# [Physics] Physical meaning of linear combination of possible states in infinite well

hilbert-spacequantum mechanicsschroedinger equationsuperpositionwavefunction

The solution of infinite well, positioned from $x=0$ $x=l$, is
$$\Psi_n(x,t)= \sqrt{\frac{2}{l}}\sin\left(\frac{n\pi}{l}x\right)e^{iE_nt}$$
But the most general solution of this problem is :
$$\Psi(x,t)= \displaystyle\sum_{n=1}^{\infty} C_n\sqrt{\frac{2}{l}}\sin\left(\frac{n\pi}{l}x\right)e^{iE_nt}$$
and
$$\Psi(x,0)=\displaystyle\sum_{n=1}^{\infty}C_n\Psi_n(x,0)$$
summing up linearly all possible states give me the general solution. But what does it mean physically? Does the summation mean all possible states constitute a wave packet in the well?

Actually I can go further and using Fourier transform show that ( shortly I will say $\Psi(x)$ for $\Psi(x,0)$ )
$$\int\Psi(x)=\displaystyle\sum_{n=1}^{\infty} \int C_n\Psi_n(x)$$
$$\int \Psi_m(x)^*\Psi(x)dx = \displaystyle\sum_{n=1}^{\infty} C_n\int \Psi_m(x)^*\Psi_n(x)dx$$
$$\int \Psi_m(x)^*\Psi_n(x)dx =\delta_{mn}$$
$$\int \Psi_m(x)^*\Psi(x)dx = C_m$$
This equation makes me not much sense. If you integrate an energy eigenstate with the general state, you get the probability of this eigenstate, if you take the square of constant $C_m$. Why?