Question

Quantum Mechanics II

Consider the linear potential V = al.]. Use a Gaussian = exp(-Bx?) as the trial wave function, and calculate the ground state energy with the variational principle. De- termine the parameter B which minimizes the energy, and find Emin Express Emin = f x (hop/2m)1/3, and give the numerical value of the factor f. This is the upper bound of the true ground state energy, E Compare Emin with the exact result, E. = 1.019 (1?o?/2m)/3, and find the relative error between them. Comment on the result.

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

$\int_{-\infty}^{\infty}\psi^{*}\psi dx=2A^{2}\int_{0}^{\infty}exp(-\beta x^{2})dx=1$

$2A^{2}\frac{1}{2}\left(\frac{\pi}{2\beta} \right )^{1/2}=1\Rightarrow A^{2}=\sqrt{\frac{2\beta}{\pi}}$

$<V> = 2\alpha A^{2}\int_{0}^{\infty }xe^{-2\beta x^{2}}dx = 2\alpha A^{2}\left(-\frac{1}{4\beta}exp(-2\beta x^{2}) \right )|_{0}^{\infty}=\frac{\alpha A^{2}}{2\beta}$

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

$<H> = <T> &plus;<V> = \frac{\hbar^{2}\beta}{2m}&plus;\frac{\alpha}{\sqrt{2\beta \pi}}$

to determine $\beta$

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

$\beta^{3/2}=\frac{\alpha}{\sqrt{2\pi}}\frac{m}{\hbar^{2}}\Rightarrow \beta=\left(\frac{m\alpha}{\sqrt{2\pi}\hbar^{2}} \right )^{2/3}$

$E_{min}=<H>_{min}=\frac{\hbar^{2}}{2m}\left(\frac{m\alpha}{\sqrt{2\pi}\hbar^{2}} \right )^{2/3}&plus;\frac{\alpha}{\sqrt{2\pi}}\left(\frac{\sqrt{2\pi}\hbar^{2}}{m\alpha} \right )^{1/3}$

$=\frac{\alpha^{2/3}\hbar^{2/3}}{m^{1/3}(2\pi)^{1/3}}\left(\frac{1}{2}&plus;1 \right )=\frac{3}{2}\left(\frac{\alpha^{2}\hbar^{2}}{2\pi m} \right )^{1/3}$

$=1.024\left(\frac{\hbar^{2}\alpha^{2}}{2m} \right )^{1/3}$

f = 1.024