Quantum Waves

Section 3.2 Quantum Waves

Let's consider a particle moving with a well-defined momentum $p\text{.}$ According to de Broglie, this particle can also be thought of as a wave with a wavelength $\lambda = h/p\text{.}$ We can represent this graphically as a simple sine wave, as shown in Figure 3.1(a).¹ But a wave doesn't have to be a simple sine wave. The wavefunction $\psi(x)$ describing a particle's wave-like nature can take any shape. Figure 3.1(b) shows a wavepacket-shaped wavefunction, and Figure 3.1(c) shows a pulse-like wavefunction. Later in this chapter, we will explain how the specific wavefunction can be determined for different physical situations.

Figure 3.1. Wavefunctions $\psi(x)$ discussed in the text: (a) a pure sine wave, (b) a partially localized function, and (c) a localized blip.

But first: what exactly is the wave here? An answer to that question was provided in 1926 by Max Born who argued that the wavefunction contains information about the probability for finding the particle. Specifically, given a wavefunction $\psi(x)$ that describes the wave properties of a particle, Born argued that $\psi(x)$ represents probability amplitude, and taking the magnitude squared of the wavefunction produces the probability density $P(x)$ — probability per unit distance for a one-dimensional problem — for finding the particle:

\begin{equation} P(x) = |\psi(x)|^2\text{.}\label{eq_probdensity}\tag{3.1} \end{equation}

This tells us that according to quantum physics, a particle has a range of possible positions, described by a probability density. We don't know where the particle is located — there is a possibility that it could be found anywhere where the wavefunction $\psi(x)$ is non-zero. But it isn't just a statement that we don't know where the particle is; rather, the particle doesn't HAVE a position until we make a measurement and locate it.

The interpretation of matter waves as probability amplitudes has tremendous implications, which we will explore later in this chapter and in the next few chapters. But before we explore those implications, we need to develop the mathematics of probability densities.

Subsection 3.2.1 Probability Densities

The concept of probability density can be tricky — it's not the same as probability! In one spatial dimension we write the probability density as $P(x)\text{,}$ since it can vary with the position $x\text{.}$ To get probability from $P(x)\text{,}$ you must multiply by a length interval since it is really probability per unit length. Specifically,

\begin{equation} P(x)\, dx = \left(\begin{array}{l} \text{ Probability of finding the particle within } \\ \text{ an interval from $x$ to $x+dx$. } \end{array} \right)\text{.}\tag{3.2} \end{equation}

If the probability density is uniform, then it is easy to find probability for any interval. To illustrate, suppose you are playing golf on a dark night (don't ask why) and after driving the ball you know it landed somewhere along the left edge of the fairway, between 200 to 250 meters from the tee.² You want to find it and learn about probability density at the same time.

The total probability of finding your ball between $=200\Xunits{m}$ and $x = 250\Xunits{m}$ is 1 — it is definitely somewhere in that interval. Furthermore, the probability density is uniform. That is, the ball is equally likely to be found anywhere in the interval. Then the probability density $P(x)$ is given by

\begin{equation} P(x) = \frac{\text{ Total Probability } }{\text{ Total Length of Interval } } \hspace{0.2in}\text{ (uniform $P(x)$ only) }\text{,}\tag{3.3} \end{equation}

so in this case

\begin{equation} P(x) = \frac{1}{(250\Xunits{m}-200\Xunits{m})} = \frac{1}{50}\Xunits{m$^{-1}$}\text{.}\tag{3.4} \end{equation}

If you limit your search to the region between $x_1 = 210\Xunits{m}$ and $x_2 = 215\Xunits{m}\text{,}$ then your probability of finding the ball is

\begin{align*} \text{ Probability } \amp = (\text{ Probability Density } )\times (\text{ Length of Interval of Interest } )\\ \amp = \frac{1}{50\Xunits{m}} \left(215\Xunits{m}-210\Xunits{m}\right) = 0.1 \text{ or 10\% } \text{.} \end{align*}

On the other hand, if the probability density is not uniform, you have to integrate:

\begin{equation} \text{ Probability of finding ball between $x_1$ and $x_2$ } = \int_{x_1}^{x_2} P(x)\, dx\text{.}\tag{3.5} \end{equation}

This is the general relation you should use to determine probabilities for 1-D problems.

Finally, we note a useful graphical connection to probability. Since integrals have the nice interpretation of being area under a curve, lots of understanding of these concepts can be gained by plotting $P(x)$ vs. $x\text{,}$ and simply calculating areas.

Example 3.2. Probability Density I.

A plot of a non-uniform probability density for finding your ball near a different hole is shown below, in Figure 3.3.

Figure 3.3. Probability density for location of a golf ball.

Show that the total probability of finding your ball is 1.
Use the plot to find the probability of finding your ball between 150 and 200 m from the tee.

Solution.

Since the total probability is the sum over the probabilities of being at all possible positions, we must integrate the probability density over all possible values of $x\text{,}$ i.e., find the total area under the graph. The total net area is a triangle with a base of $100\Xunits{m}$ and a height of $0.02\Xunits{m$^{-1}$}\text{,}$ so

\begin{equation} \text{ Probability } = \text{ Area } = {\textstyle\frac{1}{2}}\times 100 \times 0.02 = 1\text{..}\tag{3.6} \end{equation}
The probability is equal to $\int_{150}^{200}P(x)\, dx\text{,}$ which is the area of the shaded trapezoid. The trapezoid can be considered as a rectangle (of base $b = 50$ and height $h = 0.01$) plus a triangle (of base $b = 50$ and height $h = 0.01$). Therefore the total probability is

\begin{align*} \text{ Probability } = \text{ Area under curve } \amp = bh + {\textstyle\frac{1}{2}b h}\\ \amp = 50 (0.01) + {\textstyle\frac{1}{2}}50 (0.02 - 0.01)\\ \amp = 0.75 \text{ or 75\% } \text{.} \end{align*}

Subsection 3.2.2 Probabilities and Wavefunctions

While a general wavefunction may be very complicated, the prescription for its interpretation is always the same: from $\psi(x)\text{,}$ calculate $|\psi(x)|^2\text{,}$ its magnitude squared. Then the probability density is given by $P(x)=|\psi(x)|^2\text{.}$

One of the complications of a general wavefunction is that it may be complex-valued, involving factors of the imaginary number $i = \sqrt{-1}\text{.}$ When this occurs $|\psi(x)|^2$ does not just mean “square the wavefunction” to get the magnitude squared. Rather you should follow this three-step process:

1.
Write the complex conjugate of the wavefunction, $\psi^*\text{,}$ by replacing every occurrence of “$i$” with “$-i$” in the wavefunction. For example if $\psi(x) = 2 e^{i3x} - 5 i x^2\text{,}$ then $\psi^*(x) = 2 e^{-i3x} + 5 i x^2\text{.}$
2.
Multiply $\psi(x)$ times $\psi^*(x)\text{.}$
3.
Replace any occurrences of “$i^2$” with “$-1$” and simplify. Your result should be real (no left-over “$i$” s) and non-negative for all values of $x\text{,}$ as required for a probability density.

Example 3.4. Calculating probability density.

A particle in a region of space with $x > 0$ is represented by the wavefunction

\begin{equation} \psi(x) = \sqrt{5} e^{ikx} \sin(2x) e^{-x}\tag{3.7} \end{equation}

where $k$ is a real constant. Calculate the probability density for this particle.

Solution: According to the prescription given above, the probability density for this particle is calculated to be

\begin{align*} \left| \psi(x) \right| ^2 =\mathstrut \amp \psi(x) \cdot \psi^*(x)\\ =\mathstrut \amp \left( \sqrt{5} e^{ikx} \sin(2x) e^{-x} \right) \left( \sqrt{5} e^{-ikx} \sin(2x) e^{-x} \right)\\ =\mathstrut \amp 5 \sin^2(2x) e^{-2x}\text{,} \end{align*}

where we have used the fact that

\begin{align*} (e^{ikx}) (e^{ikx})^* =\mathstrut \amp (e^{ikx}) (e^{-ikx})\\ =\mathstrut \amp e^{(ikx-ikx)} = e^0 = 1\text{.} \end{align*}

Figure 3.5 is the graph of a probability density for a particle in some quantum state. The quantity $P(x_0)\, dx$ represents the probability for finding the particle within a narrow interval from $x_0$ to $x_0+dx\text{,}$ shown as the shaded area in the figure. If we would like to know the probability of finding the particle anywhere between the two positions $x_a$ and $x_b\text{,}$ then we need to integrate the probability density:

Figure 3.5. Determining the probability over a range of positions.

\begin{equation} \left( \begin{array}{l} \mbox{Probability of finding the particle} \\ \mbox{in the region } \end{array} \right) = \int_{x_a}^{x_b} P(x)\, dx = \int_{x_a}^{x_b} |\psi(x)|^2\, dx\text{,}\label{eq_IntegralprobDensity}\tag{3.8} \end{equation}

which is the area under the curve from $x_a$ to $x_b\text{.}$ (3.8) brings up an important property of wavefunctions and probability densities. What if the region we are interested in is the entire $x$-axis, that is, $[x_a, x_b] = (-\infty, +\infty)\text{?}$ Another way of saying this is, ‘What is the probability of finding the particle anywhere?’ Certainly the answer to this question is $100\%$ ! This implies that the wavefunction must be given such that

\begin{equation} \int_{-\infty}^{+\infty} |\psi(x)|^2\, dx = 1\text{.}\label{eq_normalization}\tag{3.9} \end{equation}

When a wavefunction satisfies this requirement, it is said to be normalized. Normalization is usually accomplished by multiplying the wavefunction by a suitable constant factor, as we will see in the following examples.

Example 3.6. Probability of finding a particle I.

Figure 3.7 shows the wavefunction for a certain particle, where $A$ is a positive constant. (a) Determine a value for the constant $A$ such that the wavefunction is properly normalized. (b) What is the probability of finding the particle in the region between $x = 2\Xunits{nm}$ and $x = 4\Xunits{nm}\text{?}$

Figure 3.7. Wavefunction for Example 3.6.

Solution: (a) To determine probabilities we must first determine $|\psi(x)|^2$ for the given wavefunction. We do this by computing the value of $|\psi(x)|^2$ at every position on the graph which results in the graph of Figure 3.8.

Figure 3.8. Probability density for Example 3.6.

The total probability of finding the particle anywhere is the total area under the curve of Figure 3.8:

\begin{align*} \int_{-\infty}^{+\infty} |\psi(x)|^2\, dx =\mathstrut \amp A^2 (2 - 1) + \left( - \frac{A}{2} \right)^2 (4 - 2)\\ =\mathstrut \amp A^2 + \frac{A^2}{2} = \frac{3}{2} A^2 \text{.} \end{align*}

For the wavefunction to be properly normalized, this area must equal one. Therefore, the value for the constant $A$ must be

\begin{equation} \frac{3}{2} A^2 = 1 \hspace{0.3in} \longrightarrow \hspace{0.3in} A = \sqrt{\frac{2}{3}}\text{.}\tag{3.10} \end{equation}

(b) Now that we know the value for the constant $A$ that properly normalizes the wavefunction, the probability of finding the particle in the region between $x = 2\Xunits{nm}$ and $x = 4\Xunits{nm}$ is just the area under the curve of the probability density between $x = 2\Xunits{nm}$ and $x = 4\Xunits{nm}\text{:}$

\begin{equation} \text{ Prob } ([2, 4]) = \frac{A^2}{4} (4 - 2) = \frac{A^2}{2} = \frac{1}{3}\text{.}\tag{3.11} \end{equation}

Actually, a state with well-defined momentum has a complex wavefunction $\psi(x) = Ae^{2\pi ipx/h}$ whose real and imaginary parts are simple cosine and sine waves.

This is physics, so golf is played using the metric system.