39.6.1 What Does It Mean?

Heisenberg’s uncertainty principle is a statement about our knowledge of the properties of a particle. If we want to know where a particle is located, we measure its position $x$ . That measurement is not absolutely perfect but has some uncertainty $Δ x$ . Likewise, if we want to know how fast the particle is going, we need to measure its velocity $v_{x}$ or, equivalently, its momentum $p_{x}$ . This measurement also has some uncertainty $Δ p_{x}$ .

Uncertainties are associated with all experimental measurements, but better procedures and techniques can reduce those uncertainties. Newtonian physics places no limits on how small the uncertainties can be. A Newtonian particle at any instant of time has an exact position $x$ and an exact momentum $p_{x}$ , and with sufficient care we can measure both $x$ and $p_{x}$ with such precision that the product $Δ x Δ p_{x} \to 0$ . There are no inherent limits to our knowledge about a classical, or Newtonian, particle.

Heisenberg, however, made the bold and original statement that our knowledge has real limitations. No matter how clever you are, and no matter how good your experiment, you cannot measure both $x$ and $p_{x}$ simultaneously with arbitrarily good precision. Any measurements you make are limited by the condition that $Δ x Δ p_{x} \geq h / 2$ . Our knowledge about a particle is inherently uncertain.

Why? Because of the wave-like nature of matter. The “particle” is spread out in space, so there simply is not a precise value of its position $x$ . Similarly, the de Broglie relationship between momentum and wavelength implies that we cannot know the momentum of a wave packet any more exactly than we can know its wavelength or frequency. Our belief that position and momentum have precise values is tied to our classical concept of a particle. As we revise our ideas of what atomic particles are like, we will also have to revise our old ideas about position and momentum.

EXAMPLE 39.4 The uncertainty of a dust particle

A $1 - μ m-diameter$ dust particle $(m \approx 10^{- 15} kg)$ is confined within an $8 - μ m-long$ box. Can we know with certainty if the particle is at rest? If not, within what range is its velocity likely to be found?

MODEL All matter is subject to the Heisenberg uncertainty principle.

SOLVE If we know for sure that the particle is at rest, then $p_{x} = 0$ with no uncertainty. That is, $Δ p_{x} = 0$ . But then, according to the uncertainty principle, the uncertainty in our knowledge of the particle’s position would have to be $Δ x \to \infty$ . In other words, we would have no knowledge at all about the particle’s position—it could be anywhere! But that is not the case. We know the particle is somewhere in the box, so the uncertainty in our knowledge of its position is $Δ x = L = 8 μ m$ . With a finite $Δ x$ , the momentum uncertainty $Δ p_{x}$ cannot be zero. We cannot know with certainty if the particle is at rest inside the box. No matter how hard we try to bring the particle to rest, the uncertainty in our knowledge of the particle’s momentum will be $Δ p_{x} \approx h / (2 Δ x) = h / 2 L$ . We’ve assumed the most accurate measurements possible so that the $\geq$ in Heisenberg’s uncertainty principle becomes $\approx$ . Consequently, the range of possible velocities is

Δ v_{x} = \frac{Δ p_{x}}{m} \approx \frac{h}{2 m L} \approx 4 \times 10^{- 14} m / s

This range of possible velocities will be centered on $v_{x} = 0 m / s$ if we have done our best to have the particle be at rest. Thus all we can know with certainty is that the particle’s velocity is somewhere within the interval $- 2 \times 10^{- 14} m/s \leq v_{x} \leq 2 \times 10^{- 14} m / s$ .

REVIEW For practical purposes you might consider this to be a satisfactory definition of “at rest.” After all, a particle moving with a speed of $2 \times 10^{- 14} m / s$ would need $5 \times 10^{10} s$ to move a mere 1 mm. That is about 2000 years! Nonetheless, we can’t know if the particle is “really” at rest.

NOTE One-significant-figure calculations are appropriate to use with uncertainties.

EXAMPLE 39.5 The uncertainty of an electron

What range of velocities might an electron have if confined to a 0.1-nm-wide region, about the size of an atom?

MODEL Electrons are subject to the Heisenberg uncertainty principle.

SOLVE The analysis is the same as in Example 39.4. If we know that the electron’s position is located within an interval $Δ x \approx 0.1 nm,$ then the best we can know is that its velocity is within the range

Δ v_{x} = \frac{Δ p_{x}}{m} \approx \frac{h}{2 m L} \approx 4 \times 10^{6} m / s

Because the average velocity is zero, the best we can say is that the electron’s velocity is somewhere in the interval $- 2 \times 10^{6} m/s \leq v_{x} \leq 2 \times 10^{6} m / s$ . It is simply not possible to know the electron’s velocity any more precisely than this.

REVIEW Unlike the situation in Example 39.4, where $Δ v_{x}$ was so small as to be of no practical consequence, our uncertainty about the electron’s velocity is enormous—about 1% of the speed of light!

Once again, we see that even the smallest of macroscopic objects behaves very much like a classical Newtonian particle. Perhaps a $1 - μ m-diameter$ particle is slightly fuzzy and has a slightly uncertain velocity, but it is far beyond the measuring capabilities of even the very best instruments to detect this wave-like behavior. In contrast, the effects of the uncertainty principle at the atomic scale are stupendous. We are unable to determine the velocity of an electron in an atom-size container to any better accuracy than about 1% of the speed of light.

CHAPTER 39 CHALLENGE EXAMPLE The probability of finding a particle

A particle is described by the wave function

ψ (x) = {\begin{matrix} 0 & x < 0 \\ c e^{- x / L} & x \geq 0 \end{matrix}

where $L = 1 nm$ .

Determine the value of the constant $c$ .
Draw graphs of $ψ (x)$ and the probability density $P (x)$ .
If $10^{6}$ particles are detected, how many are expected to be found in the region $x \geq 1 nm ?$

MODEL The probability of finding a particle is determined by the probability density $P (x)$ .

SOLVE

The wave function is an exponential $ψ (x) = c e^{- x / L}$ that extends from $x = 0$ to $x = + \infty$ . Equation 39.18, the normalization condition, is

$1 = \int_{- \infty}^{\infty} {| ψ (x) |}^{2} d x = c^{2} \int_{0}^{\infty} e^{- 2 x / L} d x = {- \frac{c^{2} L}{2} e^{- 2 x / L} |}_{0}^{\infty} = \frac{c^{2} L}{2}$

We can solve this for the normalization constant $c$ :

$c = \sqrt{\frac{2}{L}} = \sqrt{\frac{2}{1 nm}} = 1.414 {nm}^{- 1 / 2}$
The probability density is

$P (x) = | ψ (x) |^{2} = (2.0 {nm}^{- 1}) e^{- 2 x / (1.0 nm)}$

The wave function and the probability density are graphed in FIGURE 39.17.

FIGURE 39.17 The wave function and probability density.

FIGURE 39.17 Full Alternative Text
The probability of finding a particle in the region $x \geq 1 nm$ is the shaded area under the probability density curve in Figure 39.17. We must use Equation 39.17 and integrate to find a numerical value. The probability is

$\begin{array}{l} Prob (x \geq 1 nm) & = & \int_{1 nm}^{\infty} {| ψ (x) |}^{2} d x \\ = & (2.0 {nm}^{- 1}) \int_{1 nm}^{\infty} e^{- 2 x / (1.0 nm)} d x \\ = & {(2.0 {nm}^{- 1}) (- \frac{1.0 nm}{2}) e^{- 2 x / (1.0 nm)} |}_{_{1 nm}}^{\infty} \\ = & e^{- 2} = 0.135 = 13.5 % \end{array}$

The number of particles expected to be found at $x \geq 1 nm$ is

$N_{detected} = N \times Prob (x \geq 1 nm) = (10^{6}) (0.135) = 135, 000$

REVIEW There is a 13.5% chance of detecting a particle beyond 1 nm and thus an 86.5% chance of finding it within the interval $0 \leq x \leq 1 nm$ . Unlike classical physics, we cannot make an exact prediction of a particle’s position.