1.2.2 Uncertainty and Incompatibility

1.2.2 Uncertainty and Incompatibility#

Prompts

  • What is the commutator \([\hat{A}, \hat{B}]\), and what does it tell you about two observables?

  • Compute \([\hat{X}, \hat{Z}]\) directly from the Pauli matrices. What do you notice?

  • If two observables commute, what does that imply about their simultaneous measurability?

  • State the Heisenberg uncertainty principle. Why does the size of the commutator matter?

  • For a qubit in state \(\vert+\rangle\) (eigenstate of \(\hat{X}\)), what are \(\Delta\hat{X}\) and \(\Delta\hat{Z}\)? Do they violate the uncertainty relation?

Lecture Notes#

Overview#

Quantum mechanics places fundamental limits on the joint knowledge we can have of non-commuting observables. This section explores the commutator, proves that commuting observables share an eigenbasis, and derives the Heisenberg uncertainty principle.

Expectation Value and Variance#

For a measurement outcome that is random, we characterize the distribution by its expectation value and variance. If observable \(\hat{O}\) has eigenstates \(\vert O=m \rangle\) with eigenvalues \(m\), and state \(\vert\psi\rangle = \sum_m \psi_m \vert O=m \rangle\), then:

\[ \langle\hat{O}\rangle_\psi = \sum_m m \, P(m \vert \psi, \hat{O}) \]

where \(P(m \vert \psi, \hat{O}) = |\psi_m|^2\) is the probability of outcome \(m\).

Expectation Value and Variance

For a Hermitian operator \(\hat{O}\) and state \(\vert\psi\rangle\):

\[ \langle\hat{O}\rangle_\psi = \langle\psi\vert\hat{O}\vert\psi\rangle \]
\[ \mathrm{Var}(\hat{O}) = \langle\hat{O}^2\rangle - \langle\hat{O}\rangle^2 \]
\[ \Delta\hat{O} = \sqrt{\mathrm{Var}(\hat{O})} \]

The uncertainty \(\Delta\hat{O}\) is the standard deviation of outcomes.

Derivation: \(\langle\hat{O}\rangle_\psi = \langle\psi\vert\hat{O}\vert\psi\rangle\)

Start from the spectral decomposition \(\hat{O} = \sum_m m \, \hat{P}_{O=m} = \sum_m m \vert O{=}m\rangle\langle O{=}m\vert\) and the Born rule \(P(m\vert\psi) = \vert\psi_m|^2\):

\[ \langle\hat{O}\rangle_\psi = \sum_m m \, P(m\vert\psi) = \sum_m m \, \vert\psi_m|^2 \]

Expand \(\vert\psi\rangle = \sum_{m'} \psi_{m'} \vert O{=}m'\rangle\) and compute \(\langle\psi\vert\hat{O}\vert\psi\rangle\) directly:

\[ \langle\psi\vert\hat{O}\vert\psi\rangle = \sum_{m',m''} \psi_{m'}^* \psi_{m''} \langle O{=}m'\vert \hat{O} \vert O{=}m''\rangle = \sum_{m',m''} \psi_{m'}^* \psi_{m''} \, m'' \, \delta_{m'm''} = \sum_m m \, \vert\psi_m|^2 \]

which is the same expression. \(\square\)

The Commutator#

When two operators \(\hat{A}\) and \(\hat{B}\) applied in different orders yield different results, we measure this non-commutativity by the commutator:

Commutator

\[ [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} \]

If \([\hat{A}, \hat{B}] = 0\), the operators commute.

Key Properties of the Commutator

  • Linearity: \([\hat{A}, \hat{B} + \hat{C}] = [\hat{A}, \hat{B}] + [\hat{A}, \hat{C}]\)

  • Antisymmetry: \([\hat{A}, \hat{B}] = -[\hat{B}, \hat{A}]\)

  • Jacobi identity: \([\hat{A}, [\hat{B}, \hat{C}]] + [\hat{B}, [\hat{C}, \hat{A}]] + [\hat{C}, [\hat{A}, \hat{B}]] = 0\)

  • Product rule: \([\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}]\)

Intuition: Commute = Independent

Think of getting dressed. Putting on socks then shoes is different from shoes then socks—they don’t commute. But putting on a hat is independent of the socks-shoes order—hat and socks commute. Non-commutativity signals a constraint: we cannot simultaneously sharpen knowledge of both observables.

Commuting Operators Share Eigenbasis#

A central theorem relates commutativity to shared eigenstates:

Theorem: Commuting Operators

Two Hermitian operators \(\hat{A}\) and \(\hat{B}\) commute, i.e. \([\hat{A}, \hat{B}] = 0\), if and only if they share a complete set of simultaneous eigenstates.

Equivalently: If \([\hat{A}, \hat{B}] = 0\), we can measure both \(\hat{A}\) and \(\hat{B}\) simultaneously with arbitrary precision on a single copy of the state.

Proof: commuting operators share eigenstates

Suppose \([\hat{A}, \hat{B}] = 0\) and \(\vert u\rangle\) is an eigenstate of \(\hat{A}\) with eigenvalue \(a\):

\[ \hat{A}\vert u\rangle = a\vert u\rangle \]

Then:

\[ \hat{A}(\hat{B}\vert u\rangle) = \hat{B}\hat{A}\vert u\rangle = \hat{B}(a\vert u\rangle) = a(\hat{B}\vert u\rangle) \]

So \(\hat{B}\vert u\rangle\) is also an eigenstate of \(\hat{A}\) with the same eigenvalue \(a\). If the eigenvalue \(a\) is non-degenerate, then \(\hat{B}\vert u\rangle = b\vert u\rangle\) for some \(b\), making \(\vert u\rangle\) an eigenstate of both. For degenerate eigenvalues, \(\hat{B}\) acts within the degenerate subspace, and we can diagonalize it there. Iterating this construction yields a basis diagonalizing both \(\hat{A}\) and \(\hat{B}\) simultaneously.

Conversely, if \(\vert u_j\rangle\) are simultaneous eigenstates with \(\hat{A}\vert u_j\rangle = a_j\vert u_j\rangle\) and \(\hat{B}\vert u_j\rangle = b_j\vert u_j\rangle\), then on this basis \(\hat{A}\) and \(\hat{B}\) are diagonal matrices, which commute.

Heisenberg Uncertainty Principle#

The physical consequence of non-commutativity is the uncertainty principle:

Robertson Uncertainty Relation

For any state \(\vert\psi\rangle\) and any two observables \(\hat{A}\), \(\hat{B}\):

\[ \Delta\hat{A} \cdot \Delta\hat{B} \geq \frac{1}{2}\vert \langle[\hat{A}, \hat{B}]\rangle\vert \]

where \(\Delta\hat{O} = \sqrt{\langle\hat{O}^2\rangle - \langle\hat{O}\rangle^2}\) and \(\langle[\hat{A}, \hat{B}]\rangle = \langle\psi\vert[\hat{A}, \hat{B}]\vert\psi\rangle\).

When two observables do not commute, the product of their uncertainties has a positive lower bound set by their commutator.

Repeated identical measurements

The uncertainty principle describes repeated measurements on identical copies of the same prepared state, not sequential measurement of a single copy. If you measure \(\hat{A}\) first and then \(\hat{B}\) on the same particle, you collapse the state and \(\hat{B}\)’s outcome depends on \(\hat{A}\)’s result—that is a different story (back-action). Here we imagine preparing the state fresh many times and analyzing the distribution across those trials.

Proof: Robertson Uncertainty Relation

Define \(\Delta\hat{A} = \hat{A} - \langle\hat{A}\rangle\) and similarly for \(\hat{B}\). These are centered operators with zero expectation value.

By Cauchy-Schwarz inequality,

\[ \|\Delta\hat{A}\vert\psi\rangle\|^2 \cdot \|\Delta\hat{B}\vert\psi\rangle\|^2 \geq \vert \langle\psi\vert(\Delta\hat{A})^\dagger\Delta\hat{B}\vert\psi\rangle\vert^2 \]

Since \(\hat{A}\) and \(\hat{B}\) are Hermitian, so are \(\Delta\hat{A}\) and \(\Delta\hat{B}\), and:

\[ \langle\psi\vert(\Delta\hat{A})^\dagger\Delta\hat{B}\vert\psi\rangle = \langle\psi\vert\Delta\hat{A}\Delta\hat{B}\vert\psi\rangle \]

Decompose into symmetric and antisymmetric parts:

\[ \Delta\hat{A}\Delta\hat{B} = \frac{1}{2}\{\Delta\hat{A}, \Delta\hat{B}\} + \frac{1}{2}[\Delta\hat{A}, \Delta\hat{B}] \]

where the anticommutator \(\{\hat{A}, \hat{B}\} = \hat{A}\hat{B} + \hat{B}\hat{A}\) has real expectation value, and the commutator is anti-Hermitian. Thus:

\[ \vert\langle\psi\vert\Delta\hat{A}\Delta\hat{B}\vert\psi\rangle|^2 \geq \vert\langle\psi\vert\frac{1}{2}[\Delta\hat{A}, \Delta\hat{B}]\vert\psi\rangle|^2 = \frac{1}{4}\vert\langle[\hat{A}, \hat{B}]\rangle\vert^2 \]

Therefore:

\[ (\Delta\hat{A})^2(\Delta\hat{B})^2 \geq \frac{1}{4}\vert \langle[\hat{A}, \hat{B}]\rangle\vert^2 \]

Taking square roots gives the Robertson relation.

Example: Uncertainty Relation for Pauli Operators

For a qubit with Pauli operators:

\[ [\hat{X}, \hat{Z}] = -2\mathrm{i}\hat{Y} \]

In state \(\vert\psi\rangle = \vert 0\rangle\) (eigenstate of \(\hat{Z}\)):

  • \(\langle\hat{Z}\rangle = 1\), so \(\Delta\hat{Z} = 0\)

  • \(\langle\hat{X}\rangle = 0\), and \(\langle\hat{X}^{2}\rangle = 1\), so \((\Delta\hat{X})^2 = 1\)

  • The bound from Robertson: \(\Delta\hat{X} \cdot \Delta\hat{Z} \geq \frac{1}{2}\vert\langle[\hat{X}, \hat{Z}]\rangle\vert = \frac{1}{2}\vert\langle -2\mathrm{i}\hat{Y}\rangle\vert = 0\)

The bound is not violated because \(\Delta\hat{Z} = 0\) saturates the lower bound.

Discussion: disturbance vs intrinsic uncertainty

Is the uncertainty principle about measurement disturbance or an intrinsic property of quantum states?

Heisenberg’s original argument (1927) used a thought experiment: a gamma-ray microscope that disturbs the electron’s momentum when measuring its position. But the Robertson relation is derived purely from the mathematical structure of quantum states—no measurement apparatus is mentioned.

  • Does this mean uncertainty is a property of the state itself, not of any particular measurement?

  • If so, why did Heisenberg’s disturbance argument give the right answer?

Poll: Non-commuting observables

A qubit in state \(\vert 0\rangle\) is measured: Z gives \(+1\), then X gives \(+1\), then Z again. What is the probability of \(Z = +1\) on the final measurement?

(A) 100% (the state was \(\vert 0\rangle\) before the X measurement).

(B) 50% (the X measurement randomized the state).

(C) 0% (the state became \(\vert -\rangle\) after the X measurement).

(D) 25% (the probabilities compound non-additively).

Summary#

  • The expectation value \(\langle\hat{O}\rangle = \langle\psi\vert\hat{O}\vert\psi\rangle\) and variance \(\mathrm{Var}(\hat{O}) = \langle\hat{O}^2\rangle - \langle\hat{O}\rangle^2\) quantify measurement distributions.

  • The commutator \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\) measures whether operator order matters.

  • Commuting observables share a simultaneous eigenbasis and can be measured together with arbitrary precision.

  • The Robertson uncertainty relation \(\Delta\hat{A} \cdot \Delta\hat{B} \geq \frac{1}{2}\vert \langle[\hat{A}, \hat{B}]\rangle\vert\) is a fundamental constraint on non-commuting observables.

See Also

Homework#

1. Most general operator commuting with Z. Recall from 1.1.3 Problem 5 that every Hermitian operator on a qubit has the Pauli decomposition \(\hat A = a_0\hat I + a_x\hat X + a_y\hat Y + a_z\hat Z\) with \(a_0, a_x, a_y, a_z \in \mathbb{R}\).

(a) Compute \([\hat A, \hat Z]\) using the commutators \([\hat I, \hat Z] = 0\), \([\hat X, \hat Z] = -2\mathrm{i}\hat Y\), \([\hat Y, \hat Z] = 2\mathrm{i}\hat X\), \([\hat Z, \hat Z] = 0\).

(b) Set \([\hat A, \hat Z] = 0\) and find the constraints on \((a_0, a_x, a_y, a_z)\). Write the most general \(\hat A\) that commutes with \(\hat Z\).

(c) By the lecture theorem on commuting operators, \(\hat A\) and \(\hat Z\) share a complete simultaneous eigenbasis. Identify that basis directly from your answer to (b), and write down the eigenvalues of \(\hat A\) on each basis state.

(d) The full Hermitian operator algebra on a qubit is \(4\)-dimensional (the Pauli basis). What fraction of it commutes with \(\hat Z\)? Explain in one sentence what this fraction means for the joint-measurability of qubit observables.

2. Robertson relation on a specific state. For the state \(\vert\psi\rangle = \vert+\rangle = \tfrac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\), the \(+1\) eigenstate of \(\hat X\):

(a) Compute \(\langle\hat{X}\rangle\), \(\langle\hat{Y}\rangle\), \(\langle\hat{Z}\rangle\).

(b) Compute \(\Delta\hat{X}\), \(\Delta\hat{Y}\), \(\Delta\hat{Z}\) using \((\hat\sigma^i)^2 = \hat I\) and the lecture’s variance formula.

(c) Check the Robertson uncertainty relation for the pair \((\hat X, \hat Z)\): verify that \(\Delta\hat{X}\cdot\Delta\hat{Z} \geq \tfrac{1}{2}\vert\langle[\hat{X}, \hat{Z}]\rangle\vert\). Is the inequality saturated?

3. Spin commutator along arbitrary axes. For unit vectors \(\boldsymbol{n}_1, \boldsymbol{n}_2 \in \mathbb{R}^3\), define the spin observables along those axes,

\[ \hat A = \boldsymbol{n}_1\cdot\hat{\boldsymbol\sigma}, \qquad \hat B = \boldsymbol{n}_2\cdot\hat{\boldsymbol\sigma}. \]

(a) Using the Pauli multiplication law \(\hat\sigma^i\hat\sigma^j = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\) (1.1.3), compute \([\hat A, \hat B]\). Show that

\[ [\hat A, \hat B] = 2\mathrm{i}\,(\boldsymbol{n}_1\times\boldsymbol{n}_2)\cdot\hat{\boldsymbol\sigma}. \]

(b) State the geometric condition on \(\boldsymbol{n}_1, \boldsymbol{n}_2\) under which \(\hat A\) and \(\hat B\) commute. Interpret on the Bloch sphere.

(c) Apply the Robertson uncertainty relation: for a state with Bloch vector \(\boldsymbol n\),

\[ \Delta\hat A\cdot\Delta\hat B \geq \bigl\vert(\boldsymbol{n}_1\times\boldsymbol{n}_2)\cdot\boldsymbol n\bigr\vert. \]

For what configurations of \(\boldsymbol{n}_1, \boldsymbol{n}_2, \boldsymbol n\) does this lower bound vanish? Give two geometrically distinct scenarios.

4. Total Pauli uncertainty for a pure qubit. Show that for any pure qubit state with Bloch vector \(\boldsymbol n\) (so \(\vert\boldsymbol n\vert = 1\)),

\[ (\Delta\hat X)^2 + (\Delta\hat Y)^2 + (\Delta\hat Z)^2 = 2. \]

(a) For each Pauli component, compute \((\Delta\hat\sigma^i)^2\) using \(\langle\hat\sigma^i\rangle = n_i\) and \(\langle(\hat\sigma^i)^2\rangle = 1\).

(b) Sum the three variances and apply \(\vert\boldsymbol n\vert^2 = 1\) to obtain the identity.

(c) Interpret physically: a pure qubit state cannot have all three Pauli observables simultaneously sharp — the total uncertainty budget is exactly \(2\), independent of the state’s location on the Bloch sphere. For each state, identify the unique direction in which the uncertainty is zero.

(d) Verify the identity explicitly on \(\vert+\rangle\): compute each \((\Delta\hat\sigma^i)^2\) and confirm the sum equals \(2\). Which observable is sharp, and how is the total uncertainty distributed among the other two?

5. Sharp observable does not violate Heisenberg. Explain why the Heisenberg/Robertson uncertainty relation does not prevent measuring \(\hat Z\) with zero uncertainty on the eigenstate \(\vert 0\rangle\). Does the deterministic outcome violate the relation? Demonstrate by computing both sides of \(\Delta\hat Z\cdot\Delta\hat X \geq \tfrac{1}{2}\vert\langle[\hat Z,\hat X]\rangle\vert\) on \(\vert 0\rangle\).

6. Saturation and maximum of the Robertson bound. The Robertson relation for the pair \((\hat X, \hat Z)\) reads \(\Delta\hat X\cdot\Delta\hat Z \geq \vert\langle\hat Y\rangle\vert\) (using \([\hat X,\hat Z] = -2\mathrm{i}\hat Y\)).

(a) Over all pure qubit states (Bloch vector \(\boldsymbol n\) with \(\vert\boldsymbol n\vert = 1\)), find the state that maximises the lower bound \(\vert\langle\hat Y\rangle\vert\). Compute the maximum value.

(b) Using the Bloch parametrization with \(\Delta\hat X = \sqrt{1 - n_x^2}\) and \(\Delta\hat Z = \sqrt{1 - n_z^2}\), find the condition on the Bloch vector for which the Robertson inequality is saturated (equality \(\Delta\hat X\cdot\Delta\hat Z = \vert\langle\hat Y\rangle\vert\)). Identify the states geometrically.

(c) Find the pure states that both (i) maximise the lower bound from (a) AND (ii) saturate the inequality from (b). Identify them on the Bloch sphere. What spin direction is sharp for these states?

7. Maximising the Z-uncertainty. A qubit is prepared in \(\vert\psi\rangle = \cos\alpha\vert 0\rangle + \sin\alpha\vert 1\rangle\) (real \(\alpha\)). Find the value(s) of \(\alpha\) that maximise \(\Delta\hat Z\). Identify the maximising states on the Bloch sphere and interpret the result via Problem 4’s total-uncertainty identity.

8. Minimum of the X-Z uncertainty product. Apply the Robertson uncertainty relation to the pair \((\hat X, \hat Z)\). Over all pure qubit states, find the minimum value of \(\Delta\hat X\cdot\Delta\hat Z\) and identify the minimising states. Verify that the Robertson inequality is consistent with this minimum.