Answer to Question #203739 in Quantum Mechanics for Azam

Question

Calculate the ground state energy for delta function potential by making use of variational principle.

Answer 1

$V=\alpha/x/$

gaussian trial function

$\psi=exp(-\beta x^2)$

Since this wave function is not normalized , first we will normalize this wave function.

$\int ^\infty_{-\infty}\psi*\psi dx=2A^2 \int^\infty_0 exp(-\beta x^2)dx=1$

$2A^2 \frac{1}{2}(\frac{\pi}{2\beta})^\frac{1}{2}=1=A^2=\sqrt{\frac{2\beta}{\pi}}$

$<V>=2\alpha A^2\int^\infty_0 xe^{-2\beta x^2}dx==2\alpha A^2(-\frac{1}{4\beta}exp(-2\beta x^2))/_0^\infty=\frac{\alpha A^2}{2\beta}$

$=\frac{\alpha}{2\beta} \sqrt{\frac{2\beta}{\pi}}=\frac{\alpha}{\sqrt{2\beta \pi}}$

$<H>=<T>+<V>= \frac{h^2\beta}{2m}+\frac{\alpha}{\sqrt{2\beta\pi}}$

To determine $\beta$

$\frac{\delta <H>}{\delta \beta}=\frac{h^2}{2m}-\frac{1}{2}\frac{\alpha}{\sqrt{2\pi }} \beta ^{-3/2}=0$

$\beta^{3/2}=\frac{\alpha}{\sqrt{2\pi}} \frac{m}{h^2}=\beta=(\frac{m\alpha}{\sqrt{2\pi h^2}})^{2/3}$

$E_{min}=<H>_{min}=\frac{h^2}{2m} (\frac{m\alpha}{\sqrt{2\pi h^2}})^{2/3} +\frac{\alpha}{\sqrt{2\pi}}(\frac{\sqrt{2\pi h^2}}{m\alpha}) ^{1/3}$