Let's say we have three observables $A$, $B$ and $C$. If $A$ is a compatible observable with $B$, and $B$ is a compatible observable with $C$, then is it true that $A$ is compatible with $C$? I've tried to demonstrate that $\begin{cases} [A,B]=0 \\ [B,C]=0 \end{cases} \iff \begin{cases} AB=BA \\ BC=CB \end{cases}$ implies that $[A,C]=0 \iff AC=CA$ but I arrive at nothing. Is it simply that this isn't true or am I missing something?
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4$\begingroup$ No, it is not true (and for sure a duplicate question here). Hint: Consider e.g. $B=I$ and $A=\sigma_z$, $B=\sigma_x$. $\endgroup$Tobias Fünke– Tobias Fünke2025-06-28 20:12:52 +00:00Commented Jun 28 at 20:12
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2$\begingroup$ This question is similar to: How can the commutator operation not be transitive?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$Tobias Fünke– Tobias Fünke2025-06-29 07:54:28 +00:00Commented Jun 29 at 7:54
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1 Answer
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One way to see that, if $[A,B]=0$ and $[B,C]=0$, we can still have $[A,C]\neq 0$, is to consider them in terms of their eigenvectors. In general, two operators commute if their eigenvectors are the same. However, there can be degeneracies, in which case the choice for the eigenvectors of the associated operator can be redefined in terms of linear superpositions the degenerate eigenvectors.
So, say $B$ has two degenerate eigenvectors $|a\rangle$ and $|b\rangle$. For $A$, the associated eigenvectors are not degenerate, but instead given by two orthonormal linear combinations $|a\rangle\alpha_1+|b\rangle\alpha_2$ and $|a\rangle\beta_1+|b\rangle\beta_2$. At the same time, the associated eigenvectors of $C$ are two orthonormal linear combinations $|a\rangle\mu_1+|b\rangle\mu_2$ and $|a\rangle\nu_1+|b\rangle\nu_2$, where all the coefficients are different from those of $A$. Then clearly $A$ and $C$ don't have the same eigenvectors and thus do not commute.
