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Chapter 7: Problem 64

Find the expectation value \(\left\langle x^{2}\right\rangle\) of the square of the position for a quantum harmonic oscillator in the ground state. Note: \(\int_{-\infty}^{+\infty} d x x^{2} e^{-a x^{2}}=\sqrt{\pi}\left(2 a^{3 / 2}\right)^{-1}\)

Short Answer

Expert verified
The expectation value of the square of the position for a quantum harmonic oscillator in the ground state is \(\frac{\hbar}{2m\omega}\).

Step by step solution

01

State the ground-state wave function

For a one-dimensional harmonic oscillator, the ground state wave function, \(\psi(x)\), is given by \(\psi(x)=\left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^{2}}\), where \(m\omega/\hbar\) is \(a/2\) in this case. Thus, \(a=2m\omega/\hbar\).
02

Write the expectation value of \(x^{2}\) for the oscillator

The expectation value of \(x^{2}\) for the oscillator is defined as \( \left\langle x^{2}\right\rangle = \int_{-\infty}^{\infty} \psi^{*}(x) x^{2} \psi(x) dx\). Substituting \(\psi(x)\) in the equation, it's \(\left\langle x^{2}\right\rangle = \int_{-\infty}^{\infty} \left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{2}} e^{-\frac{m\omega}{\hbar}x^{2}}x^{2} dx\). Notice the integrand matches the given integral with \(a=2m\omega/\hbar)\).
03

Use the given integral formula to compute the expectation value

Substituting \(a=2m\omega/\hbar\) into the provided integral formula, we get \(\int_{-\infty}^{\infty} dx x^{2} e^{-\frac{m\omega}{\hbar}x^{2}} = \sqrt{\pi}((4m^{2}\omega^{2}/\hbar^{2})^{3/2})^{-1}\). Plug this result into the expectation value, we have \(\left\langle x^{2}\right\rangle = \left(\frac{m\omega}{\pi \hbar}\right)^{\frac{1}{2}} \times \sqrt{\pi}\left(\frac{\hbar^{2}}{4m^{2}\omega^{2}}\right)^{\frac{3}{2}}\).
04

Simplify the final result

Also, note that you should keep in mind that simplifying this results in \(\left\langle x^{2}\right\rangle = \frac{\hbar}{2m\omega}\). To reach this simplified result, remember to apply the properties of exponents and roots, and perform algebraic simplification.

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Most popular questions from this chapter

Use Heisenberg's uncertainty principle to estimate the ground state energy of a particle oscillating on an spring with angular frequency, \(\omega=\sqrt{k / m},\) where \(k\) is the spring constant and \(m\) is the mass.

Suppose an electron is confined to a region of length \(0.1 \mathrm{nm}\) (of the order of the size of a hydrogen atom) and its kinetic energy is equal to the ground state energy of the hydrogen atom in Bohr's model (13.6 eV). (a) What is the minimum uncertainty of its momentum? What fraction of its momentum is it? (b) What would the uncertainty in kinetic energy of this electron be if its momentum were equal to your answer in part (a)? What fraction of its kinetic energy is it?

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