Question 41.4: Average Position for a Particle in an Infinite One-Dimension......
Average Position for a Particle in an Infinite One-Dimensional Well, Quantum Approach
Find the expectation value of x for a particle in an infinite well of width L. This time apply the solution we found to Schrödinger’s equation for this situation.
INTERPRET and ANTICIPATE
This problem is the same as Example 41.3, but here we take a quantum-mechanical approach. In the end we’ll compare our results.
SOLVE
Start with Equation 41.12 for the expectation value of x. Change the limits of integration, since the particle cannot be found outside the box.
Substitute |\psi|^2=\frac{2}{L} \sin ^2 k_n x (Eq. 41.18), which came from our solution to Schrödinger’s equation.
\langle x\rangle=\int_0^L x\left(\frac{2}{L} \sin ^2 k_n x\right) d xThis integral can be solved by parts.
\langle x\rangle=\left.\frac{2}{L}\left(\frac{x^2}{4}-\frac{x \sin 2 k_n x}{4 k_n}-\frac{\cos 2 k_n x}{8 k_n^2}\right)\right|_0 ^LSubstitute the limits and use k_n=n \pi / L (Eq. 41.15). Notice that sin 2nπ = 0 and cos 2nπ = 1. (In quantum mechanics the energy is quantized.)
\begin{aligned}& \langle x\rangle=\left.\frac{2}{L}\left(\frac{L^2}{4}-\frac{\cos 2 k_n L}{8 k_n^2}+\frac{1}{8 k_n^2}\right)\right|_0 ^L \\& \langle x\rangle=\frac{2}{L}\left(\frac{L^2}{4}-\frac{\cos 2 n \pi}{8 k_n^2}+\frac{1}{8 k_n^2}\right) \\& \langle x\rangle=\frac{L}{2}\end{aligned}CHECK and THINK
Many times quantum mechanics seems counterintuitive, but not this time. Here the average position is in the center of the box, just as you’d expect (and found in Example 41.3).