Equation for probability amplitude of a free particle given a mean position, a mean velocity, and the mass of the free particle?

The uncertainty principle can be expressed using the equation $$\sigma_x\sigma_p\geq\frac{h}{4\pi}$$ with $$\sigma_x$$ being the uncertainty in position, $$\sigma_p$$ being the uncertainty in momentum, and $$h$$ being the plank constant. The uncertainty in velocity would be given by the equation $$\sigma_v=\frac{\sigma_p}{m}$$ with $$\sigma_v$$ being the uncertainty in velocity and $$m$$ being the mass. So the uncertainty principle could also be expressed using the equation $$\frac{\sigma_x\sigma_v}{m}\geq\frac{h}{4\pi}$$.

Assuming that both the uncertainty in position and uncertainty in momentum are both at a minimum what is the equation for the probability amplitude of a free particle at a chosen position and momentum, given the mean position, and mean momentum of that particle?

A usual definition of the wavefunction of a free particle is simply $$\psi(x) = \exp(-ikx)$$. This of course corresponds to a particle with definite momentum that is infinitely 'smeared out' in space ($$\delta p = 0$$, $$\delta x = \infty$$). I assume you're asking about the wavefunction of a free particle with the constraint that it's localized in space. A unique localized particle wavefunction does not exist because it depends on how you write down your Hamiltonian. You can get for instance a 'wave packet'-like wavefunction, defined as (in 1D):
$$\Psi(x,t) = \left({a \over a + i\hbar t/m}\right)^{3/2} e^{- {x^2\over 2(a + i\hbar t/m)} }.$$
Which gives a Gaussian wave packet, with width $$a$$.
• You can take the Fourier transform to get the wavefunction in momentum space, it turns out to have a very similar form, but with the width being replaced by $\frac{a}{\delta x}$. Refer to this wiki page. – Al Nejati Sep 22 '18 at 6:23
• Do you mean the one with $\psi(k,t)$ – Anders Gustafson Sep 22 '18 at 7:05