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If we have an initial state given by $ \Psi(x,0) $ and we want to find $ \Psi(x,t) $, we would expand the function in the basis of eigenstates of the Hamiltonian, $\{\psi_n\}$:

$ \Psi(x,t)=\sum _nC_n \psi _n(x)e^{-i E_nt/\hbar}$, with $C_n=(\psi_n(x), \Psi(x,0))$.

However, in the case of a free particle, the eigenstates of the Hamiltonian are

$\psi _k=Ae^{ikx}+Be^{-ikx}$

So, now, the basis is not discrete. Then, how could we find out the time-dependent state, $ \Psi=(x,t) $? How could we expand the wavefunction in the basis of eigenfunctions of the free particle?

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    $\begingroup$ Instead of a sum over $n$ you've got an integral over $k$, i.e., a Fourier transform. $\endgroup$ – Javier Jan 16 at 19:08
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With a continuous spectrum, we can take the integral, i.e. $$\Psi(x, t) = \int_{-\infty}^{\infty} \phi(k) e^{ikx - i\frac{\hbar k^2}{2m} t} dk,$$ where $\phi(k)$ are analogous to the coefficients $C_n$. This is also the well-known Fourier transform.

Actually, this is pretty hand-wavey, because the eigenfunctions of the free particle aren't normalizable, so can't be the "basis" of our Hilbert space.

See, for example, Rigged Hilbert space and QM

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  • $\begingroup$ Ok, thanks. And how would we determine those $ \phi (k)$? $\endgroup$ – Quaerendo Jan 17 at 13:54
  • $\begingroup$ Would it be done this way? $\Psi(x, 0) = \int_{-\infty}^{\infty} \phi(k) e^{ikx} dk$, therefore $\phi(k) = \int_{-\infty}^{\infty} \Psi(x, 0) e^{-ikx} dk\space$? $\endgroup$ – Quaerendo Jan 17 at 14:03

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