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?

## Best Answer

There are three answers posted but so far nobody has posted the obvious physical interpretation. The energy eigenstates all have the particle spread out within the box, stationary in time. If you want the particle to bounce back and forth between the walls of the box, then you do this by combining eigenstates. The simplest case is just to mix the ground with the first excited state. If you look carefully at the resulting function, you should see that it bounces back and forth between the left and right hand sides of the box.

In this example, the wave function still isn't very precisely localized at any instant, but if you want to do better you just add more eigenstates.