Physical Chemistry 2 Quiz
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A student performs a variational calculation for the ground state of a particle in a 1D box using ψt(x)=Nsin⁡(πx/L)cos⁡(πx/L)\psi_t(x) = N\sin(\pi x/L)\cos(\pi x/L) for 0≤x≤L0 ≤ x ≤ L. After computing ⟨E⟩\langle E \rangle, the result is found to be exactly 2π2ℏ2mL2\frac{2\pi^2\hbar^2}{mL^2}. How does this compare to the exact ground state energy, and what does it reveal about the trial function?
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