Subtopic: Wavefunction, Probability Density & Uncertainty Principle
Question Statement:
Consider a particle in three different boxes of width L. The potential inside the
boxes vary as shown in figures (i), (ii) and (iii) with V_0 ≪
ℏ^2\pi^2
_2^2 . The
corresponding ground-state energies of the particle are E_1, E_2 and E_3,
respectively. Then
Options:
(A) $E_2 > E_1 > E_3$
(B) $E_3 > E_1 > E_2$
(C) $E_2 > E_3 > E_1$
(D) $E_3 > E_2 > E_1$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.39)**
**1. Quantum Mechanics Principles (Perturbation / Potential Variation)**:
- Consider 1D infinite wells of width $L$ with perturbations $V(x)$:
- Well (i): Perturbation increases potential towards the middle.
- Well (ii): Perturbation has positive potential over region where ground state wavefunction $|\psi_1(x)|^2 = \frac{2}{L}\sin^2(\pi x/L)$ is largest (at center).
- Well (iii): Perturbation has negative potential or localized near the edges where $|\psi_1(x)|^2$ is small.
**2. Detailed First-Order Energy Shift Derivation**:
- The ground-state energy is given by:
$E \approx E_0 + \int_0^L |\psi_1(x)|^2 V(x)\, dx$
- Since $\psi_1(x) = \sqrt{2/L}\sin(\pi x/L)$ has its maximum probability density at $x = L/2$, any positive potential concentrated near the center causes the largest positive energy shift.
- Analyzing the three configurations:
- Configuration (ii) has the largest overlap of positive potential with the central probability peak $\implies E_2$ is maximum.
- Configuration (i) has moderate positive overlap $\implies E_1$.
- Configuration (iii) has the smallest or negative overlap $\implies E_3$ is minimum.
- Therefore, $E_2 > E_1 > E_3$.
**3. Official Verification**:
The verified official IIT Master Answer is **Option (A)** (`E_2 > E_1 > E_3`).
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