So the set of solutions for the particle in a box is given by $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L}).$$
In Dirac notation $<\psi_i|\psi_j>=\delta_{ij}$ assuming $|\psi_i>$ is orthonormal. My question is, are $\psi_i$ and $\psi_j$ simply corresponding to different values of $n$ for the above set of solutions?
For instance would $$\psi_1(x) = \sqrt{\frac{2}{L}}\sin(\frac{1\pi x}{L})$$ and $$\psi_2(x) = \sqrt{\frac{2}{L}}\sin(\frac{2\pi x}{L})?$$
The labels $i,j$ can correspond to any label that labels the set of orthonormal states $\{| \psi_i \rangle\}_i$. In your case, your states are energy eigenstates with different eigenvalues, so they are indeed orthogonal (by some linear algebra theorem), and I guess you've normalized them properly, so they are one possible choice. But you can change basis orthogonally
$$|u_i \rangle \equiv O_{ij} |\psi_j \rangle$$
and you'd still have the relation $\langle u_i|u_j \rangle = \delta_{ij}.$
For example, on a finite lattice you have momentum eigenstates $|\mathbf{p} \rangle$, but you can also Fourier transform to get position eigenstates $|\mathbf{x} \rangle.$ The latter aren't energy eigenstates, but are still orthogonal.
Your equation is the solution to Schrodinger's equation that describes a particle in a 1-D "box" of length L. The particular solution you have written represents the odd states. The integer-valued parametre, n, labels the quantum levels at which the particle can be inside the box. In each of these quantum states, 1, 2, 3,...$i$,...$j,...n,...$ the particle has different amount of energy, which are given by the equation
$E_n=\frac{h^2}{2mL^2}n^2$ where $n=1,2,3,...,i,...j,...n,...$.
A quantum state corresponding to the energy level $j$ is sybolically indicated by $|j>$ and is given by your equation with replacing $n$ by $j$, as you have done for $n=1$ and $n=2$. The particular expression you have written, $<\psi_i|\psi_j>$, represents the transition probability amplitude for the particle to jump from quantum state $|\psi_j>$ to $|\psi_i>$. The latter is zero since these wave-functions form an orthonormal set of quantum states, and you are showing this with the $\delta_{ij}$ symbole.
