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The Lüders rule specifies the ideal quantum instrument for a projective measurement. Conditional on an outcome with nonzero probability, it givesfor a density operator and a pure state, respectively. The second formula applies only to a pure state input. If the result is discarded, the output is the nonselective projective measurement ${\sum }_i{P}_i\rho P_i$.

The crucial feature for a degenerate eigenvalue is preservation of quantum coherence within its eigenspace. For a qutrit with orthonormal basis $|0⟩,|1⟩,|2⟩$, considerOn $|\chi ⟩=(|0⟩+|1⟩)/\sqrt{2}$, outcome zero has probability one and the Lüders rule leaves $|\chi ⟩$ unchanged. Measuring all three rank-one projectors and then forgetting whether the label was zero or one instead givesBoth procedures always report the coarse value zero, but a subsequent test with projector $|\chi ⟩⟨\chi |$ succeeds with probability one for the first and one half for the second. Resolving extra information destroyed coherence. This explains why a quantum nondemolition measurement of a degenerate quantity must specify what information the apparatus actually records.

Gerhart Lüders proposed this state-change rule in 1951, explicitly discussing degenerate quantities and distinguishing selected subensembles from the aggregate measured ensemble. Kirkpatrick's translation was first posted in 2004; equations (7) and (8) give the unnormalized selective update and the nonselective update. ^[ref].

Repeatability follows briefly from $P_j{P}_i={\delta }_{ij}{P}_i$: after outcome $i$, $\mathrm{Tr}(P_j{\rho }_i)={\delta }_{ij}$. This assumes a second application of the same ideal instrument without intervening dynamics. An instrument with the same effects can rotate or reset the output and need not satisfy this conclusion; see POVM does not determine the post-measurement state.

For the three given qutrit vectors, $⟨{\varphi }_0|=(⟨0|-i⟨2|)/\sqrt{2}$ and $⟨{\varphi }_2|=(⟨0|+i⟨2|)/\sqrt{2}$. Taking the complex conjugate in the bra is essential. Their squared norms are $1,1,1$, andThus they are an orthonormal set of three vectors in the three-dimensional inner product space ${\mathbb{C}}^3$. Such a set is linearly independent and spans the space, so it is an orthonormal basis.

Measure the first qutrit in the computational basis and the second in the orthonormal basis $\{|{\varphi }_0⟩,|{\varphi }_1⟩,|{\varphi }_2⟩\}$ from Question 1(a), with those labels as outcomes. For first-party outcome 0, the unnormalized conditional vector of the second party is $|0⟩/\sqrt{3}$, so this outcome occurs with probability $1/3$ and leaves the normalized state $|0⟩$.

The second-party Born rule probabilities are then $|⟨{\varphi }_0|0⟩{|}^2=1/2$, $|⟨{\varphi }_1|0⟩{|}^2=0$, and $|⟨{\varphi }_2|0⟩{|}^2=1/2$. HenceThe relevant joint probabilities are $1/6,0,1/6$, each divided by $\mathrm{Pr}(A=0)=1/3$. This distinguishes the conditional statement from an unconditional outcome probability. More generally, conjugate-state steering from a maximally entangled pair explains how the first local measurement determines the second party's conditional vector; the basis choice here makes that vector especially simple.

Use three-way qutrit sign-flip discrimination. Prepare the qutrit in the uniform superposition stateFor an unknown label $j\in \{0,1,2\}$, the supplied unitary operator is $U_j=I-2|j⟩⟨j|$. Apply it once and then apply the known inversion about the meanSince $D^{†}=D$ and $D^2=I$, this is an allowed unitary transformation. Using $⟨s|j⟩=1/\sqrt{3}$ givesPerform a computational-basis measurement and guess that its outcome is the unknown label. The Born rule givesEach incorrect outcome has probability $1/27$, verifying normalization. The success probability is $25/27$ separately for every label, so no assumption about their prior probabilities is needed.

Let $\omega =e^{2\pi i/3}$. The positive-exponent quantum Fourier transform gives $|\xi ⟩=3^{-1/2}{\sum }_{y=0}^2{\omega }^{2y}|y⟩$. Reindexing the cyclic shift operator givesThus $|\xi ⟩$ is an eigenstate with eigenvalue $\omega$, with this sign fixed by the printed quantum Fourier transform convention.

Prepare the first two qutrits in $({\mathrm{QFT}}_3|0⟩{)}^{\otimes 2}$ and the answer qutrit in $|\xi ⟩$. The modular-addition quantum oracle applies $S^{f(x_1,x_2)}$ to the answer. Its eigenvalue produces quantum phase kickback, leavingApply ${\mathrm{QFT}}_3^{-1}$ to each input qutrit, then measure them in the computational basis. The result isThe preparation, inverse transforms and measurements are independent of $f$, and there is exactly one use of $U_f$. This is qutrit linear-function identification, the ternary version of Bernstein-Vazirani phase kickback.

A modular-addition quantum oracle for a linear function over the three-element finite field reveals its two coefficients in one query. Prepare the answer qutrit in ${\mathrm{QFT}}_3|2⟩$, whose unit shift eigenvalue is $e^{2\pi i/3}$. Quantum phase kickback produces a product of Fourier states labelled by the coefficients; inverse quantum Fourier transforms read both coefficients exactly.

For the three qutrit unitary operators $U_j=I-2|j⟩⟨j|$, prepare $|s⟩=(|0⟩+|1⟩+|2⟩)/\sqrt{3}$, query once, apply the known inversion about the mean $D=2|s⟩⟨s|-I$, and use a computational-basis measurement to guess $j$. Since $D^2=I$, the processing is a unitary transformation. Direct multiplication gives $D{U}_j|s⟩=(5|j⟩-{\sum }_{k\ne j}|k⟩)/(3\sqrt{3})$, establishing success probability $25/27$ for each label. This calculation establishes the achieved probability, without requiring an optimality claim.