# Time Independent Schrödinger Equation - Infinite Square Well

I'm studying Quantum Mechanics for one week and I don't understand one thing about wave function harmonics. If we have a particle in an infinite square well with an initial wave function $\psi(x,0),$ we can calculate the time evolution($\psi(x,t)$), knowing all the $\psi_n(x)$ do a Fourier series.

My question is, why i don't simply calculate $$\psi(x,t) = \sum c_n \psi(x,0) \phi(t)$$ and we must calculate $\psi(x)$ which in infinite square Well is $$\psi_n(x)=A \sin\left(n \frac{\pi}{a}x\right),$$ $A$ is normalization constant and a the well length.

Why $\psi(x,0)$ is different than $\psi_n(x)\,.$

• What do you mean with $\phi(t)$ in your first equation? – FrodCube Sep 17 '16 at 19:47
• What is ϕ(t) ? Is it exp(i𝜔t) ? – freecharly Sep 17 '16 at 19:52
• When You solve the Schrodinger equation with separete variables you got solution like $\psi(x,t) =\psi(x)\phi(t)$ – Tiago Portela Sep 17 '16 at 20:25
• @freecharly yes – Tiago Portela Sep 17 '16 at 20:27

Yo,

So you can't directly write $$\Psi= \Sigma c_n \psi(x,0) \phi(t)$$ because its really $\phi_n (t)$ for each $n$.

So usually when people write $\psi(x,0) =f(x)$ what they mean to say is that the initial wave function is some combination of states, $\psi_n (x)$ at $t=0$.

Now by finding the constants: $c_n$ of each of the states $n$ , we can write

$$\Psi= \Sigma c_n \psi(x)_n \phi(t)_n$$

we use the Fourier trick: $c_n = \int_{- \infty}^{\infty} f (x,0) \psi_n(x)dx$ , and that's the answer.

N.B: $\phi_n (t) = e^{i\omega_n t}$ where $\omega_n = {E_n \over \hbar}$

• Hope I have answered your question. – Haru Fujimura Sep 17 '16 at 23:37