# Zero uncertainity in components of angular momentum in Hydrogen atom

It is given that L and Lz,Lx,Ly commute.(L is total angular momentum, Lx is angular momentum along x axis). So, I can simultaneously know the value of let's say L and Lz. But, if I perform huge no of measurements and in a certain measurement, I get the value of L = Lz, then I know for certain that Lx and Ly are 0. But, according to the uncertainty principle, I can't know the exact values of any two of Lx, Ly and Lz. So, where did I go wrong?

You can't get

$$\left\lVert\vec L\right\rVert = L_z$$

for non-zero $$l$$, since:

$$\left\lVert\vec L\right\rVert = \hbar\sqrt{l(l+1)}$$

while the maximum value of $$L_z$$ is

$$(L_z)_{\mathrm{max}} = \hbar l$$

Also: The maximal state is:

$$Y_l^l(\theta, \phi) \propto \sin^l{\theta}e^{il\phi}$$

which is not an eigenstate of $$L_x$$, nor $$L_z$$.

JEB's answer is correct: you can't do a huge number of measurements so that out of luck in one of them you will find

$$||\vec L|| = L_z$$

You could try another way: since the reference frame is arbitrary, you could just choose it having the z-axis parallel to the angular momentum vector. This won't work either, because for doing so you would have to know where the angular momentum vector is pointing, and this would require simultaneous knowledge of its three components.

• This answer is wrong, since there is no single value of the vector $L$. The state is a superposition of various eigenstates, none of which has a value of $L$. Indeed, since definition of a measurable property is one for which there exists an operator and eigenstates, and there are no eigenstates of $L$, this is not a property a quantum system can have.
– Eddy
Nov 24, 2018 at 17:08
• @Eddy what do you mean there is no single value of $L$? As JEB has pointed out, spherical harmonics are eigenfunctions of $L^2$ with eigenvalue $\hbar l(l+1)$, so you can measure $L$. What you cannot do is measure simultaneously its three components, since they not commute and therefore do not share simultaneous eigenstates. Nov 24, 2018 at 17:22
• As I stated and you also just stated, there is no single value of the vector $L$. However, in your answer, you propose setting up a coordinate system parallel to this vector, which can't be done because it cannot have a well defined value for any state.
– Eddy
Nov 24, 2018 at 17:25
• There is a measurable value for the lenght of the vector $L$. That which you cannot measure is its orientation, and the impossibility of setting up such a coordinate system is precisely my point. Nov 24, 2018 at 17:36
• Guys, I get that mathematically the value of $L_z$ can never attain the value of $L$, but what I don't understand is what would go wrong if we got the same value for $L_z and L$. I mean will any principle of QM be violated by it? Dec 1, 2018 at 18:26

L does not commute with $$L_x$$, $$L^2$$ does. Also, one is a scalar operator, the other is a vector.