Simultaneous Diagonalization of Commuting Hermitian Operators

Commuting Operators and Common Eigenbases

The relationship between commuting operators and a shared eigenbasis is a two-way street. Let's first prove the more straightforward direction: if two operators have a common eigenbasis, they must commute.

Suppose operators $A$ and $B$ are simultaneously diagonalizable. This means there exists a complete orthonormal basis of vectors $\{|{\psi }_n⟩\}$ that are eigenvectors of both $A$ and $B$:

where $a_n$ and $b_n$ are the corresponding eigenvalues.

To check if the operators commute, we can act their commutator, $[A,B]=AB-BA$, on an arbitrary basis vector $|{\psi }_n⟩$:
\begin{align*}
[A, B]|\psi_n\rangle &= (AB - BA)|\psi_n\rangle \
&= AB|\psi_n\rangle - BA|\psi_n\rangle \
&= A(b_n|\psi_n\rangle) - B(a_n|\psi_n\rangle) \
&= b_n(A|\psi_n\rangle) - a_n(B|\psi_n\rangle) \
&= b_n a_n |\psi_n\rangle - a_n b_n |\psi_n\rangle \
&= (a_n b_n - b_n a_n)|\psi_n\rangle = 0
\end{align*}
Since the commutator annihilates every vector in a complete basis, the operator itself must be the null operator. Therefore, $[A,B]=0$.

The following video segment provides a clear summary of this part of the proof.

The Converse: Commutation Implies a Common Eigenbasis

Now we tackle the more intricate direction: if two Hermitian operators $A$ and $B$ commute, they possess a common eigenbasis. The proof strategy depends on whether the eigenvalues of the operators are degenerate.

The Non-Degenerate Case

Let's start with the simple case where the operator $A$ has a non-degenerate spectrum (all its eigenvalues are distinct). Let $|{\psi }_n⟩$ be an eigenvector of $A$ with eigenvalue $a_n$:

Now, let's consider the vector $B|{\psi }_n⟩$ and see how $A$ acts on it. Using the fact that $[A,B]=0$, or $AB=BA$:

This shows that the vector $B|{\psi }_n⟩$ is also an eigenvector of $A$ with the same eigenvalue $a_n$. Since we assumed the eigenvalue $a_n$ is non-degenerate, any eigenvector corresponding to it must be a scalar multiple of $|{\psi }_n⟩$. Therefore,

for some scalar $b_n$. This is precisely the eigenvalue equation for $B$! It shows that $|{\psi }_n⟩$ is also an eigenvector of $B$. Since this holds for all eigenvectors of $A$, the eigenbasis of $A$ is also an eigenbasis of $B$.

The Degenerate Case

The situation becomes more interesting when $A$ has degenerate eigenvalues. Let $V_{a_n}$ be the eigenspace corresponding to a degenerate eigenvalue $a_n$, with degeneracy $d_n>1$. This means any vector $|\psi ⟩\in V_{a_n}$ satisfies $A|\psi ⟩=a_n|\psi ⟩$.

From the argument above, we know that if $|\psi ⟩\in V_{a_n}$, then the vector $B|\psi ⟩$ is also an eigenvector of $A$ with the same eigenvalue $a_n$. This has a crucial implication: the eigenspace $V_{a_n}$ is an invariant subspace under the action of $B$. That is, $B$ maps vectors from $V_{a_n}$ to other vectors within $V_{a_n}$.

Because $B$ is a Hermitian operator on the full Hilbert space, its restriction to the subspace $V_{a_n}$ can be treated as a Hermitian operator on that $d_n$-dimensional space. By the spectral theorem, this restricted operator can be diagonalized. This means we can find an orthonormal basis of $d_n$ vectors within the subspace $V_{a_n}$ that are eigenvectors of $B$.

Let's call this basis for $V_{a_n}$ $\{|{\psi }_{n,i}⟩{\}}_{i=1}^{d_n}$.

• Each $|{\psi }_{n,i}⟩$ is an eigenvector of $B$ by construction.
• Each $|{\psi }_{n,i}⟩$ is also an eigenvector of $A$, because every vector in the space $V_{a_n}$ is an eigenvector of $A$ with eigenvalue $a_n$.

By repeating this procedure for every eigenspace of $A$, we can construct a complete orthonormal basis for the entire Hilbert space, where every basis vector is a simultaneous eigenvector of both $A$ and $B$.

The following document from MIT OpenCourseWare provides a formal proof that walks through these steps with mathematical precision.

The Block-Diagonal Picture

The argument for the degenerate case has a very clear visual interpretation in terms of matrices. If we choose an eigenbasis of $A$ where we group the basis vectors according to their degenerate eigenvalues, the matrix for $A$ is diagonal.

Here, $I_{d_n}$ is the $d_n\times d_n$ identity matrix for the eigenspace $V_{a_n}$.

Because $[A,B]=0$, the matrix for $B$ in this same basis must be block-diagonal, with blocks corresponding to the degenerate eigenspaces of $A$.

The procedure of "diagonalizing $B$ within each eigenspace of $A$" is equivalent to diagonalizing each of these smaller sub-matrices. A unitary transformation that diagonalizes a block only affects the basis vectors within that corresponding subspace. Since all vectors in that subspace are eigenvectors of $A$ with the same eigenvalue, this transformation leaves the diagonal form of $A$ unchanged. The result is a new basis where both $A$ and $B$ are fully diagonal.

The following video provides an excellent summary of this block-diagonal approach.

Complete Set of Commuting Observables (CSCO)

This theorem has a profound physical consequence. If an observable $A$ has a degenerate spectrum, measuring the value $a_n$ is not sufficient to uniquely determine the state of the system, as there is a whole subspace of states with that eigenvalue.

To resolve this ambiguity, we need to perform another measurement of an observable $B$ that commutes with $A$. If we are lucky, the eigenvalues of $B$ will be different for the different states within the degenerate eigenspace of $A$. This "lifts the degeneracy." The pair of eigenvalues, $(a_n,b_{n,i})$, can now serve as a unique label for the state $|{\psi }_{n,i}⟩$.

A set of commuting observables whose eigenvalues are sufficient to uniquely specify all basis states of a system is called a Complete Set of Commuting Observables (CSCO). The Hamiltonian $H$, angular momentum operators $L^2$ and $L_z$, and spin are classic examples from atomic physics that you've encountered. Their eigenvalues—the quantum numbers $n,l,m_l,m_s$—form a CSCO for the electron in a hydrogen atom.

Conclusion

In this lesson, we have established one of the most crucial theorems in the formalism of quantum mechanics.

Key Takeaways:

• Two Hermitian operators $A$ and $B$ can be simultaneously diagonalized (i.e., they possess a common orthonormal eigenbasis) if and only if they commute, $[A,B]=0$.
• Physically, this means that two observables are compatible—can be measured simultaneously to arbitrary precision—if and only if their corresponding operators commute.
• For an operator with a degenerate spectrum, we can use other commuting operators to lift the degeneracy. A set of such operators that uniquely labels all basis states is known as a Complete Set of Commuting Observables (CSCO).

We now have a clear criterion for when observables can and cannot be known simultaneously. In our next lesson, we will begin to explore the consequences for operators that famously do not commute, namely position ($\hat{x}$) and momentum ($\hat{p}$). This will lead us to the celebrated Heisenberg uncertainty principle and the deep connection between the position and momentum representations, which are linked via the Fourier transform.