Infinite Square Well, finding Psi(x,t)

[Cogswell]

Homework Statement

A particle in the infinite square well has the initial wave function $\Psi (x, 0) = Ax(a-x), (0 \le x \le a)$, for some constant A. Outside the well, of course, $\Psi = 0$.
Find ## \Psi (x,t)

2. Homework Equations :
Equation [1.0]:$\displaystyle c_n = \sqrt{\dfrac{2}{a}} \int_0^a \sin \left( \dfrac{n \pi x}{a}\right) \Psi (x,0)dx$3. The answer (that is given):
First, determine A by normalising $\Psi (x,0)$

$\displaystyle 1 = \int_0^a | \Psi(x,t)|^2 dx$

Therefore $A = \sqrt{\dfrac{30}{a^5}}$

The nth coefficient is (Equation [1.0])

$\displaystyle c_n = \sqrt{\dfrac{2}{a}} \int_0^a \sin \left( \dfrac{n \pi x}{a}\right) \sqrt{\dfrac{30}{a^5}}x(a-x)dx$

... skipped working...

$\displaystyle c_n = \dfrac{4 \sqrt{15}}{(n \pi)^3}[\cos(0) - \cos(n \pi)]$

Therefore:

$c_n = 0$ if n is even

$c_n = \dfrac{8 \sqrt{15}}{(n \pi)^3}$ if n is oddThus:

$\displaystyle \Psi (x,t) = \sqrt{\dfrac{30}{a}} \left( \dfrac{2}{a} \right) \sum \limits_{n=1, 3, 5...} \dfrac{1}{n^3} \sin \left( \dfrac{n \pi x}{a} \right) e^{(-in^2 \pi^2 \hbar t) / (2ma^2)}$
4. The attempt at understanding it:

I get the normalisation part, and I get how they do the nth coefficient (hence I omitted the working) but I don't get why.

I thought all I had to do was normalise the $\Psi (x,0)$ and then add the time dependent factor to it, so shouldn't I get:

$\Psi (x,t) = \sqrt{\dfrac{30}{a^5}} x (a-x) \cdot e^{(-in^2 \pi^2 \hbar t) / (2ma^2)}$

Can someone explain this to me (in simple terms)?

 

[TSny]

Hello, Cogswell.

If you look at your proposed solution, you can see a difficulty. What value of $n$ are you going to use in the time dependent factor?

Only the energy eigenstates $\sin \left( \dfrac{n \pi x}{a}\right)$ evolve in time according to an exponential factor Exp$(-iE_n t/\hbar)$. Each energy eigenstate has its own time factor. So, the idea is to write the initial wavefunction as a superposition of energy eigenstates and then let each of those eigenstates evolve in time with its own time factor.