# 7.1 Wave functions  (Page 7/22)

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## Expectation value (part ii)

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

$\psi \left(x,t\right)=A{e}^{\text{−}i\omega t}\text{sin}\left(\text{π}x\text{/}L\right)$

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 wave function is zero, consistent with the boundary conditions.) Calculate the expectation values of position, momentum, and kinetic energy.

## Strategy

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

## Solution

Computation of the normalization constant:

$1=\underset{0}{\overset{L}{\int }}dx{\psi }^{*}\left(x\right)\psi \left(x\right)=\underset{0}{\overset{L}{\int }}dx\left(A{e}^{+i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)\left(A{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)={A}^{2}\underset{0}{\overset{L}{\int }}dx\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\frac{\pi x}{L}={A}^{2}\frac{L}{2}\phantom{\rule{0.5em}{0ex}}⇒\phantom{\rule{0.5em}{0ex}}A=\sqrt{\frac{2}{L}}.$

The expectation value of position is

$⟨x⟩=\underset{0}{\overset{L}{\int }}dx{\psi }^{*}\left(x\right)x\psi \left(x\right)=\underset{0}{\overset{L}{\int }}dx\left(A{e}^{+i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)x\left(A{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)={A}^{2}\underset{0}{\overset{L}{\int }}dxx\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\frac{\pi x}{L}={A}^{2}\frac{{L}^{2}}{4}=\frac{L}{2}.$

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 wave function $\psi \left(x\right)$ :

$\text{−}i\hslash \frac{d}{dx}\psi \left(x\right)=\text{−}i\hslash \frac{d}{dx}A{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}=\text{−}i\frac{Ah}{2L}{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\text{ }.$

Therefore, the expectation value of momentum is

$⟨p⟩=\underset{0}{\overset{L}{\int }}dx\left(A{e}^{+i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)\left(\text{−}i\frac{Ah}{2L}{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)=\text{−}i\frac{{A}^{2}h}{4L}\underset{0}{\overset{L}{\int }}dx\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{2\pi x}{L}=0.$

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

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

$-\frac{{\hslash }^{2}}{2m}\phantom{\rule{0.2em}{0ex}}\frac{{d}^{2}}{d{x}^{2}}\psi \left(x\right)=-\frac{{\hslash }^{2}}{2m}\phantom{\rule{0.2em}{0ex}}\frac{{d}^{2}}{d{x}^{2}}A{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}=-\frac{{\hslash }^{2}}{2m}A{e}^{\text{−}i\omega t}\frac{{d}^{2}}{d{x}^{2}}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}=\frac{A{h}^{2}}{2m{L}^{2}}{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}.$

Thus, the expectation value of the kinetic energy is

$\begin{array}{}\\ \\ \hfill ⟨K⟩& =\underset{0}{\overset{L}{\int }}dx\left(A{e}^{+i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)\left(\frac{A{h}^{2}}{2m{L}^{2}}{e}^{\text{−}i\omega t}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{\pi x}{L}\right)\hfill \\ & \hfill =\frac{{A}^{2}{h}^{2}}{2m{L}^{2}}\underset{0}{\overset{L}{\int }}dx\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\frac{\pi x}{L}=\frac{{A}^{2}{h}^{2}}{2m{L}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{L}{2}=\frac{{h}^{2}}{2m{L}^{2}}\text{ }.\end{array}$

## 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

$|\psi {|}^{\text{ }2}=\left(2\text{/}L\right){\text{sin}}^{2}\left(\pi x\text{/}L\right).$

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.

Check Your Understanding For the particle in the above example, find the probability of locating it between positions 0 and L /4

$\left(1\text{/}2-1\text{/}\text{π}\right)\text{/}2=9\text{%}$

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.

## Summary

• In quantum mechanics, the state of a physical system is represented by a wave function.
• In Born’s interpretation, the square of the particle’s wave function represents the probability density of finding the particle around a specific location in space.
• Wave functions must first be normalized before using them to make predictions.
• The expectation value is the average value of a quantity that requires a wave function and an integration.

## Conceptual questions

What is the physical unit of a wave function, $\text{Ψ}\left(x,t\right)?$ What is the physical unit of the square of this wave function?

$1\text{/}\sqrt{L},$ where $L=\text{length}$ ; 1/ L , where $L=\text{length}$

Can the magnitude of a wave function $\left({\text{Ψ}}^{*}\left(x,t\right)\phantom{\rule{0.2em}{0ex}}\text{Ψ}\left(x,t\right)\right)$ be a negative number? Explain.

What kind of physical quantity does a wave function of an electron represent?

The wave function does not correspond directly to any measured quantity. It is a tool for predicting the values of physical quantities.

What is the physical meaning of a wave function of a particle?

What is the meaning of the expression “expectation value?” Explain.

The average value of the physical quantity for a large number of particles with the same wave function.

## Problems

Compute $|\text{Ψ}\left(x,t\right){|}^{\text{ }2}$ for the function $\text{Ψ}\left(x,t\right)=\psi \left(x\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\omega \text{ }\text{ }t$ , where $\omega$ is a real constant.

$|\psi \left(x\right){|}^{2}{\text{sin}}^{2}\omega t$

Given the complex-valued function $f\left(x,y\right)=\left(x-iy\right)\text{/}\left(x+iy\right)$ , calculate $|f\left(x,y\right){|}^{\text{ }2}$ .

Which one of the following functions, and why, qualifies to be a wave function of a particle that can move along the entire real axis? (a) $\psi \left(x\right)=A{e}^{\text{−}{x}^{\text{ }2}}$ ;
(b) $\psi \left(x\right)=A{e}^{\text{−}x}$ ; (c) $\psi \left(x\right)=A\phantom{\rule{0.2em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}x$ ;
(d) $\psi \left(x\right)=A\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x\right)\text{/}x$ ; (e) $\psi \left(x\right)=A{e}^{\text{−}|x|}$ .

(a) and (e), can be normalized

A particle with mass m moving along the x -axis and its quantum state is represented by the following wave function:

$\text{Ψ}\left(x,t\right)=\left\{\begin{array}{cc}\hfill 0,& x<0,\hfill \\ \hfill Ax{e}^{\text{−}\alpha \text{ }x}{e}^{\text{−}i\text{ }E\text{ }t\text{/}\hslash },& x\ge 0\text{,}\hfill \end{array}$

where $\alpha =2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{ }10}{\text{m}}^{-1}$ . (a) Find the normalization constant. (b) Find the probability that the particle can be found on the interval $0\le x\le L$ . (c) Find the expectation value of position. (d) Find the expectation value of kinetic energy.

A wave function of a particle with mass m is given by

$\psi \left(x\right)=\left\{\begin{array}{cc}\hfill A\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\alpha x,& -\frac{\pi }{2\alpha }\le x\le +\frac{\pi }{2\alpha },\hfill \\ \hfill 0,\hfill & \text{otherwise,}\hfill \end{array}$

where $\alpha =1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\text{/}\text{m}$ . (a) Find the normalization constant. (b) Find the probability that the particle can be found on the interval $0\le x\le 0.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\text{m}$ . (c) Find the particle’s average position. (d) Find its average momentum. (e) Find its average kinetic energy $-0.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\text{m}\le x\le +0.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\text{m}$ .

a. $A=\sqrt{2\alpha \text{/}\pi }$ ; b. $\text{probability}=29.3\text{%}$ ; c. $⟨x⟩=0$ ; d. $⟨p⟩=0$ ; e. $⟨K⟩={\alpha }^{2}{\hslash }^{2}\text{/}2m$

in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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