- #1
Hart
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Homework Statement
The solution of Schrodinger’s equation for a free particle can be written in the form:
[tex] \psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-wt)}dk [/tex]
[Q1]: Explain why the function [tex] \phi(k) [/tex] is given by:
[tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{-ikx}dx [/tex]
[Q2]: A particle is described by a wavefunction such that:
[tex] \psi(x,0) = Ne^{-\alpha|x|}[/tex] , where N is a normalisation constant.
i) Find [tex] \phi(k) [/tex] and find [tex] | \phi(k) |^{2} [/tex]
ii) By thinking of the value of k at which [tex] | \phi(k) |^{2} [/tex] begins to rapidly decrease with k, estimate the most probable range of momentum values for this state.
Homework Equations
(Within the first section and solution attempts)
The Attempt at a Solution
[Q1]: [tex] \psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-wt)}dk [/tex]
t=0 --> [tex] \psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{ikx}dk [/tex] which is the inverse Fourier transform of [tex]\phi(k)[/tex]
Therefore [tex][\phi(k)][/tex] is the Fourier transform of [tex]\psi(x,0)[/tex]
So: [tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{-ikx}dx [/tex]
.. correct method?!
[Q2]:
i)
[tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{-ikx}dx [/tex]
and:
[tex] \psi(x,0) = Ne^{-\alpha|x|}[/tex]
therefore:
[tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}Ne^{-\alpha|x|}e^{-ikx}dx [/tex]
[tex] \phi(k) = \frac{N}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{\alpha x}e^{-ikx}dx + \int_{0}^{\infty}e^{- \alpha x}e^{-ikx}dx[/tex]
.. after some calculations ..
[tex] \phi(k) = \frac{2N}{\sqrt{2\pi}}\left( \frac{\alpha}{\alpha^{2} + k^{2}} \right)[/tex]
Hence:
[tex] | \phi(k) |^{2} = \frac{4N^{2}}{2\pi}}\left( \frac{\alpha}{\alpha^{2} + k^{2}} \right)^{2}[/tex]
ii) I have no idea how to attempt this part of the question at the moment.