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.

$\hat{O}_1(\hat{O}_2\psi)=\hat{O}_1(\Omega_2\psi)=\Omega_2\hat{O}_1\psi=\Omega_2\Omega_1\psi$

$\hat{O}_2(\hat{O}_1\psi)=\hat{O}_2(\Omega_1\psi)=\Omega_1\hat{O}_2\psi=\Omega_1\Omega_2\psi$

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 1^st 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.

Commuting operators

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