advanced31 min studyLesson 364 in Physics

Compatible observables and complete bases

Resolve degeneracy using commuting observables and distinguish a shared eigenbasis from an arbitrary set of state coordinates.

On this page

What you’ll understand

  • Construct a common eigenbasis for commuting Hermitian matrices.
  • Explain how additional eigenvalue labels resolve degeneracy.
  • Transform states and observables consistently under a passive unitary basis change.

One measured eigenvalue need not identify a unique state. If an observable has a degenerate eigenspace, an additional compatible observable can distinguish directions inside it. A complete set of commuting observables supplies enough joint labels to identify basis rays. This finite-dimensional construction clarifies why commuting operators can be measured jointly, while also showing why neither an arbitrary basis change nor a vanishing formal differential commutator is sufficient to settle every infinite-dimensional measurement question.

Commutation preserves eigenspaces

For Hermitian matrices A and B that commute, applying B to an A eigenvector keeps the result inside the same A eigenspace. When that eigenspace is one-dimensional, the vector is automatically also a B eigenvector. When it is degenerate, B can mix its vectors, but its Hermitian restriction can be diagonalized within that subspace. Repeating this operation constructs an orthonormal basis of simultaneous eigenvectors without disturbing the already diagonal first observable.

Degeneracy does not imply that every eigenvector of A is an eigenvector of B. It guarantees that a suitable basis can be chosen inside each degenerate eigenspace. If two listed joint eigenvalue tuples remain identical, the chosen commuting set has not resolved that remaining degeneracy. Complete here means enough joint labels to identify basis rays, not enough information to determine an arbitrary unknown superposition from one measurement on one copy.

[A,B]=0,A∣a⟩=a∣a⟩⟹A(B∣a⟩)=a(B∣a⟩),p(a,b)=⟨ψ∣PaQb∣ψ⟩[A,B]=0,\quad A|a\rangle=a|a\rangle\quad\Longrightarrow\quad A(B|a\rangle)=a(B|a\rangle),\qquad p(a,b)=\langle\psi|P_aQ_b|\psi\rangle
The first statement is finite-dimensional linear algebra for Hermitian A and B. Pa and Qb are their commuting spectral projectors, and the second formula gives dimensionless joint Born probabilities for a normalized state. Observable eigenvalues carry whatever units their quantities require.↔ Scroll wide equations horizontally.

A unitary coordinate change preserves every paired prediction

Arrange the new orthonormal basis vectors as columns of a unitary matrix U in the old coordinates. The new coefficient vector is U dagger times the old one, and an observable matrix is conjugated by the same coordinate change. Transforming only the vector while leaving the observable matrix fixed usually describes an active change relative to a fixed apparatus. Confusing these operations can manufacture an apparent disagreement between two equivalent descriptions.

Unitarity preserves inner products and normalization, whereas mere invertibility does not. A general invertible coordinate system can still be used with its proper Gram matrix, but the simple sum of squared coefficients is then no longer automatically the physical norm. In the orthonormal bases used here, no such correction is needed. A matrix with determinant one can stretch some vectors and shrink others, so determinant preservation alone does not establish a valid unitary change between orthonormal quantum coordinates.

c′=U†c,A′=U†AU,U†U=I,c′†A′c′=c†Acc'=U^\dagger c,\qquad A'=U^\dagger AU,\qquad U^\dagger U=I,\qquad c'^\dagger A'c'=c^\dagger Ac
c is a dimensionless normalized coefficient column, U is dimensionless and unitary, and A has observable units. U's columns are the new basis vectors expressed in the old basis. These are passive coordinate changes; both state and observable representations transform.↔ Scroll wide equations horizontally.

Spectral compatibility is the reliable general statement

In finite dimension, commuting Hermitian matrices have commuting projectors and a common eigenbasis. Infinite-dimensional self-adjoint observables require a stronger spectral formulation: their spectral projections must commute for a joint sharp observable. A formal commutator vanishing on some common test functions does not automatically establish this strong commutation. Domains and continuous spectra prevent an unrestricted transfer of the finite-matrix proof to arbitrary differential operators.

Even when a compatible measurement exists, its implementation matters for disturbance inside degenerate eigenspaces. An ideal joint projective measurement of the refined labels is a specific instrument, not the only possible apparatus returning the same coarse A outcome. The joint probabilities follow from the commuting projectors, while statements about the state afterward need an update model. This separation becomes essential when generalized measurements and quantum instruments are introduced later in the module.

Resolve a twofold degeneracy

In a three-dimensional space let A=diag⁡(1,1,2)A=\operatorname{diag}(1,1,2) and B=(010100003)B=\begin{pmatrix}0&1&0\\1&0&0\\0&0&3\end{pmatrix}, both dimensionless. Find their common eigenvectors and the joint probabilities for ∣ψ⟩=(1,0,1)T/2|\psi\rangle=(1,0,1)^T/\sqrt2.

  1. A is constant on its first two coordinates, and B mixes only those coordinates, so AB=BA. Diagonalize B in this A=1 subspace using (1,1,0)T/2(1,1,0)^T/\sqrt2 and (1,−1,0)T/2(1,-1,0)^T/\sqrt2, with B eigenvalues +1 and −1.

  2. The remaining vector (0,0,1) has A=2 and B=3. The three joint labels are therefore (1,+1), (1,−1), and (2,3), all distinct and sufficient to identify the chosen basis rays.

  3. Project ψ onto the common basis. Its coefficients are 1/2, 1/2, and 1/21/\sqrt2, giving probabilities 1/4, 1/4, and 1/2. They sum to one and imply ⟨A⟩=3/2\langle A\rangle=3/2 and ⟨B⟩=3/2\langle B\rangle=3/2.

The result

The symmetric and antisymmetric first-coordinate pairs resolve A's degeneracy. The joint outcome probabilities are 1/4, 1/4, and 1/2; an A-only measurement would merge the first two outcomes.

Explore the model

Interactive lab

Interactive lab · matrix

Invertibility is weaker than unitarity

Reset and set both controls to Diagonal scale a=1 and Shear b=0. Predict determinant 1 and both singular values 1. For a fresh comparison, Reset and set both controls to a=1 and b=1. The determinant stays 1, but the singular values become approximately 1.618 and 0.618, so lengths are not all preserved. Readouts describe the selected final matrix throughout playback.

Invertibility is weaker than unitarityThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA shear is invertible and area preserving but is not a unitary change between orthonormal state bases. This positive triangular matrix family is not a general observable diagonalizer or a quantum evolution simulator.

Check your understanding

What follows for two commuting Hermitian matrices when the first observable has a degenerate eigenspace?

Try it yourself

In a two-dimensional space let A=2I and B=diag(4,5). A normalized state has coefficients (3,i)T/2(\sqrt3,i)^T/2. Find joint outcome probabilities and ⟨B⟩, and explain why measuring only A does not distinguish the two basis states.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The coordinate basis already diagonalizes both matrices. Keep the imaginary coefficient when forming its modulus squared.

Compare with the solution

The probabilities are 3/4 for (A,B)=(2,4) and 1/4 for (2,5). Hence ⟨B⟩=4(3/4)+5(1/4)=17/4=4.25\langle B\rangle=4(3/4)+5(1/4)=17/4=4.25. Both basis states have the same A eigenvalue 2, so A alone gives no label distinguishing them. The pair (A,B) resolves that degeneracy, although one joint measurement still samples only one outcome and does not reconstruct the unknown relative phase of a general prepared state.

Take it with you

  • Commuting Hermitian matrices admit a common eigenbasis after diagonalizing within degeneracies.
  • Passive basis changes transform state coefficients and observable matrices together.
  • Infinite-dimensional compatibility concerns commuting spectral projectors, including domain issues.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.