Gaussian Wavepackets At $t=0$

Question

The momentum eigenstates of a free particle are of the form $$\Psi_k(x,t) = Ae^{ikx - iE_k t/\hbar}$$ where $$E_k = \frac{k^2 \hbar^2}{2m}.$$ Suppose we assume that our quantum state is a continuous superposition of these eigenstates $$\Psi(x,t) = \int_{-\infty}^\infty g(k)\Psi_k(x,t)dk$$ where $g(k)$ is a shape distribution that is assumed to peaked around some $k=k_0$. Suppose we further assume that $g$ is a Gaussian density function $$g(k) = \frac{1}{2\pi \sigma^2}\exp\left(-\frac{(k-k_0)^2}{2\sigma^2}\right).$$ From this, we find $$\Psi(x,0) = \frac{Ae^{ik_0 x}}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty e^{ix(k-k_0)-(k-k_0)^2/2\sigma^2}dk = Ae^{ik_0x}e^{-x^2\sigma^2/2}.$$ Thus, at time $t=0$, the probability density of finding the particle at a given $x$ is $$|\Psi(x,0)|^2 = |A|^2 e^{-x^2\sigma^2}$$ which is the shape of a normal distribution centered around $x=0$. If we consider this wavepacket to correspond to a particle with an imprecisely specified momentum, this suggests that its position at $t=0$ is around $x=0$. Mathematically, all the steps here make sense. But, my problem is that as far as I can see, nowhere did we make any assumption that the particle must start at $x=0$ at time $t=0$. All we did was assume that our particle was in a superposition of momentum eigenstates (each of which specify no initial position) and assumed our shape function had a particular form that is peaked around an arbitrary number $k=k_0$. I was hoping someone could point what I am missing here and where I inherently made an assumption resulting in the particle starting at $x=0$ at time $t=0$.

Answer 1

Your wavepacket is centered around $x=0$ because you assumed that your density distribution $g(k)$ is real. Let me remind you a property of Fourier transform: if the latter is defined as $$g(x)=\int g(k)e^{ikx}dk$$ then the Fourier transform of $e^{ikx_0}g(k)$ is $$\int e^{ikx_0}g(k)e^{ikx}dk =\int g(k)e^{ik(x+x_0)}dk =g(x+x_0)$$ As a consequence, adding a phase factor to $g(k)$, $${1\over\sqrt{2\pi\sigma^2}}e^{-(k-k_0)^2/2\sigma^2}e^{-ikx_0}$$ leads to a wavepacket centered around $x_0$.