Projective Measurement

A projective measurement is a sharp quantum measurement whose mutually exclusive outcomes are represented by mutually orthogonal projectors resolving the identity. For an outcome $a$ with projector $P_a$, the Born probability is $\operatorname{Tr}(\rho P_a)$, and the standard ideal conditional update is the Lüders projection $P_a\rho P_a/\operatorname{Tr}(\rho P_a)$.

Projective measurements are the cleanest setting in which outcome subspaces, probabilities, conditioning, and repeatability fit together. They describe basis measurements, yes–no tests, ideal measurements of discrete observables, and spectral events such as finding position in a region. They are also an idealization: a realistic detector need not be sharp, perfectly efficient, repeatable, or minimally disturbing.

Helpful background. Projectors explains orthogonal subspaces; the Born rule assigns probabilities to them.

Projection-Valued Measurements

A projection-valued measure (PVM) fixes outcome probabilities. It does not determine every possible physical state change with those probabilities. Here an ideal projective measurement means a PVM together with the Lüders update; a PVM or sharp observable refers to the statistical object alone.

Let $\mathcal H$ be the system Hilbert space. Pure states are represented by normalized vectors $\lvert\psi\rangle$, up to an overall phase, and general states by density operators $\rho$ satisfying

$$\rho\geq 0, \qquad \operatorname{Tr}\rho=1.$$

For the main development, the outcome set is finite or countable. The label $a$ may be a numerical value, a detector record, or any other classical symbol. The sum of the outcome projectors converges strongly to the identity when the set is countably infinite: its partial sums converge on each Hilbert-space vector. Sums defining an unbounded observable additionally require its domain.

From an Observable to a PVM

Let $A$ be self-adjoint with a pure point spectral measure: its mutually orthogonal eigenspaces span the Hilbert space. Label its distinct eigenvalues by $a$. Its spectral decomposition is

$$A = \sum_a aP_a,$$

where $P_a$ projects onto the eigenspace

$$\mathcal H_a = \ker(A-aI).$$

The spectral projectors obey

$$P_aP_b = \delta_{ab}P_a, \qquad \sum_aP_a = I.$$

For infinitely many unbounded eigenvalues, the weighted sum for $A\psi$ converges on

$$\mathcal D(A)=\left\{\psi:\sum_a a^2\|P_a\psi\|^2<\infty\right\}.$$

The projectors themselves remain bounded on the whole space. An ideal measurement of $A$ therefore gives a PVM whose outcome projector for the value $a$ is $P_a$. If $A$ is nondegenerate, $P_a$ has rank one. If $A$ is degenerate, $P_a$ projects onto the full eigenspace rather than onto an arbitrarily selected eigenvector.

PVMs do not require numerical labels

The projectors are the quantum part of the measurement data. Numerical eigenvalues are useful when the records represent values of an observable, but the same projector family can be labeled by words, bit strings, detector ports, or other classical records.

Conversely, assigning distinct real numbers $\alpha_a$ to a finite PVM defines a self-adjoint operator

$$B = \sum_a\alpha_aP_a.$$

Changing the distinct numerical labels changes the reported quantity but not the decomposition into alternatives. A one-to-one relabeling preserves the PVM structure. A many-to-one relabeling merges alternatives and produces a coarser PVM.

This is why saying only “measure the operator” can hide relevant operational information. The spectral projectors identify the outcome subspaces; the displayed numbers specify how those outcomes are recorded as values.

Geometry of the Outcome Subspaces

Define

$$\mathcal H_a = \operatorname{Ran}P_a.$$

Orthogonality and completeness give the direct-sum decomposition

$$\mathcal H = \bigoplus_a\mathcal H_a.$$

Every state vector has a unique decomposition

$$\lvert\psi\rangle = \sum_a\lvert\psi_a^{\mathrm{un}}\rangle, \qquad \lvert\psi_a^{\mathrm{un}}\rangle = P_a\lvert\psi\rangle.$$

The superscript “un” indicates that the branch vector is generally unnormalized. Distinct branch vectors are orthogonal:

$$\left\langle \psi_a^{\mathrm{un}} \middle| \psi_b^{\mathrm{un}} \right\rangle = \langle\psi\rvert P_aP_b\lvert\psi\rangle = 0 \qquad (a\neq b).$$

The Pythagorean identity then gives

$$\sum_a \left\lVert \lvert\psi_a^{\mathrm{un}}\rangle \right\rVert^2 = \lVert\psi\rVert^2 = 1.$$

Projective measurement turns this orthogonal decomposition into a probability distribution: the squared norm of each component is the probability of its corresponding outcome.

Events and coarse graining

An event $S$ is a set of outcome labels. Its projector is

$$P(S) = \sum_{a\in S}P_a.$$

These event projectors satisfy

$$\begin{aligned} P(\varnothing) &= 0, \\ P(\Omega) &= I, \\ P(S\cap T) &= P(S)P(T), \\ P(S^{\mathrm c}) &= I-P(S), \end{aligned}$$

where $\Omega$ is the full outcome set. If $S$ and $T$ are disjoint, then

$$P(S\cup T) = P(S)+P(T).$$

Suppose a classical readout reports $\alpha=f(a)$ rather than the fine label $a$. The coarse projector for $\alpha$ is

$$Q_\alpha = \sum_{a:f(a)=\alpha}P_a.$$

The family $\{Q_\alpha\}$ is again a PVM. It answers a coarser question by grouping orthogonal outcome subspaces. This statement concerns the reported probabilities: performing a fine measurement and forgetting labels need not realize the Lüders update of the coarser PVM.

Outcome Probabilities

The Born rule assigns a probability to each outcome projector. Projective measurement supplies a particularly transparent realization of the probability axioms.

Pure states

For normalized $\lvert\psi\rangle$,

$$\begin{aligned} p(a) &= \langle\psi\rvert P_a\lvert\psi\rangle \\ &= \langle\psi\rvert P_a^\dagger P_a\lvert\psi\rangle \\ &= \left\lVert P_a\lvert\psi\rangle \right\rVert^2. \end{aligned}$$

The probability is the squared length of the component of the state in $\mathcal H_a$. If $P_a\lvert\psi\rangle=0$, the state has no component in that outcome subspace and $p(a)=0$. If $P_a\lvert\psi\rangle=\lvert\psi\rangle$, the state already lies in that subspace and $p(a)=1$.

Mixed states

For a density operator $\rho$,

$$p(a) = \operatorname{Tr}(\rho P_a).$$

Using cyclicity of the trace and $P_a^2=P_a$,

$$\operatorname{Tr}(\rho P_a) = \operatorname{Tr}(P_a\rho P_a).$$

The operator $P_a\rho P_a$ is positive and generally has trace less than one. Its trace is the probability of the branch; after division by that trace, it becomes the ideal conditional density operator.

If

$$\begin{aligned} \rho &= \sum_jw_j \lvert\psi_j\rangle\langle\psi_j\rvert, \\ w_j &\geq 0, \qquad \sum_jw_j=1, \end{aligned}$$

then

$$p(a) = \sum_jw_j \langle\psi_j\rvert P_a\lvert\psi_j\rangle.$$

The prediction depends only on $\rho$, not on which ensemble decomposition is used to represent it.

Why the probabilities are consistent

Positivity follows because $P_a\geq0$:

$$p(a) = \operatorname{Tr}(\rho P_a) \geq0.$$

Normalization follows from completeness:

$$\sum_ap(a) = \operatorname{Tr} \left( \rho\sum_aP_a \right) = \operatorname{Tr}\rho = 1.$$

For an event $S$,

$$p(S) = \operatorname{Tr}\!\left(\rho P(S)\right) = \sum_{a\in S}p(a).$$

Thus orthogonal sums of projectors implement classical additivity for mutually exclusive records.

Ideal Conditional State

The unnormalized branch associated with outcome $a$ is

$$\mathcal I_a(\rho) = P_a\rho P_a.$$

Its trace is the outcome probability:

$$\operatorname{Tr}\mathcal I_a(\rho) = p(a).$$

The maps $\mathcal I_a$ form the Lüders instrument associated with the PVM. They encode both the classical outcome probabilities and the ideal quantum output states.

Pure-state update

For $p(a)>0$, normalizing the projected vector gives

$$\lvert\psi\rangle \longmapsto \lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle} {\sqrt{ \langle\psi\rvert P_a\lvert\psi\rangle }}.$$

The normalization check is

$$\begin{aligned} \langle\psi_a\vert\psi_a\rangle &= \frac{ \langle\psi\rvert P_a^2\lvert\psi\rangle }{ \langle\psi\rvert P_a\lvert\psi\rangle } \\ &= 1. \end{aligned}$$

Density-operator update

For a general input state,

$$\rho \longmapsto \rho_a = \frac{P_a\rho P_a} {\operatorname{Tr}(\rho P_a)}, \qquad p(a)>0.$$

The conditional state has support inside the outcome subspace:

$$P_a\rho_aP_a = \rho_a.$$

If the measurement occurs but the record is ignored, the ideal nonselective state is

$$\rho' = \sum_aP_a\rho P_a.$$

The selective/nonselective distinction and its relation to conditional inference are developed in State Update Rule.

Zero-probability outcomes

If $p(a)=0$, then

$$P_a\rho P_a = 0.$$

There is no normalized conditional state for that outcome because the normalization denominator vanishes. Writing the update formula with $p(a)=0$ is not a valid operation. A regularized or approximate detector model can assign a small nonzero probability, but that is a different measurement model.

Probability data do not determine disturbance

The PVM $\{P_a\}$ determines the probabilities $\operatorname{Tr}(\rho P_a)$. Other instruments can have the same effects $P_a$ while applying additional outcome-dependent transformations inside or after the outcome subspace. For example,

$$\widetilde{\mathcal I}_a(\rho) = U_aP_a\rho P_aU_a^\dagger$$

has the same outcome probability when $U_a$ is unitary:

$$\operatorname{Tr} \widetilde{\mathcal I}_a(\rho) = \operatorname{Tr}(\rho P_a).$$

Unless $U_a$ preserves $\operatorname{Ran}P_a$, this instrument need not be repeatable. Even when it preserves the subspace, it can disturb degrees of freedom within a degenerate sector. The Lüders update is the standard minimally refining ideal choice, not a consequence of probability data alone.

Rank-One Measurements

A nondegenerate discrete observable has rank-one spectral projectors

$$P_a = \lvert a\rangle\langle a\rvert,$$

where the eigenvectors form an orthonormal basis:

$$\langle a\vert b\rangle = \delta_{ab}, \qquad \sum_a \lvert a\rangle\langle a\rvert = I.$$

Writing

$$\lvert\psi\rangle = \sum_ac_a\lvert a\rangle, \qquad c_a = \langle a\vert\psi\rangle,$$

gives

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

If outcome $a$ occurs,

$$P_a\lvert\psi\rangle = c_a\lvert a\rangle.$$

After normalization, the output ray is the eigenstate ray represented by $\lvert a\rangle$. The phase $c_a/\lvert c_a\rvert$ has no physical significance for that isolated conditional state:

$$\lvert\psi_a\rangle \sim \lvert a\rangle.$$

For a density operator, the same rank-one conclusion is

$$\rho_a = \lvert a\rangle\langle a\rvert$$

whenever $p(a)>0$. Thus the ideal conditional output of a rank-one measurement is independent of the input state except through which outcomes can occur and with what probabilities.

Degenerate Outcomes

For a degenerate eigenvalue $a$, choose any orthonormal basis $\{\lvert a,\mu\rangle\}$ of the eigenspace. The spectral projector is

$$P_a = \sum_{\mu=1}^{g_a} \lvert a,\mu\rangle \langle a,\mu\rvert,$$

where $g_a=\operatorname{rank}P_a>1$. A state can be expanded as

$$\lvert\psi\rangle = \sum_{a,\mu} c_{a\mu}\lvert a,\mu\rangle.$$

The probability and projected component are

$$p(a) = \sum_{\mu=1}^{g_a} \lvert c_{a\mu}\rvert^2$$

and

$$P_a\lvert\psi\rangle = \sum_{\mu=1}^{g_a} c_{a\mu}\lvert a,\mu\rangle.$$

The Lüders update preserves amplitude ratios and relative phases within the selected eigenspace. It does not select one basis vector $\lvert a,\mu\rangle$ unless the apparatus actually resolves the additional label $\mu$.

A measurement that first resolves $\mu$ and later forgets it can have the same coarse probability $p(a)$ but a different conditional state. That distinction is the central topic of Degenerate Measurements and Lüders Rule.

Ideal Repeatability

An ideal projective measurement with the Lüders update is repeatable in the following conditional sense: if outcome $a$ is obtained and the same measurement is immediately repeated with no intervening dynamics, the second result is $a$ with probability one.

For a pure conditional state,

$$P_a\lvert\psi_a\rangle = \lvert\psi_a\rangle.$$

Therefore

$$\begin{aligned} p(b\mid a) &= \langle\psi_a\rvert P_b\lvert\psi_a\rangle \\ &= \frac{ \langle\psi\rvert P_aP_bP_a \lvert\psi\rangle }{p(a)} \\ &= \delta_{ab}. \end{aligned}$$

The density-operator calculation is the same:

$$\begin{aligned} p(b\mid a) &= \operatorname{Tr}(\rho_aP_b) \\ &= \frac{ \operatorname{Tr} \left( P_a\rho P_aP_b \right) }{p(a)} \\ &= \delta_{ab}. \end{aligned}$$

Repeatability does not say that the first outcome was predictable. It says that, after conditioning on the first result under this ideal update, the state lies entirely in the corresponding outcome subspace.

The claim also assumes:

• the second measurement has the same projectors;
• no intervening evolution moves the state out of the selected subspace;
• the apparatus implements the ideal repeatable instrument;
• the first measurement does not destroy or remove the system.

Most real measurements are not exactly repeatable, and exact repeatability is especially subtle for continuous observables. The ideal statement is a property of this mathematical model, not a universal feature of laboratory readout.

Worked Example: A Yes–No Test

Every orthogonal projector $P$ defines a two-outcome PVM

$$P_{\mathrm{yes}} = P, \qquad P_{\mathrm{no}} = I-P.$$

The two projectors are orthogonal because

$$P(I-P) = P-P^2 = 0,$$

and they sum to $I$. For a state $\rho$,

$$\begin{aligned} p(\mathrm{yes}) &= \operatorname{Tr}(\rho P), \\ p(\mathrm{no}) &= 1-\operatorname{Tr}(\rho P). \end{aligned}$$

If “yes” occurs and its probability is nonzero, the ideal conditional state is

$$\rho_{\mathrm{yes}} = \frac{P\rho P} {\operatorname{Tr}(\rho P)}.$$

If “no” occurs,

$$\rho_{\mathrm{no}} = \frac{(I-P)\rho(I-P)} {1-\operatorname{Tr}(\rho P)}.$$

This model asks whether the state lies in the subspace $\operatorname{Ran}P$. It need not distinguish any basis vectors within that subspace.

Worked Example: Qubit Along an Axis

Let a qubit state have Bloch representation

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

For a unit vector $\boldsymbol n$, the projectors for spin along $\boldsymbol n$ are

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

Using

$$\left( \boldsymbol n\cdot\boldsymbol\sigma \right)^2 = I,$$

one verifies

$$\begin{aligned} P_\pm^2 &= P_\pm, \\ P_+P_- &= 0, \\ P_++P_- &= I. \end{aligned}$$

The probabilities are

$$\begin{aligned} p(\pm) &= \operatorname{Tr}(\rho P_\pm) \\ &= \frac12 \left( 1\pm\boldsymbol r\cdot\boldsymbol n \right). \end{aligned}$$

Because each projector has rank one, the conditional states are

$$\rho_\pm = P_\pm$$

whenever the corresponding outcome has nonzero probability. The ideal unread measurement gives

$$\begin{aligned} \rho' &= P_+\rho P_+ + P_-\rho P_- \\ &= \frac12 \left[ I+ \left( \boldsymbol r\cdot\boldsymbol n \right) \boldsymbol n\cdot\boldsymbol\sigma \right]. \end{aligned}$$

Geometrically, the nonselective map removes the component of the Bloch vector perpendicular to $\boldsymbol n$. In general this averaged state differs from either conditional output $P_+$ or $P_-$. If the input already has a certain outcome, the average equals that one branch.

Worked Example: A Degenerate Three-Level Observable

In the orthonormal basis

$$\{ \lvert0\rangle, \lvert1\rangle, \lvert2\rangle \},$$

consider

$$\begin{aligned} A &= a \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) + b\lvert2\rangle\langle2\rvert, \\ a &\neq b. \end{aligned}$$

The spectral projectors are

$$\begin{aligned} P_a &= \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert, \\ P_b &= \lvert2\rangle\langle2\rvert. \end{aligned}$$

Prepare

$$\lvert\psi\rangle = \frac{ 2\lvert0\rangle +i\lvert1\rangle +\lvert2\rangle }{\sqrt6}.$$

The probabilities are

$$p(a) = \frac56, \qquad p(b) = \frac16.$$

Conditional on outcome $a$,

$$\lvert\psi_a\rangle = \frac{ 2\lvert0\rangle+i\lvert1\rangle }{\sqrt5}.$$

The measurement has established that the state lies in the two-dimensional $a$ eigenspace, but it has not distinguished $\lvert0\rangle$ from $\lvert1\rangle$. Their relative phase remains available to later interference-sensitive measurements within that subspace.

Conditional on outcome $b$,

$$\lvert\psi_b\rangle = \lvert2\rangle.$$

An immediate repetition returns $a$ or $b$, respectively, with certainty.

Continuous Outcomes and Spectral Measures

The finite formulas generalize to observables with continuous spectra through a projection-valued measure $E$. For each Borel set $\Delta\subseteq\mathbb R$, $E(\Delta)$ is a projector satisfying

$$E(\varnothing)=0, \qquad E(\mathbb R)=I,$$

and

$$E(\Delta\cap\Gamma) = E(\Delta)E(\Gamma).$$

For pairwise disjoint sets $\{\Delta_k\}$, countable additivity means

$$E\!\left( \bigcup_k\Delta_k \right) = \sum_kE(\Delta_k),$$

with convergence in the strong-operator topology. The probability that the measured value lies in $\Delta$ is

$$p(\Delta) = \operatorname{Tr} \left( \rho E(\Delta) \right).$$

For position on the line, one writes symbolically

$$E_X(\Delta) = \int_\Delta \lvert x\rangle\langle x\rvert\,dx,$$

so that a pure wavefunction gives

$$\begin{aligned} p(X\in\Delta) &= \langle\psi\rvert E_X(\Delta) \lvert\psi\rangle \\ &= \int_\Delta \lvert\psi(x)\rvert^2\,dx. \end{aligned}$$

The generalized kets $\lvert x\rangle$ are not normalizable Hilbert-space vectors, so an exact point outcome should not be treated as an ordinary rank-one projector $\lvert x\rangle\langle x\rvert$. The rigorous projectors correspond to measurable sets; for real-line position, $E_X(\{x\})=0$ for every singleton. A two-outcome region test uses $E_X(\Delta)$ and $I-E_X(\Delta)$. Its ideal sharp spatial cutoff can create discontinuities and infinite kinetic energy, so it should not be confused with a finite-resolution detector model. See Born Rule for Continuous Spectra for the probability-density formulation.

What Projective Measurement Idealizes

A PVM describes sharp alternatives. Its effects are idempotent:

$$P_a^2 = P_a.$$

This distinguishes it from a general positive effect $E_a$, which need only satisfy

$$0\leq E_a\leq I.$$

The ideal projective model packages several assumptions that should not be silently transferred to every apparatus:

• outcome alternatives are represented by orthogonal subspaces;
• the projectors sum to the identity, so one listed outcome occurs;
• calibration is exact at the level of the model;
• the Lüders instrument supplies the conditional disturbance;
• immediate repetition is repeatable;
• detector inefficiency, noise, resolution, and readout dynamics are omitted.

An ideal nondemolition interaction is sometimes summarized by

$$U \left( \lvert a,\mu\rangle \otimes \lvert M_0\rangle \right) = \lvert a,\mu\rangle \otimes \lvert M_a\rangle,$$

where $\lvert M_a\rangle$ are mutually orthogonal pointer records and $\mu$ labels states inside a possibly degenerate eigenspace. This relation motivates repeatability because the system remains in the same $a$ sector. It does not by itself derive a unique observed outcome from unitary dynamics, nor does every real detector implement this interaction.

Noisy, inefficient, unsharp, or deliberately nonorthogonal readouts require generalized measurements and POVMs. Detailed apparatus models and quantum instruments belong in Measurement and Open Quantum Systems.

Common Pitfalls

A degenerate outcome selects its full eigenspace. Replacing its projector by one arbitrarily chosen eigenvector changes the probabilities. Sum squared coefficients over an orthonormal basis of the eigenspace.

Calculate the branch probability before normalizing. The squared norm of the projected ket is the probability. Once a nonzero branch is selected, divide by its norm to obtain the conditional ket.

The same sharp statistics can accompany different disturbance. Outcome-dependent unitaries leave the PVM probabilities unchanged. Only suitable updates are repeatable; the Lüders choice preserves the state within the selected subspace.

A point in a continuous spectrum is not a normalizable output ket. Use spectral projectors for measurable events and specify an instrument for continuous readout. The discrete rank-one update cannot simply be copied to $|x\rangle$.

Exercises

Exercise 1: A three-outcome PVM

In $\mathbb C^3$, define

$$\begin{aligned} \lvert u_0\rangle &= \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \\ \lvert u_+\rangle &= \frac{1}{\sqrt2} \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \\ \lvert u_-\rangle &= \frac{1}{\sqrt2} \begin{pmatrix} 0\\1\\-1 \end{pmatrix}. \end{aligned}$$

Let $P_j=\lvert u_j\rangle\langle u_j\rvert$. Verify that $\{P_0,P_+,P_-\}$ is a PVM. For

$$\lvert\psi\rangle = \frac{1}{\sqrt3} \begin{pmatrix} 1\\1\\i \end{pmatrix},$$

find all three outcome probabilities.

Solution

The three vectors are normalized and mutually orthogonal. Therefore

$$P_jP_k = \delta_{jk}P_j$$

and, because the vectors form an orthonormal basis,

$$P_0+P_++P_- = I.$$

The amplitudes are

$$\begin{aligned} \langle u_0\vert\psi\rangle &= \frac1{\sqrt3}, \\ \langle u_+\vert\psi\rangle &= \frac{1+i}{\sqrt6}, \\ \langle u_-\vert\psi\rangle &= \frac{1-i}{\sqrt6}. \end{aligned}$$

Hence

$$p(0) = p(+) = p(-) = \frac13.$$

The probabilities sum to one, as required.

Exercise 2: Probability from orthogonal components

Let $\{P_a\}$ be a PVM and define $\lvert\psi_a^{\mathrm{un}}\rangle=P_a\lvert\psi\rangle$. Prove directly that

$$\lvert\psi\rangle = \sum_a \lvert\psi_a^{\mathrm{un}}\rangle$$

and

$$\sum_a \left\lVert \lvert\psi_a^{\mathrm{un}}\rangle \right\rVert^2 = 1$$

for a normalized input state.

Solution

Completeness gives

$$\sum_a \lvert\psi_a^{\mathrm{un}}\rangle = \sum_aP_a\lvert\psi\rangle = \lvert\psi\rangle.$$

For $a\neq b$,

$$\left\langle \psi_a^{\mathrm{un}} \middle| \psi_b^{\mathrm{un}} \right\rangle = \langle\psi\rvert P_aP_b\lvert\psi\rangle = 0.$$

The components are therefore orthogonal. Their squared norms satisfy

$$\begin{aligned} \sum_a \left\lVert \lvert\psi_a^{\mathrm{un}}\rangle \right\rVert^2 &= \sum_a \langle\psi\rvert P_a^2\lvert\psi\rangle \\ &= \langle\psi\rvert \left( \sum_aP_a \right) \lvert\psi\rangle \\ &= 1. \end{aligned}$$

These squared norms are precisely the projective outcome probabilities.

Exercise 3: A subspace yes–no test

Let

$$P = \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert$$

in a three-dimensional Hilbert space. For

$$\lvert\psi\rangle = \frac{ \lvert0\rangle +2\lvert1\rangle +i\lvert2\rangle }{\sqrt6},$$

find the yes and no probabilities and the two possible Lüders conditional states.

Solution

The yes branch is

$$P\lvert\psi\rangle = \frac{ \lvert0\rangle+2\lvert1\rangle }{\sqrt6},$$

so

$$p(\mathrm{yes}) = \frac{1+4}{6} = \frac56.$$

The normalized yes state is

$$\lvert\psi_{\mathrm{yes}}\rangle = \frac{ \lvert0\rangle+2\lvert1\rangle }{\sqrt5}.$$

The complementary branch is

$$(I-P)\lvert\psi\rangle = \frac{i\lvert2\rangle}{\sqrt6}.$$

Thus

$$p(\mathrm{no}) = \frac16, \qquad \lvert\psi_{\mathrm{no}}\rangle \sim \lvert2\rangle.$$

The phase $i$ is irrelevant to the final ray.

Exercise 4: Rank-one conditional outputs

Let $P_a=\lvert a\rangle\langle a\rvert$ be rank one. Show that, for every density operator $\rho$ with $\langle a\rvert\rho\lvert a\rangle>0$,

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

Explain what information about the input state remains in the measurement statistics.

Solution

Insert the rank-one projector on both sides:

$$\begin{aligned} P_a\rho P_a &= \lvert a\rangle \langle a\rvert\rho\lvert a\rangle \langle a\rvert \\ &= \langle a\rvert\rho\lvert a\rangle \lvert a\rangle\langle a\rvert. \end{aligned}$$

Its trace is

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

Dividing gives the claimed conditional state. The input state still determines the set of probabilities

$$p(a) = \langle a\rvert\rho\lvert a\rangle,$$

but, conditional on a fixed nonzero-probability outcome, the ideal rank-one output is the same projector for every input.

Exercise 5: Qubit dephasing along an axis

Starting from

$$\rho = \frac12 \left( I+\boldsymbol r\cdot\boldsymbol\sigma \right)$$

and

$$P_\pm = \frac12 \left( I\pm\boldsymbol n\cdot\boldsymbol\sigma \right),$$

derive the nonselective update

$$\sum_{s=\pm}P_s\rho P_s = \frac12 \left[ I+ \left( \boldsymbol r\cdot\boldsymbol n \right) \boldsymbol n\cdot\boldsymbol\sigma \right].$$

What happens when $\boldsymbol r\perp\boldsymbol n$?

Solution

Write

$$N = \boldsymbol n\cdot\boldsymbol\sigma.$$

Because $P_\pm=(I\pm N)/2$,

$$\begin{aligned} \sum_{s=\pm}P_s\rho P_s &= \frac14 (I+N)\rho(I+N) + \frac14 (I-N)\rho(I-N) \\ &= \frac12 \left( \rho+N\rho N \right). \end{aligned}$$

The Pauli identity gives

$$N \left( \boldsymbol r\cdot\boldsymbol\sigma \right) N = \left[ 2(\boldsymbol r\cdot\boldsymbol n) \boldsymbol n-\boldsymbol r \right] \cdot\boldsymbol\sigma.$$

Substitution yields

$$\sum_{s=\pm}P_s\rho P_s = \frac12 \left[ I+ (\boldsymbol r\cdot\boldsymbol n) \boldsymbol n\cdot\boldsymbol\sigma \right].$$

If $\boldsymbol r\perp\boldsymbol n$, then $\boldsymbol r\cdot\boldsymbol n=0$ and the output is

$$\rho' = \frac I2.$$

The two outcomes are equally likely, and ignoring the record leaves the maximally mixed state.

Exercise 6: Degenerate coherence

Let

$$\begin{aligned} P &= \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert, \\ \lvert\psi\rangle &= \frac{ \lvert0\rangle +e^{i\phi}\lvert1\rangle +\lvert2\rangle }{\sqrt3}. \end{aligned}$$

Compare the state conditional on the coarse outcome $P$ for:

1. a Lüders measurement of $\{P,I-P\}$;
2. a rank-one measurement that distinguishes $\lvert0\rangle$ and $\lvert1\rangle$, followed by forgetting which of those two results occurred.

Solution

The coarse outcome has probability

$$p(P) = \frac23.$$

The Lüders conditional state is the pure state

$$\lvert\psi_P\rangle = \frac{ \lvert0\rangle +e^{i\phi}\lvert1\rangle }{\sqrt2},$$

whose density operator contains off-diagonal terms:

$$\rho_P^{\mathrm L} = \frac12 \bigl( \lvert0\rangle\langle0\rvert +e^{-i\phi}\lvert0\rangle\langle1\rvert +e^{i\phi}\lvert1\rangle\langle0\rvert +\lvert1\rangle\langle1\rvert \bigr).$$

The refined measurement produces

$$\rho_P^{\mathrm{ref}} = \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right).$$

Both procedures have the same probability for the coarse record, but only the Lüders measurement preserves coherence within $\operatorname{Ran}P$.

Exercise 7: Repeatability for mixed states

Let

$$\rho_a = \frac{P_a\rho P_a} {\operatorname{Tr}(\rho P_a)}$$

with $\operatorname{Tr}(\rho P_a)>0$. Prove that a second measurement of the same PVM gives $a$ with probability one and every $b\neq a$ with probability zero.

Solution

For any second outcome $b$,

$$\begin{aligned} p(b\mid a) &= \operatorname{Tr}(\rho_aP_b) \\ &= \frac{ \operatorname{Tr} \left( P_a\rho P_aP_b \right) }{ \operatorname{Tr}(\rho P_a) }. \end{aligned}$$

Using $P_aP_b=\delta_{ab}P_a$,

$$p(b\mid a) = \delta_{ab} \frac{ \operatorname{Tr}(P_a\rho P_a) }{ \operatorname{Tr}(\rho P_a) } = \delta_{ab}.$$

The final ratio equals one because

$$\operatorname{Tr}(P_a\rho P_a) = \operatorname{Tr}(\rho P_a).$$

Exercise 8: Coarse graining a PVM

Let $\{P_1,P_2,P_3,P_4\}$ be a PVM. Define

$$Q_{\mathrm L} = P_1+P_2, \qquad Q_{\mathrm R} = P_3+P_4.$$

Prove that $\{Q_{\mathrm L},Q_{\mathrm R}\}$ is a PVM and show that

$$p(\mathrm L) = p(1)+p(2).$$

Does measuring the four fine outcomes and forgetting whether $1$ or $2$ occurred always produce the same conditional state for $\mathrm L$ as a Lüders measurement of $\{Q_{\mathrm L},Q_{\mathrm R}\}$?

Solution

Orthogonality of the fine projectors gives

$$\begin{aligned} Q_{\mathrm L}^2 &= (P_1+P_2)^2 = P_1+P_2 = Q_{\mathrm L}, \\ Q_{\mathrm R}^2 &= Q_{\mathrm R}, \\ Q_{\mathrm L}Q_{\mathrm R} &= 0. \end{aligned}$$

Completeness gives

$$Q_{\mathrm L}+Q_{\mathrm R} = \sum_{j=1}^4P_j = I.$$

Therefore the two coarse projectors form a PVM. Their probability is additive:

$$\begin{aligned} p(\mathrm L) &= \operatorname{Tr}(\rho Q_{\mathrm L}) \\ &= \operatorname{Tr}(\rho P_1) + \operatorname{Tr}(\rho P_2) \\ &= p(1)+p(2). \end{aligned}$$

The conditional state need not be the same. A Lüders measurement of the coarse outcome gives

$$\rho_{\mathrm L}^{\mathrm L} = \frac{ Q_{\mathrm L}\rho Q_{\mathrm L} }{p(\mathrm L)},$$

which retains the cross terms $P_1\rho P_2$ and $P_2\rho P_1$. Measuring the fine outcomes and forgetting the result gives

$$\rho_{\mathrm L}^{\mathrm{fine}} = \frac{ P_1\rho P_1 + P_2\rho P_2 }{p(\mathrm L)},$$

which removes those cross terms. For $p(\mathrm L)>0$, the two states are equal exactly when $P_1\rho P_2=P_2\rho P_1=0$. A restricted set of later observables may fail to distinguish them even when the density operators differ.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016. Publisher.
• E. B. Davies and J. T. Lewis, “An Operational Approach to Quantum Probability,” Communications in Mathematical Physics 17, 239–260 (1970). DOI.
• G. Lüders, “Über die Zustandsänderung durch den Meßprozeß,” Annalen der Physik 8, 322–328 (1951). Translation and discussion by K. A. Kirkpatrick.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.