A question about the general solution to the infinite square well

hilbert-spacehomework-and-exercisesquantum mechanicsschroedinger equationwavefunction

I was working through Griffiths' Introduction to Quantum Mechanics, specifically the part about the 1D infinite square well potential (situated between $x = 0$ and $x = a$). To my understanding, this allows for multiple wave functions, each associated with a discrete level of energy:

$$\Psi_n(x, t) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\,e^{-i\frac{E_n}{\hbar}t}$$

where:

$$E_n = \frac{n^2\pi^2\hbar^2}{2a^2m}.$$

This is where it starts to get confusing to me. Does this mean that only one of these wave functions describes the particle state? Or is it that a general solution can be obtained by combining all the above possible wave functions to get the following one:

$$\Psi(x, t) = \sqrt{\frac{2}{a}}\sum_{n=1}^{+\infty}C_n\, \sin\left(\frac{n\pi}{a}x\right)\,e^{-i\frac{E_n}{\hbar}t}$$

This is how the textbook says the general solution is determined, but what's confusing me, in this case, are the $C_n$'s (the equation was taken directly from the book). How did they get here, even though they were not present in the first equations? And how are we to determine them?

Best Answer

I have now understood that the general solution is not simply a sum of the stationary states, but rather a LINEAR COMBINATION of the stationary states, thus the presence of the $C_n$'s.

I also understood that the way to compute the $C_n$'s is to use the orthogonality of the stationary states to get:

$$C_n = \int_0^a\Psi(x,0)\psi_n^*(x)dx$$