Chapter 5: Q65E (page 191)

equation (5-33). The twosolutionsare added in equal amounts. Show that if we instead added a different percentage of the two solutions. It would not change the important conclusion related to the oscillation frequency of the charge density.

Short Answer

Expert verified

The frequency of the time-dependent wave function is the same as it was before

Step by step solution

Identification of given data

The sum of the two different solutions is $A ψ_{n} (x) e^{- i (E_{n} / ℏ) t} + B ψ_{m} e^{- i (E_{m} / ℏ) t}$

Concept/significance of oscillation frequency

The frequency of an oscillation is the number of oscillations that occur once per unit time or once per second. The frequency of an oscillation can be calculated as the inverse of the time of oscillation.

Explanation of the adding different percentages of the two solutions

The wave function of the particle is mathematically presented as:

$ψ (x, t) = A ψ_{n} (x) e^{- i (E_{n} / ℏ) t} + B ψ_{m} e^{- i (E_{m} / ℏ) t}$

Multiply the wave function with its complex conjugate as shown below.

$\begin{array}{rcl} <ψ (x, t) ψ^{*} (x, t)> & = & (A ψ_{n} (x) e^{- i (E_{n} / ℏ) t} + B ψ_{m} (x) e^{- i (E_{m} / ℏ) t}) A ψ_{n} (x) e^{i (E_{n} / ℏ) t} + B ψ_{m} e^{i (E_{m} / ℏ) t} \\ = & A^{2} ψ_{n}^{2} (x) + B^{2} ψ_{m}^{2} (x) + (e^{- i (E_{n} - E_{m} / ℏ) t} + e^{- i (E_{m} - E_{n} / ℏ) t}) \\ = & A^{2} ψ_{n}^{2} (x) + B^{2} ψ_{m}^{2} (x) + 2 A B ψ_{n} (x) ψ_{m} (x) \cos (\frac{(E_{n} - E_{m}) t}{ℏ}) \end{array}$