1.2.3 Measurement Operators#

Prompts

Define a measurement operator (projection operator) 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: is collapse subjective?

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.

Poll: Projective measurement and collapse

A qubit state \(\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\) is measured with projector \(\hat{P} = \vert 0\rangle\langle 0\vert\). If outcome 0 occurs (probability \(\vert\alpha\vert^2\)), what is the post-measurement state?

(A) \(\vert\psi\rangle\) (unchanged).

(B) \(\vert 0\rangle\) (collapsed to the eigenstate).

(D) The post-measurement state cannot be defined (measurement is non-deterministic).

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. Pauli measurement projectors. 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. Measurement probability via projectors. A qubit is prepared in the state \(\vert\psi\rangle = \frac{3\vert 0\rangle + 4\vert 1\rangle}{5}\). Calculate the probabilities of measuring \(+1\) and \(-1\) when measuring \(\hat{Z}\), and write the post-measurement states for each outcome.

3. Universal Pauli projector pattern. 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. Two-qubit measurement. Consider measuring \(\hat{Z}\) on a two-qubit state \(\vert\Psi\rangle = \frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 01\rangle)\). 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 \(+1\).

(s) for outcome \(+1\). Is it a superposition or a single eigenstate?

5. Y operator projectors. 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. Collapse as projection. A qubit is prepared in \(\vert\psi\rangle = \frac{1}{\sqrt{3}}\vert 0\rangle + \sqrt{\frac{2}{3}}\vert 1\rangle\). You measure \(\hat{Z}\) and obtain \(+1\).

(a) Write down the post-measurement state using the projection rule \(\vert\psi'\rangle = \hat{P}_{+1}\vert\psi\rangle / \|\hat{P}_{+1}\vert\psi\rangle\|\). Verify it is normalized.

(b) You now measure \(\hat{X}\) on the post-measurement state. What are the probabilities of obtaining \(+1\) and \(-1\)?

(c) Suppose instead you had measured \(\hat{X}\) first (without measuring \(\hat{Z}\)). Are the \(\hat{X}\) outcome probabilities the same as in (b)? Explain why measurement of \(\hat{Z}\) changes subsequent \(\hat{X}\) statistics — this is the physical content of “collapse.”

7. Completeness of projectors. 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. Expectation via spectral decomposition. 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.