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 \([\hat{\sigma}^x, \hat{\sigma}^z]\) 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 \(|+\rangle\) (eigenstate of \(\hat{\sigma}^x\)), what are \(\Delta\hat{\sigma}^x\) and \(\Delta\hat{\sigma}^z\)? Do they violate the uncertainty relation?

Lecture Notes#

Overview#

Quantum mechanics places fundamental limits on the joint knowledge we can have of two observables. These limits arise not from measurement error, but from the non-commutativity of the corresponding operators. This section defines variance and expectation value, introduces 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 \(P(m\vert\psi) = |\psi_m|^2\):

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

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

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

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.

Physical meaning: The commutator measures whether operator application order matters. In classical mechanics, all observables commute. In quantum mechanics, non-commutativity reflects a deep constraint: we cannot simultaneously sharpen our knowledge of both observables.

Analogy: Commute = Independent

Think of getting dressed. Let

  • \(\hat{A}\) = put on socks,

  • \(\hat{B}\) = put on shoes,

  • \(\hat{C}\) = put on a hat.

Then \(\hat{A}\hat{B} \neq \hat{B}\hat{A}\) (socks then shoes is different from shoes then socks), but \(\hat{A}\hat{C} = \hat{C}\hat{A}\) (hat doesn’t care about the socks-shoes order). The hat and socks/shoes commute because they are independent.

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

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

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 \(\hat{A}\), \(\hat{B}\):

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

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

Interpretation: When two observables do not commute, the product of their uncertainties has a positive lower bound set by their commutator. The more strongly they fail to commute, the larger the lower bound on the product of uncertainties.

Important: 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,

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

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:

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

Therefore:

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

Taking square roots gives the Robertson relation.

Example: Uncertainty Relation for Pauli Operators

For a qubit with Pauli operators:

\[ [\hat{\sigma}^x, \hat{\sigma}^z] = -2\mathrm{i}\hat{\sigma}^y \]

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 \(\Delta\hat{\sigma}^z = 0\) saturates the lower bound.

Discussion

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. This suggests uncertainty is caused by the act of measurement.

But the Robertson relation \(\Delta\hat{A} \cdot \Delta\hat{B} \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|\) is derived purely from the mathematical structure of quantum states — no measurement apparatus is mentioned. The uncertainties \(\Delta\hat{A}\) and \(\Delta\hat{B}\) refer to the spread of outcomes in repeated measurements on identically prepared ensembles.

  • 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?

  • Can you design an experiment that distinguishes “intrinsic uncertainty” from “measurement disturbance”? (Hint: look up Ozawa’s inequality.)

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

See Also

Homework#

1. Compute \([\hat{\sigma}^x, \hat{\sigma}^z]\) directly by matrix multiplication and verify that it equals \(-2\mathrm{i}\hat{\sigma}^y\).

2. For the state \(\vert\psi\rangle = \frac{1}{\sqrt{2}}(\vert0\rangle + \vert1\rangle)\) (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: \(\Delta\hat{X} \cdot \Delta\hat{Z} \geq \frac{1}{2}|\langle[\hat{X}, \hat{Z}]\rangle|\).

3. 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. Verify all three Pauli commutation relations: (a) \([\hat{\sigma}^x, \hat{\sigma}^y] = 2\mathrm{i}\hat{\sigma}^z\) (b) \([\hat{\sigma}^y, \hat{\sigma}^z] = 2\mathrm{i}\hat{\sigma}^x\) (c) \([\hat{\sigma}^z, \hat{\sigma}^x] = 2\mathrm{i}\hat{\sigma}^y\)

5. Explain why the uncertainty principle does not prevent measuring \(\hat{Z}\) with zero uncertainty on the eigenstate \(\vert0\rangle\). Does this violate the Robertson relation? Why or why not?

6. 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. A qubit is prepared in state \(\vert\psi\rangle = \cos\alpha\vert0\rangle + \sin\alpha\vert1\rangle\) (real \(\alpha\)). Find the value(s) of \(\alpha\) that maximize \(\Delta\hat{Z}\). Interpret your result.

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