A particle of mass $m$ is subjected to a potential, $V(x, y) = \frac{1}{2} m\omega^2 (x^2 + y^2)$, $- \infty \le x \le \infty$, $- \infty \le y \le \infty$ The state with energy $4\hbar\omega$ is $g$-fold degenerate. The value of $g$ is ________.
Harmonic Oscillator System Identification
The problem involves a particle of mass $m$ in a 2-dimensional potential $V(x, y) = \frac{1}{2} m\omega^2 (x^2 + y^2)$. This represents a 2D isotropic quantum harmonic oscillator.
Energy Level Formula
The energy eigenvalues for a 2D isotropic harmonic oscillator are given by:
$E_{n_x, n_y} = \hbar\omega (n_x + n_y + 1)$
where $n_x, n_y$ are the quantum numbers for the x and y directions, respectively ($n_x, n_y = 0, 1, 2, \dots$).
Let $N = n_x + n_y$. The energy depends only on the sum $N$, so the energy levels can be written as:
$E_N = \hbar\omega (N + 1)$
Calculating Degeneracy
We are asked to find the degeneracy ($g$) of the energy level $E = 4\hbar\omega$. Degeneracy refers to the number of different quantum states that share the same energy level.
- Equate the given energy to the energy formula:
$E_N = \hbar\omega (N + 1) = 4\hbar\omega$
- Solve for the principal quantum number $N$:
$N + 1 = 4$ $N = 3$
- The degeneracy $g$ for a given $N$ in a 2D isotropic harmonic oscillator is the number of ways the sum $N = n_x + n_y$ can be achieved with non-negative integers $n_x, n_y$. This is given by the formula $g = N + 1$.
- Calculate the degeneracy for $N=3$:
$g = 3 + 1 = 4$
The specific states $(n_x, n_y)$ corresponding to $N=3$ are $(0, 3), (1, 2), (2, 1),$ and $(3, 0)$. There are 4 such states, confirming the degeneracy is 4.
Result
The degeneracy $g$ for the energy state $4\hbar\omega$ is 4.
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
- A particle of mass $ m $ is in a potential $ V(x) = \frac{1}{2}m\omega^2x^2 $ for $ x > 0 $ and $ V(x) = \infty $ for $ x \leq 0 $, where $ \omega $ is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
- 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 $\Psi(x, t)$ is
