How Does a Free Particle's Wave Function Evolve Over Time?

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SUMMARY

The discussion focuses on the evolution of a free particle's wave function in one dimension, specifically for a particle of mass m with an initial state represented by \(\Psi(x,0) = \sin(k_{0}x)\). The correct approach involves using the Fourier transform to express the wave function over time, given by \(\Psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} b(k)e^{i(kx-\omega t)} dk\). The transformation of the sine function into exponential form, \(\sin(kx) = \frac{1}{2i}(e^{ikx}-e^{-ikx})\), is crucial for solving the integral. The final solution must satisfy the Schrödinger equation and the initial normalization condition.

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Homework Statement



free particle of mass m moving in 1d
state: [tex]\Psi(x,0) = sin(k_{0}x)[/tex]

Homework Equations




[tex]\Psi(x,t) = \stackrel{1}{\overline{\sqrt{2\pi}}}\overline{}[/tex][tex]\int^{\infty}_{-\infty}b(k)e^{i(kx-\omega t)}[/tex]

The Attempt at a Solution



b(k)=[tex]\stackrel{1}{\overline{\sqrt{2\pi}}}\overline{}[/tex][tex]\int^{\infty}_{-\infty}sin(k_{0}x)e^{-ikx}[/tex]
 
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The problem here is that you have sine instead of [tex]e^{ikx}[/tex] factor.
Use the fact that:
[tex]sin(kx) = \frac{1}{2i}(e^{ikx}-e^{-ikx})[/tex]

Second your answer should be:
[tex]\Psi(x,t) = ...[/tex]
not
[tex]b(k) = ...[/tex]
Look up the source of your "relevant equation" to see what role [tex]b(k)[/tex] plays...also you should give the variable of integration which appears to be k.

Your attempted solution is I suppose integrated over x? The integral you give has a known solution in terms of Dirac delta functions (which relates to my point above)

Remember ultimately you are looking for a solution to the Schrödinger equation which a.) is properly normalized and b.) satisfies the initial condition given.
 

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