GATE PH 2022 • Q34 NAT • 1 Mark

Section 2: Classical Mechanics

Small oscillations

Subtopic: Stable Equilibrium & Normal Coordinates

Question Statement:

The wavefunction of a particle in a one-dimensional infinite well of size 2 a at a certain time is 𝜓(x) = 1 √8 √2sin ( \pi ) +√3cos ( \pi 2) + cos ( ^3\pi 2 ). Probability of finding the particle in n = 2 state at that time is ______% (Round off to the nearest integer)

Official Answer Key:

33 to 34

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.34)** **1. Quantum Mechanics Formulation (Infinite Square Well)**: - Potential well of width $2a$ centered at $x = 0$ ($-a \le x \le a$). - Stationary state wavefunctions: - Even states ($n = 1, 3, \dots$): $\psi_n(x) = \frac{1}{\sqrt{a}} \cos\left(\frac{n\pi x}{2a}\right)$ - Odd states ($n = 2, 4, \dots$): $\psi_n(x) = \frac{1}{\sqrt{a}} \sin\left(\frac{n\pi x}{2a}\right)$ - For the first excited state ($n = 2$): $\psi_2(x) = \frac{1}{\sqrt{a}} \sin\left(\frac{2\pi x}{2a}\right) = \frac{1}{\sqrt{a}} \sin\left(\frac{\pi x}{a}\right)$ **2. Detailed Step-by-Step Derivation**: - The given state at time $t$ is: $\psi(x) = \frac{1}{\sqrt{6a}} \left[ \sqrt{2}\sin\left(\frac{\pi x}{a}\right) + \sqrt{3}\cos\left(\frac{\pi x}{2a}\right) + \cos\left(\frac{3\pi x}{2a}\right) \right]$ - Express $\psi(x)$ in terms of the orthonormal basis $\{\psi_n(x)\}$: $\psi(x) = \frac{\sqrt{2}}{\sqrt{6}} \left[\frac{1}{\sqrt{a}}\sin\left(\frac{\pi x}{a}\right)\right] + \frac{\sqrt{3}}{\sqrt{6}} \left[\frac{1}{\sqrt{a}}\cos\left(\frac{\pi x}{2a}\right)\right] + \frac{1}{\sqrt{6}} \left[\frac{1}{\sqrt{a}}\cos\left(\frac{3\pi x}{2a}\right)\right]$ $\psi(x) = \frac{1}{\sqrt{3}} \psi_2(x) + \frac{1}{\sqrt{2}} \psi_1(x) + \frac{1}{\sqrt{6}} \psi_3(x)$ - Check normalization: $\sum |c_n|^2 = \left|\frac{1}{\sqrt{3}}\right|^2 + \left|\frac{1}{\sqrt{2}}\right|^2 + \left|\frac{1}{\sqrt{6}}\right|^2 = \frac{1}{3} + \frac{1}{2} + \frac{1}{6} = \frac{2 + 3 + 1}{6} = 1$ - Probability of finding the particle in the $n = 2$ state: $P(n=2) = |c_2|^2 = \left|\frac{1}{\sqrt{3}}\right|^2 = \frac{1}{3} \approx 33.33\%$ - Rounding to the nearest integer yields **33%**. **3. Official Verification**: The verified official IIT Master Answer range is **33 to 34**.

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