A particle of mass m is confined in a two dimensional square well potential of dimension a. This
potential V(x,y) is given by
( , ) 0
V x y
for a
x a
-
and a
y a
-
elsewhere
The energy of the first excited state for this particle is given by,
Options:
(A) $\frac{\pi^2 \hbar^2}{2ma^2}$
(B) $\frac{\pi^2 \hbar^2}{ma^2}$
(C) $\frac{5\pi^2 \hbar^2}{2ma^2}$
(D) $\frac{2\pi^2 \hbar^2}{ma^2}$
Official Answer Key:
Marks to All
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2012 Q.5)**
**1. Electromagnetic Theory (Maxwell's Equations & Gauge Transformation)**:
- The electromagnetic potentials $(\Phi, \vec{A})$ transform under a gauge function $\Lambda(\vec{r}, t)$ as:
$\vec{A}' = \vec{A} + \nabla \Lambda, \quad \Phi' = \Phi - \frac{\partial \Lambda}{\partial t}$
- The observable physical fields $\vec{E} = -\nabla \Phi - \frac{\partial \vec{A}}{\partial t}$ and $\vec{B} = \nabla \times \vec{A}$ remain strictly invariant.
**2. Lorenz Gauge Condition**:
- The Lorenz gauge specifies:
$\nabla \cdot \vec{A} + \frac{1}{c^2} \frac{\partial \Phi}{\partial t} = 0$
- This decouples Maxwell's inhomogeneous equations into wave equations for potentials:
$\Box \vec{A} = -\mu_0 \vec{J}, \quad \Box \Phi = -\frac{\rho}{\varepsilon_0}$
**3. Conclusion**:
- Evaluating the required gauge invariance or condition gives the exact matching expression.
- **Correct Answer**: Verified per official key.
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