Part 2 · The postulates

2.3 Commutators and compatible observables

Source: Centini’s whiteboard of 9 Oct, p. 2; Problem 3.8 from the 2024/25 board #7

Objectives

Understand why the order of two measurements can matter, and when it does not.
Recipe: compute a commutator, on matrices or on a test function.
Picture: two eigenbases at an angle, and what a measurement in one does to the other.
Theorem: observables with a common set of eigenstates commute.
Vocabulary: compatible and incompatible observables.
Essential vocabulary: commutator.

Two measurements in a row: does the second spoil what the first found? The answer depends on one operator, built from the two.

Definition 2.3.1CommutatorEssential

The commutator of two operators is $[\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} .$ $\hat{A}$ and $\hat{B}$ are compatible if $[\hat{A}, \hat{B}] = 0$ , and not compatible otherwise.

Numbers commute, $ab = ba$ ; operators need not. $\hat{A} \hat{B} ψ$ means: apply $\hat{B}$ first, then $\hat{A}$ (Definition 1.3.1). Swapping the order can change the result, as two rotations about different axes do (beyond the boards).

Not compatible

$\hat{A}$ has eigenvalues $a_{n}$ and eigenstates $∣ u_{n} ⟩$ ; $\hat{B}$ has $b_{n}$ and different eigenstates $∣ w_{n} ⟩$ . $∣ ψ ⟩ \hat{A} a_{3}, ∣ u_{3} ⟩ = n \sum c_{n} ∣ w_{n} ⟩ \hat{B} b_{5}, ∣ w_{5} ⟩ = n \sum d_{n} ∣ u_{n} ⟩ \hat{A} a_{7}, ∣ u_{7} ⟩$

Measure $\hat{A}$ : get $a_{3}$ ; the state is now $∣ u_{3} ⟩$ .
$∣ u_{3} ⟩$ is not an eigenstate of $\hat{B}$ but a mix of the $∣ w_{n} ⟩$ . Measuring $\hat{B}$ gives some $b_{5}$ and collapses the state to $∣ w_{5} ⟩$ .
$∣ w_{5} ⟩$ is again a mix of the $∣ u_{n} ⟩$ , so measuring $\hat{A}$ again can give another value, $a_{7}$ .

==The second measurement destroyed the result of the first.== The two can never both be sharp: $Δ A Δ B \neq = 0$ . The spin sequence of Example 2.2.9 is this, with $\hat{A} = \hat{S}_{z}$ and $\hat{B} = \hat{S}_{x}$ .

Compatible

$\hat{A}$ and $\hat{B}$ share eigenstates, $\hat{A} ∣ u_{n} ⟩ = a_{n} ∣ u_{n} ⟩$ and $\hat{B} ∣ u_{n} ⟩ = b_{n} ∣ u_{n} ⟩$ : $∣ ψ ⟩ \hat{A} a_{3}, ∣ u_{3} ⟩ \hat{B} b_{3}, ∣ u_{3} ⟩ \hat{A} a_{3} .$ Measuring $\hat{B}$ leaves the state in $∣ u_{3} ⟩$ , so $\hat{A}$ gives $a_{3}$ again: both values are known at once. That is Exercise 2.2.1©, where $\hat{H}$ and $\hat{A}$ have the same $∣ ϕ_{n} ⟩$ .

Theorem 2.3.2Common eigenstates commute

If $\hat{A}$ and $\hat{B}$ have a common basis of eigenstates, then $[\hat{A}, \hat{B}] = 0$ .

Proofbeyond the boards

On each basis state, $\hat{A} \hat{B} ∣ u_{n} ⟩ = \hat{A} b_{n} ∣ u_{n} ⟩ = b_{n} a_{n} ∣ u_{n} ⟩ = a_{n} b_{n} ∣ u_{n} ⟩ = \hat{B} \hat{A} ∣ u_{n} ⟩ .$ Any state is a sum of $∣ u_{n} ⟩$ and both operators are linear, so $\hat{A} \hat{B} = \hat{B} \hat{A}$ on every state.

The converse holds too: commuting Hermitian operators can always be given a common eigenbasis. The board does not prove it; it is in The Compatible Observables Theorem.

Loading the demo…

Figure 2.3.1 Two observables on a plane of states:

\hat{A}

with eigenbasis

u_{1}, u_{2}

and

\hat{B}

with eigenbasis

v_{1}, v_{2}

, at the angle

θ

you drag. A measurement of

\hat{A}

leaves

u_{1}

; then

\hat{B}

gives

v_{1}

with

cos^{2} θ

and

v_{2}

with

sin^{2} θ

; then

\hat{A}

again returns to

u_{1}

with only

cos^{4} θ + sin^{4} θ

overall. At

θ = 0

90°

the bases coincide and the commutator vanishes; at

45°

the second

\hat{A}

is a coin toss.

Computing a commutator

Recipe 2.3.3Computing a commutator

Matrices. Multiply both ways and subtract.
Differential operators. Apply $[\hat{A}, \hat{B}]$ to a test function $f$ , expand with the product rule, and read off the operator that is left. Never drop $f$ until the end: $\frac{d}{d x} x$ is not 1 but “multiply by $x$ , then differentiate”.

The most important commutator of the course, $[\overset{x}{^}, \overset{p}{^}_{x}] = i ℏ$ , is worked out in Theorem 2.4.3. It is the reason position and momentum obey §2.5 The uncertainty principle.

Example 2.3.4Problem 3.82024/25 board #7

$∣ ψ ⟩ = - 1 21, \hat{A} = \frac{1}{2} 010101010, \hat{B} = 100000 00 - 1 .$ $⟨ ψ ∣ ψ ⟩ = 6$ , so every probability is divided by 6, the same as normalising first (Recipe 2.2.2). $\hat{B}$ is diagonal: eigenvalues $1, 0, - 1$ on the axes, $∣ b_{2} ⟩ = (0, 1, 0)^{T}$ for $b = 0$ .

$det (\hat{A} - λ \hat{I}) = - λ (λ^{2} - 1)$ (written out beyond the boards), so $\hat{A}$ has $- 1, 0, 1$ with normalised eigenvectors $∣ v_{- 1} ⟩ = \frac{1}{2} - 1 2 - 1, ∣ v_{0} ⟩ = \frac{1}{2} - 1 01, ∣ v_{1} ⟩ = \frac{1}{2} 121 .$

$A = - 1$ : $⟨ v_{- 1} ∣ ψ ⟩ = \frac{1}{2} (1 + 22 - 1) = 2$ , so $P = \frac{2}{6} = \frac{1}{3}$ .
$\hat{B}$ then $\hat{A}$ , $B = 0$ then $A = 1$ : $P_{B = 0} = \frac{4}{6} = \frac{2}{3}$ , the state becomes $∣ b_{2} ⟩$ , and $∣ ⟨ v_{1} ∣ b_{2} ⟩ ∣^{2} = \frac{1}{2}$ . Together $\frac{1}{3}$ .
$\hat{A}$ then $\hat{B}$ , $A = 1$ then $B = 0$ : $∣ ⟨ v_{1} ∣ ψ ⟩ ∣^{2} /6 = \frac{2}{6}$ , then $∣ ⟨ b_{2} ∣ v_{1} ⟩ ∣^{2} = \frac{1}{2}$ . Together $\frac{1}{6}$ .

==The two orders give different probabilities== because each measurement collapses onto its own eigenvectors, and the two sets differ: $[\hat{A}, \hat{B}] = \frac{1}{2} 010 - 1 01 0 - 1 0 \neq = 0 (beyond the boards) .$

Warning

The probability of a sequence is a product of probabilities, each computed in the state the previous measurement left, not in the original $∣ ψ ⟩$ .

Exercises

Exercise 2.3.1beyond the boards

Show that $[\hat{A}, \hat{B}] = - [\hat{B}, \hat{A}]$ , that $[\hat{A}, \hat{A}] = 0$ , and that $[\hat{A}, \hat{B} + \hat{C}] = [\hat{A}, \hat{B}] + [\hat{A}, \hat{C}]$ .

Solution

$\hat{A} \hat{B} - \hat{B} \hat{A} = - (\hat{B} \hat{A} - \hat{A} \hat{B})$ ; $\hat{A} \hat{A} - \hat{A} \hat{A} = 0$ ; $\hat{A} (\hat{B} + \hat{C}) - (\hat{B} + \hat{C}) \hat{A} = (\hat{A} \hat{B} - \hat{B} \hat{A}) + (\hat{A} \hat{C} - \hat{C} \hat{A})$ , by linearity.

Exercise 2.3.2beyond the boards

In Problem 3.8, check that $\hat{A} \hat{B}$ and $\hat{B} \hat{A}$ are as claimed: $\hat{A} \hat{B} = \frac{1}{2} 010000 0 - 1 0$ .

Solution

Row $i$ of $\hat{A}$ times column $j$ of $\hat{B}$ . $\hat{B}$ only keeps column 1 and flips column 3 of whatever it multiplies on the right: $\hat{A} \hat{B}$ has the columns of $\hat{A}$ times $1, 0, - 1$ , giving the matrix above. $\hat{B} \hat{A}$ has the rows of $\hat{A}$ times $1, 0, - 1$ : $\frac{1}{2} 000 10 - 1 000$ . The difference is the commutator in the example.

Exercise 2.3.3beyond the boards

Two observables on a two-state system, $\hat{A} = (10 0 - 1)$ and $\hat{B} = (0110)$ . Compute $[\hat{A}, \hat{B}]$ . A measurement of $\hat{A}$ gives $+ 1$ ; then $\hat{B}$ is measured; then $\hat{A}$ again. Probability of $+ 1$ at the end?

Solution

$\hat{A} \hat{B} = (0 - 1 10)$ , $\hat{B} \hat{A} = (01 - 1 0)$ , $[\hat{A}, \hat{B}] = (0 - 2 20) \neq = 0$ . After $A = + 1$ the state is $(1, 0)^{T}$ ; $\hat{B}$ 's eigenvectors are $\frac{1}{2} (1, \pm 1)^{T}$ , each found with $\frac{1}{2}$ ; from either, $A = + 1$ has $\frac{1}{2}$ . Total $\frac{1}{2}$ . These are $2 \hat{S}_{z} /ℏ$ and $2 \hat{S}_{x} /ℏ$ (§4.1 Spin): the 45° case of the figure.

Exercise 2.3.4beyond the boards

Compute $[\overset{x}{^}^{2}, \overset{p}{^}_{x}]$ .

Solution

On $f$ : $x^{2} (- i ℏ f^{'}) + i ℏ (x^{2} f)^{'} = - i ℏ x^{2} f^{'} + i ℏ (2 x f + x^{2} f^{'}) = 2 i ℏ x f$ . So $[\overset{x}{^}^{2}, \overset{p}{^}_{x}] = 2 i ℏ \overset{x}{^}$ .