Discussion
The Hamiltonian \(H = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} - \alpha\delta(x)\) is given for a particle. Using the trial wave function \(\psi_{trial}(x) = Ae^{-bx^{2}}\) (with b as the variational parameter), the bound on the ground state energy is :
(Given : \(\int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^{2}}\,dx\) = \(\Gamma(n+1/2)/\alpha^{n+1/2}\))
Answer: C
The bound is \(-m\alpha^{2}/\pi\hbar^{2}\).
Normalising the Gaussian gives \(|A|^{2} = \sqrt{2b/\pi}\). The kinetic and potential terms are:
\[ \begin{aligned} \langle T \rangle &= \frac{\hbar^{2}b}{2m} \\ \langle V \rangle &= -\alpha|\psi(0)|^{2} = -\alpha\sqrt{\frac{2b}{\pi}} \end{aligned} \]
Setting \(dE/db = 0\) gives \(\sqrt{b} = \frac{m\alpha}{\hbar^{2}}\sqrt{\frac{2}{\pi}}\), and then:
\[ \begin{aligned} E_{min} &= \frac{m\alpha^{2}}{\pi\hbar^{2}} - \frac{2m\alpha^{2}}{\pi\hbar^{2}} \\ &= -\frac{m\alpha^{2}}{\pi\hbar^{2}} \end{aligned} \]
The option \(-m\alpha^{2}/2\hbar^{2}\) is the exact ground-state energy of the delta well. The variational bound lies above it, as it must, since \(1/\pi \lt 1/2\). The positive option cannot describe a bound state.
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