91影视

The trial wave function has the form:

$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

The harmonic oscillator potential has the form

$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m蝇}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}$

$鉄�\mathrm{蠄}\left(\mathrm{x}\right)鈭�\mathrm{蠄}\left(\mathrm{x}\right)鉄�={鈭�}_{-鈭�}^{鈭�}{\mathrm{蠄}}^{*}\left(\mathrm{x}\right)\mathrm{蠄}\left(\mathrm{x}\right)\mathrm{dx}={鈭�}_{-鈭�}^{鈭�}{\frac{{\mathrm{A}}^2}{{\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)}^2}}^{}\mathrm{dx}=1$

We know that $\mathrm{x}=\mathrm{b}\tan \mathrm{胃}$

role="math" localid="1656045400533" $$\begin{gathered} \mathrm{dx}={\mathrm{bsec}}^2\mathrm{胃诲胃} \\ {鈭�}_{-鈭�}^{鈭�}\frac{{\mathrm{A}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx}=2{鈭�}_0^{鈭�}\frac{{\mathrm{A}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx} \\ \frac{2{\mathrm{A}}^2}{{\mathrm{b}}^3}{鈭�}_0^{鈭�}{\cos }^2\mathrm{胃诲胃}=2{\mathrm{A}}^2\frac{\mathrm{蟺}}{4{\mathrm{b}}^3}=1 \\ \mathrm{A}=\sqrt{\frac{2{\mathrm{b}}^3}{\mathrm{蟺}}} \end{gathered}$$

The expectation value of kinetic energy.

$$\begin{gathered} 鉄�\mathrm{T}鉄�=\frac{-{鈩�}^2}{2\mathrm{m}}{\mathrm{A}}^2{鈭�}_{-鈭�}^{鈭�}\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}\mathrm{dx} \\ \frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}=\frac{8{\mathrm{x}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^3}-\frac{2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}=\frac{6{\mathrm{x}}^2-2{\mathrm{b}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^3} \\ 鉄�\mathrm{T}鉄�=\frac{-{鈩�}^2}{2\mathrm{m}}2{\mathrm{A}}^2{鈭�}_0^{鈭�}\frac{6{\mathrm{x}}^2-2{\mathrm{b}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^4}\mathrm{dx}=\frac{{鈩�}^2}{4{\mathrm{mb}}^2} \end{gathered}$$

Now find the value of V expectation potential energy.

$鉄�\mathrm{V}鉄�=\frac{1}{2}{\mathrm{m蝇}}^{2}2{\mathrm{A}}^2{鈭�}_0^{鈭�}\frac{{\mathrm{x}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx}={\mathrm{m蝇}}^2\frac{2{\mathrm{b}}^3}{\mathrm{蟺}}\frac{\mathrm{蟺}}{4\mathrm{b}}=\frac{{\mathrm{m蝇}}^2{\mathrm{b}}^2}{2}$

Expectation value $鉄�H鉄�=鉄�T鉄�+鉄�V鉄�$

$=\frac{{鈩�}^2}{4{\mathrm{mb}}^2}+\frac{{\mathrm{m蝇}}^2{\mathrm{b}}^2}{2}$

Now minimize H

role="math" localid="1656046908842" $$\begin{gathered} \frac{鈭�鉄�\mathrm{H}鉄�}{鈭�\mathrm{b}}=\frac{-{鈩�}^2}{2{\mathrm{mb}}^3}+{\mathrm{m蝇}}^2\mathrm{b} \\ =0 \\ \mathrm{b}={\left(\frac{{鈩�}^2}{2{\mathrm{m}}^2{\mathrm{蝇}}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \\ 鉄�\mathrm{H}鉄�\frac{{鈩�}^2}{4\mathrm{m}}\frac{1}{{\left(\frac{{鈩�}^2}{2{\mathrm{m}}^2{\mathrm{蝇}}^2}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{{\mathrm{m蝇}}^2}{2}{{\left(\frac{{鈩�}^2}{2{\mathrm{m}}^2{\mathrm{蝇}}^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}_{\min } \\ =\frac{\sqrt{2}}{2}鈩�\mathrm{蝇} \\ =0.707鈩�\mathrm{蝇} \end{gathered}$$

Which is bigger than the expected value of the ground state energy of the harmonic oscillator$=0.707鈩�蝇$