A pair of * commuting operators* that are Hermitian can have a common complete set of eigenfunctions.

Let and be two different operators, with observables and respectively.

So, . If this is so, we say that the two operators commute. The short notation for is , where in the case of two commuting operators, .

When the effect of two operators depends on their order, we say that they do not commute, i.e. . If this is the case, we say that the observables and are * complementary*.

One important concept in quantum mechanics is that we can select a common complete set of eigenfunctions for a pair of commuting Hermitian operators. The proof is as follows:

Let and be the complete sets of eigenfunctions of and respectively, such that and . If the two operators have a common complete set of eigenfunctions, we can express as a linear combination of :

For example, the eigenfunction is:

Since some of the eigenfunctions may describe degenerate states (i.e. some are associated with the same eigenvalue ), we can rewrite as:

where and represents distinct eigenvalues of the complete set of eigenfunctions of .

For example, if the linear combination of in eq6 has and describing the same eigenstate with eigenvalue , and and describing another common eigenstate with eigenvalue ,

where , , and so on.

In other words, eq7 is a sum of eigenfunctions with distinct eigenvalues of . Since a linear combination of eigenfunctions describing a degenerate eigenstate is an eigenfunction of , we have

i.e. is an eigenfunction of . Furthermore, the set is complete, which is deduced from eq7, where the set is complete.

From , we have:

Substituting eq7 in the above equation, we have

By operating on the 1^{st} term of the summation in the above equation with , and using the fact that commute with ,

Substituting eq8, where in the above equation,

Repeating the operation of on the remaining terms of the summation in eq9, we obtain equations similar to eq11 and we can write:

i.e. is an eigenfunction of with distinct eigenvalues . Since is Hermitian and are associated with distinct eigenvalues, the eigenfunctions are orthogonal and therefore linearly independent. Consequently, each term in the summation in eq9 must be equal to zero:

This implies that , which is a complete set as mentioned earlier, is also an eigenfunction of . Therefore, we can select a common complete set of eigenfunctions for a pair of commuting Hermitian operators. Conversely, if two Hermitian operators do not commute, eq10 is no longer valid and we cannot select a common complete set of eigenfunctions for them.