Special case of angular momentum components simultaneous knowledge

[LostConjugate]

Lets say we measure $$L_z$$ and $$L^2$$ simultaneously and $$L^2$$ happens to be equal to $$(L_z)^2$$. This requires a zero value for the x and y components of angular momentum.

Does this mean, in this special case, we know the values of $$L_x L_y L_z & L^2$$ simultaneously?

 

[SpectraCat]

Lets say we measure $$L_z$$ and $$L^2$$ simultaneously and $$L^2$$ happens to be equal to $$(L_z)^2$$. This requires a zero value for the x and y components of angular momentum.

Does this mean, in this special case, we know the values of $$L_x L_y L_z & L^2$$ simultaneously?

That is not possible, precisely for the reason that you describe. The angular momentum of a quantum system is never precisely aligned with a particular spatial axis, except in the limit of macroscopic systems in accordance with the Bohr correspondence principle (i.e. large mass, high quantum number).

 

[LostConjugate]

That is not possible, precisely for the reason that you describe. The angular momentum of a quantum system is never precisely aligned with a particular spatial axis, except in the limit of macroscopic systems in accordance with the Bohr correspondence principle (i.e. large mass, high quantum number).

Ok good, so that state does not exist. As is intuitive, since an exact direction in space is not logical even classically.

 

[espen180]

The eigenvalues of $L^2$ are $\hbar ^2 j(j+1)$, $j\in \mathbb{N}$, while the eigenvalues of $L_z$ are $\hbar m$, $m\in [-j,j]$.

If all of the angular momentum were to be directed along the (arbitrarily chosen) z-axis, we would require $\hbar^2 j^2$ to be an eigenvalue of $L^2$, which it isn't.

 

[LostConjugate]

The eigenvalues of $L^2$ are $\hbar ^2 j(j+1)$, $j\in \mathbb{N}$, while the eigenvalues of $L_z$ are $\hbar m$, $m\in [-j,j]$.

If all of the angular momentum were to be directed along the (arbitrarily chosen) z-axis, we would require $\hbar^2 j^2$ to be an eigenvalue of $L^2$, which it isn't.

Ah that proves that the state does not exist as well. It seems that this is more of a property of space than of angular momentum.

I can't remember but doesn't p_x commute with the other 2 components of momentum? Now that would not be logical.

 

[espen180]

No, it is a property of angular momentum. This result arises in a purely algebraic context. It is only later applied to a physical theory.

You are correct, the three orthogonal components of the linear momentum commute with each other. A way to see this in the position representation is Clairaut's theorem, assuming the second partial derivatives are continous. (Momentum operators are differential operators in the position representation).
(See http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem).

 

[SpectraCat]

Ah that proves that the state does not exist as well. It seems that this is more of a property of space than of angular momentum.

I can't remember but doesn't p_x commute with the other 2 components of momentum? Now that would not be logical.

No, it's fine for the 3 components of linear momentum to commute with each other. One way to appeciate how the non-commutativity of the angular momentum operators arises is by noticing that angular momentum is the cross product of the non-commuting operators for position and momentum. Thus, it is just another manifestation of the non-commutativity of momentum and position, which is (usually) taken to be one of the postulates of QM.

 

[LostConjugate]

Thus, it is just another manifestation of the non-commutativity of momentum and position, which is (usually) taken to be one of the postulates of QM.

Wow! And it was right in front of me the whole time. px - py

That really clears things up.