Wave packet width given a wave function

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To find the wave packet Ψ(x, t) given the wave function φ(k) defined within specific bounds, the solution involves an integral that accounts for the contributions of different wave numbers. The integral expression for Ψ(x, t) is derived based on the system's dispersion relation, ω = vk. The discussion highlights confusion regarding the meaning of wave packet width and the necessity of the integral in the solution. Clarification is sought on whether the wave equation should be derived from the dispersion relation. Understanding these concepts is crucial for determining the wave packet's width effectively.
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Homework Statement


Find the wave packet Ψ(x, t) if φ(k) = A for k0 − ∆k ≤ k ≤ k0 + ∆k and φ(k) = 0 for all other k. The system’s dispersion relation is ω = vk, where v is a constant. What is the wave packet’s width?

Homework Equations


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I solved for Ψ(x, t):

$$\Psi(x,t) = \frac{1}{\pi\sqrt{K_0+\Delta K}} \int_{-\infty}^{\infty} \frac{sin(k(k_0 + \Delta K))}{k} e^{i(kx-hk^2/2m)t}dk$$

The Attempt at a Solution



How would I go about finding the wave packet's width. I'm not even sure what this means. Thank you for any guidance.
 
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I don't understand why you have an integral there. You should be able to find an analytical form for Ψ.
 
DrClaude said:
I don't understand why you have an integral there. You should be able to find an analytical form for Ψ.

Am I supposed to be deriving the the wave equation from the dispersion equation?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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