# How do you expand a wavefunction in the basis of eigenfunctions of the free particle?

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?

• Instead of a sum over $n$ you've got an integral over $k$, i.e., a Fourier transform. – Javier Jan 16 '19 at 19:08

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.
• Ok, thanks. And how would we determine those $\phi (k)$? – Quaerendo Jan 17 '19 at 13:54
• Would it be done this way? $\Psi(x, 0) = \int_{-\infty}^{\infty} \phi(k) e^{ikx} dk$, therefore $\phi(k) = \int_{-\infty}^{\infty} \Psi(x, 0) e^{-ikx} dk\space$? – Quaerendo Jan 17 '19 at 14:03