Discussion
Consider a particle in a two dimensional infinite square well potential of side \(L\), with \(0 \le x \le L\) and \(0 \le y \le L\). The wavefunction of the particle is zero only along the line \(y = \frac{L}{2}\), apart from the boundaries of the well. If the energy of the particle in this state is \(E\), what is the energy of the ground state?
Answer: B
The ground-state energy is \(\frac{2}{5}E\).
In a square well of side \(L\), \(\psi = \sin\frac{n_{x}\pi x}{L}\sin\frac{n_{y}\pi y}{L}\) and \(E = E_{0}(n_{x}^{2} + n_{y}^{2})\) with \(E_{0} = \pi^{2}\hbar^{2}/2mL^{2}\). A node only along \(y = L/2\), and none along any line of constant \(x\), means \(n_{x} = 1\) and \(n_{y} = 2\):
\[ \begin{aligned} E &= E_{0}(1 + 4) = 5E_{0} \\ E_{ground} &= E_{0}(1 + 1) = 2E_{0} = \frac{2}{5}E \end{aligned} \]
1/4 comes from comparing only the \(y\) parts (1 against 4), and 1/2 and 3/8 match no pair of states.
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