The free particle

  • Thread starter E92M3
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  • #1
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Homework Statement


[tex]\psi(x,0)=Ae^{-ax^2}[/tex]
Normalize and find:
[tex]\psi(x,t)[/tex]


Homework Equations


[tex]1=\int_{-\infty}^\infty\psi^*\psi dx[/tex]


The Attempt at a Solution


[tex]1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx[/tex]
let:
[tex]u=x\sqrt{2a}[/tex]

[tex]1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}[/tex]
Therefore:
[tex] A=(\frac{2a}{\pi})^{1/4}[/tex]

[tex]\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}[/tex]
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.
 

Answers and Replies

  • #2
gabbagabbahey
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[tex] A=(\frac{2a}{\pi})^{1/4}[/tex]

[tex]\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}[/tex]
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.

Your value of [itex]A[/itex] is correct; which leaves you with

[tex]\psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}[/tex]

How did you go from that, to your equation for [itex]\psi(x,t)[/itex]?
 

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