Question

Knowing precisely all components of a nonzero $\overrightarrow{L}$would violate the uncertainty principle, but knowingthat $\overrightarrow{L}$is precisely zerodoes not. Why not?

(Hint:For $l=0$ states, the momentum vector $\overrightarrow{p}$ is radial.)

Short answer

Knowingthat $\overrightarrow{L}$is precisely zerodoes not violate uncertainty principle because the components commute in this case making them measurable simultaneously.

Step-by-step solution

Angular momentum commutation relation

The angular momentum components follow the commutation relation

$$\begin{gathered} \left[L_i,L_j\right]=i\hslash L_k \\ \left[L_j,L_k\right]=i\hslash L_i \\ \left[L_k,L_i\right]=i\hslash L_j \end{gathered}$$ ..... (I)

Here, $\hslash$ is the reduced Planck's constant.

Step 2:Determining the commutation for zero angular momentum

Observables which do not commute cannot be measured simultaneously. Thus it is evident from equation (I) that components of angular momentum cannot be measured simultaneously. But if all the components are measured to be zero, the right hand sides of equations (I) become zero and the components commute. Thus simultaneous measurement of angular momentum components if the values are zero does not violate uncertainty principle.