2.32 Consider a harmonic oscillator of mass \(m\) undergoing harmonic motion in two dimensions \(x\) and \(y .\) The potential energy is given by \(V(x, y)=\frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2}\). (a) Write down the expression for the hamiltonian operator for such a system. (b) What is the general expression for the allowable energy levels of the two- dimensional harmonic oscillator? (c) What is the energy of the ground state (the lowest energy state)? Hint. The hamiltonian operator can be written as a sum of operators. 2.33 Consider a particle of mass \(1.00 \times 10^{-25} \mathrm{g}\) moving freely in a three-dimensional cubic box of side \(10.00 \mathrm{nm}\) The potential energy is zero inside the box and is infinite at the walls and outside of the box. (a) Evaluate the zero-point energy of the particle. (b) Consider the energy level that has an energy 9 times greater than the zero-point energy. What is the degeneracy of this level? Identify all the sets of quantum numbers that correspond to this energy. (The energy levels of the cubic box were deduced in Problem \(2.18 .\) (c) Compute the wavelength, frequency, and wavenumber of the photon responsible for the transition from the ground state of the particle to the energy level of part (b).

Step by step solution

01

Solving 2.32 - Part (a): Hamiltonian operator for harmonic oscillator

The Hamiltonian Operator \(\hat H\) represents the total energy of a quantum system. This is expressed as the sum of kinetic energy \(T\) and potential energy \(V\). Here, potential energy \(V\) is provided in the question and kinetic energy \(T\) of a particle is represented as \(\frac{p^2}{2m}\), where \(p\) is momentum and \(m\) is mass.\nSo, Hamiltonian Operator, \(\hat H = T + V = -\frac{\hbar^2}{2m} \Delta + \frac{1}{2} k_{x} x^{2} + \frac{1}{2} k_{y} y^{2}\)

02

Solving 2.32 - Part (b): Energy levels of 2D Harmonic Oscillator

The general energy formula for a 2D harmonic oscillator with quantum numbers \(n_x\) and \(n_y\) can be found by finding the energy of the system in x and y dimensions separately and then adding them together: \(E = (n_x + \frac{1}{2})\hbar\omega_x + (n_y + \frac{1}{2})\hbar\omega_y\)

03

Solving 2.32 - Part (c): Ground State Energy of 2D Harmonic Oscillator

The ground state is the lowest energy state of a quantum mechanical system which occurs when the quantum number is at its minimum value '0'. Therefore, the energy of ground state can be found by substituting \(n_x = 0, n_y = 0\) into the energy formula:\n\(E_{ground} = (0 + \frac{1}{2})\hbar\omega_x + (0 + \frac{1}{2})\hbar\omega_y = \frac{1}{2}\hbar\omega_x + \frac{1}{2}\hbar\omega_y\)

04

Solving 2.33 - Part (a): Zero-point energy

Zero-point energy is the lowest possible energy that a quantum mechanical physical system may have. For a cubic box, the zero-point energy is equal to the energy of the first excited state. As our box is three-dimensional, the energy is expressed as:\n\(E_1 = \frac{h^2}{8mL^2}(1^2+1^2+1^2) = \frac{3h^2}{8mL^2}\), where \(L = 10x10^{-9} m\) is the side of the box and the mass \(m = 1x10^-25 g = 1x10^-28 kg\).

05

Solving 2.33 - Part (b): Energy level and degeneracy

An energy level is said to have degeneracy if it corresponds to two or more different measurable states of a quantum system. In our case, the energy level that has an energy 9 times greater than the zero-point energy will have quantum numbers (n,l,m) combinations for n=3. The sets are (3,0,0), (2,1,-1), (2,1,0), (2,1,1), (1,2,-2), (1,2,-1), (1,2,0), (1,2,1), (1,2,2). So, the degeneracy of this level is 9.

06

Solving 2.33 - Part (c): Wavelength, Frequency, and Wavenumber

The energy of the photon responsible for the transition can be calculated using the energy of the two states. The difference in energy is given by \(\Delta E = E_n - E_1 = E_n - 9E_1\). The wavelength of the photon responsible for this transition can be found using the energy-frequency-wavelength relation in quantum mechanics. Wavelength, \(\lambda = \frac{hc}{\Delta E}\). With the wavelength, we can then find the frequency, \(f = \frac{c}{\lambda}\), and the wavenumber, \(\nu = \frac{1}{\lambda}\).

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