Simultaneous Eigenstates

A simultaneous eigenstate is a nonzero state vector that is an eigenvector of several observables. For two observables $A$ and $B$,

$$A\lvert\psi\rangle = a\lvert\psi\rangle, \qquad B\lvert\psi\rangle = b\lvert\psi\rangle.$$

After normalization, this state has sharp $A$ value $a$ and sharp $B$ value $b$. In the ideal projective formalism, measuring either observable returns its listed value with probability one.

This state-level statement must be separated from two stronger claims:

1. a simultaneous eigenbasis is a complete orthonormal basis whose vectors are eigenvectors of both operators;
2. a complete set of commuting observables has joint labels that identify each basis ray uniquely, with no residual degeneracy.

One common vector need not imply a common basis, and a common basis need not make the chosen labels complete.

Helpful background. Eigenvalues and eigenstates gives the sharp-value criterion; projectors describes eigenspaces geometrically.

Common Eigenvectors

The algebraic common-basis arguments below use finite-dimensional self-adjoint operators. Infinite-dimensional examples need the additional spectral qualifications discussed later.

For a normalized common eigenvector,

$$\langle A\rangle_\psi=a, \qquad \langle B\rangle_\psi=b,$$

and

$$\Delta_\psi A = \Delta_\psi B = 0.$$

The corresponding probability distributions are point masses at $a$ and $b$. Multiplying $\lvert\psi\rangle$ by a nonzero scalar does not change the eigenvalue equations, and multiplying a normalized representative by a phase does not change the physical ray.

For a family $A_1,\ldots,A_r$, write

$$A_k \lvert\boldsymbol\alpha,\mu\rangle = \alpha_k \lvert\boldsymbol\alpha,\mu\rangle, \qquad k=1,\ldots,r.$$

The tuple

$$\boldsymbol\alpha = (\alpha_1,\ldots,\alpha_r)$$

contains the simultaneous eigenvalues. The extra label $\mu$ records any degeneracy that remains after every $\alpha_k$ has been specified.

What the Commutator Says on One State

If $\lvert\psi\rangle$ is a simultaneous eigenvector of $A$ and $B$, then

$$\begin{aligned} AB\lvert\psi\rangle &= ab\lvert\psi\rangle, \\ BA\lvert\psi\rangle &= ba\lvert\psi\rangle. \end{aligned}$$

Therefore

$$[A,B]\lvert\psi\rangle=0.$$

This is a necessary condition on that vector. It is not a statement that

$$[A,B]=0$$

on the whole Hilbert space.

An accidental common eigenvector

Let

$$\mathcal H = \operatorname{span}\lbrace\lvert0\rangle\rbrace \oplus \mathbb C^2$$

and define the self-adjoint operators

$$\begin{aligned} A &= a_0\lvert0\rangle\langle0\rvert \oplus\sigma_z, \\ B &= b_0\lvert0\rangle\langle0\rvert \oplus\sigma_x. \end{aligned}$$

The vector $\lvert0\rangle$ is a common eigenvector with eigenvalues $a_0$ and $b_0$, but

$$[A,B] = 0\oplus2i\sigma_y \ne0.$$

Thus noncommuting observables can share a special eigenvector while failing to admit a simultaneous eigenbasis.

The converse also fails: $[A,B]\lvert\psi\rangle=0$ does not prove that $\lvert\psi\rangle$ is an eigenvector of either operator. If $A$ and $B$ commute globally, the commutator annihilates every vector, including arbitrary superpositions that are not eigenvectors of $A$ or $B$.

From Commutation to a Common Basis

For self-adjoint operators on a finite-dimensional Hilbert space,

$$[A,B]=0$$

if and only if an orthonormal basis of simultaneous eigenvectors exists.

The key step is invariance of eigenspaces. If

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

then commutation gives

$$A(B\lvert\phi\rangle) = a(B\lvert\phi\rangle).$$

Hence $B$ maps the $a$ eigenspace of $A$ into itself. Diagonalizing the restriction of $B$ inside each $A$ eigenspace produces a common basis.

The global equivalences among commutation, spectral projectors, common refinements, and joint ideal measurements belong to Compatible Observables. Here the focus is what an individual common vector and its labels mean.

Joint Eigenspaces

Let the finite-dimensional spectral decompositions be

$$A = \sum_a aP_a, \qquad B = \sum_b bQ_b.$$

When $A$ and $B$ commute, their spectral projectors commute. Define the joint projector

$$R_{ab} = P_aQ_b = Q_bP_a.$$

Its range is the intersection

$$\mathcal H_{ab} = \operatorname{Ran}P_a \cap \operatorname{Ran}Q_b.$$

Equivalently,

$$\mathcal H_{ab} = \operatorname{Ker}(A-aI) \cap \operatorname{Ker}(B-bI).$$

Every nonzero vector in $\mathcal H_{ab}$ is a simultaneous eigenvector with eigenvalues $(a,b)$, and every such eigenvector lies in this subspace. The occurring joint spectrum is therefore

$$\operatorname{spec}(A,B) = \left\lbrace (a,b):R_{ab}\ne0 \right\rbrace.$$

The nonzero joint projectors are mutually orthogonal and resolve the identity:

$$\begin{aligned} R_{ab}R_{a'b'} &= \delta_{aa'}\delta_{bb'}R_{ab}, \\ \sum_{a,b}R_{ab} &= I. \end{aligned}$$

This decomposition organizes the Hilbert space into sectors of simultaneous sharp values:

$$\mathcal H = \mathop{\bigoplus}_{(a,b)} \mathcal H_{ab}.$$

Degeneracy and Residual Labels

The joint eigenvalues need not identify a unique ray. Define

$$g_{ab} = \dim\mathcal H_{ab} = \operatorname{rank}R_{ab}.$$

If $g_{ab}>1$, choose an orthonormal basis

$$\left\lbrace \lvert a,b,\mu\rangle: \mu=1,\ldots,g_{ab} \right\rbrace.$$

Every normalized vector in the joint subspace has the same sharp $A$ and $B$ values. The label $\mu$ is not an eigenvalue of either listed observable; it records information those observables leave unresolved.

This is why the statement “commuting observables have the same eigenvectors” is too loose. If $A$ has a degenerate eigenspace, an arbitrary basis chosen inside that space need not diagonalize $B$. Commutation guarantees that a common basis can be chosen by diagonalizing $B$ inside the degenerate block.

If degeneracy remains after $B$, another commuting observable may refine the joint subspaces further. The detailed treatment of when every surviving block has rank one is Complete Sets of Commuting Observables.

Nondegenerate Observables and Redundant Labels

For a finite-dimensional nondegenerate $A$ with $[A,B]=0$, each one-dimensional $A$ eigenspace is invariant under $B$. Thus

$$B\lvert a_n\rangle = b_n\lvert a_n\rangle.$$

The $A$ eigenbasis is automatically a common eigenbasis. In finite dimension, one can define a function on the spectrum by $f(a_n)=b_n$, giving

$$B=f(A).$$

In this setting, $B$ adds no independent label: knowing $a_n$ already fixes $b_n$. A second commuting observable is useful for labeling when it resolves a degeneracy left by the first, not merely because it commutes.

Likewise, if a compatible family is already complete, adding $C=f(A_1,\ldots,A_r)$ preserves completeness but supplies a redundant quantum number.

Measurement Meaning

For a globally compatible pair, the projectors $R_{ab}$ define the joint projective measurement. In a normalized state $\lvert\psi\rangle$, its outcome probabilities are

$$p(a,b) = \langle\psi|R_{ab}|\psi\rangle.$$

If $\lvert\psi\rangle\in\mathcal H_{a_0b_0}$, then

$$p(a,b) = \delta_{aa_0}\delta_{bb_0}.$$

Both outcomes are sharp, and ideal Lüders measurements of either observable leave the state unchanged because

$$P_{a_0}\lvert\psi\rangle = Q_{b_0}\lvert\psi\rangle = \lvert\psi\rangle.$$

If $R_{a_0b_0}$ has rank greater than one, obtaining $(a_0,b_0)$ does not identify the ray within that subspace. A Lüders measurement of the coarse projectors preserves coherence inside the unresolved joint block. A finer measurement followed by forgetting its extra labels can destroy that coherence even though it reports the same coarse outcomes. The state-update details belong to Degenerate Measurements and Lüders Rule.

For an accidental common eigenvector of globally noncommuting observables, the two individual measurements are still deterministic in that state and leave it unchanged under their ideal projectors. What is absent is one global joint PVM that works as a joint measurement for arbitrary input states. State-specific sharpness is weaker than observable compatibility.

Two-Qubit Example

On two qubits, let

$$A = \sigma_z\otimes I, \qquad B = I\otimes\sigma_z.$$

They commute because they act on different tensor factors. Their simultaneous eigenvectors are the computational basis:

$$\begin{array}{c|cc} \text{state} & A & B \\ \hline \lvert00\rangle & +1 & +1\\ \lvert01\rangle & +1 & -1\\ \lvert10\rangle & -1 & +1\\ \lvert11\rangle & -1 & -1 \end{array}.$$

Each observable alone has two-dimensional eigenspaces. Together, the ordered pair of eigenvalues distinguishes all four basis rays.

The product

$$C=AB=\sigma_z\otimes\sigma_z$$

has eigenvalue equal to the product of the first two labels. Adding $C$ does not further refine the basis; it is a redundant compatible observable.

A superposition such as

$$\frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}$$

is not a simultaneous eigenstate of $A$ and $B$, even though the operators commute. Compatibility supplies a common basis, not sharp labels for every state.

Angular-Momentum Labels

Angular momentum satisfies

$$[J^2,J_z]=0,$$

so one chooses common eigenvectors

$$J^2\lvert j,m,\alpha\rangle = \hbar^2j(j+1) \lvert j,m,\alpha\rangle$$

and

$$J_z\lvert j,m,\alpha\rangle = \hbar m \lvert j,m,\alpha\rangle.$$

The label $\alpha$ is included because a Hilbert space may contain several copies of the same angular-momentum representation. Within one fixed irreducible spin-$j$ space, no multiplicity label is needed: $J^2$ is constant and $m=-j,-j+1,\ldots,j$ labels the standard basis.

The Cartesian components cannot all be sharp in a common basis because

$$[J_x,J_y] = i\hbar J_z$$

and the cyclic permutations are nonzero in a nontrivial spin sector. In the spin-zero representation all three components vanish, so that sector is an exception. The commuting pair $J^2,J_z$ replaces the classically tempting triple $J_x,J_y,J_z$.

The ladder construction and representation theory belong to Angular Momentum Algebra.

Hydrogenic Bound States

For the ideal spinless Coulomb Hamiltonian,

$$H = \frac{\mathbf p^2}{2m} - \frac{\kappa}{r},$$

rotational invariance gives

$$[H,L^2] = [H,L_z] = [L^2,L_z] = 0.$$

The conventional bound states are simultaneous eigenstates

$$\lvert n,\ell,m\rangle$$

with

$$H\lvert n,\ell,m\rangle = E_n\lvert n,\ell,m\rangle,$$ $$L^2\lvert n,\ell,m\rangle = \hbar^2\ell(\ell+1) \lvert n,\ell,m\rangle,$$

and

$$L_z\lvert n,\ell,m\rangle = \hbar m \lvert n,\ell,m\rangle.$$

The Coulomb energy depends only on $n$, so $H$ alone leaves a large degeneracy. At fixed $n$,

$$\ell=0,1,\ldots,n-1, \qquad m=-\ell,\ldots,\ell.$$

The number of spatial states is

$$\sum_{\ell=0}^{n-1}(2\ell+1) = n^2.$$

The labels $\ell$ and $m$ resolve these spinless bound-state alternatives. Thus $H,L^2,L_z$ form a complete labeling family on the ideal spinless bound sector.

The qualification matters. Including electron spin adds an unresolved two-dimensional factor unless a spin label is supplied. Fine structure, external fields, and relativistic corrections change which commuting operators are most useful. The dynamics and degeneracies are developed in Hydrogen Atom.

Symmetry and Good Quantum Numbers

In finite dimension, if a self-adjoint symmetry observable $Q$ commutes with a time-independent Hamiltonian,

$$[H,Q]=0,$$

each energy eigenspace is invariant under $Q$. One may choose energy eigenstates that are also $Q$ eigenstates. The corresponding eigenvalue is a good quantum number. For unbounded operators, use strong spectral commutation; normalizable common eigenvectors additionally require the relevant spectral sectors to have eigenvectors. Strong commutation alone does not turn a continuous spectrum into a discrete basis.

Degeneracy is often a sign that a symmetry acts nontrivially inside an energy eigenspace. For an Abelian family of symmetry generators, simultaneous eigenvalues label the sectors directly. For a non-Abelian symmetry, not all generators commute. One instead chooses commuting Casimir operators and a maximal commuting subset, as $J^2$ and $J_z$ do for rotations.

A symmetry can also produce repeated copies of the same joint labels, leaving a multiplicity index. Good Quantum Numbers develops the symmetry and conservation viewpoint.

Continuous Spectra and Generalized Eigenstates

In an infinite-dimensional Hilbert space, a compatible family need not possess normalizable eigenvectors. On the line, the free Hamiltonian

$$H = \frac{P^2}{2m}$$

commutes with momentum. The generalized momentum kets satisfy

$$P\lvert p\rangle = p\lvert p\rangle$$

and

$$H\lvert p\rangle = \frac{p^2}{2m} \lvert p\rangle.$$

They are simultaneous generalized eigenstates, not vectors of $L^2(\mathbb R)$. For each positive energy, $H$ alone leaves the two momentum labels $p$ and $-p$ degenerate; momentum resolves that degeneracy.

The rigorous replacement for a common discrete eigenbasis is a joint spectral measure or direct-integral decomposition. For self-adjoint unbounded operators, the appropriate compatibility condition is strong commutativity:

$$P_A(\Delta)P_B(\Gamma) = P_B(\Gamma)P_A(\Delta)$$

for all Borel sets $\Delta$ and $\Gamma$. A formal equation $[A,B]\psi=0$ on a small common domain is not enough to guarantee a joint spectral representation.

The spectral theorem and generalized basis language are developed in Spectral Decomposition.

Constructing Common Eigenvectors in Practice

For commuting self-adjoint matrices, construct a common basis by resolving degeneracy:

1. diagonalize one self-adjoint operator $A$;
2. group its equal eigenvalues and construct each eigenspace projector;
3. restrict $B$ to each degenerate eigenspace;
4. diagonalize those restricted blocks;
5. repeat with further commuting observables only inside residual joint blocks.

After constructing a candidate $\lvert a,b,\mu\rangle$, verify the residuals

$$\left\lVert A\lvert\psi\rangle-a\lvert\psi\rangle \right\rVert$$

and

$$\left\lVert B\lvert\psi\rangle-b\lvert\psi\rangle \right\rVert.$$

Numerical eigensolvers are free to return arbitrary orthonormal combinations inside an exactly degenerate eigenspace. Such output can fail to diagonalize a second commuting matrix until the second matrix is diagonalized within the block.

For floating-point data, equality and commutation require tolerances tied to matrix norms and spectral gaps. A small commutator norm alone does not always guarantee a well-conditioned nearby common basis when degeneracies or tiny gaps are present.

Common Pitfalls

Zero commutator action does not mean sharp values. A superposition of different common basis vectors can satisfy $[A,B]\psi=0$ while having nonzero variances in both observables.

Commutation guarantees a choice of common basis. An eigensolver can return a different basis inside a degenerate block. Diagonalize the second observable within that block before assigning joint labels.

Labels are complete only in the stated space. Orbital labels leave spin unresolved; angular-momentum labels leave repeated representation copies unresolved. State the sector before claiming that a tuple identifies one ray.

Exercises

Exercise 1: Necessary is not sufficient

Show that a simultaneous eigenvector obeys

$$[A,B]\lvert\psi\rangle=0.$$

Then give a reason why this equation alone does not imply that $\lvert\psi\rangle$ is a simultaneous eigenvector.

Solution

If

$$A\lvert\psi\rangle=a\lvert\psi\rangle, \qquad B\lvert\psi\rangle=b\lvert\psi\rangle,$$

then

$$\begin{aligned} [A,B]\lvert\psi\rangle &= AB\lvert\psi\rangle - BA\lvert\psi\rangle \\ &= ab\lvert\psi\rangle - ba\lvert\psi\rangle \\ &=0. \end{aligned}$$

For a counterexample, take $A=B=\sigma_z$ and $|\psi\rangle=|+x\rangle$. Their commutator is zero on every vector, but $|+x\rangle$ is not an eigenvector of either operator. Thus the kernel of the commutator can be much larger than the set of common eigenvectors.

Exercise 2: A common state without global compatibility

On

$$\mathcal H = \operatorname{span}\lbrace\lvert0\rangle\rbrace \oplus\mathbb C^2,$$

let

$$A=0\oplus\sigma_z, \qquad B=0\oplus\sigma_x.$$

Verify that $\lvert0\rangle$ is a common eigenvector but that $A$ and $B$ do not commute.

Solution

Both operators annihilate the one-dimensional first summand:

$$A\lvert0\rangle = B\lvert0\rangle = 0.$$

Hence $\lvert0\rangle$ is a simultaneous eigenvector with eigenvalues $(0,0)$. On the second summand,

$$[\sigma_z,\sigma_x] = 2i\sigma_y,$$

so

$$[A,B] = 0\oplus2i\sigma_y \ne0.$$

The common eigenvector is a state-specific fact, not a global compatibility theorem.

Exercise 3: Joint-projector characterization

Let $P_a$ and $Q_b$ be commuting spectral projectors. Prove that

$$R_{ab}=P_aQ_b$$

is an orthogonal projector onto $\operatorname{Ran}P_a\cap\operatorname{Ran}Q_b$.

Solution

Because the projectors commute,

$$\begin{aligned} R_{ab}^2 &= P_aQ_bP_aQ_b \\ &= P_a^2Q_b^2 \\ &= P_aQ_b = R_{ab}. \end{aligned}$$

Also,

$$R_{ab}^\dagger = Q_bP_a = P_aQ_b = R_{ab}.$$

Thus $R_{ab}$ is an orthogonal projector. If $\lvert\phi\rangle=R_{ab}\lvert\chi\rangle$, then $P_a\lvert\phi\rangle=Q_b\lvert\phi\rangle=\lvert\phi\rangle$, so its range lies in the intersection. Conversely, if both projectors fix $\lvert\phi\rangle$, then

$$R_{ab}\lvert\phi\rangle = P_aQ_b\lvert\phi\rangle = \lvert\phi\rangle.$$

Hence the range equals the intersection.

Exercise 4: Resolving a four-dimensional degeneracy

Consider

$$A = \begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$

and

$$B = \begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&2&1\\ 0&0&1&2 \end{pmatrix}.$$

Show that they commute and find a simultaneous eigenbasis.

Solution

$A$ is constant on each two-dimensional coordinate block, while $B$ acts only within those blocks. Therefore $AB=BA$.

In the first block, the normalized eigenvectors of $B$ are

$$\lvert0,\pm\rangle = \frac{\lvert e_1\rangle\pm\lvert e_2\rangle}{\sqrt2},$$

with joint eigenvalues $(0,\pm1)$. In the second block,

$$\lvert1,\pm\rangle = \frac{\lvert e_3\rangle\pm\lvert e_4\rangle}{\sqrt2},$$

with $A$ eigenvalue $1$ and $B$ eigenvalues $3$ and $1$, respectively. These four vectors form an orthonormal simultaneous eigenbasis.

Exercise 5: Two-qubit redundancy

For

$$\begin{aligned} A &= \sigma_z\otimes I, \\ B &= I\otimes\sigma_z, \\ C &= AB. \end{aligned}$$

find the eigenvalue of $C$ on each simultaneous eigenstate of $A$ and $B$. Explain why $C$ adds no independent label.

Solution

If a common eigenstate has

$$A\lvert\psi\rangle = a\lvert\psi\rangle, \qquad B\lvert\psi\rangle = b\lvert\psi\rangle,$$

then

$$C\lvert\psi\rangle = AB\lvert\psi\rangle = ab\lvert\psi\rangle.$$

Thus the $C$ values on $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$ are $+1,-1,-1,+1$. Each is already determined by the ordered pair $(a,b)$, so $C$ does not split any joint eigenspace further.

Exercise 6: Angular-momentum multiplicity

Why can the labels $j,m$ fail to identify a unique state if the Hilbert space contains two copies of the same spin-$j$ representation?

Solution

Both copies have identical eigenvalues

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

and

$$J_z = \hbar m$$

for each $m$. Therefore the joint eigenspace for a fixed $(j,m)$ has dimension two. A multiplicity label $\alpha=1,2$, or an additional commuting observable that distinguishes the copies, is required. Completeness is relative to the full Hilbert space, not merely to one irreducible sector.

Exercise 7: Hydrogenic degeneracy count

For fixed principal quantum number $n$, use

$$\ell=0,\ldots,n-1, \qquad m=-\ell,\ldots,\ell$$

to count the spinless spatial states.

Solution

For each $\ell$, there are $2\ell+1$ values of $m$. Hence

$$\begin{aligned} g_n &= \sum_{\ell=0}^{n-1}(2\ell+1) \\ &= 2\frac{n(n-1)}2+n \\ &= n^2. \end{aligned}$$

$H$ fixes $n$ but leaves all $n^2$ states degenerate in the ideal Coulomb problem. $L^2$ and $L_z$ supply the labels $\ell$ and $m$ that distinguish the spinless bound-state basis.

Exercise 8: Generalized free-particle labels

In one dimension, show that a generalized momentum eigenket is also an energy eigenket of

$$H=\frac{P^2}{2m}.$$

Why does energy alone not distinguish the two propagation directions at positive energy?

Solution

If

$$P\lvert p_0\rangle = p_0\lvert p_0\rangle,$$

then

$$\begin{aligned} H\lvert p_0\rangle &= \frac{P^2}{2m} \lvert p_0\rangle \\ &= \frac{p_0^2}{2m} \lvert p_0\rangle. \end{aligned}$$

For any $E>0$, the values

$$p_0 = \pm\sqrt{2mE}$$

produce the same energy. The momentum label resolves the right-moving and left-moving generalized eigenstates. These kets are distribution-normalized, not ordinary vectors of $L^2(\mathbb R)$.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013. Publisher.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.