Finding $\psi(x,t)$ for a free particle starting from a Gaussian wave profile $\psi(x)$ Consider a free-particle with a Gaussian wavefunction,
$$\psi(x)~=~\left(\frac{a}{\pi}\right)^{1/4}e^{-\frac12a x^2},$$
find $\psi(x,t)$.
The wavefunction is already normalized, so the next thing to find is coefficient expansion function ($\theta(k)$), where:
$$\theta(k)=\int_{-\infty}^{\infty}  \psi(x)e^{-ikx} \,dx.$$
But this equation seems to be impossible to solve without error function (as maple 16 tells me).
Is there any trick to solve this?
 A: Your question seems rather confused,


*

*First you ask for the time evolution of the wavefunction. For this you will need to use the Schrödinger equation $i \partial \psi/\partial t= \hat H \psi $ and thus will need to know the Hamiltonian ($\hat H$).

*Second you seem to want to work out the Fourier transform of the wavefunction. This will not give you the wavefunction as a function of time but will give you the wavefunction in momentum space. The integral you want to calculate is the Fourier transform of a Gaussian which is itself a Gaussian:
$$\int_{-\infty}^{\infty} e^{-ax^2/2}e^{-i k x} \, dx \\
= \int_{-\infty}^{\infty} e^{-ax^2/2}\left(\cos{kx} - i \sin{kx} \right) \, dx .$$
The second term in the above integral is odd so will give zero. The first term is a known integral and gives
$$=\sqrt{\frac{2\pi}{a}} e^{-k^2/2 a} , $$
a Gaussian as promised with width inversey proportional to the original.


I am pretty certain Maple should also be able to calculate the integral for you as it is written in my fist line (Mathematica can), so I imagine you are just not entering it correctly.
Edit: Apologies for the first comment above. I had not seen that you had written this was for a free particle, so indeed you know the Hamiltonian, the potential is $V(x,t)=0$, and so from Schrödinger's equation we know the time evolution of the energy Eigenstates is $\psi(x,t)=e^{-i \omega t}\psi(x)$. For the free particle we have $\omega=k^2/2m$ and so you know the time evolution of the Fourier transform.
So taking the Fourier transform given above, applying the time evolution, and transforming back to position space we have
$$\psi(x,t)=\int_{-\infty}^{\infty} e^{-k^2/2 a}e^{-i\omega t}e^{ikx} \, dk
\\
=\int_{-\infty}^{\infty} e^{-\frac{k^2}{2 a}(1+iat/m)}e^{ikx}\, dk
\\
\sim e^{\frac12 \frac{x^2}{1/a+imt}}$$
as #Ron pointed out in his comment. This shows how the wavepacket spreads out with time.
