D.18 Eigen­func­tions of com­mut­ing op­er­a­tors

The fact that two op­er­a­tors that com­mute have a com­mon set of eigen­func­tions can be seen as fol­lows: as­sume that $\alpha$ is an eigen­func­tion of $A$ with eigen­value $a$. Then since $A$ and $B$ com­mute, $AB\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $BA\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $aB\alpha$. Com­par­ing start and end, $B\alpha$ must be an eigen­func­tion of $A$ with eigen­value $a$ just like $\alpha$ it­self is. If there is no de­gen­er­acy of the eigen­value, that must mean that $B\alpha$ equals $\alpha$ or is at least pro­por­tional to it. That is the same as say­ing that $\alpha$ is an eigen­func­tion of $B$ too. (In the spe­cial case that $B\alpha$ is zero, $\alpha$ is still an eigen­func­tion of $B$, with eigen­value zero.)

If there is de­gen­er­acy, the eigen­func­tions of $A$ are not unique and you can mess with them un­til they all do be­come eigen­func­tions of $B$ too. That can be shown as­sum­ing that the prob­lem has been ap­prox­i­mated by a fi­nite-di­men­sion­al one. Then $A$ and $B$ be­come ma­tri­ces and the eigen­func­tion be­come eigen­vec­tors. Con­sider each eigen­value of $A$ in turn. There will be more than one eigen­vec­tor cor­re­spond­ing to a de­gen­er­ate eigen­value $a$. Now by com­plete­ness, any eigen­vec­tor $\beta$ can be writ­ten as a com­bi­na­tion of the eigen­vec­tors of $A$, and more par­tic­u­larly as $\beta$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\beta_n+\beta_a$ where $\beta_a$ is a com­bi­na­tion of the eigen­vec­tors of $A$ with eigen­value $a$ and $\beta_n$ a com­bi­na­tion of the eigen­vec­tors of $A$ with other eigen­val­ues.

The vec­tors $\beta_n$ and $\beta_a$ sep­a­rately are still eigen­vec­tors of $B$ if nonzero, since as noted above, $B$ con­verts eigen­vec­tors of $A$ into eigen­vec­tors with the same eigen­value or zero. (For ex­am­ple, if $B\beta_a$ was not $b\beta_a$, $B\beta_n$ would have to make up the dif­fer­ence, and $B\beta_n$ can only pro­duce com­bi­na­tions of eigen­vec­tors of $A$ that do not have eigen­value $a$.) Now re­place the eigen­vec­tor $\beta$ by ei­ther $\beta_a$ or $\beta_n$, whichever is in­de­pen­dent of the other eigen­vec­tors of $B$. Do­ing this for all eigen­vec­tors of $B$ you achieve that the re­place­ment eigen­vec­tors of $B$ are ei­ther com­bi­na­tions of the eigen­vec­tors of $A$ with eigen­value $a$ or of the other eigen­vec­tors of $A$. The set of new eigen­vec­tors of $B$ that are com­bi­na­tions of the eigen­vec­tors of $A$ with eigen­value $a$ can now be taken as the re­place­ment eigen­vec­tors of $A$ with eigen­value $a$. They are also eigen­vec­tors of $B$. Re­peat for all eigen­val­ues of $A$.

Sim­i­lar ar­gu­ments can be used re­cur­sively to show that more gen­er­ally, a set of op­er­a­tors that all com­mute have a com­mon set of eigen­vec­tors.

The op­er­a­tors do not re­ally have to be Her­mit­ian, just di­ag­o­nal­iz­able: they must have a com­plete set of eigen­func­tions.

The above de­riva­tion as­sumed that the prob­lem was fi­nite-​di­men­sion­al, or dis­cretized some way into a fi­nite-di­men­sion­al one like you do in nu­mer­i­cal so­lu­tions. The lat­ter is open to some sus­pi­cion, be­cause even the most ac­cu­rate nu­mer­i­cal ap­prox­i­ma­tion is never truly ex­act. Un­for­tu­nately, in the in­fi­nite-​di­men­sion­al case the de­riva­tion gets much trick­ier. How­ever, as the hy­dro­gen atom and har­monic os­cil­la­tor eigen­func­tion ex­am­ples in­di­cate, typ­i­cal in­fi­nite sys­tems in na­ture do sat­isfy the re­la­tion­ship any­way.