 # If Operators A and B Commute, Then A and B Are Simultaneously Well-Defined

purpose of this video is to prove the theorem that if the operators a and B commute then there exists a set of eigenfunctions the sets I n which are assigned with hi-c huggin functions of a and B meaning that both a and B are simultaneously well-defined what this means is that that the a and B operators representing that we are allowed to know simultaneously and precisely in other words not subject to the Heisenberg uncertainty principle limitations and so we're gonna prove that sy is enough this on sy is a non degenerate eigen function of a operator with eigenvalue a this is getting to be our standard way of dealing with these various theorems just write down our friend eigenvalue equation itself and then if we go ahead and operate on that eigenvalue equation with the second operator B operator and so we've done that to the left-hand side B operator is now acting on a operator acting on sy and then on the right-hand side B operator acts on the eigenvalue a times sy and then we can pull the eigenvalue a out in front of our expression because a of course being scalar value is not actually affected by b operator no matter what B operator is and so the right-hand side becomes eigenvalue a times the result of the B operator acting on the function side since we're looking at the case where a and B are known to commute with each other what that means is that we can reverse the order of operation for our left hand side term so we can instead of having a operate on sy and then operate on that result with B in the second line and the third line we've reversed it so that B operates on site first followed by a operating on that particular result and then our right hand side is not change because we just have to be operator on the right hand side but we can group that our terms on the right hand side in a suggestive way that we can say well really be operate on sy is some other function and that would be an eigen function because this certainly looks like an eigenvalue equation since the right-hand side has an eigenvalue term in front of it and so by doing this we recognize then that the be operator acting on psy is in fact itself an eigen functions and eigen function of a with eigenvalue a if we just look at the equation that we generated all right when we act with them with a on b x side we get b x side be acting on side back again with eigenvalue a since the wave function psy is not degenerate there's only one eigen function that corresponds to that eigenvalue a and so what this means then is that be operating on psy has two equal sign itself or just differs by psy from by some particular constant it all this means is that of course that b is following its own eigenvalue equation when ax by itself on side and so we're gonna call that constant term that relates the result of B acting outside and call that little B eigen value B so if we have an eigenvalue equation for the a operator and we know that a and B commute then we know that when B operates on sy that that leads to an eigenvalue equation as well and the only difference between the eigenvalue equation for the operator and like an equation for the B operator is that the eigen value itself could be different but the wave function that satisfies that eigenvalue equation for the two cases that is exactly the same and the reason why this is significant then is that what this means is that both the both the properties a and B we're interested in that we represented with the operators a and B are simultaneously well-defined mean we can know we can know the values at the same time then we can know them infinitely precisely we're not bound by any kind of Heisenberg uncertainty principle limit and the reason why we want to have this commuter test available to us in the future is we're gonna have more complicated cases of operators where it's not going to be obvious to us like it was in the case of momentum because for the the one-dimensional particle that the Heisenberg uncertainty principle apply that we're gonna want to just use the commutator test to figure out whether we can know about two properties a and B simultaneously or not 