Question

H is the Hamiltonian, $\vec{L}$ the orbital angular momentum, and $L_z$ is the z-component of $\vec{L}$. The 1S state of the hydrogen atom in the non-relativistic formalism is an eigenfunction of which one of the following sets of operators?

• A. $H, L^{2}$ and $L_{z}$
• B. $H, \vec{L}, L^{2}$ and $L_{z}$
• C. $L^{2}$ and $L_{z}$ only
• D. $H$ and $L_{z}$ only

Hint:

For spherically symmetric states like the 1S state of hydrogen, the quantum numbers $l$ and $m_l$ completely define the state, and only $L^2$ and $L_z$ are relevant operators.

Answer 1

The Correct Option is C

The 1S state of the hydrogen atom refers to the quantum state with $n=1$, $l=0$, and $m_l=0$.
- In this state, the orbital angular momentum $L$ is zero, meaning it is an eigenfunction of $L^2$ and $L_z$ with eigenvalues $l(l+1) = 0$ and $m_l = 0$, respectively.
- The Hamiltonian $H$ is also involved in the description, but the 1S state is an eigenfunction of $L^2$ and $L_z$ only, because $L$ is zero in this state. Therefore, the correct answer is (C).