The problem describes a particle in a one-dimensional infinite potential well with a parabolic potential for \(x > 0\) and an infinite wall at \(x \leq 0\). This is essentially a half harmonic oscillator. Let's solve this step-by-step to find the ratio of the energies between the lowest and the next higher energy levels.
The potential is given as:
\(V(x) = \begin{cases} \frac{1}{2}m\omega^2x^2, & x > 0 \\ \infty, & x \leq 0 \end{cases}\)
Due to the boundary condition at \(x = 0\), only odd energy levels are allowed in this scenario. This results from the requirement that the wave function must be zero at the boundary. Let's break down the energy levels:
The energy levels for a standard harmonic oscillator are given by:
\(E_n = \left(n + \frac{1}{2}\right)\hbar\omega\)
In our modified case (a semi-infinite well), the energy levels correspond to the odd levels of a full harmonic oscillator, i.e., \(n = 1, 3, 5, \ldots\):
- Lowest energy level: \(E_1 = \left(1 + \frac{1}{2}\right)\hbar\omega = \frac{3}{2}\hbar\omega\)
- Next higher energy level: \(E_3 = \left(3 + \frac{1}{2}\right)\hbar\omega = \frac{7}{2}\hbar\omega\)
The ratio of the energies of these two levels is:
\(\frac{E_1}{E_3} = \frac{\frac{3}{2}\hbar\omega}{\frac{7}{2}\hbar\omega} = \frac{3}{7}\)
Therefore, the correct answer is \(\frac{3}{7}\).
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
- The energy $E$ and degeneracy $d$ of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency $\omega$ are
- Young's double slit experiment is performed using a beam of $C_{60}$ (fullerene) molecules, each molecule being made up of 60 carbon atoms. When the slit separation is 50 nm, fringes are formed on a screen kept at a distance of 1 m from the slits. Now, the experiment is repeated with $C_{70}$ molecules with a slit separation of 92.5 nm. The kinetic energies of both the beams are the same. The position of the 4th bright fringe for $C_{60}$ will correspond to the $n^{th}$ bright fringe for $C_{70}$. What is the value of $n$ (rounded off to the nearest integer) ?
- Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \le x \le L$ and $0 \le y \le L$. The wavefunction of the particle is zero only along the line $y = \frac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ isA particle is subjected to a potential
$V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$
Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?
