# Orthogonality of the particle in a box model

I found that for the particle in the box model, since the solutions represent the wave functions $\psi_n=A\sin{\frac{n\pi x}{L}}$ and $\psi_{n+k}=A\sin{\frac{\left(n+k\right)\pi x}{L}}$, hence the integral shown below $$\int_{-\infty}^{+\infty}A^2\sin{\frac{n\pi x}{L}}\sin{\frac{\left(n+k\right)\pi x}{L}}\mathrm{d}x=\frac{A^2}{2}\int_{-\infty}^{+\infty}\left(\cos{\frac{k\pi x}{L}}-\cos{\frac{\left(2n+k\right)\pi x}{L}}\right)\mathrm{d}x=0$$ From this I imply that $$\langle\psi_n|\psi_{n+k}\rangle =\int_{-\infty}^{+\infty}{\psi_n}^*\psi_{n+k}\mathrm{d}x=0.$$ Is this valid, only for the particle in the box model or for all potential shapes as well? But, this means that all the $\psi$ waves are vectors orthogonal to each other and electronic transition from one eigen value level $E_n$ to the other $E_{n+k}$ would not be deemed as a possibility. Yet electronic transitions do exist; am I doing something wrong or am I going wrong somewhere?

• @ubuntu_noob Be careful that you don't assume that all particles are in energy eigenstates. The state function of a particle is a superposition of eigenstates, and an energy measurement will collapse the state function into an energy eigenstate. A position or momentum measurement will change the state function. – Bill N Dec 18 '16 at 13:51
• Ahh, @AccidentalFourierTransform, sorry for that; I thought the query emanated from that specific problem; that's why I tagged it with homework-&-exercises; it's a valid conceptual query, for sure. You can indeed remove the tag for you can evaluate the post much better than me. – user36790 Dec 18 '16 at 13:56

To calculate electronic transition you need to have some time dependent potential then at least to first order the probability of transition between eigenstates is $|\int dt <\psi_k|V(t)|\psi_{k'}>e^{iE_k -E_{k'}t}|^2$ this is very different from $\int \psi_k^*\psi_{k'}dx$.