Find the best bound on ${\mathbf{E}}_{\mathbf{gs}}$for the one-dimensional harmonic oscillator using a trial wave function of the form role="math" localid="1656044636654" $\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\mathbf{.}\mathbf{,}$where A is determined by normalization and b is an adjustable parameter.

The trial wave function has the form:

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

The harmonic oscillator potential has the form

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

$⟨\mathrm{\psi }\left(\mathrm{x}\right)\mid \mathrm{\psi }\left(\mathrm{x}\right)⟩={\int }_{-\infty }^{\infty }{\mathrm{\psi }}^{*}\left(\mathrm{x}\right)\mathrm{\psi }\left(\mathrm{x}\right)\mathrm{dx}={\int }_{-\infty }^{\infty }{\frac{{\mathrm{A}}^2}{{\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)}^2}}^{}\mathrm{dx}=1$

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

role="math" localid="1656045400533" $$\begin{gathered} \mathrm{dx}={\mathrm{bsec}}^2\mathrm{\theta d\theta } \\ {\int }_{-\infty }^{\infty }\frac{{\mathrm{A}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx}=2{\int }_0^{\infty }\frac{{\mathrm{A}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx} \\ \frac{2{\mathrm{A}}^2}{{\mathrm{b}}^3}{\int }_0^{\infty }{\cos }^2\mathrm{\theta d\theta }=2{\mathrm{A}}^2\frac{\mathrm{\pi }}{4{\mathrm{b}}^3}=1 \\ \mathrm{A}=\sqrt{\frac{2{\mathrm{b}}^3}{\mathrm{\pi }}} \end{gathered}$$

The expectation value of kinetic energy.

$$\begin{gathered} ⟨\mathrm{T}⟩=\frac{-{\hslash }^2}{2\mathrm{m}}{\mathrm{A}}^2{\int }_{-\infty }^{\infty }\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}\mathrm{dx} \\ \frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{1}{({\mathrm{x}}^2+{\mathrm{b}}^2)}=\frac{8{\mathrm{x}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^3}-\frac{2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}=\frac{6{\mathrm{x}}^2-2{\mathrm{b}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^3} \\ ⟨\mathrm{T}⟩=\frac{-{\hslash }^2}{2\mathrm{m}}2{\mathrm{A}}^2{\int }_0^{\infty }\frac{6{\mathrm{x}}^2-2{\mathrm{b}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^4}\mathrm{dx}=\frac{{\hslash }^2}{4{\mathrm{mb}}^2} \end{gathered}$$

Now find the value of V expectation potential energy.

$⟨\mathrm{V}⟩=\frac{1}{2}{\mathrm{m\omega }}^{2}2{\mathrm{A}}^2{\int }_0^{\infty }\frac{{\mathrm{x}}^2}{({\mathrm{x}}^2+{\mathrm{b}}^2{)}^2}\mathrm{dx}={\mathrm{m\omega }}^2\frac{2{\mathrm{b}}^3}{\mathrm{\pi }}\frac{\mathrm{\pi }}{4\mathrm{b}}=\frac{{\mathrm{m\omega }}^2{\mathrm{b}}^2}{2}$

Expectation value $⟨H⟩=⟨T⟩+⟨V⟩$

$=\frac{{\hslash }^2}{4{\mathrm{mb}}^2}+\frac{{\mathrm{m\omega }}^2{\mathrm{b}}^2}{2}$

Now minimize H

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

Which is bigger than the expected value of the ground state energy of the harmonic oscillator$=0.707\hslash \omega$