Question 41.3: Average Position for a Particle in an Infinite One-Dimension......
Average Position for a Particle in an Infinite One-Dimensional Well, Classical Approach
Consider the hockey puck trapped between the walls of the one-dimensional box in Figure 41.1. As described in Section 41-2, when the puck encounters a wall, it reverses direction instantaneously and there are no dissipative forces. So the puck’s kinetic energy is constant and nonzero. Use classical mechanics to find the (average position) expectation value of x for a puck in an infinite well of width L. (In Example 41.4, we will take a quantum mechanical approach to answer the same question.)
INTERPRET and ANTICIPATE
The expectation value is another way of saying the average value. Because the puck slides back and forth without losing speed, you might expect that its average position is in the center of the box.
SOLVE
The puck doesn’t change speed, and so it is equally likely to be anywhere inside the box. So the probability density |\psi|^2 is a constant between the walls of the box and zero outside the box as shown in Figure 41.7.
Of course, the puck must be located somewhere along the x axis inside the box. To express this commonsense idea, we apply the normalization condition (Eq. 41.11). In this case, we know |\psi|^2 is a constant between x = 0 and L and zero elsewhere.
Now we are ready to find the expectation value of x using Equation 41.12. (Keep in mind that the integrals from x = – ∞ to 0 and from x = L to ∞ are zero because |\psi|^2 is zero in these regions. These two integrals are not shown here.)
\begin{aligned}& \langle x\rangle=\int_{-\infty}^{\infty} x|\psi(x)|^2 d x \quad \quad (41.12)\\& \langle x\rangle=\int_0^L x|\psi(x)|^2 d x=\int_0^L x \frac{1}{L} d x=\left.\frac{1}{L}\left(\frac{x^2}{2}\right)\right|_0 ^L \\& \langle x\rangle=\frac{L}{2}\end{aligned}CHECK and THINK
We found that the average position is in the center of the box as expected. In Example 41.4, we’ll find the same thing using quantum mechanics.
