[L^2,L(x)], [L^2,L(y)] and [L^2,L(z)] are pairs corresponding to good quantum numbers ,that is, each pair has common eigen states. But L(x),L(y) and L(z) do not commute in a pair wise manner and so they do not have common eigenstates.(adsbygoogle = window.adsbygoogle || []).push({});

Now,

L(x)^2 + L(y)^2 + L(z)^2 = L^2

therefore,

[L(x)^2 + L(y)^2 + L(z)^2 ] |phi(ml)= L^2 |phi(ml)

Where,

phi(ml) is an eigenstate common to L and L(z)

We have,

[L(x)^2 + L(y)^2] |phi(ml)+m^2 phi(ml)=l^2 phi(ml)

Or,

[L(x)^2 + L(y)^2] |phi(ml)=(l^2-m^2)phi(ml)

This means that the operator

[L(x)^2+L(y)^2] has common eigen states with L^2 and L(z)^2

Physically this means that if we consider the resolution of angular momentum into two-dimensional rectangular parts they should have common eigenstates.That is, we can measure the original angular momentum vector and the two resolved parts simultaneously.

Is this really true?Do L(x) and L(y) commute for a two dimensional rectangular resolution?

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