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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?

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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 $a$.

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  • $\begingroup$ Is there a similar wavefunction involving the particles momentum? $\endgroup$ – Anders Gustafson Sep 22 '18 at 6:10
  • $\begingroup$ 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. $\endgroup$ – Al Nejati Sep 22 '18 at 6:23
  • $\begingroup$ Do you mean the one with $\psi(k,t)$ $\endgroup$ – Anders Gustafson Sep 22 '18 at 7:05
  • $\begingroup$ correct, both momentum and position space solutions are given. $\endgroup$ – Al Nejati Sep 22 '18 at 8:04

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