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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 $$\psi_2(x) = \sqrt{\frac{2}{L}}\sin(\frac{2\pi x}{L})?$$

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Yes, n is the quantum number of your problem, it labels the eigenvalue of the energy. In dimension without spin, I don't what more you could expect. – Learning is a mess Feb 19 '13 at 9:17

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.

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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.

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