Question

A particle of mass 'm' is located at the vector position r and has a linear momentum p. The vector r and p are nonzero.if the particle moves only in y-z plane.Prove that Ly=Lz=0 and Lx is not equal to zero.

Answer 1

Answer on Question #67644- Physics / Quantum Mechanics

A particle of mass 'm' is located at the vector position $r$ and has a linear momentum $p$. The vector $r$ and $p$ are nonzero if the particle moves only in $y-z$ plane. Prove that $Ly = Lz = 0$ and $Lx$ is not equal to zero.

Solution:

In the case when particle moves only in $y-z$ plane

$$\mathbf{r} = (0, y, z),$$ $$\mathbf{p} = (0, p_y, p_z).$$

By definition the angular momentum is given by

$$\mathbf{L} = [\mathbf{r} \times \mathbf{p}]$$

The components of angular momentum

$$L_x = y p_z - z p_y \neq 0,$$ $$L_y = z p_x - x p_y = z \cdot 0 - 0 \cdot p_y = 0,$$ $$L_z = x p_y - y p_x = 0 \cdot p_y - y \cdot 0 = 0.$$

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