Electron in a one-dimensional box: Uncertainty relation (25%) Setup: One electron, mass me , charge e, in a one-dimensional box with length L. Box range: 0 ≤ x ≤ L. Use the same dimensionless units as in problem 1, where ~ = 1, me = 1. 1
(a) For the two lowest-energy states, n = 1 and n = 2, calculate 1 the expectation value for the position of the particle, hxi = hψn |xψn i. Then draw a sketch of ψ ∗ nψn for some of the states with higher n and argue why hxi is the same for each state.
(b) Calculate hx 2 i for a state with quantum number n: hx 2 i = Z L 0 ψ ∗ n (x) x 2 ψn (x) dx (2) Some help: Rπ 0 y 2 sin2 (ny) d y = π 3 6 - π 4n2 . What is the limit of the result for states with large quantum number n ? If you have difficulties with the integral, consider that for high quantum numbers n the probability density of finding the electron in a one-dimensional box of length L is ψ ∗ (x)ψ(x) and rapidly oscillating around the average 1/L. Then use the constant average value as an approximation to calculate hx 2 i for large n. (The result is the same as taking the limit n → ∞ of the exact expectation value.)
(c) Calculate the position uncertainty ?x for states with high n. For this you need (?x) 2 = hx 2 i - hxi 2 . Use the approximate formula for large n from part (b).
(d) Calculate the momentum uncertainty as a function of n. Some help: The energy is En = n 2 π 2 2L2 and purely kinetic, because the potential is zero. The average momentum hpx i is zero because the electron is as likely to move to the right as it is to move to the left. We then have (?px ) 2 = hp 2 x i - hpx i 2 = hp 2 x i = 2hEn i = 2En . (Remember that the energy has a sharply defined value for each of the quantum states.) The correct position uncertainty is, of course, also obtained if you calculate the expectation values of pˆx and pˆ 2 x = pˆxpˆx explicitly with the wavefunction for state n.
(e) Using the results of c) and d), calculate ?x?px . This is called the uncertainty product. How does it depend on n? Hint: If your result contains a square root of 12 somewhere you are on the right track.