1.2.3 Measurement Operators#

Prompts

Define a measurement operator \(\hat{P}_{O=m}\) for an observable \(\hat{O}\) with outcome \(m\). What are the key properties (Hermiticity, idempotence, completeness)?
For \(\hat{\sigma}^z\) on a qubit, write down the two projectors \(\hat{P}_0\) and \(\hat{P}_1\) explicitly as matrices and verify idempotence: \(\hat{P}^2 = \hat{P}\).
When you measure a state and get outcome \(m\), how does the state change? What is the formula for the post-measurement state?
What happens when an observable has a degenerate eigenvalue (many eigenstates with the same eigenvalue)? How does the projector change?
How is measurement mathematically similar to Bayesian updating in probability?

Lecture Notes#

Overview#

The measurement postulate (§1.2.1) tells us what happens when we measure: outcomes are eigenvalues, probabilities follow the Born rule, and the state collapses. Here we introduce the measurement operator (projector) — the mathematical object that encodes all of this in a single formula. We tabulate the projectors for all Pauli observables, interpret collapse as Bayesian updating, and extend to degenerate spectra.

Measurement Operators as Projectors#

Measurement Operator (Projector)

For an observable \(\hat{O}\) with eigenvalue \(m\) and eigenstate \(\vert O{=}m\rangle\), the measurement operator is:

\[\hat{P}_{O=m} = \vert O{=}m \rangle \langle O{=}m \vert\]

Key properties:

Hermitian: \(\hat{P}^\dagger = \hat{P}\)
Idempotent: \(\hat{P}^2 = \hat{P}\) (projecting twice is the same as projecting once)
Positive semi-definite: \(\langle \psi \vert \hat{P} \vert \psi \rangle \geq 0\) for all \(\vert \psi \rangle\)
Completeness: \(\sum_m \hat{P}_{O=m} = \hat{I}\) (outcomes are exhaustive)

Spectral Decomposition

The projectors decompose the observable:

\[\hat{O} = \sum_m m \, \hat{P}_{O=m}\]

Knowing all projectors is equivalent to knowing the observable.

Usage in measurement (recap from §1.2.1):

Probability: \(P(m \mid \psi) = \langle \psi \vert \hat{P}_{O=m} \vert \psi \rangle\)
Post-measurement state: \(\vert \psi \rangle \to \hat{P}_{O=m} \vert \psi \rangle \,/\, \sqrt{P(m \mid \psi)}\)

Pauli Measurement Operators#

For Pauli operators with eigenvalues \(m = \pm 1\), the projectors follow a universal pattern:

\[\hat{P}_{O=m} = \frac{1}{2}\big(\hat{I} + m\,\hat{O}\big)\]

Observable	\(m\)	Eigenstate	Projector	Matrix
\(\hat{X}\)	\(+1\)	\(\vert +\rangle\)	\(\frac{1}{2}(\hat{I}+\hat{X})\)	\(\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\)
\(\hat{X}\)	\(-1\)	\(\vert -\rangle\)	\(\frac{1}{2}(\hat{I}-\hat{X})\)	\(\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\)
\(\hat{Y}\)	\(+1\)	\(\vert \mathrm{i}\rangle\)	\(\frac{1}{2}(\hat{I}+\hat{Y})\)	\(\frac{1}{2}\begin{pmatrix} 1 & -\mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}\)
\(\hat{Y}\)	\(-1\)	\(\vert \bar{\mathrm{i}}\rangle\)	\(\frac{1}{2}(\hat{I}-\hat{Y})\)	\(\frac{1}{2}\begin{pmatrix} 1 & \mathrm{i} \\ -\mathrm{i} & 1 \end{pmatrix}\)
\(\hat{Z}\)	\(+1\)	\(\vert 0\rangle\)	\(\frac{1}{2}(\hat{I}+\hat{Z})\)	\(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\)
\(\hat{Z}\)	\(-1\)	\(\vert 1\rangle\)	\(\frac{1}{2}(\hat{I}-\hat{Z})\)	\(\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\)

Example: \(\hat{Z}\) Projectors

Idempotence:

\[\begin{split}\hat{P}_0^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \hat{P}_0 \quad \checkmark\end{split}\]

\[\begin{split}\hat{P}_1^2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \hat{P}_1 \quad \checkmark\end{split}\]

Completeness:

\[\begin{split}\hat{P}_0 + \hat{P}_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \hat{I} \quad \checkmark\end{split}\]

Spectral decomposition:

\[\begin{split}(+1)\hat{P}_0 + (-1)\hat{P}_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \sigma^z \quad \checkmark\end{split}\]

State Collapse as Bayesian Updating#

Measurement can be understood as updating our knowledge in response to new information:

Before measurement: the state \(\vert \psi \rangle\) encodes a probability distribution over outcomes via \(P(m) = \langle \psi \vert \hat{P}_{O=m} \vert \psi \rangle\).
After obtaining outcome \(m\): the state becomes \(\vert O{=}m \rangle\), representing certainty that the observable has value \(m\).

This is formally analogous to Bayesian updating: prior (initial state) \(\to\) posterior (collapsed state) upon receiving evidence (measurement outcome).

Collapse is Knowledge Update, Not Physical Disturbance

The collapse \(\vert \psi \rangle \to \hat{P}_{O=m} \vert \psi \rangle / \sqrt{P(m \mid \psi)}\) is an update of our information, not a physical process acting on the particle. The qubit does not “change” in response to measurement; rather, we refine what we know about its state given the result.

Discussion

If collapse is “just” updating our knowledge, does that mean quantum mechanics is subjective? Two observers with different information about the same qubit would assign different states — is this a problem, or a feature?

Degenerate Measurements#

So far each eigenvalue \(m\) had a unique eigenstate. When an eigenvalue is degenerate (multiple orthonormal eigenstates \(\vert O{=}m,i\rangle\) share the same eigenvalue), the projector generalizes to a sum over the eigenspace:

\[\hat{P}_{O=m} = \sum_{i=1}^{d_m} \vert O{=}m, i \rangle \langle O{=}m, i \vert\]

where \(d_m\) is the degeneracy. All four properties (Hermitian, idempotent, PSD, completeness) still hold.

Upon measuring outcome \(m\), the state collapses to:

\[\vert \psi' \rangle = \frac{\hat{P}_{O=m} \vert \psi \rangle}{\sqrt{P(m \mid \psi)}}\]

This is a superposition within the eigenspace — the measurement tells us the eigenvalue but does not resolve which basis state within the degenerate subspace.

Example: Two-Qubit Degenerate Measurement

Setup. Measure \(\hat{Z} \otimes \hat{I}\) on a two-qubit system. The eigenvalue \(-1\) eigenspace is spanned by \(\{\vert 10\rangle, \vert 11\rangle\}\) (both have first qubit \(\vert 1\rangle\)). The projector is:

\[\begin{split}\hat{P}_{-1} = \vert 1 \rangle \langle 1 \vert \otimes \hat{I} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{split}\]

Given state. \(\vert \Psi \rangle = \frac{1}{\sqrt{2}}\vert 00\rangle + \frac{1}{2}\vert 10\rangle + \frac{1}{2}\vert 11\rangle\).

Probability. \(P(-1) = \langle \Psi \vert \hat{P}_{-1} \vert \Psi \rangle = \vert 1/2\vert^2 + \vert 1/2\vert^2 = 1/2\).

Post-measurement state. Apply the projector and normalize:

\[\hat{P}_{-1}\vert \Psi \rangle = \frac{1}{2}\vert 10\rangle + \frac{1}{2}\vert 11\rangle, \quad \vert \Psi' \rangle = \frac{\hat{P}_{-1}\vert \Psi \rangle}{\sqrt{P(-1)}} = \frac{1}{\sqrt{2}}(\vert 10\rangle + \vert 11\rangle)\]

The first qubit is now definitely \(\vert 1\rangle\), but the second qubit remains in the superposition \(\frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) — the measurement collapsed the state into the degenerate eigenspace preserving the coherence within the eigenspace.

Summary#

A measurement operator \(\hat{P}_{O=m} = \vert O{=}m \rangle \langle O{=}m \vert\) is a projector satisfying Hermiticity, idempotence, PSD, and completeness.
For Pauli operators: \(\hat{P}_{O=m} = \tfrac{1}{2}(\hat{I} + m\,\hat{O})\) for \(m = \pm 1\).
Collapse is best understood as a Bayesian knowledge update, not a physical disturbance.
For degenerate eigenvalues, the projector sums over the eigenspace, and collapse leaves the state as a superposition within that subspace.

Homework#

1. For \(\hat{X}\), write the two projectors \(\hat{P}_{+}\) and \(\hat{P}_{-}\) as explicit \(2 \times 2\) matrices. Verify idempotence (\(\hat{P}^2 = \hat{P}\)) for each and show they sum to the identity.

2. A qubit is prepared in the state \(\vert\psi\rangle = \frac{3\vert0\rangle + 4\vert1\rangle}{5}\). Calculate the probabilities of measuring \(+1\) and \(-1\) when measuring \(\hat{Z}\), and write the post-measurement states for each outcome.

3. Verify the universal pattern for Pauli projectors: \(\hat{P}_{O=m} = \frac{\hat{I} + m\hat{O}}{2}\) for \(m = \pm 1\). (Hint: Use the fact that \(\hat{O}^2 = \hat{I}\) for any Pauli matrix.)

4. Consider measuring \(\hat{Z}\) on a two-qubit state \(\vert\Psi\rangle = \frac{1}{\sqrt{2}}(\vert00\rangle + \vert01\rangle)\). The first qubit’s \(\hat{Z}\) has a degenerate \(+1\) eigenspace. (a) Write down the projector onto the \(+1\) eigenspace. (b) Compute the measurement probability of outcome \(+1\). (c) Write down the post-measurement state(s) for outcome \(+1\). Is it a superposition or a single eigenstate?

5. For \(\hat{Y}\), write down both projectors explicitly (as \(2 \times 2\) matrices) using the pattern \(\hat{P}_{\pm} = (\hat{I} \pm \hat{Y})/2\). Verify idempotence and completeness.

6. Explain in 2–3 sentences: Why is the collapse of a quantum state after measurement best understood as an update of knowledge rather than a physical disturbance to the particle?

7. Show that the completeness relation \(\sum_m \hat{P}_{O=m} = \hat{I}\) follows directly from the orthonormality of eigenstates. (Generalize from the case of non-degenerate spectrum.)

8. Use the spectral decomposition \(\hat{O} = \sum_m m \, \hat{P}_{O=m}\) to show that for an eigenstate \(\vert O=m'\rangle\), measuring \(\hat{O}\) returns outcome \(m'\) with probability 1 and the state remains unchanged.