How to superpose an wavefunction for an infinite potential well / interval - $-d/2 <x<d/2$

Question

I know that two functions which describe the state of a particle in an infinite square (on interval $-d/2<x<d/2$) well are like:

\begin{align} \psi_{even}&= \sqrt{\frac{2}{d}}\sin\left(\frac{N\pi x}{d}\right)\quad N=2,4,6\dots\\ \psi_{odd} &= \sqrt{\frac{2}{d}}\cos\left(\frac{N\pi x}{d}\right)\quad N=1,3,5\dots \end{align}

Q1: I am a bit unsure, but correct me if I am wrong. I assume that if I want to calculate the ground state $N=1$ I have to take the odd solution and set $N=1$ like this:

$$ \psi = \sqrt{\frac{2}{d}}\cos\left(\frac{1\pi x}{d}\right) $$

but if I want to calculate for the state $N=2$ i must take the even function and set $N=2$ like this:

$$ \psi = \sqrt{\frac{2}{d}}\sin\left(\frac{2\pi x}{d}\right) $$

Is this correct? Is there any need to superpose the odd and even functions or anything like that?

Q2: Now lets say I have to calculate $\langle x^2\rangle$ for the ground state. Do I do it like this?

$$\int\limits_{-d/2}^{d/2}\sqrt{\frac{2}{d}}\cos\left(\frac{1\pi x}{d}\right) x^2\sqrt{\frac{2}{d}}\cos\left(\frac{1\pi x}{d}\right) dx$$

Or for the 1st excited state:

$$\int\limits_{-d/2}^{d/2}\sqrt{\frac{2}{d}}\sin\left(\frac{2\pi x}{d}\right) x^2\sqrt{\frac{2}{d}}\sin\left(\frac{2\pi x}{d}\right) dx$$

Q3: Are $\langle x^2\rangle$ in general the same for the intervals $-d/2<x<d/2$ and $0<x<d$?

Thank you!

Answer 1

A1: Correct. No superposition.

A2: Yes.

A3: In general no. An analogy, $\langle x \rangle=0$ for the ground state in $[−d/2,d/2]$ well; $\langle x \rangle=d/2$ for the ground state in $[0,d]$ well.

Answer 2

First Question

These are already pure energy eigenstates. There is no need to superpose them to get the ground state. Think of it like this: if the ground state was a superposition of energy eigenstates with different energies, then you could make a state with lower energy by removing some of the population from the higer-energy state.

Second Question

Those are the correct integrals.

Third Question

$\langle x^2 \rangle$ will be different. When you go from $[-L,L]$ to $[0,L]$, you're basically swapping an area with lower values of $|x|$ for one with higher values of $|x|$. This makes sense: the origin is changing, so $\langle x^2 \rangle$ should change, just like $\langle x \rangle$. It's a bit like saying the distance to the north pole is different depending on where you measure it from.

That said, the variance $\sigma_x^2=\langle x^2 \rangle-\langle x \rangle^2$ won't change when you shift the origin.