91影视

The variational principle asserts that the ground-state energy is always less than or equal to the expected value calculated using the trial wavefunction: i.e., the wavefunction and energy of the ground-state can be approximated by varying until the expected value is minimized.

Determine the value of$鉄�V鉄�$in the following way.

role="math" localid="1658998746850" $\begin{array}{c}鉄�V鉄�=2伪{螒}^2\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}x{e}^{鈭�2b{x}^2}dx\\ =2伪{螒}^2{[鈭�\frac{1}{4b}{e}^{鈭�2b{x}^2}]}_0^{\mathrm{鈭�}}\\ =\frac{伪{螒}^2}{2b}\\ =\frac{伪}{2b}\sqrt{\frac{2b}{蟺}}\\ =\frac{伪}{\sqrt{2b蟺}}\end{array}$

Determine the value of$鉄�H鉄�$in the following way.

$\begin{array}{c}鉄�H鉄�=\frac{{\mathrm{鈩�}}^{2}b}{2m}+\frac{伪}{\sqrt{2b蟺}}鈰�\frac{鈭�鉄�H鉄�}{鈭�b}\\ =\frac{{\mathrm{鈩�}}^2}{2m}鈭�\frac{1}{2}\frac{伪}{\sqrt{2蟺}}{b}^{鈭�3/2}\end{array}$

Equate the above equation to 0 and find the value of $b$.

$\begin{array}{c}\frac{{\mathrm{鈩�}}^2}{2m}鈭�\frac{1}{2}\frac{伪}{\sqrt{2蟺}}{b}^{鈭�3/2}=0\\ b^{3/2}=\frac{伪}{\sqrt{2蟺}}\frac{m}{鈭�{\mathrm{鈩�}}^2}\\ b={\left(\frac{m伪}{\sqrt{2蟺}{\mathrm{鈩�}}^2}\right)}^{2/3}\end{array}$

Determine the value of${鉄�H鉄�}_{\min }$ in the following way.

${鉄�H鉄�}_{\min }=\frac{{\mathrm{鈩�}}^{2}b}{2m}+\frac{伪}{\sqrt{2b蟺}}$

Substitute${\left(\frac{m伪}{\sqrt{2蟺}{\mathrm{鈩�}}^2}\right)}^{2/3}$for$b$ in the above expression.

role="math" localid="1658998933836" $\begin{array}{c}{鉄�H鉄�}_{\min }=\frac{{\mathrm{鈩�}}^2}{2m}{\left(\frac{m伪}{\sqrt{2蟺}{\mathrm{鈩�}}^2}\right)}^{2/3}+\frac{a}{\sqrt{2蟺}}{\left(\frac{\sqrt{2蟺}{\mathrm{鈩�}}^2}{m伪}\right)}^{1/3}\\ =\frac{{伪}^{2/3}{\mathrm{鈩�}}^{2/3}}{m^{1/3}{\left(2蟺\right)}^{1/3}}(\frac{1}{2}+1)\\ =\frac{3}{2}{\left(\frac{{伪}^2{\mathrm{鈩�}}^2}{\left(2蟺m\right)}\right)}^{1/3}\end{array}$

Thus, the lowest upper bound for the given linear potential is $\frac{3}{2}{\left(\frac{{伪}^2{\mathrm{鈩�}}^2}{\left(2蟺m\right)}\right)}^{1/3}$.

Determine the value of $鉄�V鉄�$in the following way.

$\begin{array}{c}鉄�V鉄�=2伪{螒}^2\underset{0}{\overset{\mathrm{鈭�}}{鈭�}}{x}^4{e}^{鈭�2b{x}^2}dx\\ =2伪{螒}^2\frac{3}{8{\left(2b\right)}^2}\sqrt{\frac{蟺}{2b}}\\ =\frac{3伪}{16b^2}\sqrt{\frac{蟺}{2b}}\sqrt{\frac{2b}{蟺}}\\ =\frac{3伪}{16b^2}\end{array}$

Determine the value of$鉄�H鉄�$in the following way.

$\begin{array}{c}鉄�H鉄�=\frac{{\mathrm{鈩�}}^{2}b}{2m}+\frac{3}{16b^2}鈰�\frac{鈭�鉄�H鉄�}{鈭�b}\\ =\frac{{\mathrm{鈩�}}^2}{2m}鈭�\frac{3伪}{8b^3}\end{array}$

Equate the above equation to 0 and find the value of $b$.

$\begin{array}{c}\frac{{\mathrm{鈩�}}^2}{2m}鈭�\frac{3伪}{8b^3}=0\\ b^3=\frac{3伪m}{4{\mathrm{鈩�}}^2}\\ b={\left(\frac{3伪m}{4{\mathrm{鈩�}}^2}\right)}^{1/3}\end{array}$

Determine the value of${鉄�H鉄�}_{\min }$in the following way.

${鉄�H鉄�}_{\min }=\frac{{\mathrm{鈩�}}^{2}b}{2m}+\frac{3伪}{16b^2}$

Substitute${\left(\frac{3伪m}{4{\mathrm{鈩�}}^2}\right)}^{1/3}$for $b$in the above expression.

role="math" localid="1658999135976" $\begin{array}{c}{鉄�H鉄�}_{\min }=\frac{{\mathrm{鈩�}}^2}{2m}{\left(\frac{3伪m}{4{\mathrm{鈩�}}^2}\right)}^{1/3}+\frac{3伪}{16}{\left(\frac{4{\mathrm{鈩�}}^2}{3伪m}\right)}^{2/3}\\ =\frac{{伪}^{1/3}{\mathrm{鈩�}}^{4/3}}{m^{2/3}}{3}^{1/3}{4}^{鈭�1/3}(\frac{1}{2}+\frac{1}{4})\\ =\frac{3}{4}{\left(\frac{3伪{\mathrm{鈩�}}^4}{4m^2}\right)}^{1/3}\end{array}$

Thus, the lowest upper bound for the given quartic potential is $\frac{3}{4}{\left(\frac{3伪{\mathrm{鈩�}}^4}{4m^2}\right)}^{1/3}$.