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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)\,.$

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

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

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