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. Pauli commutator calculation. Compute \([\hat{X}, \hat{Z}]\) directly by matrix multiplication and verify that it equals \(-2\mathrm{i}\hat{Y}\).

2. Commutation relations and states. For the state \(\vert\psi\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) (eigenstate of \(\hat{X}\) with eigenvalue \(+1\)), compute:

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

(b) \(\Delta\hat{X}\), \(\Delta\hat{Y}\), \(\Delta\hat{Z}\)

(c) Check the uncertainty relation: \(\Delta\hat{X} \cdot \Delta\hat{Z} \geq \frac{1}{2}\vert \langle[\hat{X}, \hat{Z}]\rangle\vert\).

3. Commutator product rule. Prove the product rule for commutators: \([\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}]\).

4. Pauli matrices. Verify all three Pauli commutation relations:

(a) \([\hat{X}, \hat{Y}] = 2\mathrm{i}\hat{Z}\)

(b) \([\hat{Y}, \hat{Z}] = 2\mathrm{i}\hat{X}\)

(c) \([\hat{Z}, \hat{X}] = 2\mathrm{i}\hat{Y}\)

5. Simultaneous measurement. Explain why the uncertainty principle does not prevent measuring \(\hat{Z}\) with zero uncertainty on the eigenstate \(\vert 0\rangle\). Does this violate the Robertson relation? Why or why not?

6. Commuting operator zero expectation. For the Pauli operators, show that \([\hat{\sigma}^i, \hat{\sigma}^j] = 0\) if and only if \(i = j\). Use this to explain why the three Pauli operators cannot all be measured simultaneously with arbitrary precision.

7. Uncertainty principle application. A qubit is prepared in state \(\vert\psi\rangle = \cos\alpha\vert 0\rangle + \sin\alpha\vert 1\rangle\) (real \(\alpha\)). Find the value

(s) of \(\alpha\) that maximize \(\Delta\hat{Z}\). Interpret your result.

8. Robertson relation. Apply the Robertson uncertainty relation to the pair \((\hat{X}, \hat{Z})\). For which state does the product of uncertainties \(\Delta\hat{X} \cdot \Delta\hat{Z}\) achieve its minimum value? Compute that minimum.