Measurement in a Chosen Basis

A basis measurement is the standard finite-dimensional projective measurement whose outcomes correspond to the vectors of an orthonormal basis. Computational readout, polarization analysis in a chosen pair of modes, and an ideal spin-$1/2$ measurement along a specified axis all fit this pattern.

The calculation is compact, but three operations must remain distinct:

1. rewriting coordinates in a new basis,
2. physically rotating a state before a fixed readout, and
3. performing the measurement, including its conditional or unread state update.

Helpful background. Change of basis supplies the column-vector convention, and probability in different bases explains how amplitudes become probabilities. Here we use that calculation to design a measurement and identify its output state.

The Basis Projectors

Let $\mathcal H$ have finite dimension $d$, and let

$$\mathcal F = \left\lbrace \lvert f_0\rangle,\ldots,\lvert f_{d-1}\rangle \right\rbrace$$

be an orthonormal basis. The associated rank-one projectors are

$$P_a = \lvert f_a\rangle\langle f_a\rvert, \qquad a=0,\ldots,d-1.$$

They obey

$$P_aP_b = \delta_{ab}P_a, \qquad \sum_{a=0}^{d-1}P_a = I.$$

Those relations make $\{P_a\}$ a projection-valued measure, or PVM. The projectors specify the outcome probabilities; the basis kets represent their one-dimensional ranges. A state-change rule requires an instrument, which below is the ideal Lüders instrument unless stated otherwise.

For a density operator $\rho$, the probability of outcome $a$ is

$$p(a) = \operatorname{Tr}(P_a\rho) = \langle f_a\rvert\rho\lvert f_a\rangle.$$

If the state is pure, $\rho=\lvert\psi\rangle\langle\psi\rvert$, this becomes

$$p(a) = \left\lvert \langle f_a\rvert\psi\rangle \right\rvert^2.$$

For the ideal rank-one Lüders instrument, a selected outcome with $p(a)>0$ gives

$$\rho_a = \frac{P_a\rho P_a}{p(a)} = P_a.$$

Thus the post-measurement ray is represented by $\lvert f_a\rangle$, independent of the input state. If the outcome is not retained, the output is

$$\mathcal D_{\mathcal F}(\rho) = \sum_a P_a\rho P_a.$$

This unread channel removes coherence between distinct basis vectors. State Update Rule develops selective and nonselective updates for general instruments.

Outcomes are labels, not necessarily eigenvalues

The index $a$ is an outcome label. If the measurement is presented as an observable

$$A = \sum_a \alpha_a P_a,$$

then outcome label $a$ is displayed as $\alpha_a$. If two coefficients coincide and only that common number is recorded, the result is a coarse outcome with their projectors summed. A computational-basis detector may instead report a bit string, and a spin apparatus may report $+\hbar/2$ or $-\hbar/2$. The same PVM can therefore be paired with different display conventions or numerical calibrations.

Basis phases do not change the measurement

Replacing each ket by

$$\lvert f_a'\rangle = e^{i\chi_a}\lvert f_a\rangle$$

leaves every projector unchanged:

$$\lvert f_a'\rangle\langle f_a'\rvert = P_a.$$

Permuting the basis vectors only relabels the outcomes. This is why a basis measurement is specified physically by its projectors, not by arbitrary ket phases or by the order used to print a matrix.

The Matrix Recipe

Suppose a state and density matrix are represented in a reference basis

$$\mathcal E = \left\lbrace \lvert e_0\rangle,\ldots,\lvert e_{d-1}\rangle \right\rbrace.$$

Place the chosen measurement vectors in the columns of a matrix $V$. Its entries are

$$V_{ja} = \langle e_j\rvert f_a\rangle.$$

Equivalently,

$$V = \begin{pmatrix} \vert & & \vert \\ [f_0]_{\mathcal E} & \cdots & [f_{d-1}]_{\mathcal E} \\ \vert & & \vert \end{pmatrix}.$$

Orthonormality means

$$V^\dagger V = VV^\dagger = I.$$

If

$$c_j = \langle e_j\rvert\psi\rangle$$

are the reference-basis amplitudes, then the measurement-basis amplitudes are

$$d = V^\dagger c, \qquad d_a = \langle f_a\rvert\psi\rangle.$$

The probability vector has entries

$$p(a) = \lvert d_a\rvert^2.$$

For a mixed state represented by $\rho_{\mathcal E}$, form

$$\rho_{\mathcal F} = V^\dagger\rho_{\mathcal E}V.$$

Then

$$p(a) = (\rho_{\mathcal F})_{aa}.$$

Only the diagonal of $\rho_{\mathcal F}$ is needed for the outcome distribution. The off-diagonal entries still matter for later coherent operations and are removed by an unread ideal basis measurement:

$$\mathcal D_{\mathcal F}(\rho_{\mathcal E}) = V\, \operatorname{diag}\!\left( V^\dagger\rho_{\mathcal E}V \right) V^\dagger.$$

Here $\operatorname{diag}(M)$ means the matrix obtained from $M$ by setting its off-diagonal entries to zero.

A reliable computational workflow

For hand calculations or code:

1. Put the chosen orthonormal basis vectors in the columns of $V$.
2. Verify $V^\dagger V=I$ to the intended numerical tolerance.
3. Compute $d=V^\dagger c$ for a pure state, or $\rho_{\mathcal F}=V^\dagger\rho_{\mathcal E}V$ for a mixed state.
4. Read probabilities from $\lvert d_a\rvert^2$ or $(\rho_{\mathcal F})_{aa}$.
5. Check that every probability is real and nonnegative within tolerance.
6. Check $\sum_a p(a)=1$.
7. State explicitly whether the result requested is only a distribution, a selected output, or an unread output.

The placement of $V^\dagger$ is a common source of errors. The columns of $V$ carry $\mathcal F$-components into the reference coordinates, so $V^\dagger$ carries reference-coordinate amplitudes into the measurement basis.

Rotated Readout in a Fixed Apparatus

Many devices directly measure only a fixed reference basis $\mathcal E$. A measurement in $\mathcal F$ can then be implemented by applying $V^\dagger$ and measuring in $\mathcal E$:

$$\rho \longmapsto V^\dagger\rho V \longmapsto \text{reference-basis readout}.$$

The probability of reference outcome $a$ is

$$\langle e_a\rvert V^\dagger\rho V \lvert e_a\rangle = \langle f_a\rvert\rho\lvert f_a\rangle,$$

because $V\lvert e_a\rangle=\lvert f_a\rangle$. Thus the outcome statistics agree with the chosen-basis PVM.

There is, however, an output-state distinction:

• after $V^\dagger$ followed by reference-basis readout, the selected state in the apparatus frame is $\lvert e_a\rangle$;
• the Lüders output of the $\mathcal F$-basis measurement is $\lvert f_a\rangle$.

To implement the full Lüders instrument on a system that will be used again, apply $V$ after the readout:

$$\begin{aligned} \rho &\longmapsto V^\dagger\rho V \longmapsto \lvert e_a\rangle\langle e_a\rvert,\\ \lvert e_a\rangle\langle e_a\rvert &\longmapsto V\lvert e_a\rangle\langle e_a\rvert V^\dagger = P_a. \end{aligned}$$

If the measured system is discarded or the readout is terminal, only the outcome probabilities may matter. If subsequent operations use the system, “rotate and measure” is incomplete unless the intended output frame is stated.

Measurement in circuits applies this distinction to circuit diagrams, destructive readout, and repeated experimental shots.

Computational-Basis Measurement

For one qubit, the computational basis is

$$\mathcal Z = \left\lbrace \lvert0\rangle,\lvert1\rangle \right\rbrace.$$

For

$$\lvert\psi\rangle = \alpha\lvert0\rangle + \beta\lvert1\rangle, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1,$$

the probabilities are

$$p(0) = \lvert\alpha\rvert^2, \qquad p(1) = \lvert\beta\rvert^2.$$

The relative phase between $\alpha$ and $\beta$ is invisible to this distribution. Before measurement, a rotation can convert the input coherence into a population difference. After an unread computational-basis Lüders measurement, that coherence has been removed and cannot be recovered simply by rotating the output.

For $n$ qubits, the computational basis is indexed by bit strings:

$$\left\lbrace \lvert x\rangle : x\in\{0,1\}^n \right\rbrace.$$

If

$$\lvert\Psi\rangle = \sum_{x\in\{0,1\}^n} c_x\lvert x\rangle,$$

then

$$p(x) = \lvert c_x\rvert^2.$$

A readout of only part of the register is a coarse-grained projective measurement. For a decomposition $x=(y,z)$, the probability of observed substring $y$ is

$$p(y) = \sum_z \lvert c_{yz}\rvert^2.$$

The ordering of qubits and the mapping between displayed strings and tensor factors are conventions that must be documented. Bits, Qubits, Qudits, and Modes develops the encoding language.

Hadamard-Basis Measurement

The Hadamard, or Pauli-$X$, basis is

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}, \qquad \lvert-\rangle = \frac{ \lvert0\rangle-\lvert1\rangle }{\sqrt2}.$$

Its basis matrix in computational coordinates is

$$V_X = H = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$

Because $H^\dagger=H$, applying a Hadamard gate and then measuring computationally implements the $X$-basis outcome statistics.

For $\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle$,

$$\begin{aligned} p(+) &= \frac12 \left\lvert \alpha+\beta \right\rvert^2,\\ p(-) &= \frac12 \left\lvert \alpha-\beta \right\rvert^2. \end{aligned}$$

Expanding the moduli gives

$$\begin{aligned} p(+) &= \frac12 \left( 1+2\operatorname{Re}(\alpha^*\beta) \right),\\ p(-) &= \frac12 \left( 1-2\operatorname{Re}(\alpha^*\beta) \right). \end{aligned}$$

The interference term contains relative-phase information that computational readout misses.

For a density matrix

$$\rho = \begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix},$$

the same result is

$$p(\pm) = \frac12 \left( 1\pm2\operatorname{Re}\rho_{01} \right) = \frac12(1\pm r_x),$$

where $r_x=\operatorname{Tr}(\rho\sigma_x)$ is the $x$ component of the Bloch vector.

The Y Basis and Complex Coherence

Computational and Hadamard measurements do not, by themselves, reveal the imaginary part of a qubit’s coherence. Use the Pauli-$Y$ basis

$$\lvert+y\rangle = \frac{ \lvert0\rangle+i\lvert1\rangle }{\sqrt2}, \qquad \lvert-y\rangle = \frac{ \lvert0\rangle-i\lvert1\rangle }{\sqrt2}.$$

The corresponding amplitudes are

$$\begin{aligned} d_{+y} &= \frac{\alpha-i\beta}{\sqrt2},\\ d_{-y} &= \frac{\alpha+i\beta}{\sqrt2}. \end{aligned}$$

Therefore

$$p(\pm y) = \frac12(1\pm r_y), \qquad r_y = \operatorname{Tr}(\rho\sigma_y).$$

In computational coordinates,

$$r_y = -2\operatorname{Im}\rho_{01}.$$

The basis matrix is $V_Y=SH$, where

$$S = \begin{pmatrix} 1 & 0\\ 0 & i \end{pmatrix}.$$

Consequently, a $Y$-basis measurement can be implemented by applying

$$V_Y^\dagger = HS^\dagger$$

before computational readout. Gate order matters: $S^\dagger$ acts first, then $H$.

Spin Along an Arbitrary Axis

Let

$$\hat{\mathbf n} = ( \sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta )$$

be a unit vector. For spin $1/2$, define

$$\sigma_{\hat n} = \hat{\mathbf n}\cdot\boldsymbol{\sigma}.$$

Its spectral projectors are

$$P_\pm(\hat{\mathbf n}) = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol{\sigma} \right).$$

Using $\lvert0\rangle=\lvert+z\rangle$ and $\lvert1\rangle=\lvert-z\rangle$, one convenient eigenket convention is

$$\lvert+\hat n\rangle = \cos\frac{\theta}{2}\lvert0\rangle + e^{i\phi} \sin\frac{\theta}{2}\lvert1\rangle,$$

and

$$\lvert-\hat n\rangle = -e^{-i\phi} \sin\frac{\theta}{2}\lvert0\rangle + \cos\frac{\theta}{2}\lvert1\rangle.$$

Other phase conventions describe the same projectors. For a qubit state written in Bloch form,

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right), \qquad \lVert\mathbf r\rVert\leq1,$$

the probabilities are

$$p(\pm\hat{\mathbf n}) = \operatorname{Tr} \left[ \rho P_\pm(\hat{\mathbf n}) \right] = \frac12 \left( 1\pm\mathbf r\cdot\hat{\mathbf n} \right).$$

This formula applies to both pure and mixed states. If a pure spin is prepared along $\hat{\mathbf m}$ and $\gamma$ is the angle between $\hat{\mathbf m}$ and $\hat{\mathbf n}$, then

$$\begin{aligned} p(+\hat{\mathbf n}) &= \cos^2\frac{\gamma}{2},\\ p(-\hat{\mathbf n}) &= \sin^2\frac{\gamma}{2}. \end{aligned}$$

For the physical observable

$$S_{\hat n} = \frac{\hbar}{2}\sigma_{\hat n},$$

the numerical outcomes are $+\hbar/2$ and $-\hbar/2$. The geometry and phase conventions are developed further in Spin-1/2 Hilbert Space and The Bloch Sphere.

A Qutrit Fourier-Basis Example

Basis measurement is not limited to qubits. Let

$$\omega = e^{2\pi i/3},$$

and define the qutrit Fourier basis

$$\lvert f_k\rangle = \frac{1}{\sqrt3} \sum_{j=0}^{2} \omega^{jk}\lvert j\rangle, \qquad k=0,1,2.$$

The basis matrix has entries

$$V_{jk} = \frac{\omega^{jk}}{\sqrt3}.$$

For a state with computational amplitudes $c_j$, the chosen-basis amplitudes are the inverse discrete Fourier transform:

$$d_k = \frac{1}{\sqrt3} \sum_{j=0}^{2} \omega^{-jk}c_j.$$

Every computational basis state gives a uniform Fourier-basis distribution:

$$\left\lvert \langle f_k\rvert j\rangle \right\rvert^2 = \frac13.$$

By contrast, preparing $\lvert f_{k_0}\rangle$ makes outcome $k_0$ certain. This pair of bases is mutually unbiased: certainty in one basis becomes maximal uncertainty in the other.

Degenerate Outcomes and Coarse Graining

A detector may group several basis labels into one reported outcome. If the sets $A_m$ partition the fine labels into disjoint groups covering the full basis, define

$$Q_m = \sum_{a\in A_m}P_a.$$

Then

$$p(m) = \operatorname{Tr}(Q_m\rho) = \sum_{a\in A_m}p(a).$$

The probability is obtained by summing fine probabilities, but the state update depends on what the apparatus physically resolves.

For $p(m)>0$, an ideal coarse Lüders measurement gives

$$\rho_m^{\mathrm{coarse}} = \frac{Q_m\rho Q_m}{p(m)}.$$

A fine basis measurement followed by deletion of the fine label gives

$$\rho_m^{\mathrm{fine\ then\ hidden}} = \frac{ \displaystyle \sum_{a\in A_m}P_a\rho P_a }{ p(m) }.$$

These states need not agree. The coarse Lüders update preserves coherence inside the subspace $Q_m\mathcal H$, whereas the unresolved fine measurement destroys it. Equal outcome probabilities therefore do not establish equal measurement dynamics. Degenerate measurements and Lüders rule develops this distinction.

What Does Not Count as a Basis PVM

The vectors must be orthonormal and complete. A collection of normalized but nonorthogonal kets generally fails because

$$P_aP_b \neq 0 \quad \text{for some }a\neq b,$$

and its rank-one projectors generally do not sum to the identity.

A system prepared in nonorthogonal signal states can still be measured projectively; the measurement cannot distinguish those inputs perfectly in one shot. To use nonorthogonal rank-one operators as the measurement effects themselves, choose suitable weights and any additional outcomes so that the full collection is a POVM. Likewise, noisy readout, inconclusive outcomes, and overcomplete measurements are naturally described by POVMs and instruments. See Generalized Measurements Overview and POVMs: First Encounter.

Basis Change Versus Measurement

The same matrix can appear in several mathematically related procedures, but the physical statements differ.

• Passive coordinate rewrite: the same abstract state receives new components. There is no random outcome and no physical disturbance.
• Active unitary $V^\dagger$: the state physically becomes $V^\dagger\rho V$. There is no random outcome yet.
• Reference readout after $V^\dagger$: an outcome is sampled with the chosen-basis probabilities, and the selected output in the apparatus frame is a reference-basis projector.
• Full chosen-basis Lüders implementation: rotate by $V^\dagger$, read out, and rotate by $V$. The selected output in the original frame is $P_a$.

For example, writing

$$\lvert0\rangle = \frac{ \lvert+\rangle+\lvert-\rangle }{\sqrt2}$$

does not perform an $X$ measurement. It merely exposes the amplitudes that would determine an $X$-basis measurement.

Common Pitfalls

Using $V$ instead of $V^\dagger$. With the basis kets in columns, $V^\dagger c$ gives the measurement amplitudes. Real symmetric examples such as Hadamard hide this mistake; the complex $Y$ basis exposes it.

Rotating probabilities instead of amplitudes. Combine the complex amplitudes first, then take absolute squares. Relative phases can change the new distribution even when the old populations are identical.

Confusing a new coordinate list with a new experiment. A passive rewrite changes no physical state. Rotating before readout changes the apparatus’s effective measurement, and adding the inverse rotation afterward changes the output instrument.

Selecting a single ket after a coarse outcome. A coarse record identifies a subspace. The surviving coherence depends on the physical measurement, as in the qutrit example below.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Apply the column-matrix convention

In the reference basis $\{\lvert0\rangle,\lvert1\rangle,\lvert2\rangle\}$, consider

$$\begin{aligned} \lvert f_0\rangle &= \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2},\\ \lvert f_1\rangle &= \frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2},\\ \lvert f_2\rangle &= \lvert2\rangle. \end{aligned}$$

For

$$\lvert\psi\rangle = \frac{\lvert0\rangle+i\lvert1\rangle}{\sqrt2},$$

construct $V$, compute $V^\dagger c$, and find all three probabilities.

Solution

The basis vectors form the columns

$$V = \begin{pmatrix} 1/\sqrt2 & 1/\sqrt2 & 0\\ 1/\sqrt2 & -1/\sqrt2 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$

The input coordinate vector is

$$c = \begin{pmatrix} 1/\sqrt2\\ i/\sqrt2\\ 0 \end{pmatrix}.$$

Because $V^\dagger=V$ here,

$$d = V^\dagger c = \begin{pmatrix} (1+i)/2\\ (1-i)/2\\ 0 \end{pmatrix}.$$

Hence

$$p(0) = \frac12, \qquad p(1) = \frac12, \qquad p(2) = 0.$$

The probabilities sum to one.

2. Reveal a phase in the Hadamard basis

Let

$$\lvert\psi_\varphi\rangle = \frac{ \lvert0\rangle+e^{i\varphi}\lvert1\rangle }{\sqrt2}.$$

Find the computational-basis and Hadamard-basis distributions. For which phases is each Hadamard outcome certain?

Solution

Computational readout gives

$$p(0) = p(1) = \frac12$$

for every $\varphi$. In the Hadamard basis,

$$\begin{aligned} p(+) &= \frac14 \left\lvert 1+e^{i\varphi} \right\rvert^2 = \frac12(1+\cos\varphi),\\ p(-) &= \frac14 \left\lvert 1-e^{i\varphi} \right\rvert^2 = \frac12(1-\cos\varphi). \end{aligned}$$

Outcome $+$ is certain when $\varphi=0$ modulo $2\pi$, and outcome $-$ is certain when $\varphi=\pi$ modulo $2\pi$.

3. Reveal the imaginary coherence

For the same state $\lvert\psi_\varphi\rangle$, compute the Pauli-$Y$ basis probabilities. Identify the phases that make $+y$ and $-y$ certain.

Solution

The Bloch component is

$$r_y = 2\operatorname{Im} \left( \frac{e^{i\varphi}}{2} \right) = \sin\varphi.$$

Therefore

$$\begin{aligned} p(+y) &= \frac12(1+\sin\varphi),\\ p(-y) &= \frac12(1-\sin\varphi). \end{aligned}$$

Outcome $+y$ is certain for $\varphi=\pi/2$ modulo $2\pi$. Outcome $-y$ is certain for $\varphi=-\pi/2$ modulo $2\pi$.

4. Measure a mixed spin state along a tilted axis

A qubit has Bloch vector

$$\mathbf r = \left( \frac{3}{10}, -\frac{4}{10}, \frac{5}{10} \right).$$

It is measured along

$$\hat{\mathbf n} = \left( 0, \frac35, \frac45 \right).$$

Verify that the state and axis are valid, then compute $p(+\hat{\mathbf n})$ and $p(-\hat{\mathbf n})$.

Solution

The axis is normalized:

$$\lVert\hat{\mathbf n}\rVert^2 = \frac{9}{25} + \frac{16}{25} = 1.$$

The state is valid because

$$\lVert\mathbf r\rVert^2 = \frac{9+16+25}{100} = \frac12 \leq 1.$$

Their dot product is

$$\mathbf r\cdot\hat{\mathbf n} = -\frac{4}{10}\frac35 + \frac{5}{10}\frac45 = \frac{4}{25}.$$

Thus

$$\begin{aligned} p(+\hat{\mathbf n}) &= \frac12 \left( 1+\frac{4}{25} \right) = \frac{29}{50},\\ p(-\hat{\mathbf n}) &= \frac12 \left( 1-\frac{4}{25} \right) = \frac{21}{50}. \end{aligned}$$

5. Compare terminal readout with a Lüders instrument

Let $\mathcal F$ be represented by the column matrix $V$. Starting from $\rho$, compare:

1. applying $V^\dagger$ and measuring the reference basis, and
2. applying $V^\dagger$, measuring the reference basis, and then applying $V$.

Show that the outcome probabilities agree, and identify each selected output state.

Solution

In either procedure, reference outcome $a$ has probability

$$\begin{aligned} p(a) &= \langle e_a\rvert V^\dagger\rho V \lvert e_a\rangle\\ &= \langle f_a\rvert \rho \lvert f_a\rangle. \end{aligned}$$

After the first procedure, ideal reference readout leaves

$$\rho_a^{(1)} = \lvert e_a\rangle\langle e_a\rvert.$$

The final rotation in the second procedure gives

$$\begin{aligned} \rho_a^{(2)} &= V\lvert e_a\rangle\langle e_a\rvert V^\dagger\\ &= \lvert f_a\rangle\langle f_a\rvert = P_a. \end{aligned}$$

The statistics are the same, but only the second procedure realizes the chosen-basis Lüders output in the original frame.

6. Change basis phases and ordering

Define

$$\lvert g_a\rangle = e^{i\chi_a} \lvert f_{\pi(a)}\rangle,$$

where $\pi$ is a permutation. Show how the projectors and probabilities compare with those of $\mathcal F$.

Solution

The phase cancels between ket and bra:

$$\begin{aligned} \lvert g_a\rangle\langle g_a\rvert &= e^{i\chi_a} \lvert f_{\pi(a)}\rangle \langle f_{\pi(a)}\rvert e^{-i\chi_a}\\ &= P_{\pi(a)}. \end{aligned}$$

Therefore

$$p_{\mathcal G}(a) = \operatorname{Tr} \left( P_{\pi(a)}\rho \right) = p_{\mathcal F}(\pi(a)).$$

The physical projectors are unchanged as a set. Only the outcome labels are permuted.

7. Distinguish coarse and hidden fine measurements

In a qutrit, let

$$Q = \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert,$$

and prepare

$$\lvert\psi\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

Compare the selected state for outcome $Q$ under:

1. the coarse Lüders measurement $\{Q,I-Q\}$, and
2. a computational-basis measurement whose labels $0$ and $1$ are later merged.

Solution

Outcome $Q$ has probability one. Because $Q\lvert\psi\rangle=\lvert\psi\rangle$, the coarse Lüders update leaves

$$\rho_Q^{\mathrm{coarse}} = \lvert\psi\rangle\langle\psi\rvert.$$

The fine measurement resolves $0$ from $1$. Forgetting that fine result gives

$$\rho_Q^{\mathrm{fine\ then\ hidden}} = \frac12 \lvert0\rangle\langle0\rvert + \frac12 \lvert1\rangle\langle1\rvert.$$

The first state retains the coherence

$$\frac12 \left( \lvert0\rangle\langle1\rvert + \lvert1\rangle\langle0\rvert \right),$$

while the second does not. The reported coarse statistics alone cannot distinguish the instruments.

8. Test a proposed measurement basis

Someone proposes the two qubit vectors $\lvert0\rangle$ and $\lvert+\rangle$ as a projective measurement basis. Test orthogonality and completeness of the associated rank-one projectors.

Solution

The vectors are not orthogonal because

$$\langle0\rvert+\rangle = \frac{1}{\sqrt2} \neq 0.$$

In the computational basis,

$$\begin{aligned} P_0+P_+ &= \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} + \frac12 \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}\\ &= \begin{pmatrix} 3/2 & 1/2\\ 1/2 & 1/2 \end{pmatrix} \neq I. \end{aligned}$$

Also $P_0P_+\neq0$. These projectors therefore do not form a PVM. A valid measurement involving nonorthogonal effects must instead be designed within the POVM formalism.