# On Angular Momentum

1. Jun 23, 2010

### Anamitra

[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.

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?

2. Jun 23, 2010

### msumm

Yes, phi is an eigenstate of (Lx^2 + Ly^2) if it is an eigenstate of Lz^2 and L^2, but in this case phi will not be an eigenstate of Lx nor Ly. So yes, Lx and Ly have a common set of eigenstates if you restrict yourself to the set of eigenstates of Lz, but that set is empty. Maybe I don't understand your question? I'm pretty sure I don't undrstand your last question.

3. Jun 24, 2010

### Anamitra

Lz and Lx^2+Ly^2 have common eigen states. So the z component of the angular momentum and the resultant in the x-y plane can be measured simultaneously.But this is simply a two dimensional resolution of the angular momentum. Now let us redefine the axes. We call the z axis the y' axis and the direction along which the resultant of the x and the y components acts the x' axis.Renaming the axes should not change the physical nature of the problem. So the Lx' component[previously the resultant in the x-y plane] and the Ly' component[previously the Lz component] may be measured at the same time------they correspond to good quantum numbers.
Therefore Lx' and Ly' should commute provided the resolution is two dimensional.[Vectors corresponding to L ,Lx' and Ly' should lie in the same plane]

4. Aug 11, 2010

### Anamitra

If we accept the fact that simultaneous observations[of the values of the components] can be made for two dimensional resolution of the angular momentum vector there is a problem.It is like this. We first consider the resolution of the angular momentum vector into components one along the z-axis and the other in x-y plane.They may be observed simultaneously.Now we resolve the component in the x-y plane into two parts, one along the x-axis and the other along the y-axis.They should again be simultaneously observable if the first statement of this posting is true.