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I came across the following statement:

If $A$ and $B$ commute and $B$ and $C$, do $A$ and $C$ necessarily commute. Prove or disprove.

I think it's generally untrue. Consider $A = X$, $B = P_y$ and $C = Ly$, where $X$ is the position operator, $P_y$ is the momentum operator in the $y$ direction and $L_y$ is the angular momentum operator in the $y$ direction.

If the statement is generally untrue, by virtue of the counterexample, I have the following question:

Don't we implicitly use the argument in the statement to assert for instance, adding an operator to a list of operators to complete a set of commuting observables. For example, assume $H$ is a rotationally invariant scalar operator (Hamiltonian). We know that it commutes with $L_z$. We also know that $L_z$ commutes with $L^2$ and we then assert that $H$ and $L^2$ must also commute as well and we then looking for the simultaneous eigenbasis of th two angular operator momentum operators to get the spherical harmonics.

Am I missing out on something here?

Qmechanic
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Junaid Aftab
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    No, we never use that argument, because it's wrong, for exactly the reasons you point out. Who told you it? – knzhou Sep 24 '16 at 18:28
  • @knzhou That was my impression of having read Shankar's text on rotational invariance. We know that $H$ and $L_z$ and $L_z$ and $L^2$ commute. How does one justify completing the set of commuting observables by adding $L^2$ if $H$ and $L^2$ don't commute. But they do right? We use this identity to prove a whole host of selection rules in atomic physics. – Junaid Aftab Sep 24 '16 at 18:31
  • Did Shankar ever actually say, "because $H$ and $L_z$, and $L_z$ and $L^2$ commute, then $H$ and $L^2$ commute"? That's completely wrong, and I can't imagine a good book like Shankar making such a big error. – knzhou Sep 24 '16 at 18:33
  • You have to show that $H$ and $L^2$ commute separately. This follows because the Hamiltonian is rotationally invariant. – knzhou Sep 24 '16 at 18:34
  • If Shankar did, it wouldn't be because of a general principle attempted here that's wrong. The point is $H$ commutes with $L_z^2$, and we can similar prove $H$ commutes with $L_x^2,,L_y^2$, and summing $H$ commutes with $L^2$. – J.G. Sep 24 '16 at 18:35
  • @knzhou No, that's why I said that's the impression I get. So where am I going wrong? Is it the case if $A$ and $B$ commute and $B$ and $C$ commute, then $A$, $B$ and $C$ admit a simulataneous eigenbasis but $A$ and $C$ generally don't commute? – Junaid Aftab Sep 24 '16 at 18:35
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    Possible duplicate: http://physics.stackexchange.com/q/95355/2451 – Qmechanic Sep 24 '16 at 18:36
  • For an easier counterexample take $C=P_x$. – J.G. Sep 24 '16 at 18:37

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