= Solution
For <observables> with finite <variances>, the <Heisenberg uncertainty principle> gives $\Delta A\Delta B\geq|\langle[A,B]\rangle|/2$; in particular $\Delta x\Delta p\geq\hbar/2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and <Bose-Einstein statistics>; half-integer-spin particles have antisymmetric full states and <Fermi-Dirac statistics>. The latter imply Pauli exclusion from one complete single-particle state, including <spin>.
Inside an <infinite square well>, solutions satisfying both zero boundary conditions are
$$
\boxed{\psi_n(x)=\sqrt{\frac2L}\sin\frac{n\pi x}{L},\qquad
E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}},\quad n=1,2,\ldots,
$$
with zero <wavefunction> outside. For the <ground state>, direct integration gives
$$
\langle x\rangle=\frac L2,\quad
\langle x^2\rangle=L^2\left(\frac13-\frac1{2\pi^2}\right),\quad
\langle p\rangle=0,\quad\langle p^2\rangle=\frac{\pi^2\hbar^2}{L^2}.
$$
The <infinite-well ground-state uncertainty product> follows:
$$
\boxed{\Delta x=L\sqrt{\frac1{12}-\frac1{2\pi^2}},\quad
\Delta p=\frac{\pi\hbar}{L},\quad
\Delta x\Delta p=\hbar\sqrt{\frac{\pi^2}{12}-\frac12}\simeq0.568\hbar>\frac\hbar2}.
$$
The <ground state> is not a minimum-uncertainty Gaussian. Here <momentum> <variance> is the usual kinetic-momentum second moment of the confined state; the hard-wall <momentum operator> has domain subtleties that do not alter this kinetic-energy result.
For <spin degeneracy in filling a one-dimensional well>, let $E_1=\pi^2\hbar^2/(2mL^2)$ and assume the <spin> levels are degenerate and available. Four spin-zero <bosons> all occupy the lowest orbital, giving \b[$4E_1$]. Four spin-$1/2$ <fermions> fill two <spin> states in each of $n=1,2$, giving \b[$10E_1$]. Spin-$3/2$ <fermions> have four <spin> projections, so all four can occupy $n=1$ in different complete states, giving \b[$4E_1$]. These fillings do not impose full <spin> polarization.
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