Probability of finding particle in half of a box for a given state

[trelek2]

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:$$u _{n}(x)=Ae ^{ \frac{in \pi x}{a} }$$ I'm pretty sure that this is correct.
I normalised A and got $$A= \frac{1}{ \sqrt{2a} }$$, which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
$$\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)]$$
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.

 

[ygolo]

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:$$u _{n}(x)=Ae ^{ \frac{in \pi x}{a} }$$ I'm pretty sure that this is correct.
I normalised A and got $$A= \frac{1}{ \sqrt{2a} }$$, which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
$$\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)]$$
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.

Be careful that you are meeting the boundary conditions. I assume you are talking about an infinite square well. If this is the case, the wave functions that you use have to be 0 at x=a, and x=-a. I believe this gives the solutions: $$u _{n}(x)=\sqrt{\frac{2}{a}}sin ( \frac{n \pi}{2a} (x+a))$$

Now whatever $$\psi(x)$$ you use, the probability of finding the particle in region 0<x<a is $$\int_0^a |\psi(x)|^2 dx$$