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Chapter 2: Problem 28

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.

Short Answer

Expert verified
The probability of finding the ground-state harmonic oscillator stretched to a displacement beyond the classical turning point is 84.27%.

Step by step solution

01

Define the problem

First, note that for a simple harmonic oscillator in the ground state, we need to find the probability that the oscillator will have a displacement greater than its turning point for a classical oscillator with the same energy. The probability \( P \) for a quantum mechanical particle to be found at a displacement \( x \) such that \( x > x_{0} \) in the ground state is given by \[ P = \frac{2}{\sqrt{\pi}} \int_{x_0}^{\infty} e^{-x^{2}} dx \] where \( x_{0} \) is the classical turning point. This will be the space to integrate over.
02

Use the error function to solve the integral

Now, we can connect the error function \( \operatorname{erf} (z) \), which is defined as \[ \operatorname{erf} (z) = 1 - \frac{2}{\sqrt{\pi}} \int_{z}^{\infty} e^{-y^{2}} dy \] to the equation from Step 1, so we rewrite the probability as \( P = \operatorname{erf}(x_{0}) \).
03

Evaluate the error function

Finally, we evaluate the error function using \( x_{0}=1 \), the classical turning point, and we get \( P = \operatorname{erf}(1)=0.8427 \).
04

Write the final answer

Therefore, the probability of finding a ground-state harmonic oscillator at a displacement beyond the classical turning point is \( 0.8427 \), or 84.27%.

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Most popular questions from this chapter

Evaluate the matrix elements (a) \(\langle v+1|x| v\rangle\) and (b) \(\left\langle v+2\left|x^{2}\right| v\right\rangle\) of a harmonic oscillator by using the relations given at the bottom of Table 2.1 for the Hermite polynomials.

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).

An electron is confined to a one-dimensional box of length \(L .\) What should be the length of the box in order for its zero-point energy to be equal to its rest mass energy \(\left(m_{\mathrm{e}} c^{2}\right) ?\) Express the result in terms of the Compton wavelength, \(\lambda_{\mathrm{C}}=h / m_{\mathrm{e}} c\)

Energy is required to compress the box when a particle is inside: this suggests that the particle exerts a force on the walls. (a) On the basis that when the length of the box changes by dL the energy changes by \(\mathrm{d} E=-F \mathrm{d} L,\) find an expression for the force. (b) At what length does \(F=1 \mathrm{N}\) when an electron is in the state \(n=1 ?\)

The root mean square deviation of the particle from its mean position is \(\Delta x=\left\\{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}\right\\}^{1 / 2} .\) Evaluate this quantity for a particle in a well and show that it approaches its classical value as \(n \rightarrow \infty\). Hint. Evaluate \(\left\langle x^{2}\right\rangle=\int_{0}^{L} x^{2} \psi^{2}(x) \mathrm{d} x\) In the classical case the distribution is uniform across the box, and so in effect \(\psi(x)=1 / L^{1 / 2}.\)
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