# 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?

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I am a bit confused, why are you trying to find $\psi (k)$ ? Or as you write it, $\theta (k)$ ? –  DJBunk Sep 20 '12 at 23:06
Can you put what you ran through maple 16? –  Magpie Apr 8 '13 at 1:38

• 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.

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The fourier trasnform evolves by simple phases, and a reverse fourier transform gives the time evolution, which is a spreading Gaussian, so that the a gets replaced everywhere by ${1\over {(1/a)+it}}$ –  Ron Maimon Sep 21 '12 at 6:48
Oh yeah, hadn't seen the part saying this was for a free particle (doh!). Have added an edit to the answer to complete it. Thanks for pointing that out. –  Mistake Ink Sep 21 '12 at 13:27