Use a gaussian trial function (Equation 7.2) to obtain the lowest upper bound you can on the ground state energy of (a) the linear potential $V\left(x\right)=\alpha \left|x\right|$ (b) the quartic potential:$V\left(x\right)=\alpha x^4$

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$⟨V⟩$in the following way.

role="math" localid="1658998746850" $\begin{array}{c}⟨V⟩=2\alpha {{\rm A}}^2\underset{0}{\overset{\infty }{\int }}x{e}^{-2b{x}^2}dx\\ =2\alpha {{\rm A}}^2{[-\frac{1}{4b}{e}^{-2b{x}^2}]}_0^{\infty }\\ =\frac{\alpha {{\rm A}}^2}{2b}\\ =\frac{\alpha }{2b}\sqrt{\frac{2b}{\pi }}\\ =\frac{\alpha }{\sqrt{2b\pi }}\end{array}$

Determine the value of$⟨H⟩$in the following way.

$\begin{array}{c}⟨H⟩=\frac{{\hslash }^{2}b}{2m}+\frac{\alpha }{\sqrt{2b\pi }}\cdot \frac{\partial ⟨H⟩}{\partial b}\\ =\frac{{\hslash }^2}{2m}-\frac{1}{2}\frac{\alpha }{\sqrt{2\pi }}{b}^{-3/2}\end{array}$

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

$\begin{array}{c}\frac{{\hslash }^2}{2m}-\frac{1}{2}\frac{\alpha }{\sqrt{2\pi }}{b}^{-3/2}=0\\ b^{3/2}=\frac{\alpha }{\sqrt{2\pi }}\frac{m}{-{\hslash }^2}\\ b={\left(\frac{m\alpha }{\sqrt{2\pi }{\hslash }^2}\right)}^{2/3}\end{array}$

Determine the value of${⟨H⟩}_{\min }$ in the following way.

${⟨H⟩}_{\min }=\frac{{\hslash }^{2}b}{2m}+\frac{\alpha }{\sqrt{2b\pi }}$

Substitute${\left(\frac{m\alpha }{\sqrt{2\pi }{\hslash }^2}\right)}^{2/3}$for$b$ in the above expression.

role="math" localid="1658998933836" $\begin{array}{c}{⟨H⟩}_{\min }=\frac{{\hslash }^2}{2m}{\left(\frac{m\alpha }{\sqrt{2\pi }{\hslash }^2}\right)}^{2/3}+\frac{a}{\sqrt{2\pi }}{\left(\frac{\sqrt{2\pi }{\hslash }^2}{m\alpha }\right)}^{1/3}\\ =\frac{{\alpha }^{2/3}{\hslash }^{2/3}}{m^{1/3}{\left(2\pi \right)}^{1/3}}(\frac{1}{2}+1)\\ =\frac{3}{2}{\left(\frac{{\alpha }^2{\hslash }^2}{\left(2\pi m\right)}\right)}^{1/3}\end{array}$

Thus, the lowest upper bound for the given linear potential is $\frac{3}{2}{\left(\frac{{\alpha }^2{\hslash }^2}{\left(2\pi m\right)}\right)}^{1/3}$.

Determine the value of $⟨V⟩$in the following way.

$\begin{array}{c}⟨V⟩=2\alpha {{\rm A}}^2\underset{0}{\overset{\infty }{\int }}{x}^4{e}^{-2b{x}^2}dx\\ =2\alpha {{\rm A}}^2\frac{3}{8{\left(2b\right)}^2}\sqrt{\frac{\pi }{2b}}\\ =\frac{3\alpha }{16b^2}\sqrt{\frac{\pi }{2b}}\sqrt{\frac{2b}{\pi }}\\ =\frac{3\alpha }{16b^2}\end{array}$

Determine the value of$⟨H⟩$in the following way.

$\begin{array}{c}⟨H⟩=\frac{{\hslash }^{2}b}{2m}+\frac{3}{16b^2}\cdot \frac{\partial ⟨H⟩}{\partial b}\\ =\frac{{\hslash }^2}{2m}-\frac{3\alpha }{8b^3}\end{array}$

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

$\begin{array}{c}\frac{{\hslash }^2}{2m}-\frac{3\alpha }{8b^3}=0\\ b^3=\frac{3\alpha m}{4{\hslash }^2}\\ b={\left(\frac{3\alpha m}{4{\hslash }^2}\right)}^{1/3}\end{array}$

Determine the value of${⟨H⟩}_{\min }$in the following way.

${⟨H⟩}_{\min }=\frac{{\hslash }^{2}b}{2m}+\frac{3\alpha }{16b^2}$

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

role="math" localid="1658999135976" $\begin{array}{c}{⟨H⟩}_{\min }=\frac{{\hslash }^2}{2m}{\left(\frac{3\alpha m}{4{\hslash }^2}\right)}^{1/3}+\frac{3\alpha }{16}{\left(\frac{4{\hslash }^2}{3\alpha m}\right)}^{2/3}\\ =\frac{{\alpha }^{1/3}{\hslash }^{4/3}}{m^{2/3}}{3}^{1/3}{4}^{-1/3}(\frac{1}{2}+\frac{1}{4})\\ =\frac{3}{4}{\left(\frac{3\alpha {\hslash }^4}{4m^2}\right)}^{1/3}\end{array}$

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