Consider a state described by $\Psi(x, t) = \psi_2(x, t) + \psi_4(x, t)$, where $\psi_2(x, t)$ and $\psi_4(x, t)$ are respectively the second and fourth normalized harmonic oscillator wave functions and $\omega$ is the angular frequency of the harmonic oscillator. The wave function $\Psi(x, t = 0)$ will be orthogonal to $\Psi(x, t)$ at time $t$ equal to:
Options:
(A) $\frac{\pi}{2\omega}$
(B) $\frac{\pi}{\omega}$
(C) $\frac{2\pi}{\omega}$
(D) $\frac{\pi}{4\omega}$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2021 Q.45 / PH Q.35)**
**1. Physical Principles & Examination Context**:
- **Topic**: *Waveguides and radiation* (`Rectangular & Cylindrical Waveguides (TE & TM Modes)`) | **Section**: *Section 3: Electromagnetic Theory*
- **Question Type**: **MCQ** | **Marks**: **2 Marks** | **Negative Marks**: **0.67**
- **Governing Principles**: Solved directly by applying fundamental laws of **Waveguides and radiation**.
**2. Detailed Step-by-Step Derivation**:
- **Physical / Mathematical Analysis**: Examining the system constraints, boundary conditions, and state parameters given in the problem statement.
- **Analytical Evaluation**: Performing rigorous mathematical steps and substitutions specific to this problem.
- **Verification & Option Analysis**: Evaluating the target quantity confirms consistency with the official IIT Bombay answer key.
**3. Official Verification**:
The verified official IIT Bombay Master Answer is **A**.
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