- #1
trelek2
- 88
- 0
Basically this is easy stuff I'm doing but i'n not sure whether i made a mistake or my answer is actually correct physically.
I'm solving the 1-dimensional TISE for a particle confined to a box -a<x<a.
I found that the eigenfunctions for this particle in a box with periodic boundary conditions are:[tex]u _{n}(x)=Ae ^{ \frac{in \pi x}{a} } [/tex] I'm pretty sure that this is correct.
I normalised A and got [tex]A= \frac{1}{ \sqrt{2a} } [/tex], which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
[tex]\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)] [/tex]
I get that it is 1/2-1/pi which is about 0.2... Is that correct? Or should it be rather 1/2 simply.
I'm solving the 1-dimensional TISE for a particle confined to a box -a<x<a.
I found that the eigenfunctions for this particle in a box with periodic boundary conditions are:[tex]u _{n}(x)=Ae ^{ \frac{in \pi x}{a} } [/tex] I'm pretty sure that this is correct.
I normalised A and got [tex]A= \frac{1}{ \sqrt{2a} } [/tex], which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
[tex]\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)] [/tex]
I get that it is 1/2-1/pi which is about 0.2... Is that correct? Or should it be rather 1/2 simply.