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. Spin-axis projector. For a unit vector \(\boldsymbol{n} = (n_x, n_y, n_z) \in \mathbb{R}^3\), the spin observable along \(\boldsymbol{n}\) is \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\) (1.1.3 Problem 7), with eigenvalues \(\pm 1\).

(a) Use the universal Pauli projector pattern \(\hat P_{O=m} = (\hat I + m\hat O)/2\) to write the two projectors \(\hat P_{\boldsymbol{n},\pm}\) for the eigenvalue \(\pm 1\) eigenspaces of \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\).

(b) Verify idempotence \(\hat P_{\boldsymbol{n},+}^2 = \hat P_{\boldsymbol{n},+}\) algebraically, using only \((\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \hat I\) from 1.1.3 Problem 7. (Do not compute \(2\times 2\) matrix products.)

(d) Compute the probability of obtaining \(+1\) when measuring \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\) on the state \(\vert 0\rangle\). Express your answer in terms of \(n_z\) and check the two limits \(\boldsymbol{n} = \boldsymbol{e}_z\) and \(\boldsymbol{n} = \boldsymbol{e}_x\).

2. Born rule via projector formula. A qubit is in the general Bloch state \(\vert\psi\rangle = \cos(\theta/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert 1\rangle\). Compute the \(\hat X\) measurement probabilities using the projector form \(P(m) = \langle\psi\vert\hat P_m\vert\psi\rangle\), rather than direct amplitude squaring.

(a) Write the projectors \(\hat P_{X,\pm} = (\hat I \pm \hat X)/2\).

(b) Use linearity of expectation values and the lecture’s Bloch-vector formula \(\langle\psi\vert\hat X\vert\psi\rangle = \sin\theta\cos\varphi\) to obtain \(P(\pm 1)\).

(c) Find the state that maximises \(P(+1)\), and the state that maximises \(P(-1)\). Identify both on the Bloch sphere.

3. Spectral decomposition. Recall from 1.1.3 Problem 5 the Hermitian operator

\[\begin{split} \hat O = 2\hat I + 2\hat X + \hat Y + \hat Z = \begin{pmatrix} 3 & 2 - \mathrm{i} \\ 2 + \mathrm{i} & 1 \end{pmatrix}, \end{split}\]

with Pauli coefficients \((a_0, a_x, a_y, a_z) = (2, 2, 1, 1)\).

(a) Using the parametrisation \(\hat O = a_0\hat I + \boldsymbol a\cdot\hat{\boldsymbol\sigma}\) and the result of 1.1.3 Problem 5(c), state the eigenvalues of \(\hat O\) in closed form. Compute their numerical values.

(b) Identify the unit vector \(\boldsymbol{e}_a = \boldsymbol a/\vert\boldsymbol a\vert\) — the “Bloch axis” of \(\hat O\) — and write the spectral projectors \(\hat P_\pm\) onto the eigenstates.

(c) Verify the spectral decomposition \(\hat O = E_+\hat P_+ + E_-\hat P_-\) by expanding the right side and recovering \(\hat O = 2\hat I + 2\hat X + \hat Y + \hat Z\) algebraically.

(d) Use the spectral form to compute \(\hat O^2\) in two ways: (i) by squaring the spectral expansion, and (ii) by direct \(4\times 4\) … wait, \(2\times 2\) matrix multiplication. Verify the two agree.

★ 4. Two-qubit measurement. Consider measuring \(\hat{Z}\) on the equal-weight two-qubit superposition \(\vert\Psi\rangle = \tfrac{1}{2}(\vert 00\rangle + \vert 01\rangle + \vert 10\rangle + \vert 11\rangle)\). The first qubit’s \(\hat{Z}\) — i.e. \(\hat Z \otimes \hat I\) acting on the two-qubit space — has a degenerate \(+1\) eigenspace.

(a) Write the projector \(\hat P_{+1}\) onto the \(+1\) eigenspace as an outer-product sum, and as an explicit \(4\times 4\) matrix in the ordered basis \(\{\vert 00\rangle, \vert 01\rangle, \vert 10\rangle, \vert 11\rangle\}\).

(b) Compute the measurement probability \(P(+1) = \langle\Psi\vert\hat P_{+1}\vert\Psi\rangle\).

(c) Write the post-measurement state for outcome \(+1\). Is it a single basis state, or a superposition? Factor it as a tensor product \(\vert\text{first qubit}\rangle\otimes\vert\text{second qubit}\rangle\), and identify what the second qubit “remembers” about the original state.

5. Sequential projectors and non-commutativity. Two single-qubit projectors \(\hat P_{Z,+} = \vert 0\rangle\langle 0\vert\) and \(\hat P_{X,+} = (\hat I + \hat X)/2 = \vert+\rangle\langle+\vert\) each describe a “yes” outcome of one Pauli measurement.

(a) Write each projector as an explicit \(2\times 2\) matrix.

(b) Compute both products \(\hat P_{Z,+}\hat P_{X,+}\) and \(\hat P_{X,+}\hat P_{Z,+}\) as matrices. Show that they are unequal.

(c) Apply each product to a generic state \(\vert\psi\rangle = a\vert 0\rangle + b\vert 1\rangle\). Verify that the order of projector application changes the result.

(d) Compute \((\hat P_{Z,+}\hat P_{X,+})^2\) and show that it is not equal to \(\hat P_{Z,+}\hat P_{X,+}\) — i.e., the product is not idempotent. Explain in one sentence why the product of two non-commuting projectors fails to be a projector, and what this means physically for sequential incompatible measurements.

6. Repeatability from idempotence. Show that if measuring observable \(\hat O\) on a state \(\vert\psi\rangle\) yields outcome \(m\) (with probability \(p_m > 0\)), then an immediately repeated measurement of \(\hat O\) on the post-measurement state returns the same outcome \(m\) with probability \(1\).

(a) Write the post-measurement state \(\vert\psi'\rangle\) in terms of the projector \(\hat P_{O=m}\) and \(\vert\psi\rangle\).

(b) Compute the probability \(P(m'\,\vert\,\vert\psi'\rangle)\) of obtaining outcome \(m'\) on the repeated measurement. Apply idempotence \(\hat P_{O=m}^2 = \hat P_{O=m}\) and projector orthogonality \(\hat P_{O=m}\hat P_{O=m'} = 0\) for \(m'\neq m\).

(c) Conclude that \(P(m\,\vert\,\vert\psi'\rangle) = 1\) and \(P(m'\,\vert\,\vert\psi'\rangle) = 0\) for every \(m' \neq m\).

(d) Explain in one sentence why this property — repeatability — would fail if the measurement operators were Hermitian but not idempotent (i.e., \(\hat P^2 \neq \hat P\)).

7. Operator functions via spectral decomposition. The spectral decomposition extends to operator functions: for any function \(f\),

\[ f(\hat O) = \sum_m f(m)\,\hat P_{O=m}. \]

This makes operator powers, exponentials, and other functions easy to compute once the projectors are known.

(a) Apply this to compute \(\hat Z^n\) for an integer \(n\). Distinguish the cases \(n\) even and \(n\) odd.

(b) Apply it to the Hermitian operator \(\hat H = \omega\hat X + \Delta\hat Z\) from 1.1.3 Problem 1, with eigenvalues \(E_\pm = \pm\Omega = \pm\sqrt{\omega^2 + \Delta^2}\). Express \(\hat H^n\) in spectral form, distinguishing even and odd \(n\). Confirm that \(\hat H^2 = \Omega^2\hat I\) (matching the result of 1.1.3 P1(c)).

(c) Compute the unitary operator \(\hat U(\theta) = \mathrm{e}^{-\mathrm{i}\hat Z\theta}\) for real \(\theta\). Write it as an explicit \(2\times 2\) matrix in the \(\{\vert 0\rangle, \vert 1\rangle\}\) basis, and verify directly that \(\hat U^\dagger\hat U = \hat I\).