A particle of mass m is in a potential V(x) =
1
2 m\omega2x2 for x > 0 and V(x) = ∞ for
x ≤ 0, where \omega is the angular frequency. The ratio of the energies corresponding
to the lowest energy level to the next higher level is
Options:
(A) 3
7
(B) 1
2
(C) 1
3
(D) 3
5
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.13)**
**1. Theoretical Framework (Half-Harmonic Oscillator)**:
- **Potential Profile**:
$V(x) = \begin{cases} \frac{1}{2}m\omega^2 x^2 & x > 0 \\ \infty & x \le 0 \end{cases}$
- **Boundary Condition**: The infinite potential wall at $x \le 0$ demands that the wavefunction vanishes identically at the origin:
$\psi(0) = 0$
**2. Step-by-Step Derivation**:
- The solutions to the full simple harmonic oscillator (SHO) are:
$E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n = 0, 1, 2, 3, \dots$
where even $n$ eigenfunctions have even parity (non-zero at origin $\psi_n(0) \ne 0$) and odd $n$ eigenfunctions have odd parity (satisfying $\psi_n(0) = 0$).
- Because of the infinite barrier at $x = 0$, only the **odd-parity states** ($n = 1, 3, 5, 7, \dots$) are physically admissible.
- Step 1: Lowest energy level (ground state, corresponding to full SHO $n = 1$):
$E_0' = E_1 = \left(1 + \frac{1}{2}\right)\hbar\omega = \frac{3}{2}\hbar\omega$
- Step 2: First excited energy level (corresponding to full SHO $n = 3$):
$E_1' = E_3 = \left(3 + \frac{1}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega$
- Step 3: Calculate the energy ratio:
$\frac{E_0'}{E_1'} = \frac{\frac{3}{2}\hbar\omega}{\frac{7}{2}\hbar\omega} = \frac{3}{7}$
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`3/7`).
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