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Homework Statement
Two Hermitian operators X and Y have a complete set of mutual eigenkets. Show that [X,Y]=0 and interpret this physically.
Homework Equations
[X,Y]=XY-YX
If [X,Y]=0, XY=YX
The Attempt at a Solution
I have proved that [X,Y]=0, but I'm just falling a little short of understanding the physics of what is going on, so let me write down what I have.
Suppose our system is in the state represented by the ket |Ψ> and that the complete set of mutual eigenkets is {|φn>}, with eigenvalues xn and yn for the operator X and Y respectively.
We may write our state in terms of these eigenkets, so that
|Ψ>=Σncn|φn>
where cn represents the quantum amplitudes of the eigenkets (or eigenstates).
Now XY=YX implies XY|Ψ>=YX|Ψ>.
Now this means that, if we first measure the observable corresponding to the operator X, our state collapses to some eigenstate |φn> and we observe it's corresponding eigenvalue xn with probability |cn|2. We can then measure the observable corresponding to the operator Y, with the state of the system fixed in the same eigenstate, and we measure the eigenvalue yn, but now with probability 1, because we're in an eigenstate - so the overall probability of measuring these values was |cn|2. We can do this the other way around, and we still measure the same observables with the same probability. Thus the fact these observables commute means that there is no uncertainty measured in the observables corresponding to the operators X and Y.
The bit in bold is what I don't get - I believe this link is true, but I don't see how we can associate say XY|Ψ> with first trying to measure the observable corresponding to Y (and so collapsing the state to an eigenstate) and then associate X as trying to measure the observable corresponding to X - if you mathematically do this, it doesn't collapse the state down to some eigenstage. The same applied to YX|Ψ> too. Could anybody explain this? Thankyou.