E92M3
Oct3-09, 06:57 PM
1. The problem statement, all variables and given/known data
\psi(x,0)=Ae^{-ax^2}
Normalize and find:
\psi(x,t)
2. Relevant equations
1=\int_{-\infty}^\infty\psi^*\psi dx
3. The attempt at a solution
1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx
let:
u=x\sqrt{2a}
1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}
Therefore:
A=(\frac{2a}{\pi})^{1/4}
\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}
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.
\psi(x,0)=Ae^{-ax^2}
Normalize and find:
\psi(x,t)
2. Relevant equations
1=\int_{-\infty}^\infty\psi^*\psi dx
3. The attempt at a solution
1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx
let:
u=x\sqrt{2a}
1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}
Therefore:
A=(\frac{2a}{\pi})^{1/4}
\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}
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.