39.4 Normalization

In our discussion of probability we noted that the dart has to hit the wall somewhere. The mathematical statement of this idea is the requirement that $P_{A} + P_{B} + P_{C} = 1$ . That is, the probabilities of all the mutually exclusive outcomes must add up to 1.

Similarly, a photon or electron has to land somewhere on the detector after passing through an experimental apparatus. Consequently, the probability that it will be detected at some position is 100%. To make use of this requirement, consider an experiment in which an electron is detected on the $x$ -axis. As FIGURE 39.7 shows, we can divide the region between positions $x_{L}$ and $x_{R}$ into $N$ adjacent narrow strips of width $δ x$ .

The figure illustrates dividing the entire detector into many small strips of width delta x.

FIGURE 39.7 Full Alternative Text

The probability that any particular electron lands in the narrow strip $i$ at position $x_{i}$ is

Prob(in δ x at x_{i}) = P (x_{i}) δ x

where $P (x_{i}) = | ψ (x_{i}) |^{2}$ is the probability density at $x_{i}$ . The probability that the electron lands in the strip at $x_{1}$ or $x_{2}$ or $x_{3}$ or $\dots$ is the sum

\begin{array}{l} Prob (between x_{L} and x_{R}) & = & Prob (in δ x at x_{1}) \\ + Prob(in δ x at x_{2}) + \dots \\ = & \sum_{i = 1}^{N} P (x_{i}) δ x = \sum_{i = 1}^{N} {| ψ (x_{i}) |}^{2} δ x \end{array}

(39.16)

That is, the probability that the electron lands somewhere between $x_{L}$ and $x_{R}$ is the sum of the probabilities of landing in each narrow strip.

If we let the strips become narrower and narrower, then $δ x \to d x$ and the sum becomes an integral. Thus the probability of finding the particles in the range $x_{L} \leq x \leq x_{R}$ is

Prob ({in range x}_{L} \leq x \leq x_{R}) = \int_{x_{L}}^{x_{R}} P (x) d x = \int_{x_{L}}^{x_{R}} {| ψ (x) |}^{2} d x

(39.17)

As FIGURE 39.8a shows, we can interpret $Prob(in range x_{L} \leq x \leq x_{R})$ as the area under the probability density curve between $x_{L}$ and $x_{R}$ .

The figure labeled ("a") illustrates a probability density versus position graph.

The figure labeled (b) illustrates a probability density versus position graph.

Full Alternative Text

NOTE The integral of Equation 39.17 is needed when the probability density changes over the range $x_{L}$ to $x_{R}$ . For sufficiently narrow intervals, over which $P (x)$ remains essentially constant, the expression $Prob(in δ x at x) = P (x) δ x$ is still valid and is easier to use.

Now let the detector become infinitely wide, so that the probability that the electron will arrive somewhere on the detector becomes 100%. The statement that the electron has to land somewhere on the $x$ -axis is expressed mathematically as

\int_{- \infty}^{\infty} P (x) d x = \int_{- \infty}^{\infty} {| ψ (x) |}^{2} d x = 1

(39.18)

Equation 39.18 is called the normalization condition. Any wave function $ψ (x)$ must satisfy this condition; otherwise we would not be able to interpret $| ψ (x) |^{2}$ as a probability density. As FIGURE 39.8b shows, Equation 39.18 tells us that the total area under the probability density curve must be 1.

NOTE The normalization condition integrates the square of the wave function. We don’t have any information about what the integral of $ψ (x)$ might be.

EXAMPLE 39.2 Normalizing and interpreting a wave function

FIGURE 39.9 shows the wave function of a particle confined within the region between $x = 0 nm$ and $x = L = 1.0 nm$ . The wave function is zero outside this region.

Determine the value of the constant $c$ that makes this a normalized wave function.
Draw a graph of the probability density $P (x)$ .
Draw a dot picture showing where the first 40 or 50 particles might be found.
Calculate the probability of finding the particle in a region of width $δ x = 0.01 nm$ at positions $x_{1} = 0.05 nm, x_{2} = 0.50 nm,$ and $x_{3} = 0.95 nm$ .

The figure illustrates the wave function of a particle confined with region x equal zero and x equal L.

FIGURE 39.9 Full Alternative Text

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

VISUALIZE The wave function is shown in Figure 39.9.

SOLVE

The wave function is $ψ (x) = c (1 - x / L)$ between 0 and $L$ , 0 otherwise. This is a function that decreases linearly from $ψ = c$ at $x = 0$ to $ψ = 0$ at $x = L$ . The constant $c$ is the height of this wave function. The particle has to be in the region $0 \leq x \leq L$ with probability 1, and only one value of $c$ will make it so. We can determine $c$ by using Equation 39.18, the normalization condition. The wave function is zero outside the interval from 0 to $L$ , so we need to integrate the probability density only from 0 to L. Thus

$\begin{array}{l} 1 & = & \int_{0}^{L} {| ψ (x) |}^{2} d x = c^{2} \int_{0}^{L} {(1 - \frac{x}{L})}^{2} d x \\ = & c^{2} \int_{0}^{L} (1 - \frac{2 x}{L} + \frac{x^{2}}{L^{2}}) d x \\ = & c^{2} {[x - \frac{x^{2}}{L} + \frac{x^{3}}{3 L^{2}}]}_{0}^{L} = \frac{1}{3} c^{2} L \end{array}$

The solution for $c$ is

$c = \sqrt{\frac{3}{L}} = \sqrt{\frac{3}{1.0 nm}} = 1.732 {nm}^{- 1 / 2}$

Note the unusual units for $c$ . Although these are not SI units, we can correctly compute probabilities as long as $δ x$ has units of nm. A multiplicative constant such as $c$ is often called a normalization constant.
The wave function is

$ψ (x) = (1.732 {nm}^{- 1 / 2}) (1 - \frac{x}{1.0 nm})$

Thus the probability density is

$P (x) = | ψ (x) |^{2} = (3.0 {nm}^{- 1}) (1 - \frac{x}{1.0 nm})^{2}$

This probability density is graphed in FIGURE 39.10a.

FIGURE 39.10 The probability density $P (x)$ and the detected positions of particles.

FIGURE 39.10 Full Alternative Text
Particles are most likely to be detected at the left edge of the interval, where the probability density $P (x)$ is maximum. The probability steadily decreases across the interval, becoming zero at $x = 1.0 nm$ . FIGURE 39.10b shows how a group of particles described by this wave function might appear on a detection screen.
$P (x)$ is essentially constant over the small interval $δ x = 0.01 nm,$ so we can use

$Prob (in δ x at x) = P (x) δ x = {| ψ (x) |}^{2} δ x$

for the probability of finding the particle in a region of width $δ x$ at the position $x$ . We need to evaluate $| ψ (x) |^{2}$ at the three positions $x_{1} = 0.05 nm, x_{2} = 0.50 nm,$ and $x_{3} = 0.95 nm$ . Doing so gives

$\begin{array}{l} Prob (in 0.01 nm at x_{1} = 0.05 nm) & = & c^{2} {(1 - x_{1} / L)}^{2} δ x \\ = & 0.0270 = 2.70 % \\ Prob (in 0.01 nm at x_{2} = 0.50 nm) & = & c^{2} {(1 - x_{2} / L)}^{2} δ x \\ = & 0.0075 = 0.75 % \\ Prob (in 0.01 nm at x_{3} = 0.95 nm) & = & c^{2} {(1 - x_{3} / L)}^{2} δ x \\ = & 0.00008 = 0.008 % \end{array}$

39.4 Normalization

FIGURE 39.7 Dividing the entire detector into many small strips of width $δ x$ .

FIGURE 39.8 The area under the probability density curve is a probability.

EXAMPLE 39.2 Normalizing and interpreting a wave function

FIGURE 39.9 The wave function of Example 39.2.

FIGURE 39.10 The probability density $P (x)$ and the detected positions of particles.