Which of the following option(s) is/are correct for the ground state of a hydrogen
atom?
Options:
(A) Linear Stark effect is zero
(B) It has definite parity
(C) Spin-orbit coupling is zero
(D) Hyperfine splitting is zero
Official Answer Key:
A;B;C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.24)**
**1. Quantum Mechanical Properties of Hydrogen Ground State**:
- Ground state designation: $|1s\rangle$ with quantum numbers:
$n = 1, \quad l = 0, \quad m_l = 0, \quad s = 1/2$
**2. Analysis of Each Statement (MSQ)**:
- **(A) Linear Stark effect is zero [TRUE]**:
- The first-order Stark shift is $\Delta E^{(1)} = -e \langle 1s | z | 1s \rangle$.
- Because the $n = 1$ level has only a single orbital state with $l = 0$, there are no opposite-parity degenerate states at $n=1$. The matrix element of the odd-parity dipole operator $z$ between identical parity states vanishes identically.
- **(B) It has definite parity [TRUE]**:
- The spatial parity operator eigenvalue is $\Pi = (-1)^l = (-1)^0 = +1$ (even parity). The state has well-defined even parity.
- **(C) Spin-orbit coupling is zero [TRUE]**:
- Spin-orbit Hamiltonian $H_{SO} \propto \vec{L} \cdot \vec{S}$.
- For an $s$-orbital, $l = 0 \implies \vec{L} = 0$, so $\vec{L} \cdot \vec{S} = 0$. The spin-orbit coupling vanishes identically.
- **(D) Hyperfine splitting is zero [FALSE]**:
- Hyperfine interaction arises from contact interaction between electron spin $\vec{S}$ and nuclear spin $\vec{I}$ via Fermi contact term $\propto |\psi(0)|^2 \vec{I} \cdot \vec{S}$.
- For $l = 0$, $|\psi(0)|^2 > 0$ is non-zero, giving rise to the famous $21\text{ cm}$ ground state hyperfine splitting into $F = 1$ and $F = 0$.
**3. Official Verification**:
The verified official IIT answer is **Options (A), (B), and (C)** (`['A', 'B', 'C']`).
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