Question 705038

The given potential describes a half-harmonic oscillator: $V(x) = \frac{1}{2}\omega^2 x^2$ for $x > 0$, and $\infty$ for $x \le 0$.

Because the potential is infinite for $x \le 0$, the wavefunction must be identically zero at $x = 0$ and for all negative values of $x$. This boundary condition allows only the odd-parity wavefunctions of the standard full harmonic oscillator to exist. Therefore, the allowed quantum numbers from the standard harmonic oscillator ($n_{full}$) are restricted to odd integers: $1, 3, 5, 7, \dots$

The energy levels are given by the standard formula $E = (n_{full} + \frac{1}{2})\hbar\omega$. Let's list the allowed states and their corresponding energies for this half-harmonic oscillator:

• Ground state: corresponds to $n_{full} = 1 \implies E_0 = (1 + 0.5)\hbar\omega = 1.5\hbar\omega$
• First excited state: corresponds to $n_{full} = 3 \implies E_1 = (3 + 0.5)\hbar\omega = 3.5\hbar\omega$
• Second excited state: corresponds to $n_{full} = 5 \implies E_2 = (5 + 0.5)\hbar\omega = 5.5\hbar\omega$
• Third excited state: corresponds to $n_{full} = 7 \implies E_3 = (7 + 0.5)\hbar\omega = 7.5\hbar\omega$

According to the Maxwell-Boltzmann distribution, the number of particles $N_i$ in a state with energy $E_i$ at thermal equilibrium temperature $T$ is proportional to $\exp(-\frac{E_i}{k_B T})$.

The number of particles in the second excited state ($n_2$) is: $n_2 \propto \exp(-\frac{E_2}{k_B T}) = \exp(-\frac{5.5\hbar\omega}{k_B T})$

The number of particles in the third excited state ($n_3$) is: $n_3 \propto \exp(-\frac{E_3}{k_B T}) = \exp(-\frac{7.5\hbar\omega}{k_B T})$

Now, calculate the ratio $n_2 / n_3$: $\frac{n_2}{n_3} = \frac{\exp(-\frac{5.5\hbar\omega}{k_B T})}{\exp(-\frac{7.5\hbar\omega}{k_B T})}$ $\frac{n_2}{n_3} = \exp\left( -\frac{5.5\hbar\omega}{k_B T} - \left( -\frac{7.5\hbar\omega}{k_B T} \right) \right)$ $\frac{n_2}{n_3} = \exp\left( \frac{7.5\hbar\omega - 5.5\hbar\omega}{k_B T} \right)$ $\frac{n_2}{n_3} = \exp\left( \frac{2\hbar\omega}{k_B T} \right)$

Final Answer: The ratio $n_2/n_3$ is $\exp\left(\frac{2\hbar\omega}{k_B T}\right)$. This corresponds to Option (A).