91影视

The trial wave function of the form:

$蠄\left(x\right)=\frac{1}{\sqrt{蟺b^3}}{e}^{-r/b}$

Same hydrogen but replaced a with b.

For hydrogen,

So for trial wave function, .

Let $蠄=\frac{1}{\sqrt{蟺b^3}}{e}^{-r/b}$(same as hydrogen, but with a鈫抌 adjustable).

we have $鉄�T鉄�=-E_1=\frac{{\mathrm{鈩�}}^2}{2m{a}^2}$for hydrogen, so in this case $鉄�T鉄�=\frac{{\mathrm{鈩�}}^2}{2m{b}^2}.$

$鉄�T鉄�=-E_n;鉄�V鉄�=2E_n$ (4.218).

$\begin{array}{rcl}鉄�V鉄�& =& -\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{4蟺}{蟺b^3}{鈭�}_0^{鈭�}{e}^{-2r/b}\frac{e^{-渭r}}{r}{r}^{2}dr=-\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{4}{b^3}{鈭�}_0^{鈭�}{e}^{-(渭+2/b)r}rdr\\ & =& -\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{4}{b^3}\frac{1}{{(渭+2/b)}^2}\\ & =& -\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{1}{b{\left(1+\frac{渭b}{2}\right)}^2}\end{array}$

Now, calculate ,

$鉄�H鉄�=\frac{{\mathrm{鈩�}}^2}{2m{b}^2}-\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{1}{b{\left(1+\frac{渭b}{2}\right)}^2}.$

$\begin{array}{rcl}\frac{鈭�鉄�H鉄�}{鈭�b}& =& -\frac{{\mathrm{鈩�}}^2}{m{b}^3}+\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\left[\frac{1}{b^2{(1+渭b/2)}^2}+\frac{渭}{b{(1+渭b/2)}^3}\right]\\ & =& -\frac{{\mathrm{鈩�}}^2}{m{b}^3}+\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{(1+3渭b/2)}{b^2{(1+渭b/2)}^3}\\ & =& 0\end{array}$

$$\begin{gathered} \frac{{\mathrm{鈩�}}^2}{m}\left(\frac{4\mathrm{蟺}{鈭�}_0}{e^2}\right)=b\frac{(1+3渭b/2)}{{(1+渭b/2)}^3}, \\ \text{or}b\frac{(1+3渭b/2)}{{(1+渭b/2)}^3}=a. \end{gathered}$$

This determines b, but unfortunately it鈥檚 a cubic equation. So we use the fact that 渭 is small to obtain a suitable approximate solution. If 渭 = 0 , then b = a (of course), so$渭a鈮�1;渭b鈮�1$

. We鈥檒l expand in powers of $渭b$:

$$\begin{gathered} a鈮�b\left(1+\frac{3渭b}{2}\right)\left[1-\frac{3渭b}{2}+6{\left(\frac{渭b}{2}\right)}^2\right] \\ 鈮�b\left[1-\frac{9}{4}{\left(渭b\right)}^2+\frac{6}{4}{\left(渭b\right)}^2\right] \\ =b\left[1-\frac{3}{4}{\left(渭b\right)}^2\right]. \end{gathered}$$

Since the $\frac{3}{4}(渭b{)}^2$ term is already a second-order correction, we can replace b by a:

$$\begin{gathered} b鈮�\frac{a}{\left[1-\frac{3}{4}{\left(渭b\right)}^2\right]} \\ 鈮�a\left[1+\frac{3}{4}{\left(渭a\right)}^2\right] \end{gathered}$$

$鉄�H{鉄�}_{\min }=\frac{{\mathrm{鈩�}}^2}{2m{a}^2{\left[1+\frac{3}{4}{\left(渭a\right)}^2\right]}^2}-\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{1}{a\left[1+\frac{3}{4}{\left(渭a\right)}^2\right]{\left[1+\frac{1}{2}\left(渭a\right)\right]}^2}$

$$\begin{gathered} 鈮�\frac{{\mathrm{鈩�}}^2}{2m{a}^2}\left[1-2\frac{3}{4}{\left(渭a\right)}^2\right]-\frac{e^2}{4\mathrm{蟺}{鈭�}_0}\frac{1}{a}\left[1-\frac{3}{4}{\left(渭a\right)}^2\right]\left[1-2\frac{渭a}{2}+3{\left(\frac{渭a}{2}\right)}^2\right]. \\ =-E_1\left[1-\frac{3}{2}{\left(渭a\right)}^2\right]+2E_1\left[1-渭a+\frac{3}{4}{\left(渭a\right)}^2-\frac{3}{4}{\left(渭a\right)}^2\right] \\ =E_1\left[1-2\left(渭a\right)+\frac{3}{2}{\left(渭a\right)}^2\right]. \end{gathered}$$