Commuting Observables and Simultaneous Diagonalization
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You should be comfortable with the spectral theorem for self-adjoint operators and the definition of the commutator \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\). Familiarity with unitary transformations and orthonormal bases is assumed.
Why this matters
The ability to measure two physical quantities simultaneously without disturbing the state is a cornerstone of quantum mechanics.
In classical physics, we rarely worry about the order in which we measure position and momentum; they are just numbers. In quantum mechanics, observables are operators, and the order of operations matters. If two operators do not commute, they represent incompatible observables, and the uncertainty principle limits our knowledge of both. However, if they do commute, we can find a special set of states where both observables have definite values at the same time. This lecture explains the mathematical guarantee that such states exist and how to find them.
The Intuition: Compatible Measurements
Imagine you have a box containing balls of different colours and sizes. If you want to sort them by colour, you might use a filter that only lets red balls through. If you then want to sort the red balls by size, you can do that easily because the colour filter did not change the size of the balls. The two properties, colour and size, are "compatible." You can measure both at once, or in any order, and get consistent results.
Now imagine a different scenario. Suppose you have a machine that sorts balls by their spin direction (up or down) along the x-axis. If you then try to sort them by spin along the y-axis, the first measurement disturbs the second. You cannot know both the x-spin and y-spin precisely at the same time. These properties are "incompatible."
In linear algebra, "compatible" means the operators commute. If \(\hat{A}\) and \(\hat{B}\) commute, applying \(\hat{A}\) then \(\hat{B}\) gives the same result as applying \(\hat{B}\) then \(\hat{A}\). This mathematical property ensures that the eigenvectors of one operator can be chosen to be eigenvectors of the other. In other words, there is a single list of "pure states" (basis vectors) that are perfectly defined for both observables simultaneously. This is the essence of simultaneous diagonalization.
The Commutator as a Test
The commutator \([\hat{A}, \hat{B}]\) is the primary tool for checking compatibility. If \([\hat{A}, \hat{B}] = 0\), the operators commute. This condition is necessary and sufficient for the existence of a common eigenbasis, provided the operators are self-adjoint (which they are, for physical observables).
Consider the position operator \(\hat{x}\) and the momentum operator \(\hat{p}_x\). We know from the canonical commutation relation that \([\hat{x}, \hat{p}_x] = i\hbar\). Since this is not zero, position and momentum along the same axis are incompatible. You cannot have a state that is an eigenstate of both \(\hat{x}\) and \(\hat{p}_x\). This is the mathematical root of the Heisenberg uncertainty principle.
However, consider position \(\hat{x}\) and momentum \(\hat{p}_y\). Since these involve different axes, they commute: \([\hat{x}, \hat{p}_y] = 0\). This means you can have a state with a definite x-position and a definite y-momentum. The physical interpretation is that measuring the x-position does not disturb the y-momentum.
Thus, the commutator acts as a switch. Zero commutator means "yes, we can diagonalize them together." Non-zero commutator means "no, we must choose one basis or the other, and trade off precision."
Simultaneous Diagonalization Theorem
The formal result, derived in D-084, states that if two self-adjoint operators \(\hat{A}\) and \(\hat{B}\) commute, there exists a unitary operator \(U\) that diagonalizes both simultaneously.
This means that in the basis \(\{\lvert e_1\rangle, \dots, \lvert e_n\rangle\}\) defined by the columns of \(U\), every vector \(\lvert e_k\rangle\) is an eigenvector of both \(\hat{A}\) and \(\hat{B}\):
The key insight is that the eigenvectors are shared. If \(\hat{A}\) has a non-degenerate spectrum (all \(a_k\) are distinct), the eigenvectors are uniquely determined by \(\hat{A}\), and the commutation condition forces \(\hat{B}\) to act as a scalar on these same vectors. If \(\hat{A}\) has degeneracies (some \(a_k\) are equal), the eigenspaces are subspaces, and \(\hat{B}\) must be diagonalizable within those subspaces. The theorem guarantees that we can always find a basis that works for both.
Physical Implications: Complete Sets of Observables
In quantum mechanics, a "complete set of commuting observables" (CSCO) is a set of operators that all commute with each other and whose common eigenstates form a complete basis for the Hilbert space. This is crucial because it allows us to label states uniquely by a set of quantum numbers.
For example, in the hydrogen atom, the Hamiltonian \(\hat{H}\), the square of the angular momentum \(\hat{L}^2\), and the z-component of angular momentum \(\hat{L}_z\) form a CSCO. They all commute: \([\hat{H}, \hat{L}^2] = 0\), \([\hat{H}, \hat{L}_z] = 0\), and \([\hat{L}^2, \hat{L}_z] = 0\). Therefore, we can find states \(\lvert n, l, m \rangle\) that are simultaneous eigenstates of all three. The quantum numbers \(n, l, m\) label these states uniquely.
If we tried to include \(\hat{L}_x\) in this set, it would fail because \([\hat{L}_z, \hat{L}_x] = i\hbar \hat{L}_y \neq 0\). Thus, we cannot have a state with definite \(L_z\) and definite \(L_x\) simultaneously. The CSCO concept tells us exactly which measurements we can make together and which we must choose between.
The derivation
Simultaneous Diagonalization of Commuting Operators
Let \(\hat{A}\) and \(\hat{B}\) be self-adjoint operators on a finite-dimensional complex inner-product space \(V\), \(\dim V = n\), and suppose they commute, \([\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} = 0\). Then \(V\) possesses a single orthonormal basis \(\{\lvert e_1\rangle,\dots,\lvert e_n\rangle\}\) whose every vector is simultaneously an eigenvector of \(\hat{A}\) and of \(\hat{B}\), so that one unitary \(U\) brings both to diagonal form: \(U^{\dagger}\hat{A}U = \operatorname{diag}(a_1,\dots,a_n)\) and \(U^{\dagger}\hat{B}U = \operatorname{diag}(b_1,\dots,b_n)\). The converse also holds: two self-adjoint operators sharing an orthonormal eigenbasis necessarily commute.
A common pitfall is assuming that if \([\hat{A}, \hat{B}] = 0\), then the eigenvectors of \(\hat{A}\) are automatically eigenvectors of \(\hat{B}\). This is only true if the eigenspaces of \(\hat{A}\) are one-dimensional (non-degenerate). If \(\hat{A}\) has degenerate eigenvalues, the eigenspace is multi-dimensional, and \(\hat{B}\) might mix vectors within that subspace. You must explicitly diagonalize \(\hat{B}\) within each degenerate subspace of \(\hat{A}\) to find the common eigenbasis. The theorem guarantees the existence of such a basis, but it does not mean any arbitrary basis of \(\hat{A}\)'s eigenspaces will work for \(\hat{B}\). C
Worked example
Problem. Consider a two-dimensional Hilbert space with two self-adjoint operators \(\hat{A}\) and \(\hat{B}\) given by the matrices:
Check if they commute and find a common eigenbasis.
- Compute the commutator \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\). Since \(\hat{A} = I\) (the identity matrix), \(\hat{A}\hat{B} = \hat{B}\) and \(\hat{B}\hat{A} = \hat{B}\). Thus, \([\hat{A}, \hat{B}] = \hat{B} - \hat{B} = 0\). They commute.
- Find the eigenvectors of \(\hat{B}\). The characteristic equation is \(\det(\hat{B} - \lambda I) = \lambda^2 - 1 = 0\), so \(\lambda = \pm 1\).
- For \(\lambda = 1\), solve \((\hat{B} - I)\mathbf{v} = 0\): \(\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = v_2\). Normalized eigenvector: \(\lvert + \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).
- For \(\lambda = -1\), solve \((\hat{B} + I)\mathbf{v} = 0\): \(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = -v_2\). Normalized eigenvector: \(\lvert - \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
- Check if these are eigenvectors of \(\hat{A}\). Since \(\hat{A} = I\), \(\hat{A}\lvert + \rangle = 1 \cdot \lvert + \rangle\) and \(\hat{A}\lvert - \rangle = 1 \cdot \lvert - \rangle\). Yes, they are eigenvectors of \(\hat{A}\) with eigenvalue \(a=1\).
- The common eigenbasis is \(\{\lvert + \rangle, \lvert - \rangle\}\). In this basis, \(\hat{A}\) is \(\operatorname{diag}(1, 1)\) and \(\hat{B}\) is \(\operatorname{diag}(1, -1)\).
Units check. The operators are dimensionless in this abstract example, so no physical units are required. The eigenvectors are normalized, so their norm is 1 (dimensionless).
Three quick tests
- Do the operators \(\hat{L}_x\) and \(\hat{L}_y\) commute?
Answer
No, \([\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z \neq 0\). - If \(\hat{A}\) and \(\hat{B}\) commute, can we always find a basis where both are diagonal?
Answer
Yes, provided they are self-adjoint on a finite-dimensional space. - What is the physical significance of a non-zero commutator \([\hat{A}, \hat{B}] \neq 0\)?
Answer
The observables are incompatible; they cannot be simultaneously measured with arbitrary precision.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.