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Ch 40: Quantum Mechanics I: Wave Functions
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 40: Quantum Mechanics I: Wave FunctionsProblema 3
Capitolo 40, Problema 3

An electron is moving as a free particle in the −x-x-direction with momentum that has magnitude 4.50×10−244.50\(\times\)10^{-24} kg*m/s. Let k2=3k1=3kk_2 = 3k_1 = 3k. At t=0 t = 0, the probability distribution func­tion ∣Ψ(x,t)∣2|Ψ(x, t)|^2 has a maximum at x=0x = 0.
(a) What is the smallest positive value of xx for which the probability distribution function has a maximum at time t=2πωt=\(\frac{2\pi}{\omega}\), where ω=hk2/2mω = hk^2/2m?
(b) From your result in part (a), what is the average speed with which the probability distribution is moving in the +x+x­-direction?

Guida verificata passo dopo passo
1
Step 1: Start by understanding the problem. The electron is described as a free particle, and its wavefunction Ψ(x, t) is a superposition of two plane waves with wave numbers k_1 = k and k_2 = 3k. The probability distribution |Ψ(x, t)|^2 exhibits interference, leading to maxima and minima. The goal is to find the smallest positive x where the probability distribution has a maximum at t = 2π/ω, and then calculate the average speed of the probability distribution in the +x-direction.
Step 2: Write the wavefunction Ψ(x, t) as a superposition of two plane waves: Ψ(x, t) = A[e^(i(kx - ω_1t)) + e^(i(3kx - ω_2t))], where ω_1 = ℏk^2/2m and ω_2 = ℏ(3k)^2/2m. The probability distribution is given by |Ψ(x, t)|^2, which depends on the interference between the two waves.
Step 3: To find the maxima of |Ψ(x, t)|^2, calculate the phase difference between the two waves. The phase difference is Δϕ = (3kx - ω_2t) - (kx - ω_1t) = 2kx - (ω_2 - ω_1)t. For maxima, Δϕ must be an integer multiple of 2π: 2kx - (ω_2 - ω_1)t = 2nπ, where n is an integer.
Step 4: At t = 2π/ω, substitute ω = ℏk^2/2m and calculate ω_2 - ω_1. Using ω_1 = ℏk^2/2m and ω_2 = ℏ(3k)^2/2m, we find ω_2 - ω_1 = ℏ(9k^2 - k^2)/2m = 4ℏk^2/m. Substitute this into the phase condition: 2kx - (4ℏk^2/m)(2π/ω) = 2nπ. Simplify to solve for x in terms of n, k, and other constants.
Step 5: For the smallest positive x, set n = 1 and solve for x. Once x is determined, calculate the average speed of the probability distribution. The average speed is given by v_avg = x/t, where t = 2π/ω. Substitute the expression for x and simplify to find v_avg in terms of k, ℏ, and m.

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Concetti chiave

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Momentum

Momentum is a vector quantity defined as the product of an object's mass and its velocity. In this context, the electron's momentum is given as 4.50 x 10^-24 kg*m/s, indicating its motion in the -x-direction. Understanding momentum is crucial for analyzing the motion of particles and their interactions, especially in quantum mechanics.
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Intro to Momentum

Wave Function and Probability Distribution

In quantum mechanics, the wave function Ψ(x, t) describes the quantum state of a particle, and its squared magnitude |Ψ(x, t)|^2 represents the probability distribution of finding the particle at position x at time t. The maxima of this distribution indicate the most likely positions of the particle, which is essential for solving the problem regarding the position of the electron at a given time.
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Probability Distribution Graph

Angular Frequency (ω)

Angular frequency ω is a measure of how quickly an oscillating system changes its phase, defined as ω = 2π/T, where T is the period. In this problem, ω is related to the wave number k and the mass m of the electron, influencing the time evolution of the wave function. Understanding ω is key to determining the time-dependent behavior of the probability distribution.
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