Degenerate Measurements

A degenerate measurement is a projective measurement in which at least one reported outcome corresponds to a subspace with dimension greater than one. The outcome identifies an eigenspace, not a unique vector inside that eigenspace.

This matters because the probabilities may be fixed by an observable $A$, while the state left after the measurement depends on what the apparatus actually distinguishes. An ideal Lüders measurement of a degenerate eigenvalue preserves coherence inside the reported eigenspace. A refined apparatus can erase that same coherence while reporting the same eigenvalue.

For the focused update rule, see Lüders Rule. This page explains the modeling distinction that makes the rule necessary.

Degenerate Spectral Decomposition

Let a discrete observable have spectral decomposition

$$A = \sum_a a\Pi_a,$$

where the sum is over distinct reported eigenvalues. The projectors satisfy

$$\Pi_a\Pi_b = \delta_{ab}\Pi_a, \qquad \sum_a\Pi_a=I.$$

If the eigenspace for $a$ has degeneracy $g_a$, choose an orthonormal basis $\{\lvert a,\lambda\rangle\}_{\lambda=1}^{g_a}$ inside that eigenspace:

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

The label $\lambda$ is not part of the outcome when the device measures only the value of $A$. The reported outcome is $a$, and the operational projector is the whole $\Pi_a$.

Born Probability

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

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

For a pure state

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

the same probability is

$$p(a) = \sum_{\lambda=1}^{g_a} |c_{a\lambda}|^2.$$

This probability adds all amplitudes inside the reported eigenspace. The measurement of $A$ asks “which eigenspace?”, not “which basis vector inside that eigenspace?”

Lüders Measurement

The ideal minimal-disturbance operation for outcome $a$ is

$$\mathcal I_a^{L}(\rho) = \Pi_a\rho\Pi_a.$$

If outcome $a$ is known and $p(a)$ is nonzero,

$$\rho_a^{L} = \frac{\Pi_a\rho\Pi_a} {\operatorname{Tr}(\Pi_a\rho)}.$$

If the outcome is ignored, the nonselective Lüders channel is

$$\mathcal M_L(\rho) = \sum_a \Pi_a\rho\Pi_a.$$

This channel removes coherences between different reported eigenspaces but preserves the operator block inside each eigenspace:

$$\rho = \sum_{a,b} \Pi_a\rho\Pi_b \quad \longmapsto \quad \mathcal M_L(\rho) = \sum_a \Pi_a\rho\Pi_a.$$

The diagonal block $\Pi_a\rho\Pi_a$ may still contain coherences between different $\lambda$ labels. A Lüders measurement has not measured those labels.

Refined Measurement

A physical apparatus may measure more than the coarse observable $A$. Suppose a second compatible observable or apparatus degree of freedom distinguishes orthogonal projectors $\Pi_{a\lambda}$ inside each degenerate eigenspace, with

$$\Pi_a = \sum_\lambda \Pi_{a\lambda}.$$

If the apparatus distinguishes $\lambda$ but the lab notebook reports only the coarse value $a$, the operation associated with the reported outcome is

$$\mathcal I_a^{\mathrm{ref}}(\rho) = \sum_\lambda \Pi_{a\lambda}\rho\Pi_{a\lambda}.$$

The probability for $a$ is still

$$p(a) = \operatorname{Tr}(\Pi_a\rho),$$

but the conditional state is generally different from the Lüders state:

$$\rho_a^{\mathrm{ref}} = \frac{ \sum_\lambda \Pi_{a\lambda}\rho\Pi_{a\lambda} }{ \operatorname{Tr}(\Pi_a\rho) }.$$

The refined measurement erases coherences between distinct $\lambda$ sectors. It can therefore have the same reported eigenvalue statistics as the Lüders measurement but different predictions for later measurements.

Three-Level Example

Let a three-dimensional system have observable

$$A = a \left( \lvert1\rangle\langle1\rvert + \lvert2\rangle\langle2\rvert \right) + b \lvert3\rangle\langle3\rvert, \qquad a\ne b.$$

The $a$ eigenspace is two-dimensional:

$$\Pi_a = \lvert1\rangle\langle1\rvert + \lvert2\rangle\langle2\rvert.$$

Prepare

$$\lvert\psi\rangle = \frac{ \lvert1\rangle+\lvert2\rangle }{\sqrt2}.$$

The outcome $a$ occurs with probability one. A Lüders measurement leaves the state unchanged:

$$\rho_a^{L} = \lvert\psi\rangle\langle\psi\rvert.$$

Now suppose the apparatus actually distinguishes $\lvert1\rangle$ from $\lvert2\rangle$ and then reports only $a$. The refined operation gives

$$\rho_a^{\mathrm{ref}} = \frac12 \lvert1\rangle\langle1\rvert + \frac12 \lvert2\rangle\langle2\rvert.$$

Both procedures always report $a$ for this input. They are nevertheless different measurements because they leave different states for future predictions. A later measurement in the basis

$$\left\{ \frac{\lvert1\rangle+\lvert2\rangle}{\sqrt2}, \frac{\lvert1\rangle-\lvert2\rangle}{\sqrt2} \right\}$$

distinguishes the two output states immediately.

Degeneracy and Complete Sets

Degeneracy is often caused by symmetry. If a Hamiltonian is rotationally invariant, energy levels can be degenerate in angular-momentum labels. Measuring the energy need not measure the angular-momentum projection. Measuring a complete set of commuting observables refines the description by adding compatible labels until the common eigenspaces are one-dimensional, at least in the ideal finite-dimensional case.

This is the role of a complete set of commuting observables: it converts a degenerate spectral label into a more detailed set of labels. But an apparatus implements a physical measurement, not a slogan. One must ask which compatible quantities the device actually couples to and records.

For the algebraic background, see Complete Sets of Commuting Observables and Compatible Observables. For symmetry examples, see Angular Momentum Algebra and Selection Rules.

Repeatability Is Not Enough

Both a Lüders measurement and a refined measurement can be repeatable for the coarse observable $A$: after the outcome $a$, a second ideal measurement of $A$ gives $a$ again.

Repeatability of the coarse value does not determine the internal state. The difference appears when one later measures an observable that acts nontrivially within the degenerate eigenspace. This is why the phrase “the measurement of $A$” is incomplete unless the instrument or apparatus model is specified.

Common Mistakes

• Treating a degenerate eigenvalue as if it identified a unique eigenvector.
• Replacing an eigenspace projector $\Pi_a$ by an arbitrary basis vector inside the eigenspace.
• Assuming that every apparatus reporting $a$ implements the Lüders update.
• Forgetting that a refined apparatus can destroy coherence inside a degenerate eigenspace.
• Believing that repeatability for $A$ fixes all post-measurement predictions.
• Ignoring hidden compatible labels measured by the apparatus.

Exercises

Probability in a Degenerate Eigenspace

Let

$$\Pi_a = \lvert1\rangle\langle1\rvert + \lvert2\rangle\langle2\rvert$$

and

$$\lvert\psi\rangle = \alpha\lvert1\rangle + \beta\lvert2\rangle + \gamma\lvert3\rangle.$$

What is the probability of outcome $a$?

Solution

The probability is

$$p(a) = \langle\psi\rvert\Pi_a\lvert\psi\rangle = |\alpha|^2+|\beta|^2.$$

The amplitudes inside the eigenspace add as probabilities because $\lvert1\rangle$ and $\lvert2\rangle$ are orthogonal components of the same reported outcome.

Lüders Versus Refined Output

For the state

$$\lvert\psi\rangle = \frac{\lvert1\rangle+\lvert2\rangle}{\sqrt2}$$

inside the degenerate eigenspace above, compare the Lüders output for outcome $a$ with the refined output that distinguishes $\lvert1\rangle$ and $\lvert2\rangle$ but reports only $a$.

Solution

The Lüders output is the pure state

$$\rho_a^{L} = \lvert\psi\rangle\langle\psi\rvert = \frac12 \left( \lvert1\rangle\langle1\rvert + \lvert1\rangle\langle2\rvert + \lvert2\rangle\langle1\rvert + \lvert2\rangle\langle2\rvert \right).$$

The refined output is

$$\rho_a^{\mathrm{ref}} = \frac12 \lvert1\rangle\langle1\rvert + \frac12 \lvert2\rangle\langle2\rvert.$$

The refined measurement has removed the off-diagonal coherences inside the degenerate eigenspace.

Later Measurement

For the two outputs in the previous exercise, what is the probability of the state

$$\lvert+\rangle = \frac{\lvert1\rangle+\lvert2\rangle}{\sqrt2}$$

in a later projective measurement?

Solution

For the Lüders output,

$$\operatorname{Tr} \left( \lvert+\rangle\langle+\rvert \rho_a^{L} \right) = 1.$$

For the refined output,

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

Thus the two measurement models have the same coarse outcome probability for $A$ but different later predictions.

Cross-Links

• Projective Measurements
• Lüders Rule
• State Update Rules
• Selective and Nonselective Measurements
• Measurement Backaction
• Degenerate Measurements and Lüders Rule
• Complete Sets of Commuting Observables
• Compatible Observables
• Angular Momentum Algebra
• Selection Rules

References

• G. Lüders, “Concerning the state-change due to the measurement process,” Annalen der Physik 15, 663-670 (2006 English translation of 1951 article).
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press (1955).
• P. Busch, P. Lahti, and P. Mittelstaedt, The Quantum Theory of Measurement, 2nd ed., Springer (1996).
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer Academic (1995).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010).