1
$\begingroup$

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)\,.$

$\endgroup$
4
  • $\begingroup$ What do you mean with $\phi(t)$ in your first equation? $\endgroup$
    – FrodCube
    Commented Sep 17, 2016 at 19:47
  • $\begingroup$ What is ϕ(t) ? Is it exp(i𝜔t) ? $\endgroup$
    – freecharly
    Commented Sep 17, 2016 at 19:52
  • $\begingroup$ When You solve the Schrodinger equation with separete variables you got solution like $\psi(x,t) =\psi(x)\phi(t) $ $\endgroup$ Commented Sep 17, 2016 at 20:25
  • $\begingroup$ @freecharly yes $\endgroup$ Commented Sep 17, 2016 at 20:27

1 Answer 1

1
$\begingroup$

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} $

$\endgroup$
1
  • $\begingroup$ Hope I have answered your question. $\endgroup$ Commented Sep 17, 2016 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.