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# 7.2: Wave functions

In general, an even function times an even function produces an even function. A simple example of an even function is the product $$x^2e^{-x^2}$$ (even times even is even). Similarly, an odd function times an odd function produces an even function, such as x sin x (odd times odd is even). However, an odd function times an even function produces an odd function, such as $$x^2e^{-x^2}$$ (odd times even is odd). The integral over all space of an odd function is zero, because the total area of the function above the x-axis cancels the (negative) area below it. As the next example shows, this property of odd functions is very useful.

Exercise $$\PageIndex{2}$$: Expectation Value (Part I)

The normalized wavefunction of a particle is

$\psi(x) = e^{-|x|/x_0} /\sqrt{x_0}. \nonumber$

Find the expectation value of position.

Strategy

Substitute the wavefunction into Equation \ref{7.7} and evaluate. The position operator introduces a multiplicative factor only, so the position operator need not be “sandwiched.”

Solution

First multiply, then integrate:

\begin{align*} \langle x \rangle &= \int_{-\infty}^{\infty} dx\,x|\psi(x)|^2 \nonumber \\[4pt] &= \int_{-\infty}^{\infty} dx\, x|\dfrac{e^{-|x|/x_0}}{\sqrt{x_0}}|^2 \nonumber \\[4pt] &= \dfrac{1}{x_0} \int_{-\infty}^{\infty} dx\, xe^{-2|x|/x_0} \nonumber \\[4pt] &= 0. \nonumber \end{align*}

Significance

The function in the integrand ($$xe^{-2|x|/x_0}$$) is odd since it is the product of an odd function (x) and an even function ($$e^{-2|x|/x_0}$$). The integral vanishes because the total area of the function about the x-axis cancels the (negative) area below it. The result ($$\langle x \rangle = 0$$) is not surprising since the probability density function is symmetric about $$x = 0$$.

Exercise $$\PageIndex{3}$$: Where Is the Ball? (Part II)

The time-dependent wavefunction of a particle confined to a region between 0 and L is

$\psi(x,t) = A \, e^{-i\omega t} \sin \, (\pi x/L) \nonumber$

where $$\omega$$ is angular frequency and $$E$$ is the energy of the particle. (Note: The function varies as a sine because of the limits (0 to L). When $$x = 0$$, the sine factor is zero and the wavefunction is zero, consistent with the boundary conditions.) Calculate the expectation values of position, momentum, and kinetic energy.

Strategy

We must first normalize the wavefunction to find A. Then we use the operators to calculate the expectation values.

Solution

Computation of the normalization constant:

\begin{align*} 1 &= \int_0^L dx\, \psi^* (x) \psi(x) \nonumber \\[4pt] &= \int_0^L dx \, \left(A e^{+i\omega t} \sin \, \dfrac{\pi x}{L}\right) \left(A e^{-i\omega t} \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= A^2 \int_0^L dx \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= A^2 \dfrac{L}{2} \nonumber \\[4pt] \Rightarrow A &= \sqrt{\dfrac{2}{L}}. \nonumber \end{align*}

The expectation value of position is

\begin{align*}\langle x \rangle &= \int_0^L dx \, \psi^* (x) x \psi(x) \nonumber \\[4pt] &= \int_0^L dx \, \left(A e^{+i\omega t} \sin \, \dfrac{\pi x}{L}\right) x \left(A e^{-i\omega t} \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= A^2 \int_0^L dx\,x \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= A^2 \dfrac{L^2}{4} \nonumber \\[4pt] \Rightarrow A &= \dfrac{L}{2}. \nonumber \end{align*}

The expectation value of momentum in the x-direction also requires an integral. To set this integral up, the associated operator must— by rule—act to the right on the wavefunction $$\psi(x)$$:

\begin{align*} -i\hbar\dfrac{d}{dx} \psi(x) &= -i\hbar \dfrac{d}{dx} Ae^{-i\omega t}\sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= - i\dfrac{Ah}{2L} e^{-i\omega t} \cos\, \dfrac{\pi x}{L}. \nonumber \end{align*}

Therefore, the expectation value of momentum is

\begin{align*} \langle p \rangle &= \int_0^L dx \left(Ae^{+i\omega t}sin \dfrac{\pi x}{L}\right)\left(-i \dfrac{Ah}{2L} e^{-i\omega t} cos \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= -i \dfrac{A^2h}{4L} \int_0^L dx \, \sin \, \dfrac{2\pi x}{L} \nonumber \\[4pt] &= 0. \nonumber \end{align*}

The function in the integral is a sine function with a wavelength equal to the width of the well, L—an odd function about $$x = L/2$$. As a result, the integral vanishes.

The expectation value of kinetic energy in the x-direction requires the associated operator to act on the wavefunction:

\begin{align} -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2} \psi (x) &= - \dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} Ae^{-i\omega t} \, \sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= - \dfrac{\hbar^2}{2m} Ae^{-i\omega t} \dfrac{d^2}{dx^2} \, \sin \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= \dfrac{Ah^2}{2mL^2} e^{-i\omega t} \, \sin \, \dfrac{\pi x}{L}. \nonumber \end{align} \nonumber

Thus, the expectation value of the kinetic energy is

\begin{align*} \langle K \rangle &= \int_0^L dx \left( Ae^{+i\omega t} \, \sin \, \dfrac{\pi x}{L}\right) \left(\dfrac{Ah^2}{2mL^2} e^{-i\omega t} \, \sin \, \dfrac{\pi x}{L}\right) \nonumber \\[4pt] &= \dfrac{A^2h^2}{2mL^2} \int_0^L dx \, \sin^2 \, \dfrac{\pi x}{L} \nonumber \\[4pt] &= \dfrac{A^2h^2}{2mL^2} \dfrac{L}{2} \nonumber \\[4pt] &= \dfrac{h^2}{2mL^2}. \end{align*}

Significance

The average position of a large number of particles in this state is $$L/2$$. The average momentum of these particles is zero because a given particle is equally likely to be moving right or left. However, the particle is not at rest because its average kinetic energy is not zero. Finally, the probability density is

$|\psi|^2 = (2/L) \, \sin^2 (\pi x/L). \nonumber$

This probability density is largest at location $$L/2$$ and is zero at $$x = 0$$ and at $$x = L$$. Note that these conclusions do not depend explicitly on time.

Exercise $$\PageIndex{3}$$

For the particle in the above example, find the probability of locating it between positions $$0$$ and $$L/4$$.

$$(1/2 - 1/\pi) /2 = 9\%$$

Quantum mechanics makes many surprising predictions. However, in 1920, Niels Bohr (founder of the Niels Bohr Institute in Copenhagen, from which we get the term “Copenhagen interpretation”) asserted that the predictions of quantum mechanics and classical mechanics must agree for all macroscopic systems, such as orbiting planets, bouncing balls, rocking chairs, and springs. This correspondence principle is now generally accepted. It suggests the rules of classical mechanics are an approximation of the rules of quantum mechanics for systems with very large energies. Quantum mechanics describes both the microscopic and macroscopic world, but classical mechanics describes only the latter.

## Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).