Let's say there are two observables corresponding to two operators A and B, and let's say my system is in a state Phi where with probability 1 if I measure A I get 3 (let's say 3 Joules), If I measure B I get 4 (let's say 4 m/s). If I measure A and then B I would get 3 Joules for that energy measurement and 4 m/s for the speed measurement, however, mathematically, I would write:

BA Phi=12 Phi

So the measurements kind of mixed up, I don't understand this.

This question arised from problem 3.5 of Zetilli's book

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    $\begingroup$ What exactly is the question? If $|\phi\rangle$ is an eigenstate of both $A$ and $B$ with eigenvalues 3 and 4, it's also an eigenstate of $AB$ (and also $BA$) with eigenvalue 12. What is it that bothers you about this? $\endgroup$ – Javier Feb 4 '19 at 1:32
  • $\begingroup$ The mathematical representation of measuring A and B isn't $AB \Phi = 12 Phi$. $AB \Phi $ means you are applying the operator $AB$ to the $\Phi$ state. A measurement has no actual mathematical representation. (since you can't know what you are getting, only the probabilities which are represented in $\Phi$). $\endgroup$ – IvanMartinez Feb 4 '19 at 5:13

It seems the problem is in distinguishing measuring $A$, measuring $B$ and measuring $BA$. If you have an apparatus that measures $BA$, then you’d get $12 J m/s$, and there is no way you can “separate out” the $A$ and $B$ part: presumably the apparatus to measure $BA$ would yield a single pulse of some height (or whatever other signal) from which you’d deduce the outcome is $12 J m/s$, and you’d have no way of knowing if this were $3\times 4$ or $2\times 6$.

You could separate this out if you measured $A$, recorded the outcome, then measured $B$ and recorded the outcome. That’s not quite the same as measuring $BA$, which is technically a different operator and thus would require a different setup than measuring $A$ or measuring $B$ alone. Measuring $A$ then $B$ is two measurements, whereas measuring $BA$ is a single measurement.

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  • $\begingroup$ Okey I get that, but, supposing Phi is an eigenstate of both A and B? What I'm confused with is the fact that experimentally measuring A and then B would give me and energy of 3J and a velocity of 4 m/s... however, using the formalism, I just get an eigenvalue of 12... and I do not know what information its giving me $\endgroup$ – Juan Pablo Arcila Feb 4 '19 at 1:19
  • $\begingroup$ of course you need to keep track of the units, v.g. the angular momentum operator $\hat L_z$ really have eigenvalues $m\hbar$, not just $m$. your factor here would be $12 Jm/s$, not $12$. Maybe edit your question to clarify... $\endgroup$ – ZeroTheHero Feb 4 '19 at 1:49
  • $\begingroup$ yes sure, it would be 12 Jm/s... but that is not nor the energy nor the velocity I measured, is the product of them.. so I don't know from the formalism how much is the energy or the velocity, I just know their product? $\endgroup$ – Juan Pablo Arcila Feb 4 '19 at 2:48
  • $\begingroup$ I’ve re-edited my answer from scratch in view of your comments. Hopefully this is closer to the mark. $\endgroup$ – ZeroTheHero Feb 4 '19 at 3:08

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