How many degrees of freedom does angular momentum have?

[Ahmes]

Angular momentum is a vector, so alegedly it has 3 degrees of freedom.

It has never been formally told me, but I noticed angular momentum is taken as two separate magnitudes and not three. i.e. in quantum mechanics there's an operator for $\bf{L}^2$ and for $L_z$ and this should be enough.

My question is whether knowing a vector's magnitude (it's "absolute value") and one of it's components is sufficient to determine the other two components (i.e. $L_x$ and $L_y$)
I guess that for a general vector this would be a NO. but specifically angular momentum is a cross product... and I think it matters.

Thanks in advance.

 

[Physics Monkey]

Let's talk first about classical vectors. You always need three numbers to specify a three dimensional vector. That's all there is to it. If you just draw a picture you can easily convince yourself that fixing the magnitude and one component of a vector leaves quite a lot of freedom, in other words it doesn't determine the vector completely.

Now, on to quantum mechanics. Remember the key to describing possible states is to find as many commuting operators as you can. But the components of angular momentum do not all commute with each other. In practice this means that there is no state with definite values for all three angular momentum components. In general the best you can do is to label states by their total angular momentum and by their angular momentum along a single direction. Let me emphasize again that this does not mean you know all three components of the angular momentum in such a state. You don't.

 

[Tide]

Angular momentum or rotation correspond to two degrees of freedom because, given the magnitude, of the angular momentum you can specify the direction with only two numbers - a polar angle and azimuthal angle. I.e., the object has freedom to rotate in two directions at once.

 

[Ahmes]

Ok. Two different things said here.
1. In quantum mechanics things go wacky, and as $\left|\ell,m\right\rangle$ are just the base to all states, it doesn't contradict having 3 degrees of freedom.
2. Classical angular momentum does cocorrespond to only two degrees of freedom.

I didn't understand you explanation, Tide, to why it is so. If we have a magnitude and two angles - those are three numbers.

But yet, an n-dimensional system should have 2n-1 constants of motion (right? I'm not 100% sure). And in a free body in 3D space we have 1 energy, 3 linear momenums and 3 angular momentums, thats already 7 constants when alegedly 5 is the maximum...

 

[Tide]

Yes, I am thinking "classically" but you don't consider magnitudes per se when dealing with degrees of freedom which was the context of your question. E.g., we talk about three spatial degrees of freedom with regard to translational motion. We do that because there are three orthogonal spatial dimensions in which an unconstrained particle can move but we don't ask for the magnitude of the particle's velocity or its spatial displacement in determining the number of degrees of freedom.