How Does Angular Momentum Uncertainty Affect Angular Position?

[neelakash]

Homework Statement

It is a homework question and I solved it in the following way:
To Prove that (∆θ)(∆L)~(ћ/2) ;For what uncertainty of L will the angular position of the particle would be indeterminate?

Homework Equations

The Attempt at a Solution

I considered a particle of mass m moving in a circle of constant radius r at speed v.Then,classically,L=mvr.

Clearly,from uncertainty principle, ∆s ∆p~(ћ/2)
=> (∆s/r)[(∆p)r]~(ћ/2)
=> (∆θ)(∆L)~(ћ/2)
What I am worried if this particular case of a circle will do;and the second question.I think it is ∆L=0 which makes ∆θ indeterminate...But I am not sure...

Please check if this is correct and rectify if I am wrong.

 

[Jhero]

Your solution is correct. The uncertainty principle states that the product of the uncertainty in position (∆s) and the uncertainty in momentum (∆p) must be greater than or equal to the reduced Planck's constant (ћ/2). In this case, we can rewrite the equation as (∆s/r)(∆p)r~(ћ/2) to account for the circular motion of the particle.

For the second question, you are correct in saying that ∆L=0 would make ∆θ indeterminate. This is because if there is no uncertainty in the angular momentum (∆L=0), then the uncertainty in the angular position (∆θ) would be infinite. This can be seen from the equation (∆θ)(∆L)~(ћ/2), where if ∆L=0, then (∆θ)~(ћ/2)/0, which is undefined.

I hope this helps clarify any doubts you had. Keep up the good work in your studies! (Scientist)