Commuting operators

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

Let \hat{O}_1 and \hat{O}_2 be two different operators, with observables \Omega_1 and \Omega_2 respectively.



So, \hat{O}_1(\hat{O}_2\psi)=\hat{O}_2(\hat{O}_1\psi)\; \; \; or\; \; \;\hat{O}_1(\hat{O}_2\psi)-\hat{O}_2(\hat{O}_1\psi)=0. If this is so, we say that the two operators commute. The short notation for \hat{O}_1(\hat{O}_2\psi)-\hat{O}_2(\hat{O}_1\psi) is \left [\hat{O}_1,\hat{O}_2\right ], where in the case of two commuting operators, \left [\hat{O}_1,\hat{O}_2\right ]=0.

When the effect of two operators depends on their order, we say that they do not commute, i.e. \left [\hat{O}_1,\hat{O}_2\right ]\neq 0. If this is the case, we say that the observables \Omega_1 and \Omega_2 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 \left \{ \vert a_n\rangle \right \} and \left \{ \vert b_n\rangle \right \} be the complete sets of eigenfunctions of \hat{A} and \hat{B} respectively, such that \hat{A}\vert a_n\rangle=a_n\vert a_n\rangle and \hat{B}\vert b_m\rangle=b_m\vert b_m\rangle. If the two operators have a common complete set of eigenfunctions, we can express \vert a_n\rangle as a linear combination of \vert b_m\rangle:

\vert a_n\rangle=\sum_{m=1}^{k}c_{nm}\vert b_m\rangle\; \; \; \; \; \; \; \; 5

For example, the eigenfunction is:

\vert a_1\rangle=c_{11}\vert b_1\rangle+c_{12}\vert b_2\rangle+\cdots+c_{1k}\vert b_k\rangle\; \; \; \; \; \; \; \; 6

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

\vert a_n\rangle=\sum_{i=1}^{j}\vert(a_n) b_i\rangle\; \; \; \; \; \; \; \; 7

where \vert(a_n) b_i\rangle=\sum_{m=1}^{k}d_{nm}\vert b_m\rangle\delta_{b_i,b_m} and b_i represents distinct eigenvalues of the complete set of eigenfunctions of \hat{B}.

For example, if the linear combination of \vert a_1 \rangle in eq6 has \vert b_1\rangle and \vert b_2\rangle describing the same eigenstate with eigenvalue b_1, and \vert b_4\rangle and \vert b_5\rangle describing another common eigenstate with eigenvalue b_3,

\vert a_1\rangle=\vert(a_1) b_1\rangle+\vert(a_1) b_2\rangle+\vert(a_1) b_3\rangle+\cdots+\vert(a_1) b_j\rangle

where \vert(a_1) b_1\rangle=d_{11}\vert b_1\rangle+d_{12}\vert b_2\rangle, \vert(a_1) b_2\rangle=d_{13}\vert b_3\rangle, \vert(a_1) b_3\rangle=d_{14}\vert b_4\rangle+d_{15}\vert b_5\rangle and so on.

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

\hat{B}\vert(a_n)b_i\rangle=b_i\vert(a_n)b_i\rangle\; \; \; \; \; \; \; \; 8

i.e. \vert(a_n)b_i\rangle is an eigenfunction of \hat{B}. Furthermore, the set \left \{ \vert(a_n)b_i\rangle \right \} is complete, which is deduced from eq7, where the set \left \{ \vert a_n\rangle \right \} is complete.

From \hat{A}\vert a_n\rangle=a_n\vert a_n\rangle, we have:

(\hat{A}-a_n)\vert a_n\rangle=0

Substituting eq7 in the above equation, we have

(\hat{A}-a_n)\vert a_n\rangle=\sum_{i=1}^{j}(\hat{A}-a_n) \vert (a_n)b_i\rangle=0\; \; \; \; \; \; \; \; 9

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

\hat{B}(\hat{A}-a_n)\vert (a_n)b_1\rangle=(\hat{A}-a_n)\hat{B} \vert (a_n)b_1\rangle\; \; \; \; \; \; \; \; 10

Substituting eq8, where i=1 in the above equation,

\hat{B}(\hat{A}-a_n)\vert (a_n)b_1\rangle=b_1(\hat{A}-a_n) \vert (a_n)b_1\rangle\; \; \; \; \; \; \; \; 11

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

\hat{B}(\hat{A}-a_n)\vert (a_n)b_i\rangle=b_i(\hat{A}-a_n) \vert (a_n)b_i\rangle

i.e. (\hat{A}-a_n)\vert (a_n)b_i\rangle is an eigenfunction of \hat{B} with distinct eigenvalues b_i. Since \hat{B} is Hermitian and (\hat{A}-a_n)\vert (a_n)b_i\rangle are associated with distinct eigenvalues, the eigenfunctions (\hat{A}-a_n)\vert (a_n)b_i\rangle are orthogonal and therefore linearly independent. Consequently, each term in the summation in eq9 must be equal to zero:

(\hat{A}-a_n)\vert (a_n)b_i\rangle=0\; \; \; \Rightarrow \; \; \;\hat{A}\vert (a_n)b_i\rangle=a_n\vert (a_n)b_i\rangle

This implies that \vert (a_n)b_i\rangle, which is a complete set as mentioned earlier, is also an eigenfunction of \hat{A}. Therefore, we can select a common complete set of eigenfunctions \left \{ \vert (a_n)b_i\rangle\right \} 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.



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