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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 $R$?

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  • $\begingroup$ Is this the distribution that minimizes uncertainty? On that one note, Wikipedia has this to say: "The normal distribution saturates the [entropic uncertainty principle] inequality, and it is the only distribution with this property, because it is the maximum entropy probability distribution among those with fixed variance." Just don't get mixed up between information-theoretic entropy and Fourier/quantum standard deviations. $\endgroup$ – user10851 Feb 13 '14 at 1:42
  • $\begingroup$ There is no joint probability distribution for position and momentum of a quantum particle, precisely because these two observables do not commute. The object that most closely resembles it is the so-called Wigner function (en.wikipedia.org/wiki/Wigner_quasiprobability_distribution), which is not a bona fide probability density for not being everywhere non-negative. It is also most certainly not a (complex) Gaussian in general - this happens precisely when the uncertainty principle is minimized, as observed by Schrödinger himself. The corresponding state is called a coherent state. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 13 '14 at 1:46
  • $\begingroup$ @ Pedro Does this mean that the probability distribution of a quantum particle is an uncertain formula by itself and it is impossible to have an exact formula for the distribution for a particular specific case? for example a free particle with no potential in space. $\endgroup$ – user40229 Feb 13 '14 at 2:17
  • $\begingroup$ No, it simply means that you cannot have a probability distribution for a quantum particle in phase space. Of course you have it for all possible values of a single observable (such as position or momentum). More generally, if you have $n$ observables $O_1,\ldots,O_n$ that commute (up to domain subtleties for unbounded observables that do not concern us here), you can find a joint probability distribution for their possible values. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 13 '14 at 3:26
  • $\begingroup$ @ Pedro. Thank you. If we measure the particle position to Δx giving uncertainty in momentum ΔP=ℏ/(2Δx) - then what would be the corresponding probability distributions? $\endgroup$ – user40229 Feb 13 '14 at 3:35
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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.

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  • $\begingroup$ For a free particle in space with zero potential, what formula would represent the probability distribution derived from the Schrodinger equation? $\endgroup$ – user40229 Feb 13 '14 at 2:13
  • $\begingroup$ I've updated my answer to include more detail about the free particle case. Let me know if it's still unclear. $\endgroup$ – chase Feb 13 '14 at 2:33
  • $\begingroup$ Thank you. If we measure the particle position to $\Delta x$ giving uncertainty in momentum $\Delta P = \hbar/(2 \Delta x)$ - then what would be the corresponding probability distributions? $\endgroup$ – user40229 Feb 13 '14 at 2:42
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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 $\sigma_x\sigma_p = \frac{\hbar}{2} \times \sqrt{\frac{\pi^2}{3}-2}$

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  • $\begingroup$ Thank you. For example, a gas in a box at a given temperature and pressure, what can we say about the average uncertainty in position and momentum of the particles? What would be the average probability distributions for position and momentum of the gas atoms? $\endgroup$ – user40229 Feb 13 '14 at 4:15
  • $\begingroup$ That will depend on how the particles are interacting with each other. I have not seen such a calculation before. The UP is not that interesting for such large systems as we don't really measure or talk about individual atoms when we have a gas: its macroscopic quantities like pressure and temperature that count. Anyway, if the particles don't interact (collide), then the total wave-function is just the product of the individual wave-functions and we have the same uncertainty relation for each of the gas-particles as for the single particle in a box. $\endgroup$ – Winther Feb 13 '14 at 4:30

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