Quartic potential ground state

Setup

Consider a Particle in a Potential $V(x)=\lambda x^4$. We want to find the Energy of the Ground state of a Particle in this Potential.

Solution

We will find an upper bound for the Ground state energy by calculating the Energy for the $ansatz$ given by $e^{\frac{-\alpha x^2}{2}}$. The associated Hamiltonian is given by $H=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+\lambda x^4$. We perform the Wave function energy calculation as $E(\alpha )={\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\left(-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+\lambda x^4\right)e^{\frac{-\alpha x^2}{2}}dx/{\int }_{\mathbb{R}}{e}^{-\alpha x^2}dx$.

We will split the calculations into 2 parts. We start with ${\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}(\frac{d^2}{d{x}^2})e^{\frac{-\alpha x^2}{2}}dx$:

$$\begin{array}{rll}{\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}(\frac{d^2}{d{x}^2})e^{\frac{-\alpha x^2}{2}}dx& ={\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\frac{d}{dx}\left(-\alpha x{e}^{\frac{-\alpha x^2}{2}}\right)dx& \text{Taking −α out of the integral}\\ & =-\alpha {\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\frac{d}{dx}\left(x{e}^{\frac{-\alpha x^2}{2}}\right)dx& \text{By the product rule}\\ & =-\alpha {\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\left(e^{\frac{-\alpha x^2}{2}}+x(-\alpha x)e^{\frac{-\alpha x^2}{2}}\right)dx& \text{Separating into 2 integrals}\\ & =-\alpha {\int }_{\mathbb{R}}{e}^{-\alpha x^2}dx+{\alpha }^2{\int }_{\mathbb{R}}{x}^2{e}^{-\alpha x^2}dx& \text{Setting Cα:=∫Re−αx2dx}\\ & =-\alpha {\mathcal{C}}_{\alpha }+{\alpha }^2{\int }_{\mathbb{R}}{x}^2{e}^{-\alpha x^2}dx& \\ & =-\alpha {\mathcal{C}}_{\alpha }+{\alpha }^2{\int }_{\mathbb{R}}{x}^2\frac{1}{-2\alpha x}d{e}^{-\alpha x^2}& \text{Moving constants out of the integral}\\ & =-\alpha {\mathcal{C}}_{\alpha }-\frac{\alpha }{2}{\int }_{\mathbb{R}}xd{e}^{-\alpha x^2}& \text{Integration by parts}\\ & =-\alpha {\mathcal{C}}_{\alpha }+\frac{\alpha }{2}{\int }_{\mathbb{R}}{e}^{-\alpha x^2}dx& \text{Setting Cα:=∫Re−αx2dx}\\ & =-\alpha {\mathcal{C}}_{\alpha }+\frac{\alpha }{2}{\mathcal{C}}_{\alpha }& \\ & =-\frac{\alpha }{2}{\mathcal{C}}_{\alpha }& \end{array}$$

Secondly we simplify ${\int }_{\mathbb{R}}{x}^4{e}^{-\alpha x^2}dx$:

$$\begin{array}{rll}{\int }_{\mathbb{R}}{x}^4{e}^{-\alpha x^2}dx& ={\int }_{\mathbb{R}}{x}^4\frac{-1}{2\alpha x}d{e}^{-\alpha x^2}& \\ & =-\frac{1}{2\alpha }{\int }_{\mathbb{R}}{x}^{3}d{e}^{-\alpha x^2}& \text{Integration by parts}\\ & =\frac{1}{2\alpha }{\int }_{\mathbb{R}}{e}^{-\alpha x^2}d{x}^3& \\ & =\frac{1}{2\alpha }{\int }_{\mathbb{R}}3x^2{e}^{-\alpha x^2}dx& \\ & =\frac{3}{2\alpha }{\int }_{\mathbb{R}}{x}^2{e}^{-\alpha x^2}dx& \\ & =\frac{3}{2\alpha }{\int }_{\mathbb{R}}{x}^2\frac{-1}{2\alpha x}d{e}^{-\alpha x^2}& \\ & =-\frac{3}{4{\alpha }^2}{\int }_{\mathbb{R}}xd{e}^{-\alpha x^2}& \text{Integration by parts}\\ & =\frac{3}{4{\alpha }^2}{\int }_{\mathbb{R}}{e}^{-\alpha x^2}dx& \text{Setting Cα:=∫Re−αx2dx}\\ & =\frac{3}{4{\alpha }^2}{\mathcal{C}}_{\alpha }& \end{array}$$

Hence

$$\begin{array}{rll}E(\alpha )& ={\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\left(-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+\lambda x^4\right)e^{\frac{-\alpha x^2}{2}}dx/{\int }_{\mathbb{R}}{e}^{-\alpha x^2}dx& \\ & =\left(-\frac{{\hslash }^2}{2m}{\int }_{\mathbb{R}}{e}^{\frac{-\alpha x^2}{2}}\left(\frac{d^2}{d{x}^2}\right)e^{\frac{-\alpha x^2}{2}}dx+\lambda {\int }_{\mathbb{R}}{x}^4{e}^{-\alpha x^2}dx\right)/{\mathcal{C}}_{\alpha }& \text{Substituting the results above}\\ & =\left(\frac{-{\hslash }^2}{2m}\frac{-\alpha }{2}{\mathcal{C}}_{\alpha }+\lambda \frac{3}{4{\alpha }^2}{\mathcal{C}}_{\alpha }\right)/{\mathcal{C}}_{\alpha }& \text{Cancelling out the Cα terms}\\ & =\frac{-{\hslash }^2}{2m}\frac{-\alpha }{2}+\lambda \frac{3}{4{\alpha }^2}& \\ & =\frac{{\hslash }^2\alpha }{4m}+\frac{3\lambda }{4{\alpha }^2}& \end{array}$$

= Since we want to find an upper bound for the Ground state Energy $E_0$, we want to minimize $E(\alpha )$, using the standard method:

$$\begin{array}{rll}0& =\frac{d}{d\alpha }E(\alpha )& \\ & =\frac{d}{d\alpha }\left(\frac{{\hslash }^2\alpha }{4m}+\frac{3\lambda }{4{\alpha }^2}\right)& \\ & =\frac{{\hslash }^2}{4m}-\frac{3\lambda }{2{\alpha }^3}& \\ & \Downarrow & \\ {\hslash }^2{\alpha }^3& =6\lambda m& \\ & \Downarrow & \text{Relabelling α as α0 to disguish it as a minimum}\\ {\alpha }_0& ={\left(\frac{6\lambda m}{{\hslash }^2}\right)}^{\frac{1}{3}}& \end{array}$$

We get the upper bound $E({\alpha }_0)=\frac{{\hslash }^2{\alpha }_0}{4m}+\frac{3\lambda }{4{\alpha }_0^2}=\frac{3}{8}{\left(\frac{6{\hslash }^4\lambda }{m^2}\right)}^{\frac{1}{3}}$.