Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.