Infinite Square Well

Let the position space wavefunction describing this system be given by $\Psi(x, \ t)$ . We know that we can write the wavefunction as a linear combination of energy eigenfunctions:

$\Psi(x, \ t) \ = \ \displaystyle\sum_{n} \ c_n(t) \psi_n(x)$

Now, we will have $\hat{H} \psi_n(x) \ = \ E_n \psi_n(x)$ . We also know that by the Schrödinger equation we will have:

$i\hbar \frac{d \Psi}{dt} \ = \ \hat{H} \Psi \ \Rightarrow \ i\hbar \displaystyle\sum_{n} \frac{d c_n(t)}{dt} \psi_n(x) \ = \ \displaystyle\sum_{n} c_n(t) E_n \psi_n(x)$

As each energy eigenfunction is orthonormal, we solve the general differential equation by matching up terms in the sums, getting $c_n(t) \ = \ c_n(0) e^{- i E_n t / \hbar}$ . By Born's rules, we know that the probability of finding the particle in the $n$ -th energy eigenstate is given by:

$|c_n(t)|^2 \ = \ |c_n(0) e^{- i E_n t / \hbar} |^2 \ = \ |c_n(0)|^2$

We know that at some time $t_0$ , the particle is equally likely to be found anywhere in the well. Thus, we will have $\Psi(x, \ t_0) \ = \ C$ , where $C$ is some constant. We also know that:

$\Psi(x, \ t_0) \ = \ \displaystyle\sum_{n} \ c_n(0) e^{-i E_n t_0 / \hbar} \psi_n(x)$

Thus, since each $\psi_n(x)$ is orthonormal, multiplying by $\psi^{*}_i(x)$ and integrating will cancel all terms except for the $i$ -th, where the integral of the eigenstate multiplied by its conjugate will be $1$ , leaving $c_n(0) e^{-i E_n t_0 / \hbar}$ , where we multiply by its conjugate to get the final probability $|c_n(0)|^2$ (the exponential cancels with its conjugate). Thus, we have:

$c_1(0) e^{-i E_1 t_0 / \hbar} \ = \ \displaystyle\int_{0}^{a} \ \Psi(x, \ t_0) \psi^{*}_1(x) \ dx$

By the normalization condition, we can find $C$ :

$\displaystyle\int_{0}^{a} \ |\Psi(t_0, \ x)|^2 \ dx \ = \ \displaystyle\int_{0}^{a} \ |C|^2 \ dx \ = \ 1 \ \Rightarrow \ C \ = \ \frac{1}{\sqrt{a}}$

Now, all that's left to do is to solve the time-independent Schrödinger equation to get $\psi^{*}_1(x)$ . I'm not going to go through this, but we end up getting:

$\psi_n(x) \ = \ \sqrt{\frac{2}{a}} \ \sin \Big( \frac{n \pi x}{a} \Big)$

Thus, we get:

$c_1(0) e^{-i E_1 t_0 / \hbar} \ = \ \displaystyle\int_{0}^{a} \ \frac{1}{\sqrt{a}} \sqrt{\frac{2}{a}} \ \sin \Big( \frac{\pi x}{a} \Big) \ dx \ = \ \frac{2 \sqrt{2}}{\pi}$

Thus, we have:

$|c_1(0) \ = \ \frac{8}{\pi^2} \ \approx \ 0.81056946913$