Gaussian Probability Distribution? The uncertainty principle states that,
$$\sigma _{{x}}\sigma _{{p}}\geq {\frac {\hbar }{2}}.$$
It is mentioned from many sources that the probability distribution of the particle position and momentum would follow a Gaussian distribution.
Why is it a Gaussian distribution? is this the distribution that minimizes uncertainty? Is this distribution definitely the case for the uncertainty principle or can it be different under different conditions? Has this been proven?
What are the formulas for position and momentum probability distributions of a free particle? How is this derived from the wave function? What would be the formulas of the probability distributions for the position and momentum for a system of 2 identical bosons separated by a distance $R$?
 A: I believe this can be attributed to the central limit theorem, which states that a large number of samples from a population with a well-defined variance will follow a gaussian distribution. The key idea is that because of quantum mechanics, we must treat both position and momentum as random variables; the uncertainty principle gives us a relation between the variance of the two quantities.
We cannot talk about the "formula for position" per se; however we can derive a deterministic formula for the wavefunction, which represents the probability density for these random variables. The exact form of the wavefunction is dependent on the problem, but can (in principle) generally be obtained from the Schrödinger equation.
Wikipedia has a good writeup for the free particle. The Hamiltonian for a free particle with fixed momentum $\mathbf{p}$ is $\mathcal{H} = \mathbf{p}^2/2m$ (the potential is zero). Eigenstates of this Hamiltonian are plane-waves in position-space (that is, their wavefunctions oscillate throughout space and time):
$$
\psi(\mathbf{x}, t) = Ae^{i(\mathbf{x}\cdot\mathbf{p}-Et)/\hbar}
$$
that means that the probability distribution is simply:
$$
\left|\psi(\mathbf{x},t)\right|^2 = \left|A\right|^2
$$
which is a constant independent of position $\mathbf{x}$. Note that this wavefunction cannot be renormalized to unity, but the takeaway is that the particle is equally likely to be anywhere. This is consistent with the uncertainty princple: since we specified $\mathbf{p}$ exactly ($\sigma_p=0$), the uncertainty in position is infinite.
For more complex systems, the Hamiltonian is not always exactly known; this is often the case in multi-particle systems, such as atoms. In still other cases, the Hamiltonian is known but cannot be solved analytically.
A: It is not correct that the probability distribution of $x$ and $p$ are Gaussian in general.
Take a simple system of a particle moving in some potential $V(x)$.
The probability distribution of $x$ is the square of the wave-function $\Psi(x)$ of the particle, i.e. the probability of finding your particle in $[x,x+dx]$ is $|\Psi(x)|^2dx$. 
The probability distribution of $p$ is the square of the momentum-space wave-function $\Psi(p) = \int dx \Psi(x) e^{ipx}$ (the Fourier transform of $\Psi(x)$).
It is only when the wave-function $\Psi(x)$ is a Gaussian that the uncertainty principle is minimized, i.e. $\sigma_x \sigma_p = \frac{\hbar}{2}$. See (http://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle) for a proof that $\frac{\hbar}{2}$ is the lower limit.
Now why exactly the Gaussian?  To minimize the uncertainty product we need a wave-function that is sufficiently well localized both in both real space and Fourier space. If we squeeze a function in real space and it broadens in Fourier space and vice versa. The Gaussian happens to be the unique function that maintains its 'shape' when Fourier transformed, i.e. the Fourier transform of a Gaussian (with variance $\sigma^2$) is just another Gaussian (with variance $1/(4\sigma^2)$) and the product of variance (uncertainty) remains a constant independent of $\sigma$.
Finally, there exist many systems where the uncertainty principle is not minimized. The simplest example is a 'particle in a box' (http://en.wikipedia.org/wiki/Particle_in_a_box). Here the ground state has $\sigma_x\sigma_p = \frac{\hbar}{2} \times \sqrt{\frac{\pi^2}{3}-2}$
