Degenerate Measurements and Lüders Rule

A degenerate projective measurement has at least one outcome whose projector has rank greater than one. The outcome identifies an eigenspace, not a unique eigenvector. The Lüders rule conditions on that entire eigenspace:

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

This distinction matters whenever a measured value leaves other quantum labels unresolved. An ideal Lüders measurement preserves coherence within the selected eigenspace. An apparatus that resolves additional labels and later discards them can have the same coarse outcome probabilities while producing a different state.

Helpful background. State update explains how a branch is normalized; spectral decomposition identifies its eigenspace projector. We first work in finite dimensions. The parity example uses two bounded projectors on an infinite-dimensional space, for which the same update formulas remain valid.

What Degeneracy Means

An eigenvalue $a$ is degenerate when

$$g_a = \dim\ker(A-aI) > 1.$$

The number $g_a$ is the degeneracy or multiplicity of the eigenvalue. Every vector in

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

satisfies

$$A\lvert\phi\rangle = a\lvert\phi\rangle.$$

The value $a$ alone does not specify the post-measurement state. For a repeatable measurement the output lies in $\mathcal H_a$, but the input and the instrument still determine which state in that subspace results. A general apparatus with the same outcome probabilities need not even leave its output there.

Degeneracy can arise in several ways:

• a symmetry can force several states to share one eigenvalue;
• an additional compatible quantum number can remain unmeasured;
• an “accidental” spectral coincidence can occur beyond the degeneracy required by an obvious symmetry;
• an experiment can deliberately group finer alternatives into one coarse record.

Exact mathematical degeneracy should not be confused with two distinct eigenvalues that an imperfect detector cannot resolve. Finite resolution is an apparatus property and may require coarse-grained POVM effects. Exact degeneracy is a property of the operator spectrum in the model.

Spectral Projectors and Basis Independence

Choose any orthonormal basis

$$\{ \lvert a,1\rangle, \ldots, \lvert a,g_a\rangle \}$$

for $\mathcal H_a$. The spectral projector is

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

The basis vectors are not unique. Let another orthonormal basis be related by a unitary matrix $U$:

$$\lvert a,\nu\rangle' = \sum_\mu U_{\nu\mu} \lvert a,\mu\rangle.$$

Then

$$\begin{aligned} \sum_\nu \lvert a,\nu\rangle' \langle a,\nu\rvert' &= \sum_{\nu,\mu,\lambda} U_{\nu\mu} U_{\nu\lambda}^* \lvert a,\mu\rangle \langle a,\lambda\rvert \\ &= \sum_{\mu,\lambda} \delta_{\mu\lambda} \lvert a,\mu\rangle \langle a,\lambda\rvert \\ &= P_a. \end{aligned}$$

The projector is therefore an intrinsic property of the eigenspace. The Lüders update therefore depends on the eigenspace, not on the basis chosen to describe it. An apparatus can nevertheless distinguish a particular internal basis physically; that additional measurement must be specified.

The spectral decomposition consequently sums over distinct eigenvalues:

$$A = \sum_a aP_a,$$

with each eigenspace grouped into one outcome. It is also algebraically valid to expand $P_a$ into rank-one terms and repeat the coefficient $a$; those terms must not be mistaken for separately recorded outcomes.

Decomposing a State by Eigenspaces

Insert the resolution of identity

$$\sum_aP_a = I.$$

Every state vector decomposes as

$$\lvert\psi\rangle = \sum_aP_a\lvert\psi\rangle.$$

Within each eigenspace,

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

where

$$c_{a\mu} = \langle a,\mu\vert\psi\rangle.$$

The outcome probability is the squared norm of the full eigenspace component:

$$\begin{aligned} p(a) &= \langle\psi\rvert P_a\lvert\psi\rangle \\ &= \left\lVert P_a\lvert\psi\rangle \right\rVert^2 \\ &= \sum_{\mu=1}^{g_a} \lvert c_{a\mu}\rvert^2. \end{aligned}$$

There are no interference cross terms in this sum because the chosen $\lvert a,\mu\rangle$ are orthonormal. The coherent information inside the subspace remains in the projected state, even though the coarse outcome probability is the sum of component probabilities.

For a density operator, insert identity resolutions on both sides:

$$\rho = \sum_{a,b} P_a\rho P_b.$$

The diagonal block $P_a\rho P_a$ is the unnormalized state associated with outcome $a$. Its trace is

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

Selective Lüders Rule

If outcome $a$ occurs with $p(a)>0$, the Lüders conditional state is

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

For a pure input, this becomes

$$\lvert\psi_a^{\mathrm L}\rangle = \frac{P_a\lvert\psi\rangle}{\sqrt{p(a)}}.$$

The conditional state is supported in the selected eigenspace:

$$P_a\rho_a^{\mathrm L}P_a = \rho_a^{\mathrm L}.$$

It follows that an immediate repetition of the same coarse PVM, with no intervening evolution, gives $a$ again with probability one:

$$\operatorname{Tr} \left( \rho_a^{\mathrm L}P_b \right) = \delta_{ab}.$$

Repeatability fixes the coarse value, not a unique vector within the eigenspace. A later measurement of an additional compatible observable can still have nontrivial statistics.

Zero-probability outcomes

When $p(a)=0$,

$$P_a\rho P_a = 0,$$

so no normalized conditional state is defined. Degeneracy does not alter this rule: projection onto a large subspace still yields no branch if the input has no support there.

The Precise Sense of Minimal Disturbance

Calling the Lüders rule “minimal” should be tied to a definite property rather than used as a slogan.

Suppose the input state already lies entirely in the eigenspace for $a$:

$$\rho = P_a\rho P_a.$$

Then $p(a)=1$ and the Lüders update leaves the state unchanged:

$$\rho_a^{\mathrm L} = \rho.$$

Thus the ideal Lüders instrument does not disturb any state already confined to the recorded outcome subspace. In particular, it preserves every superposition and mixture within that eigenspace.

Repeatability alone does not imply this stronger nondisturbance. Let $U_a$ map $\mathcal H_a$ unitarily onto itself and define

$$\mathcal I_a^{U}(\rho) = U_aP_a\rho P_aU_a^\dagger.$$

The branch remains in $\mathcal H_a$, so the same coarse outcome repeats with certainty. But a state inside $\mathcal H_a$ can be rotated:

$$\rho \longmapsto U_a\rho U_a^\dagger.$$

The Lüders choice corresponds to no additional outcome-dependent transformation inside the selected subspace. This is the operational content of its minimally refining character.

It does not follow that a real apparatus causes no disturbance. The statement belongs to the specified ideal instrument.

The Nonselective Lüders Channel

When the measurement occurs but its outcome is ignored, the state becomes

$$\mathcal L_A(\rho) = \sum_aP_a\rho P_a.$$

Relative to the eigenspace decomposition

$$\mathcal H = \bigoplus_a\mathcal H_a,$$

the density operator has a block form. The Lüders channel removes off-diagonal blocks $P_a\rho P_b$ with $a\neq b$ and preserves every diagonal block $P_a\rho P_a$.

The channel therefore:

• removes coherence between distinct eigenvalue sectors;
• preserves coherence within each degenerate eigenspace;
• leaves outcome probabilities unchanged;
• is idempotent;
• fixes every state commuting with all $P_a$.

The idempotence calculation is

$$\begin{aligned} \mathcal L_A^2(\rho) &= \sum_{a,b} P_aP_b\rho P_bP_a \\ &= \sum_aP_a\rho P_a \\ &= \mathcal L_A(\rho). \end{aligned}$$

The channel acts once by deleting inter-eigenspace blocks; applying it again has nothing further to delete.

Nondisturbance of Compatible Observables

Let $B$ be a bounded observable that commutes with every spectral projector of $A$:

$$[B,P_a] = 0 \qquad \text{for every }a.$$

After a nonselective Lüders measurement of $A$, its expectation value is

$$\begin{aligned} \operatorname{Tr} \left[ B\mathcal L_A(\rho) \right] &= \sum_a \operatorname{Tr}(BP_a\rho P_a) \\ &= \operatorname{Tr} \left[ \rho\sum_aP_aBP_a \right] \\ &= \operatorname{Tr}(\rho B). \end{aligned}$$

The equality holds for every input state. More generally, every spectral projector of a compatible observable is fixed by the dual Lüders map, so its full probability distribution is preserved.

Conversely, suppose

$$\operatorname{Tr} \left[ B\mathcal L_A(\rho) \right] = \operatorname{Tr}(\rho B)$$

for every density operator $\rho$. Then

$$\sum_aP_aBP_a = B.$$

Multiplying by $P_b$ on the left and $P_c$ on the right gives

$$P_bBP_c = 0 \qquad (b\neq c).$$

Therefore $B$ is block diagonal in the eigenspace decomposition and

$$[B,P_a] = 0$$

for every $a$. This is the sharp-observable form of the Lüders nondisturbance theorem: an unread ideal Lüders measurement of $A$ preserves the statistics of $B$ in every state if and only if the observables are compatible.

The statement is instrument-specific. A more disturbing measurement with the same outcome projectors can alter $B$ even when $[A,B]=0$.

Refining a Degenerate Measurement

A refinement resolves additional alternatives inside each degenerate eigenspace. Let

$$P_a = \sum_{\mu\in M_a} R_{a\mu},$$

where

$$R_{a\mu}R_{b\nu} = \delta_{ab}\delta_{\mu\nu} R_{a\mu}.$$

The refined PVM has records $(a,\mu)$. Its probabilities are

$$p(a,\mu) = \operatorname{Tr}(\rho R_{a\mu}).$$

Summing over the unresolved label gives the original coarse probability:

$$\begin{aligned} \sum_{\mu\in M_a} p(a,\mu) &= \operatorname{Tr} \left( \rho\sum_{\mu\in M_a}R_{a\mu} \right) \\ &= \operatorname{Tr}(\rho P_a) \\ &= p(a). \end{aligned}$$

Agreement on the coarse probabilities does not imply agreement on the output state.

Direct coarse Lüders measurement

If the apparatus implements the coarse Lüders instrument, the unnormalized branch is

$$\widetilde\rho_a^{\mathrm L} = P_a\rho P_a.$$

Expanding the projector gives

$$\widetilde\rho_a^{\mathrm L} = \sum_{\mu,\nu\in M_a} R_{a\mu}\rho R_{a\nu}.$$

This includes the off-diagonal terms with $\mu\neq\nu$.

Fine measurement followed by forgetting

If the apparatus resolves $(a,\mu)$ and the label $\mu$ is later discarded, the unnormalized coarse branch is

$$\widetilde\rho_a^{\mathrm{ref}} = \sum_{\mu\in M_a} R_{a\mu}\rho R_{a\mu}.$$

For $p(a)>0$, the corresponding normalized state is

$$\rho_a^{\mathrm{ref}} = \frac{ \sum_{\mu\in M_a} R_{a\mu}\rho R_{a\mu} }{p(a)}.$$

The difference from the Lüders branch is

$$\widetilde\rho_a^{\mathrm L} - \widetilde\rho_a^{\mathrm{ref}} = \sum_{\substack{\mu,\nu\in M_a\\\mu\neq\nu}} R_{a\mu}\rho R_{a\nu}.$$

These are precisely the coherences among refined alternatives inside the coarse eigenspace.

Neither formula is universally “the correct collapse” independent of the apparatus. The direct Lüders expression models a measurement that preserves every state within each coarse subspace. Merely failing to record $\mu$ does not guarantee this: other disturbances or rotations are possible. The refined expression assumes a Lüders measurement of the fine projectors, followed by discarding the fine record.

Complete Commuting Refinements

Degeneracy often signals that one observable does not provide enough labels to identify a ray. A complete set of commuting observables supplies additional compatible labels.

Suppose $A$ and $B$ commute. Their joint eigenspaces can be represented by projectors $R_{ab}$ satisfying

$$\begin{aligned} P_a &= \sum_bR_{ab}, \\ Q_b &= \sum_aR_{ab}. \end{aligned}$$

Measuring only $A$ uses the coarse projectors $P_a$. Measuring the compatible pair $(A,B)$ uses the finer joint projectors $R_{ab}$. If the joint eigenspaces are one-dimensional, the pair provides a complete basis label.

The physical distinction is not erased by saying that $A$ and $B$ commute. Compatibility means that a joint sharp measurement is possible; it does not mean that every device reporting $A$ also measures $B$.

Historical Terminology

For nondegenerate observables, projection onto the measured eigenstate is unambiguous. Degeneracy exposes an ambiguity: the recorded eigenvalue does not identify one vector in its eigenspace.

In discussions descended from von Neumann’s measurement postulate, a measurement of a degenerate observable may be represented through a finer, degeneracy-breaking observable. Lüders proposed the full-eigenspace update

$$\rho \longmapsto \frac{P_a\rho P_a}{p(a)}$$

for an ideal measurement that registers the degenerate value without resolving additional labels.

Modern measurement theory treats these as descriptions of different instruments rather than competing universal formulas detached from experimental context. A degeneracy-preserving Lüders device and a degeneracy-resolving device are both mathematically realizable; their distinction must be determined by how the apparatus couples and what information it records.

Worked Example: A Three-Level Coarse Measurement

Let

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

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

$$\lvert\psi\rangle = \frac{ \lvert0\rangle +e^{i\phi}\lvert1\rangle +\lvert2\rangle }{\sqrt3}.$$

The probability of the degenerate outcome $a$ is

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

The Lüders conditional state is

$$\lvert\psi_a^{\mathrm L}\rangle = \frac{ \lvert0\rangle +e^{i\phi}\lvert1\rangle }{\sqrt2}.$$

Its density operator is

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

Now suppose the apparatus distinguishes $\lvert0\rangle$ from $\lvert1\rangle$ and only later suppresses that label. Conditional on the coarse report $a$, the state is

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

Both instruments give $p(a)=2/3$, but a later interference measurement distinguishes them.

Define

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}, \qquad Q_+ = \lvert+\rangle\langle+\rvert.$$

For the Lüders state,

$$\begin{aligned} p_{\mathrm L}(+) &= \operatorname{Tr} \left( \rho_a^{\mathrm L}Q_+ \right) \\ &= \frac{1+\cos\phi}{2}. \end{aligned}$$

For the refined-and-forgotten state,

$$p_{\mathrm{ref}}(+) = \frac12.$$

When $\phi=0$, the Lüders state gives $+$ with certainty, while the refined state gives it with probability $1/2$. The difference is experimentally accessible through later measurements inside the degenerate subspace.

Physical Examples

Total angular momentum

For a sector with total angular momentum $j$, a measurement of $J^2$ reports

$$J^2 = \hbar^2j(j+1)$$

but does not determine the magnetic label $m$. In a multiplicity-free sector, the corresponding projector has the form

$$P_j = \sum_{m=-j}^{j} \lvert j,m\rangle \langle j,m\rvert.$$

An ideal Lüders measurement of $J^2$ preserves superpositions over $m$ within the selected $j$ sector. Measuring the compatible pair $(J^2,J_z)$ is a refinement that also resolves $m$.

Central-potential energy

In a rotationally invariant Hamiltonian, an energy eigenspace can contain several angular-momentum states. For the nonrelativistic Coulomb problem, the bound-state energy depends on the principal quantum number while multiple orbital labels share that energy.

An ideal energy measurement projects onto the full energy eigenspace. A device that also resolves $L^2$ and $L_z$ performs a more detailed compatible measurement. Whether angular coherence survives depends on which measurement is physically implemented.

Parity

Let $\Pi$ be the parity operator. The projectors

$$P_\pm = \frac12 \left( I\pm\Pi \right)$$

select the even and odd subspaces. Each subspace is generally infinite dimensional. A parity measurement reports only the sign; a Lüders update preserves arbitrary superpositions within the selected parity sector.

Subspace and syndrome readout

Quantum-information protocols often ask whether a state lies in a particular subspace rather than which basis vector it occupies. Error-syndrome measurements are designed to reveal an error sector while preserving encoded superpositions within that sector. The ideal mathematical model is a degenerate projective measurement; unwanted resolution of logical information is an additional disturbance.

How Degeneracy Is Lifted or Hidden

The mathematical eigenspace and the laboratory record should be distinguished.

• A perturbation can split an exact degeneracy into nearby eigenvalues.
• A detector with enough resolution can distinguish the split values.
• A low-resolution detector can group distinct values into one record.
• An apparatus can couple to an additional commuting observable and resolve internal labels.
• Environmental interactions can decohere states inside a nominally degenerate subspace.

These situations can produce similar coarse histograms but different post-measurement states. Spectroscopy of the first outcome distribution is therefore not always enough to identify the measurement instrument; sequential or interference-sensitive tests may be required.

Common Pitfalls

Selecting an arbitrary vector in a degenerate eigenspace. The Lüders projector includes the whole eigenspace. A different orthonormal basis gives the same projector and the same update.

Treating repeatability as absence of disturbance. A device can rotate the state within the selected subspace and still reproduce the same coarse outcome. The stronger Lüders property is that every state already in that subspace is left unchanged.

Inferring the instrument from a coarse histogram. Fine measurement followed by discarding its label and direct coarse Lüders measurement agree on the coarse probabilities. The interference test above distinguishes their outputs.

Confusing degeneracy with detector resolution. Equal eigenvalues are a property of the observable; grouping nearby eigenvalues is a property of a measurement. State explicitly which projectors or effects describe the recorded alternatives.

References

• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer, 2016 — modern treatment of Lüders instruments, repeatability, and degeneracy.
• P. Busch and J. Singh, “Lüders theorem for unsharp quantum measurements,” Physics Letters A 249, 10–12, 1998, doi:10.1016/S0375-9601(98)00704-X, arXiv:1304.0054 — sharp Lüders nondisturbance theorem and extensions.
• G. Lüders, “Über die Zustandsänderung durch den Meßprozeß,” Annalen der Physik 8, 322–328, 1951; K. A. Kirkpatrick, “Translation of Lüders’ ‘Über die Zustandsänderung durch den Messprozess’,” Annalen der Physik 15, 663–670, 2006, arXiv:quant-ph/0403007 — the full-eigenspace update and its relation to compatibility.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010 — subspace measurements and finite-dimensional examples.

Exercises

Exercise 1: Basis independence of a degenerate projector

Let $\{\lvert e_1\rangle,\lvert e_2\rangle\}$ be an orthonormal basis of a two-dimensional eigenspace. Define

$$\begin{aligned} \lvert f_1\rangle &= \cos\theta\lvert e_1\rangle + e^{i\varphi}\sin\theta\lvert e_2\rangle, \\ \lvert f_2\rangle &= -e^{-i\varphi}\sin\theta\lvert e_1\rangle + \cos\theta\lvert e_2\rangle. \end{aligned}$$

Show that

$$\lvert f_1\rangle\langle f_1\rvert + \lvert f_2\rangle\langle f_2\rvert = \lvert e_1\rangle\langle e_1\rvert + \lvert e_2\rangle\langle e_2\rvert.$$

Solution

Expanding the first projector gives

$$\lvert f_1\rangle\langle f_1\rvert = \cos^2\theta \lvert e_1\rangle\langle e_1\rvert + \sin^2\theta \lvert e_2\rangle\langle e_2\rvert + e^{-i\varphi} \sin\theta\cos\theta \lvert e_1\rangle\langle e_2\rvert + e^{i\varphi} \sin\theta\cos\theta \lvert e_2\rangle\langle e_1\rvert.$$

The second gives

$$\lvert f_2\rangle\langle f_2\rvert = \sin^2\theta \lvert e_1\rangle\langle e_1\rvert + \cos^2\theta \lvert e_2\rangle\langle e_2\rvert - e^{-i\varphi} \sin\theta\cos\theta \lvert e_1\rangle\langle e_2\rvert - e^{i\varphi} \sin\theta\cos\theta \lvert e_2\rangle\langle e_1\rvert.$$

The off-diagonal terms cancel and $\sin^2\theta+\cos^2\theta=1$, yielding the original projector.

Exercise 2: Lüders update in a qutrit

Let

$$A = 0 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right) + 3\lvert2\rangle\langle2\rvert$$

and

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

Find the probability of outcome $0$ and the Lüders conditional state.

Solution

The projector for outcome $0$ is

$$P_0 = \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert.$$

The projected vector is

$$P_0\lvert\psi\rangle = \frac{ 2\lvert0\rangle+i\lvert1\rangle }{\sqrt6}.$$

Therefore

$$p(0) = \frac{4+1}{6} = \frac56,$$

and

$$\lvert\psi_0^{\mathrm L}\rangle = \frac{ 2\lvert0\rangle+i\lvert1\rangle }{\sqrt5}.$$

The relative phase between the two internal components survives.

Exercise 3: Interference distinguishes two instruments

For

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

compare the probability of

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

with that obtained from

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

Solution

For the Lüders state,

$$\begin{aligned} p_{\mathrm L}(+) &= \left\lvert \langle+\vert\psi_a^{\mathrm L}\rangle \right\rvert^2 \\ &= \left\lvert \frac{1+e^{i\phi}}{2} \right\rvert^2 \\ &= \frac{1+\cos\phi}{2}. \end{aligned}$$

For the refined mixture,

$$\begin{aligned} p_{\mathrm{ref}}(+) &= \operatorname{Tr} \left( \rho_a^{\mathrm{ref}} \lvert+\rangle\langle+\rvert \right) \\ &= \frac12. \end{aligned}$$

The difference depends on the coherence phase. At $\phi=0$, the probabilities are $1$ and $1/2$, respectively.

Exercise 4: Repeatability does not imply Lüders minimality

Let $P_a$ be a degenerate outcome projector and let $U_a$ be a unitary satisfying

$$U_aP_a = P_aU_aP_a.$$

Consider

$$\mathcal I_a^U(\rho) = U_aP_a\rho P_aU_a^\dagger.$$

Show that the outcome $a$ repeats with certainty after conditioning on this branch, but a state initially supported in $P_a$ need not remain unchanged.

Solution

The condition on $U_a$ means that it maps $\operatorname{Ran}P_a$ into itself. Hence

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

After normalization, the conditional state $\rho_a^U$ therefore satisfies

$$\operatorname{Tr}(\rho_a^UP_a) = 1.$$

The same coarse PVM returns $a$ with certainty.

If the input already obeys

$$\rho = P_a\rho P_a,$$

the output is

$$\rho \longmapsto U_a\rho U_a^\dagger.$$

This equals $\rho$ only when $U_a$ leaves that particular state invariant. The instrument can be repeatable while rotating states inside the degenerate eigenspace; the Lüders instrument adds no such rotation.

Exercise 5: Lüders nondisturbance

Let

$$\mathcal L_A(\rho) = \sum_aP_a\rho P_a$$

and suppose $[B,P_a]=0$ for every $a$. Prove that

$$\operatorname{Tr} \left[ B\mathcal L_A(\rho) \right] = \operatorname{Tr}(B\rho)$$

for every density operator $\rho$.

Solution

Using cyclicity of the trace,

$$\begin{aligned} \operatorname{Tr} \left[ B\mathcal L_A(\rho) \right] &= \sum_a \operatorname{Tr}(BP_a\rho P_a) \\ &= \sum_a \operatorname{Tr}(P_aBP_a\rho). \end{aligned}$$

Because $B$ commutes with $P_a$,

$$P_aBP_a = BP_a.$$

Therefore

$$\begin{aligned} \sum_a \operatorname{Tr}(P_aBP_a\rho) &= \operatorname{Tr} \left( B\sum_aP_a\rho \right) \\ &= \operatorname{Tr}(B\rho). \end{aligned}$$

Exercise 6: When coarse and refined updates agree

Let

$$P_a = \sum_{\mu=1}^{g_a}R_{a\mu}.$$

Show that the unnormalized coarse Lüders branch and refined-and-forgotten branch agree if and only if

$$\sum_{\substack{\mu,\nu=1\\\mu\neq\nu}}^{g_a} R_{a\mu}\rho R_{a\nu} = 0.$$

Give a sufficient condition stated as a commutator.

Solution

Expanding the coarse branch gives

$$\begin{aligned} P_a\rho P_a &= \sum_{\mu,\nu} R_{a\mu}\rho R_{a\nu} \\ &= \sum_\mu R_{a\mu}\rho R_{a\mu} + \sum_{\mu\neq\nu} R_{a\mu}\rho R_{a\nu}. \end{aligned}$$

The first sum is the refined-and-forgotten branch. The branches agree exactly when the second sum vanishes.

A sufficient condition is

$$[\rho,R_{a\mu}] = 0$$

for every $\mu$. Then, for $\mu\neq\nu$,

$$R_{a\mu}\rho R_{a\nu} = \rho R_{a\mu}R_{a\nu} = 0.$$

Exercise 7: Parity measurement

Let $\Pi=\Pi^\dagger$ and $\Pi^2=I$, and define

$$P_\pm = \frac12(I\pm\Pi).$$

1. Verify that $\{P_+,P_-\}$ is a PVM.

2. For a pure state decomposed as

   $$\lvert\psi\rangle = \lvert\psi_+\rangle + \lvert\psi_-\rangle,$$

   with $\Pi\lvert\psi_\pm\rangle=\pm\lvert\psi_\pm\rangle$, find the outcome probabilities and conditional states.

Solution

Hermiticity of $\Pi$ gives $P_\pm^\dagger=P_\pm$. Using $\Pi^2=I$,

$$\begin{aligned} P_\pm^2 &= \frac14 \left( I\pm2\Pi+\Pi^2 \right) \\ &= P_\pm. \end{aligned}$$

Also,

$$P_+P_- = \frac14(I-\Pi^2) = 0,$$

and

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

Because $P_\pm\lvert\psi\rangle=\lvert\psi_\pm\rangle$,

$$p(\pm) = \lVert\psi_\pm\rVert^2.$$

For nonzero probability, the conditional state is

$$\lvert\psi_{\pm,\mathrm{cond}}\rangle = \frac{ \lvert\psi_\pm\rangle }{\lVert\psi_\pm\rVert}.$$

The measurement selects a parity sector without resolving a basis inside that sector.

Exercise 8: Energy measurement versus a complete refinement

Suppose an energy eigenspace is spanned by orthonormal states

$$\{ \lvert E,\lambda_1\rangle, \lvert E,\lambda_2\rangle \},$$

where $\lambda$ is the value of a compatible observable. The input is

$$\lvert\psi\rangle = \alpha\lvert E,\lambda_1\rangle + \beta\lvert E,\lambda_2\rangle + \gamma\lvert E'\rangle.$$

Compare the state conditional on energy $E$ for:

1. a Lüders energy measurement;
2. a joint measurement of energy and $\lambda$, followed by forgetting $\lambda$.

Assume $E\neq E'$, all three displayed kets are normalized and mutually orthogonal, $|\alpha|^2+|\beta|^2+|\gamma|^2=1$, and $|\alpha|^2+|\beta|^2>0$.

Solution

The energy projector is

$$P_E = \lvert E,\lambda_1\rangle \langle E,\lambda_1\rvert + \lvert E,\lambda_2\rangle \langle E,\lambda_2\rvert.$$

The energy-$E$ probability is

$$p(E) = \lvert\alpha\rvert^2 + \lvert\beta\rvert^2.$$

The Lüders conditional state is the pure state

$$\lvert\psi_E^{\mathrm L}\rangle = \frac{ \alpha\lvert E,\lambda_1\rangle + \beta\lvert E,\lambda_2\rangle }{ \sqrt{ \lvert\alpha\rvert^2+\lvert\beta\rvert^2 } }.$$

The refined-and-forgotten conditional state is

$$\begin{aligned} w_1 &= \frac{\lvert\alpha\rvert^2}{p(E)}, & w_2 &= \frac{\lvert\beta\rvert^2}{p(E)}. \end{aligned}$$

Therefore

$$\rho_E^{\mathrm{ref}} = w_1 \lvert E,\lambda_1\rangle \langle E,\lambda_1\rvert + w_2 \lvert E,\lambda_2\rangle \langle E,\lambda_2\rvert.$$

The Lüders state retains the relative phase between $\alpha$ and $\beta$; the refined state does not.