Particle in an infinite square well - interval -d/2<x<d/2

[71GA]

Homework Statement

Particle is in an infinite square well of width $L$ on an interval $-L/2<x<L/2$. The wavefunction which describes the state of this particle is of form:
$$\psi = A_0\psi_0(x) + A_1\psi_1(x)$$
where $A_1=1/2$ and where $\psi_0$ and $\psi_1$ are ground and first excited state. What are $\psi_0$ and $\psi_1$ like? Write them down. Calculate $A_0$.

Homework Equations

I allways used this equation to get normalisation factors:
$$\int\limits_{-d/2}^{d/2} \psi =1$$
If i recall corectly the eigenfunctions are orthogonal? I remember something about this and i think there is different way too solve this problem using this orthogonality. Please someone show me, because i am interested in both ways!

The Attempt at a Solution

I know that normalisation factor should be $A_0=\sqrt{3}/2$ but what i got was totaly different. I think i chose $\psi_0$ and $\psi_1$ wrong. Here is how i got the wrong result:
http://shrani.si/?f/OH/aKLhcp9/582013-01837.jpg

Is it possible that I should use $\psi_0(x) = \cos \left(\frac{(2N-1)\pi}{d}x\right)$ where $\boxed{N=1}$ and $\psi_1(x) = \sin \left(\frac{2N\pi}{d}x\right)$ where $\boxed{N=1}$?

 

[TheShrike]

You're correct that your initial eigenfunctions are wrong, but your suggested corrections are not quite right either (but really close! :) ). N labels the energy states, so that the ground state is N=1 and the first excited state is N=2.

Using that in your new eigenfunctions should result in the correct answer. I didn't go in detail through your method though.

Edit: Whoops, don't forget the \sqrt{\frac{2}{L}} factor in those new eigenfunctions also!

 

[71GA]

Whoops, don't forget the $\sqrt{2/L}$ factor in those new eigenfunctions also!

What i thought was: "Oh! $A_0$ and $A_1$ are already there and are playing the role of $\sqrt{2/L}$". I guess i missread the instructions...

I don,t understand the why N=2 though. If i plug in the N=1 in the equations i provided (i added factor of $\sqrt{2/L}$) I get $\psi_0(x) = \sqrt{2/L}\cos \left(\frac{1\pi}{d}x\right)$ (this should be the ground state) and $\psi_1(x) =\sqrt{2/L} \sin \left(\frac{2\pi}{d}x\right)$ (this should be first excited state).

 

[vela]

Typically, you label the states n=1, n=2, n=3, and so on. You don't use the same n for different states. The eigenfunctions would be given by
$$\phi_n(x) = \begin{cases}
\sqrt{\frac{2}{L}}\cos \frac{n\pi x}{L}, & n\text{ odd} \\\\
\sqrt{\frac{2}{L}}\sin \frac{n \pi x}{L}, & n\text{ even}
\end{cases}$$ for $-L/2 \le x \le L/2$ and 0 elsewhere. Following this convention for labeling the states, the energy of each state is given succinctly by
$$E_n = \frac{\hbar^2 n^2 \pi^2}{2mL^2}.$$

 

[71GA]

If i understand your point you wanted to say that my sugestion would have worked, but is not written in accordance to convention?

 

[vela]

I was just clarifying what TheShrike meant. You can use whatever convention you want as long as it works for you.

 

[71GA]

Yes i know that. But would my own convention work?

 

[71GA]

It worked and i got the solution $A_0 = \sqrt{3}/2$. Thank you all for your help.

 

[TheShrike]

What i thought was: "Oh! $A_0$ and $A_1$ are already there and are playing the role of $\sqrt{2/L}$". I guess i missread the instructions...

I don,t understand the why N=2 though. If i plug in the N=1 in the equations i provided (i added factor of $\sqrt{2/L}$) I get $\psi_0(x) = \sqrt{2/L}\cos \left(\frac{1\pi}{d}x\right)$ (this should be the ground state) and $\psi_1(x) =\sqrt{2/L} \sin \left(\frac{2\pi}{d}x\right)$ (this should be first excited state).

Oh, I see what you were doing now. A conflict of conventions, as vela stated. My mistake.

Looks like you had the answer all along.

 

[71GA]

My mistake.

Nope. The mistake is always a comunication :). If you were here and i could ask you in person we would solve the problem in first two sentances. But writing and reading is sometimes blah :). We should create some sort of a Skype Q&A site ;)