A particle in an infinite square well has an initial wavefunction
$$\psi (x,0) ~=~ Ax(a-x) \qquad \mathrm{for}\qquad 0\leq x\leq a.$$
Now the question is to calculate $\psi (x,t)$.
I have normalised it and calculated the value of A in terms of $a$. Now I do not know how to proceed.
Any help please?
P.S. I am from a non-physics background.
Best Answer
For a particle in a box, you know the behavior of special solutions:
$$ \psi_0(x)= \sin(kx)$$
Where $k={\pi n\over a}$, where n is an integer, evolves into
$$ \psi(x,t) = e^{- i\omega t} \sin(kx)$$
where $\omega = {k^2\over 2m}$ (up to $\hbar$ factors, I have set $\hbar=1$, as you should for these type of things).
You can expand your wavefunction $x(A-x)$ in the interval between 0 and A, as an infinite series of these sine waves, this is the Fourier series. Since you know how each sine wave evolves, you know how the whole thing evolves, since the Schrodinger equation is linear.
The linear property says that in a sum of initial conditions, each term in the sum time evolves independently, and then adds up to the time evolution of the sum.
I will stop here, because this looks like homework. But for a tip about how to do Fourier series efficiently, the Fourier series of a delta function is simple, and if you integrate a delta function twice (with the appropriate integration constants) you find your initial condition. You can also find the Fourier coefficients by brute force, but it takes a little bit of integration by parts.