .

Exercise 2.2.5beyond the boards

Normalise $∣ φ ⟩ = ∣ ϕ_{1} ⟩ + 2 i ∣ ϕ_{2} ⟩$ and find the probabilities of the two energies. Then show that $Δ A = 0$ in any eigenstate of $\hat{A}$ .

Solution

$⟨ φ ∣ φ ⟩ = 1 + (- 2 i) (2 i) = 5$ , so $∣ ψ ⟩ = \frac{1}{5} (∣ ϕ_{1} ⟩ + 2 i ∣ ϕ_{2} ⟩)$ : $P_{1} = \frac{1}{5}$ , $P_{2} = \frac{4}{5}$ . The $i$ changes no probability. In an eigenstate with eigenvalue $a$ , $⟨ \hat{A} ⟩ = a$ and $⟨ \hat{A}^{2} ⟩ = a^{2}$ , so $Δ A = 0$ .