Let $\sigma_x$ and $\sigma_p$ be the standard deviation of position and momentum of a particle. The ordinary uncertainty relation tells us that in general we have


Now, consider a particle in an interval of length $L$, contrained by an infinite potential well. Since the wave function of that particle is zero at the boundary, the product $\sigma_x\,\sigma_p$ should be striktly greater that $\hbar/2$ because the particle can not be in the (gaussian) state of minimum uncertainty.

Moreover, it is not to expect that the ground state of a particle in the box is the state of minimum uncertainty. So, my question is:

What is the minimum value of the product $\sigma_x\,\sigma_p$ which a particle in the box can reach?

  • $\begingroup$ Related: physics.stackexchange.com/q/88267/2451 $\endgroup$ – Qmechanic Mar 21 '18 at 19:15
  • $\begingroup$ I am rather asking for linear combinations of eigenfunctions that minimize the uncertainty instead of expectation values correponding to particular eigenfunctions discussed in that link. $\endgroup$ – kaffeeauf Mar 21 '18 at 19:58
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    $\begingroup$ Is there a reason you aren't trying to do the problem yourself? Evaluate $\langle m|x|n \rangle$, $\langle m|x^2|n \rangle$, $\langle m|p|n \rangle$, $\langle m|p^2|n \rangle$ and construct the sums over $m$ and $n$. See if there is a minimum (if certain coefficients in the linear combinations are zero). $\endgroup$ – Bill N Mar 21 '18 at 20:28
  • $\begingroup$ The matrix elements of the squared position operator are not diagonal in the standard energy representation. That makes the problem somewhat difficult. $\endgroup$ – kaffeeauf Mar 21 '18 at 20:48
  • $\begingroup$ so you want people to solve the problem for you.. or at least give you the solution? Presumably if you well is wide enough you can fit something pretty close to a Gaussian in there... $\endgroup$ – ZeroTheHero Mar 21 '18 at 22:14

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