Discussion
If a particle has the wave function \( \psi = e^{ikz} \), the z-component of its angular momentum is :
Answer: B
The \( z \)-component of angular momentum is zero.
In cylindrical coordinates \( L_{z} = -i\hbar \frac{\partial}{\partial \phi} \). The function \( e^{ikz} \) depends only on \( z \), not on the azimuthal angle \( \phi \), so
\[ \begin{aligned} L_{z}\, e^{ikz} &= -i\hbar \frac{\partial}{\partial \phi} e^{ikz} \\ &= 0 \end{aligned} \]
The state is an eigenstate of \( L_{z} \) with eigenvalue 0. \( \hbar k \) is the trap: it is the eigenvalue of the linear momentum \( p_{z} \), not of the angular momentum. The imaginary values cannot be eigenvalues of a Hermitian operator.
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