1.2.3 Measurement Operators#

Worked solutions for the homework problems in the 1.2.3 Measurement Operators lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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 \(\boldsymbol{n} = \boldsymbol{e}_x\).

Solution.

(a) Setting \(\hat O = \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\) in the pattern,

\[ \hat P_{\boldsymbol{n},\pm} = \tfrac{1}{2}\bigl(\hat I \pm \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\bigr) = \tfrac{1}{2}\bigl(\hat I \pm (n_x\hat X + n_y\hat Y + n_z\hat Z)\bigr). \]

(b) Expand the square:

\[ \hat P_{\boldsymbol{n},+}^2 = \tfrac{1}{4}(\hat I + \boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \tfrac{1}{4}\bigl(\hat I + 2\boldsymbol{n}\cdot\hat{\boldsymbol\sigma} + (\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2\bigr). \]

Substitute \((\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \hat I\) (which uses \(\vert\boldsymbol{n}\vert = 1\)):

\[ \hat P_{\boldsymbol{n},+}^2 = \tfrac{1}{4}\bigl(\hat I + 2\boldsymbol{n}\cdot\hat{\boldsymbol\sigma} + \hat I\bigr) = \tfrac{1}{4}(2\hat I + 2\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}) = \tfrac{1}{2}(\hat I + \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}) = \hat P_{\boldsymbol{n},+}. \quad\checkmark \]

The unit-norm condition is exactly what makes idempotence hold — without it, the cross term would not collapse correctly.

\[ \hat P_{\boldsymbol{n},+} + \hat P_{\boldsymbol{n},-} = \tfrac{1}{2}(\hat I + \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}) + \tfrac{1}{2}(\hat I - \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}) = \hat I. \quad\checkmark \]

(d) Born rule via projector form, with \(\hat X\vert 0\rangle, \hat Y\vert 0\rangle\) orthogonal to \(\vert 0\rangle\) (so \(\langle 0\vert\hat X\vert 0\rangle = \langle 0\vert\hat Y\vert 0\rangle = 0\)) and \(\hat Z\vert 0\rangle = \vert 0\rangle\):

\[\begin{split} \begin{split} P(+1) &= \langle 0\vert\hat P_{\boldsymbol{n},+}\vert 0\rangle \\ &= \tfrac{1}{2}\langle 0\vert(\hat I + n_x\hat X + n_y\hat Y + n_z\hat Z)\vert 0\rangle \\ &= \tfrac{1}{2}(1 + n_z) = \frac{1 + n_z}{2}. \end{split} \end{split}\]

Limits. \(\boldsymbol{n} = \boldsymbol{e}_z\): \(n_z = 1\), so \(P(+1) = 1\) — deterministic, since \(\vert 0\rangle\) is the \(+1\) eigenstate of \(\boldsymbol{e}_z\cdot\hat{\boldsymbol\sigma} = \hat Z\). \(\boldsymbol{n} = \boldsymbol{e}_x\): \(n_z = 0\), so \(P(+1) = 1/2\) — a coin flip, since \(\hat X\) is incompatible with \(\hat Z\) and \(\vert 0\rangle\) is an equal superposition of \(\hat X\) eigenstates. Both limits match the Bloch picture: \(P(+1) = (1 + \boldsymbol n_\psi\cdot\boldsymbol{n})/2\), with \(\boldsymbol n_\psi = (0,0,1)\) being the Bloch vector of \(\vert 0\rangle\).

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.

Solution.

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

(b) Using linearity,

\[ P(\pm 1) = \langle\psi\vert\hat P_{X,\pm}\vert\psi\rangle = \tfrac{1}{2}\bigl(\langle\psi\vert\hat I\vert\psi\rangle \pm \langle\psi\vert\hat X\vert\psi\rangle\bigr) = \tfrac{1}{2}\bigl(1 \pm \sin\theta\cos\varphi\bigr). \]

This is the projector-form result, equivalent to the direct amplitude-squaring calculation that would give \(P(+1) = \tfrac{1}{2}\vert\cos(\theta/2) + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert^2\) — but obtained in one line.

(c) Maximising \(P(+1) = (1 + \sin\theta\cos\varphi)/2\) requires \(\sin\theta\cos\varphi = +1\), i.e. \(\sin\theta = 1\) and \(\cos\varphi = 1\), so \(\theta = \pi/2\) and \(\varphi = 0\). The state is

\[ \vert\psi\rangle = \cos(\pi/4)\vert 0\rangle + \sin(\pi/4)\vert 1\rangle = \tfrac{1}{\sqrt 2}(\vert 0\rangle + \vert 1\rangle) = \vert+\rangle, \]

which gives \(P(+1) = 1\) — the \(\hat X = +1\) eigenstate. Bloch position: \((\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta) = (1, 0, 0)\), the \(+\boldsymbol{e}_x\) direction.

Maximising \(P(-1)\) requires \(\sin\theta\cos\varphi = -1\), i.e. \(\sin\theta = 1\) and \(\cos\varphi = -1\), so \(\theta = \pi/2\), \(\varphi = \pi\). The state is \(\vert\psi\rangle = \tfrac{1}{\sqrt 2}(\vert 0\rangle - \vert 1\rangle) = \vert-\rangle\), with Bloch position \((-1, 0, 0)\) — the \(-\boldsymbol{e}_x\) direction.

Geometric reading. Maximum \(P(+1)\) for \(\hat X\) measurement occurs at the Bloch vector that points directly along \(+\boldsymbol{e}_x\) (the measurement’s “yes” axis); maximum \(P(-1)\) at the antipodal Bloch direction \(-\boldsymbol{e}_x\). Probability is a linear function of the Bloch vector’s projection onto the measurement axis, \(P(\pm 1) = (1 \pm \boldsymbol{e}_x\cdot\boldsymbol n)/2\) — generalising the \(P(0) = (1 + n_z)/2\) identity of 1.1.2 Problem 6.

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.

Solution.

(a) From 1.1.3 P5(c), the eigenvalues of \(\hat O = a_0\hat I + \boldsymbol a\cdot\hat{\boldsymbol\sigma}\) are \(E_\pm = a_0 \pm \vert\boldsymbol a\vert\). Here \(\boldsymbol a = (2, 1, 1)\), so \(\vert\boldsymbol a\vert = \sqrt{4 + 1 + 1} = \sqrt 6\), and

\[ E_\pm = 2 \pm \sqrt 6, \]

numerically \(E_+ \approx 4.449\) and \(E_- \approx -0.449\).

(b) \(\boldsymbol{e}_a = (2, 1, 1)/\sqrt 6\). The spectral projectors are

\[ \hat P_\pm = \tfrac{1}{2}(\hat I \pm \boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}) = \tfrac{1}{2}\hat I \pm \frac{1}{2\sqrt 6}\bigl(2\hat X + \hat Y + \hat Z\bigr). \]

\[ E_+\hat P_+ + E_-\hat P_- = (2 + \sqrt 6)\tfrac{1}{2}(\hat I + \boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}) + (2 - \sqrt 6)\tfrac{1}{2}(\hat I - \boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}). \]

Collect \(\hat I\) and \(\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}\) coefficients separately:

\[ = \tfrac{1}{2}[(2+\sqrt 6) + (2-\sqrt 6)]\,\hat I + \tfrac{1}{2}[(2+\sqrt 6) - (2-\sqrt 6)]\,\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma} = 2\hat I + \sqrt 6\,\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}. \]

Substitute \(\sqrt 6\,\boldsymbol{e}_a = (2, 1, 1) = \boldsymbol a\):

\[\begin{split} \begin{split} \sqrt 6\,\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma} &= 2\hat X + \hat Y + \hat Z, \\ \Longrightarrow\;\; E_+\hat P_+ + E_-\hat P_- &= 2\hat I + 2\hat X + \hat Y + \hat Z = \hat O. \quad\checkmark \end{split} \end{split}\]

(d) Method 1 — squaring the spectral expansion. Using orthogonality \(\hat P_+\hat P_- = 0\) and idempotence,

\[ \hat O^2 = E_+^2\hat P_+ + E_-^2\hat P_- = (2 + \sqrt 6)^2\hat P_+ + (2 - \sqrt 6)^2\hat P_-. \]

Expanding: \((2 \pm \sqrt 6)^2 = 4 \pm 4\sqrt 6 + 6 = 10 \pm 4\sqrt 6\). So

\[\begin{split} \begin{split} \hat O^2 &= (10 + 4\sqrt 6)\hat P_+ + (10 - 4\sqrt 6)\hat P_- \\ &= 10(\hat P_+ + \hat P_-) + 4\sqrt 6(\hat P_+ - \hat P_-) \\ &= 10\hat I + 4\sqrt 6\,\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma} \\ &= 10\hat I + 4(2\hat X + \hat Y + \hat Z). \end{split} \end{split}\]

Hence \(\hat O^2 = 10\hat I + 8\hat X + 4\hat Y + 4\hat Z\).

Method 2 — direct matrix squaring.

\[\begin{split} \hat O^2 = \begin{pmatrix} 3 & 2 - \mathrm{i} \\ 2 + \mathrm{i} & 1 \end{pmatrix}^2 = \begin{pmatrix} 9 + (2-\mathrm{i})(2+\mathrm{i}) & 3(2-\mathrm{i}) + (2-\mathrm{i}) \\ (2+\mathrm{i})(3+1) & (2+\mathrm{i})(2-\mathrm{i}) + 1 \end{pmatrix}. \end{split}\]

Compute: \((2-\mathrm{i})(2+\mathrm{i}) = 5\). So the \((1,1)\) entry is \(9 + 5 = 14\). The \((1,2)\) entry: \(3(2-\mathrm{i}) + (2-\mathrm{i}) = 4(2-\mathrm{i}) = 8 - 4\mathrm{i}\). The \((2,1)\) entry: \((2+\mathrm{i})\cdot 4 = 8 + 4\mathrm{i}\). The \((2,2)\) entry: \(5 + 1 = 6\). So

\[\begin{split} \hat O^2 = \begin{pmatrix} 14 & 8 - 4\mathrm{i} \\ 8 + 4\mathrm{i} & 6 \end{pmatrix}. \end{split}\]

Read off Pauli coefficients via \(a_0 = \tfrac{1}{2}\operatorname{Tr}(\hat O^2)\) etc.: trace \(= 20\), so \(a_0 = 10\). Off-diagonal \(\hat X\) contribution: \(a_x = \tfrac{1}{2}(\operatorname{Tr}(\hat O^2\hat X)) = \tfrac{1}{2}((8 - 4\mathrm{i}) + (8 + 4\mathrm{i})) = 8\). Similarly \(a_y = 4\) and \(a_z = (14 - 6)/2 = 4\). So \(\hat O^2 = 10\hat I + 8\hat X + 4\hat Y + 4\hat Z\), matching Method 1. ✓

The spectral form turned a \(2\times 2\) matrix squaring into eigenvalue squaring — and the result is automatically in Pauli form, without the trace-extraction step. This is the general payoff: operator functions are scalar functions of eigenvalues, plus projectors.

★ 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.

Solution.

The observable \(\hat Z\otimes\hat I\) has eigenvalues \(\pm 1\), each with a degenerate two-dimensional eigenspace. The \(+1\) eigenspace is spanned by \(\{\vert 00\rangle, \vert 01\rangle\}\) (first qubit \(\vert 0\rangle\)), and the \(-1\) eigenspace by \(\{\vert 10\rangle, \vert 11\rangle\}\). The initial state factors as

\[ \vert\Psi\rangle = \tfrac{1}{2}(\vert 00\rangle + \vert 01\rangle + \vert 10\rangle + \vert 11\rangle) = \vert+\rangle\otimes\vert+\rangle, \]

an unentangled product with each qubit in \(\vert+\rangle\).

(a) The projector onto the \(+1\) eigenspace sums the outer products of the two spanning states (equivalently, projects the first qubit onto \(\vert 0\rangle\) and leaves the second untouched):

\[ \hat P_{+1} = \vert 0\rangle\langle 0\vert \otimes \hat I = \vert 00\rangle\langle 00\vert + \vert 01\rangle\langle 01\vert. \]

As a \(4\times 4\) matrix in the ordered basis,

\[\begin{split} \hat P_{+1} \mapsto \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{split}\]

(b) \(\hat P_{+1}\) acts component-by-component on \(\vert\Psi\rangle\): it keeps the \(\vert 00\rangle,\vert 01\rangle\) components (first qubit already \(\vert 0\rangle\)) and kills the \(\vert 10\rangle,\vert 11\rangle\) components:

\[ \hat P_{+1}\vert\Psi\rangle = \tfrac{1}{2}(\vert 00\rangle + \vert 01\rangle). \]

The probability is the squared norm,

\[ P(+1) = \langle\Psi\vert\hat P_{+1}\vert\Psi\rangle = \bigl\Vert\hat P_{+1}\vert\Psi\rangle\bigr\Vert^{2} = \left(\tfrac{1}{2}\right)^{2} + \left(\tfrac{1}{2}\right)^{2} = \tfrac{1}{2}. \]

By the same calculation \(P(-1) = 1/2\) — the two outcomes are equally likely, reflecting that the first qubit is in \(\vert+\rangle\), a balanced superposition of \(\vert 0\rangle\) and \(\vert 1\rangle\). Unlike a state that already lies in one eigenspace, this measurement is genuinely informative.

\[\begin{split} \begin{split} \vert\Psi'\rangle &= \frac{\hat P_{+1}\vert\Psi\rangle}{\sqrt{P(+1)}} = \frac{(1/2)(\vert 00\rangle + \vert 01\rangle)}{\sqrt{1/2}}\\ &= \tfrac{1}{\sqrt 2}(\vert 00\rangle + \vert 01\rangle) = \vert 0\rangle\otimes\tfrac{1}{\sqrt 2}(\vert 0\rangle + \vert 1\rangle) = \vert 0\rangle\otimes\vert+\rangle. \end{split} \end{split}\]

The post-measurement state is a superposition in the full two-qubit space (not a single basis state) — it spans the entire 2D \(+1\) eigenspace — but it factors cleanly as a tensor product. The first qubit has collapsed to \(\vert 0\rangle\) (the \(+1\) eigenvector), while the second qubit emerges in \(\vert+\rangle\) — exactly its component in the original \(\vert+\rangle\otimes\vert+\rangle\). The measurement \(\hat Z\otimes\hat I\) acts on the first-qubit factor only; because the initial state was unentangled, the second-qubit factor passes through unchanged.

Defining feature of degenerate measurement. A rank-\(r\) projector leaves the post-state somewhere inside an \(r\)-dimensional eigenspace, but does not pick out a unique direction within it; the exact post-state vector depends on the initial state. Here the post-state is \(\vert 0\rangle\otimes\vert+\rangle\). The second qubit emerges unchanged from its initial value \(\vert+\rangle\) only because the input was a product state: nothing about the first-qubit projection could feed into the second-qubit factor. For an entangled input such as \((\vert 00\rangle+\vert 11\rangle)/\sqrt 2\), the same projector \(\vert 0\rangle\langle 0\vert\otimes\hat I\) collapses the second qubit too (to \(\vert 0\rangle\), via the correlation), even though it is the same rank-\(2\) projector. Compare with non-degenerate measurement (Pauli on a single qubit): a rank-\(1\) projector pins the post-state to a unique eigenvector (up to a global phase) regardless of the initial 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.

Solution.

(a) \(\hat P_{Z,+} \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), \(\hat P_{X,+} \mapsto \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\).

(b) Compute the two products:

\[\begin{split} \hat P_{Z,+}\hat P_{X,+} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \cdot \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}, \end{split}\]

\[\begin{split} \hat P_{X,+}\hat P_{Z,+} = \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \tfrac{1}{2}\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}. \end{split}\]

These are different matrices: \(\hat P_{Z,+}\hat P_{X,+}\) has nonzero top row; \(\hat P_{X,+}\hat P_{Z,+}\) has nonzero left column. So \(\hat P_{Z,+}\hat P_{X,+} \neq \hat P_{X,+}\hat P_{Z,+}\) — projectors do not commute.

\[\begin{split} \begin{split} \hat P_{X,+}\vert\psi\rangle &= \tfrac{a + b}{\sqrt 2}\vert+\rangle = \tfrac{a + b}{2}(\vert 0\rangle + \vert 1\rangle), \\ \hat P_{Z,+}\hat P_{X,+}\vert\psi\rangle &= \tfrac{a + b}{2}\vert 0\rangle, \end{split} \end{split}\]

\[\begin{split} \begin{split} \hat P_{Z,+}\vert\psi\rangle &= a\vert 0\rangle, \\ \hat P_{X,+}\hat P_{Z,+}\vert\psi\rangle &= \tfrac{a}{2}(\vert 0\rangle + \vert 1\rangle) = \tfrac{a}{\sqrt 2}\vert+\rangle. \end{split} \end{split}\]

The two outputs differ in both their magnitude (after-norm survival probability) and their direction (post-measurement state). Order matters.

(d) Compute the square of \(\hat P_{Z,+}\hat P_{X,+}\):

\[\begin{split} (\hat P_{Z,+}\hat P_{X,+})^2 = \tfrac{1}{4}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} = \tfrac{1}{4}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} = \tfrac{1}{2}\hat P_{Z,+}\hat P_{X,+}. \end{split}\]

This is not equal to \(\hat P_{Z,+}\hat P_{X,+}\) — the square is half of the original, not equal. So the product is not idempotent, and therefore not a projector.

Physical interpretation. A projector represents the “filter” or “yes/no” outcome of a single measurement. The product of two non-commuting projectors represents the consecutive application of two such filters. After both filters have acted, the remaining amplitude is smaller (some of the state was discarded by each filter), and a third application of the same filter sequence would reduce the amplitude again — not leave it invariant. This is the algebraic counterpart of “incompatible measurements modify the state”: the sequence \(Z\)-then-\(X\) does not commute with \(X\)-then-\(Z\), and neither sequence is a fixed-point operation. The non-idempotence reflects information actively being thrown away at each stage.

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., \(\hat P^2 \neq \hat P\)).

Solution.

(a) The projection rule of the measurement postulate:

\[ \vert\psi'\rangle = \frac{\hat P_{O=m}\vert\psi\rangle}{\sqrt{p_m}}, \qquad p_m = \langle\psi\vert\hat P_{O=m}\vert\psi\rangle. \]

(b) For the repeated measurement, the probability of outcome \(m'\) is

\[ P(m'\,\vert\,\vert\psi'\rangle) = \langle\psi'\vert\hat P_{O=m'}\vert\psi'\rangle = \frac{\langle\psi\vert\hat P_{O=m}\hat P_{O=m'}\hat P_{O=m}\vert\psi\rangle}{p_m}, \]

where the projector \(\hat P_{O=m}^\dagger = \hat P_{O=m}\) (Hermitian) was used in moving from bra \(\langle\psi'\vert\) back to \(\langle\psi\vert\). The numerator contains the triple product \(\hat P_{O=m}\hat P_{O=m'}\hat P_{O=m}\), which simplifies in two cases.

Case \(m' = m\): apply idempotence twice,

\[ \hat P_{O=m}\hat P_{O=m}\hat P_{O=m} = \hat P_{O=m}^3 = \hat P_{O=m}^2 = \hat P_{O=m}. \]

So the numerator equals \(\langle\psi\vert\hat P_{O=m}\vert\psi\rangle = p_m\), giving \(P(m\,\vert\,\vert\psi'\rangle) = p_m/p_m = 1\).

Case \(m' \neq m\): projector orthogonality \(\hat P_{O=m}\hat P_{O=m'} = 0\) (eigenstates of \(\hat O\) with distinct eigenvalues are orthogonal) collapses the triple product:

\[ \hat P_{O=m}\hat P_{O=m'}\hat P_{O=m} = \hat P_{O=m}\cdot 0 = 0. \]

So the numerator vanishes and \(P(m'\,\vert\,\vert\psi'\rangle) = 0\).

(c) Combining, the repeated measurement returns \(m\) with certainty: \(P(m) = 1\), \(P(m') = 0\) for every \(m' \neq m\). This is the repeatability of measurement: once \(\hat O\) has been measured, the system sits in an \(\hat O\) eigenstate, and any further \(\hat O\) measurement merely confirms that result.

(d) Without idempotence, \(\hat P^2 \neq \hat P\), so the chain \(\hat P_{O=m}^3 = \hat P_{O=m}\) used in part (b) would fail. The numerator \(\langle\psi\vert\hat P_{O=m}^3\vert\psi\rangle\) would generally not equal \(p_m\), breaking the equality \(P(m\,\vert\,\vert\psi'\rangle) = 1\). Idempotence is the precise algebraic property that turns “projecting twice = projecting once,” and that in turn is what makes “measurement reveals a value that is then stable under re-measurement” — a non-trivial empirical claim of quantum mechanics, anchored in a one-line operator identity.

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 \(\hat U^\dagger\hat U = \hat I\).

Solution.

(a) The spectral form of \(\hat Z\): \(\hat Z = (+1)\hat P_0 + (-1)\hat P_1 = \hat P_0 - \hat P_1\), with \(\hat P_0 = \vert 0\rangle\langle 0\vert\) and \(\hat P_1 = \vert 1\rangle\langle 1\vert\). So

\[ \hat Z^n = (+1)^n\hat P_0 + (-1)^n\hat P_1 = \hat P_0 + (-1)^n\hat P_1. \]

For even \(n\): \(\hat Z^n = \hat P_0 + \hat P_1 = \hat I\) (completeness).

For odd \(n\): \(\hat Z^n = \hat P_0 - \hat P_1 = \hat Z\).

Equivalently and more memorably, \(\hat Z^2 = \hat I\), which iterates to even powers being \(\hat I\) and odd powers being \(\hat Z\) — the same pattern as \((-1)^n\).

(b) From 1.1.3 P1, \(\hat H = \Omega\hat P_+ + (-\Omega)\hat P_-\). Then

\[ \hat H^n = \Omega^n\hat P_+ + (-\Omega)^n\hat P_- = \Omega^n\bigl(\hat P_+ + (-1)^n\hat P_-\bigr). \]

Even \(n\): \(\hat H^n = \Omega^n(\hat P_+ + \hat P_-) = \Omega^n\hat I = (\omega^2 + \Delta^2)^{n/2}\hat I\). Matches 1.1.3 P1(c) at \(n=2\): \(\hat H^2 = \Omega^2\hat I = (\omega^2 + \Delta^2)\hat I\). ✓

Odd \(n\): \(\hat H^n = \Omega^n(\hat P_+ - \hat P_-) = \Omega^{n-1}\,\Omega(\hat P_+ - \hat P_-) = \Omega^{n-1}\hat H\).

\[ \hat U(\theta) = \mathrm{e}^{-\mathrm{i}\hat Z\theta} = \mathrm{e}^{-\mathrm{i}\theta}\hat P_0 + \mathrm{e}^{+\mathrm{i}\theta}\hat P_1. \]

As a matrix in the \(\{\vert 0\rangle, \vert 1\rangle\}\) basis,

\[\begin{split} \hat U(\theta) = \mathrm{e}^{-\mathrm{i}\theta}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \mathrm{e}^{+\mathrm{i}\theta}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i}\theta} & 0 \\ 0 & \mathrm{e}^{+\mathrm{i}\theta} \end{pmatrix}. \end{split}\]

Unitarity check: \(\hat U^\dagger = \begin{pmatrix} \mathrm{e}^{+\mathrm{i}\theta} & 0 \\ 0 & \mathrm{e}^{-\mathrm{i}\theta} \end{pmatrix}\), and

\[\begin{split} \hat U^\dagger\hat U = \begin{pmatrix} \mathrm{e}^{+\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta} & 0 \\ 0 & \mathrm{e}^{-\mathrm{i}\theta}\mathrm{e}^{+\mathrm{i}\theta} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \hat I. \quad\checkmark \end{split}\]

The same structure underlies the time-evolution operator \(\mathrm{e}^{-\mathrm{i}\hat H t/\hbar}\) developed in §1.3: once the Hamiltonian’s spectral form is known, time evolution reduces to scalar exponentials of eigenvalues, multiplying the corresponding projectors.