# Bizzard measurement outcome of Angular Momentum, Why?

1. Nov 11, 2009

### wam_mi

Hi there,

I am just a bit confused about the measurement of angular momentum.
If we have a state which we know for sure the outcome of the measurement of L(z), the z-component of the angular momentum L, say h. Why is it that all the other components, say L(x) would not reveal itself to have a definite value, but instead there is a probability (but not for sure) of getting one of the eigenvalues of L(x)....

This concept is very bizzard and I don't really know what I'm talking about... has it got something to do with the uncertainty principle and the commutation relations between the two observables?

Thank you!

2. Nov 11, 2009

### zenith8

As I wrote in response to some thread the other day:

The answer to this in orthodox QM is not defined, since it rejects as meaningless any attempt to understand the reality of a quantum event. You simple have to accept the mathematics and 'shut up and calculate'.

However, in the de Broglie-Bohm interpretation of QM (merely a minor shift in perspective), the answer is straightforward. Particles have trajectories, pushed around by an objectively existing wave field represented mathematically by the wave function. And orbital angular momentum really is the angular momentum of particles 'orbiting' the nucleus. Of course this is a bit difficult if you believe that particles only exist when you look at them (as in the orthodox view).

Why is it quantized? In de Broglie-Bohm theory quantities are well-defined and continuously variable for all quantum states - values for the subset of wave functions which are eigenstates of some operator have no fundamental physical significance. So this characteristic feature of QM - the existence of discrete energy levels - is due to the restriction of a basically continuous theory to motion associated with a subclass of eigenfunctions.

Now what happens with angular momentum? Let's say we prepare the system with a wave function which is an eigenfunction of the $${\hat L}_z$$ operator (where the direction of the z-axis is arbitrary). Then the z component of the angular momentum will coincide with the eigenvalue of $${\hat L}_z$$ , and the total orbital angular momentum with the eigenvalue of $${\hat L}^2$$ . Conventionally one would say that the x- and y- components are 'undefined'. However, in de Broglie-Bohm they are perfectly well-defined. In fact you have:

$$L_x= -m \hbar \cot \theta \cos \theta$$

$$L_y = m\hbar \cot \theta \sin \theta$$

$$L_z = m\hbar$$

So the difference is that along the trajectory $$L_z$$ and the total angular momentum are conserved, but $$L_x$$ and $$L_y$$ are not. Thus, although the classical force is central the motion is asymmetrical, which reflects the fact that a particular direction in space has been singled out as the quantization axis.

And if you'll forgive me correcting your English, I believe the word is 'bizarre' not 'bizzard'.