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Chapter 2: Problem 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.

Short Answer

Expert verified
The Hamiltonian operator for the system is \( \hat{H} = -\frac{\hbar^2}{2m}\left(\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}\right) + \frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2} \). The energy levels are given by \(E = (n_x + \frac{1}{2})\hbar\omega_x + (n_y + \frac{1}{2})\hbar\omega_y \)and the ground state energy is \( E_0 = \frac{1}{2}\hbar(\omega_x + \omega_y) \).

Step by step solution

01

Hamiltonian operator

The Hamiltonian operator represents the total energy of the quantum system and can be written as the sum of kinetic and potential energy. In this case, the kinetic energy operator is \(-\frac{\hbar^2}{2m}\) times the Laplacian, since we are dealing with two dimensions (x and y), and the potential energy from the problem is \(V(x, y)=\frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2}\). So the Hamiltonian operator, \( \hat{H} \), is \[ \hat{H} = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2} = -\frac{\hbar^2}{2m}\left(\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}\right) + \frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2} \] .
02

Energy Levels

The energy levels of the two-dimensional harmonic oscillator are made up of the sum of energy levels of each independent one-dimensional oscillator. Therefore, the energy of a state with quantum numbers \(n_x\) and \(n_y\) (where \(k_x\) and \(k_y\) are the force constants in the x and y directions, respectively) is: \[ E = (n_x + \frac{1}{2})\hbar\omega_x + (n_y + \frac{1}{2})\hbar\omega_y \] with \(\omega_x =\sqrt{\frac{k_x}{m}}, \omega_y =\sqrt{\frac{k_y}{m}} \) are the frequencies of the oscillators in the x and y directions.
03

Ground State Energy

The ground state energy (the lowest energy state) is found when \(n_x = 0\) and \(n_y = 0\) in the energy level expression. So, the ground state energy \(E_0\) is: \[ E_0 = (\frac{1}{2})\hbar \omega_x + (\frac{1}{2})\hbar \omega_y = \frac{1}{2}\hbar(\omega_x + \omega_y) \] where \(\omega_x =\sqrt{\frac{k_x}{m}}, \omega_y =\sqrt{\frac{k_y}{m}} \) are the frequencies of the oscillators in x and y directions.

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Most popular questions from this chapter

A particle was prepared travelling to the right with all momenta between \(\left(k-\frac{1}{2} \Delta k\right) \hbar\) and \(\left(k+\frac{1}{2} \Delta k\right) \hbar\) contributing equally to the wavepacket. Find the explicit form of the wavepacket at \(t=0,\) normalize it, and estimate the range of positions, \(\Delta x,\) within which the particle is likely to be found. Compare the last conclusion with a prediction based on the uncertainty principle. Hint. Use eqn 2.13 with \(g=B\) a constant, inside the range \(k-\frac{1}{2} \Delta k\) to \(k+\frac{1}{2} \Delta k\) and zero elsewhere, and eqn 2.12 with \(t=0\) for \(\Psi_{k} .\) To evaluate \(\int\left|\Psi_{k}\right|^{2} \mathrm{d} \tau\) (for the normalization step) use the integral \(\int_{-\infty}^{\infty}(\sin x / x)^{2} \mathrm{d} x=\pi .\) Take \(\Delta x\) to be determined (numerically) by the locations where \(|\Psi|^{2}\) falls to half its value at \(x=0\) For the last part use \(\Delta p_{x} \approx \hbar \Delta k\)

Determine the probability of finding the ground-state harmonic oscillator stretched to a displacement beyond the classical turning point. Hint. Relate the expression for the probability to the error function, erf \(z,\) defined as \\[ \operatorname{erf} z=1-\frac{2}{\pi^{1 / 2}} \int_{z}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d} y \\] and evaluate it using erf \(1=0.8427 .\) The error function is incorporated into most mathematical software packages.

Calculate the values of (a) \(\langle x\rangle\) (b) \(\left\langle x^{2}\right\rangle\) \(,(\mathrm{c})\left\langle p_{x}\right\rangle,(\mathrm{d})\left\langle p_{x}^{2}\right\rangle\) for a harmonic oscillator in its ground state by evaluation of the appropriate integrals (as in Problems \(2.13-2.15\) ). Examine the value of \(\Delta x \Delta p_{x}\) in the light of the uncertainty principle. Hint. Use the integrals \\[ \begin{array}{l} \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^{1 / 2} \\ \int_{0}^{\infty} x \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2 \alpha} \\\ \int_{-\infty}^{\infty} x^{2} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2}\left(\frac{\pi}{\alpha^{3}}\right)^{1 / 2} \end{array} \\]

Locate the nodes of the harmonic oscillator wavefunction for the state with \(v=6 .\) Hint. Use mathematical software.

Demonstrate that accidental degeneracies can exist in a rectangular infinite square-well potential provided that the lengths of the sides are in a rational proportion, that is when \(L_{1}=\lambda L_{2},\) with \(\lambda\) an integer. What is the degeneracy?
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