Discussion
Two identical spin-1/2 fermions are placed in a 1-dimensional square potential well with infinitely high walls, V = 0 for \(0 \le x \le L\), otherwise \(V = \infty\). The normalized single particle energy eigenstates are \(u_{n}(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)\); n = 1, 2, 3, ... . Let s be the quantum number corresponding to total spin \(\vec{S}\). The ground state energy of the system is given by :
Answer: C
The ground state energy is \(\frac{\hbar^{2}\pi^{2}}{mL^{2}}\) with \(s = 0\).
Both fermions can sit in the lowest level \(n = 1\) only if their total wavefunction is antisymmetric. The spatial part \(u_{1}(x_{1})u_{1}(x_{2})\) is symmetric, so the spin part must be the antisymmetric singlet, \(s = 0\). The energy is
\[ \begin{aligned} E &= 2E_{1} = 2\cdot\frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &= \frac{\pi^{2}\hbar^{2}}{mL^{2}} \end{aligned} \]
A triplet (\(s = 1\)) needs an antisymmetric spatial state, so the lowest triplet puts one particle in \(n = 1\) and one in \(n = 2\), with energy \(5\pi^{2}\hbar^{2}/2mL^{2}\). That is higher, and it belongs to \(s = 1\), not \(s = 0\) as the last option states. The option \(\pi^{2}\hbar^{2}/mL^{2}\) with \(s = 1\) breaks the Pauli principle, and \(5\pi^{2}\hbar^{2}/4mL^{2}\) has half the correct triplet energy.
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