1D Infinite Square Well: Box Suddenly Increase in Size. Why the coefficients can be derived by inner producting the wave functions?

Question

I have read the question 1D Infinite Square Well: Box Suddenly Increase in Size. How treat this?. There are two key equations in the selected answer.

$$\int_0^{2L} dx \ \psi^*_m\left(x\right) \Psi\left(x,0\right) = \sum_{n=1}^\infty c_n \int_0^{2L} dx \ \psi^*_m\left(x\right) \psi_n\left(x\right) = \sum_{n=1}^\infty c_n\delta_{mn} = c_m \tag{1}$$

$$c_n = \frac{\sqrt{2}}{L} \int_0^{L} dx \ \sin\left(\frac{n \pi x}{ 2L}\right) \sin\left(\frac{\pi x}{L}\right) \tag{2} $$ In the first equation, $\psi^*_m\left(x\right)$ and $\psi_n\left(x\right)$ are orthonormal (the wave function is mutually orthonormal), so the right hand side(R.H.S.) of the equation is equal to the left hand side(L.H.S.) of the equation. However, in the second equation, $\sin\left(\frac{n \pi x}{ 2L}\right)$ and $\sin\left(\frac{\pi x}{L}\right)$ are not orthonormal through $[0, L]$. Why $c_{n}$ still equal to the L.H.S.?

Answer 1

The system suffers a sudden perturbation when the size of the well is doubled. If the perturbation occurs at $t=0$ the wave function of the system will be $\psi(0^+)=\psi(0^-)$. This means that at $t=0^+$, the wave function will still be in the ground state of an infinite well of width $L$. However, the Hamiltonian changed during the perturbation, so you wave function will no longer be an eigenstate.

You should expand you wave function in the new basis so that you can deduce the time evolution of the wave function for $t>0$. Equation (2) gives you the coefficients of the expansion into the new basis. The formula between (1) and (2) makes use of the orthonormality relation among the eigenstates of the new Hamiltonian to define $c_n$.