Quadratic potential ground state energy via the Variational Method

Setup

We have a spinor evolving according to the one-dimensional Hamiltonian $H=-\frac{\hslash }{2m}\frac{{\partial }^2}{\partial x^2}+x^2$ , where

• $\hslash$ is planck’s constant
• $m$ is the mass

Goal

We want to find the Ground state Energy $E_0$ using the Variational Method.

Solution

We parameterise our Quantum states as $|{\psi }_{\alpha }⟩=({\alpha }_0+{\alpha }_{1}x)e^{-{\alpha }_2{x}^2}$ where

• $\alpha =({\alpha }_0,{\alpha }_1,{\alpha }_3)\in {\mathbb{R}}^3$.
• $E_{\alpha }:=\frac{⟨{\psi }_{\alpha }|H|{\psi }_{\alpha }⟩}{⟨{\psi }_{\alpha }|{\psi }_{\alpha }⟩}$ describes the expected energy.
• $I_{n,k}:={\int }_{{\mathbb{R}}^{+}}{x}^n{e}^{-k{x}^2}dx,k\in {\mathbb{R}}^{+}$. We use this to simplify the calculations below.

We can write the denominator as

$$\begin{array}{rll}⟨{\psi }_{\alpha }|{\psi }_{\alpha }⟩& ={\int }_{\mathbb{R}}({\alpha }_0+{\alpha }_{1}x{)}^2{e}^{-2{\alpha }_2{x}^2}dx& \\ & ={\int }_{\mathbb{R}}({\alpha }_0^2+2{\alpha }_0{\alpha }_{1}x+{\alpha }_1^2{x}^2)e^{-2{\alpha }_2{x}^2}dx& \text{Removing the xe−2αx2 term since it is odd}\\ & ={\int }_{\mathbb{R}}({\alpha }_0^2+{\alpha }_1^2{x}^2)e^{-2{\alpha }_2{x}^2}dx& \text{Splitting the integral}\\ & ={\alpha }_0^2{\int }_{\mathbb{R}}{e}^{-2{\alpha }_2{x}^2}dx+{\alpha }_1^2{\int }_{\mathbb{R}}{x}^2{e}^{-2{\alpha }_2{x}^2}dx& \text{Since the integrands are even}\\ & =2{\alpha }_0^2{\int }_{{\mathbb{R}}^{+}}{e}^{-2{\alpha }_2{x}^2}dx+2{\alpha }_1^2{\int }_{{\mathbb{R}}^{+}}{x}^2{e}^{-2{\alpha }_2{x}^2}dx& \text{Using In,α as defined above}\\ & =2{\alpha }_0^2{I}_{0,2{\alpha }_2}+2{\alpha }_1^2{I}_{2,2{\alpha }_2}& \end{array}$$

The 2nd derivative can be written as

$$\begin{array}{rll}\frac{{\partial }^2}{\partial x^2}{\psi }_{\alpha }& =\frac{{\partial }^2}{\partial x^2}\left(({\alpha }_0+{\alpha }_{1}x)e^{-{\alpha }_2{x}^2}\right)& \text{By the product rule for derivatives}\\ & =\frac{\partial }{\partial x}\left({\alpha }_1{e}^{-{\alpha }_2{x}^2}+({\alpha }_0+{\alpha }_{1}x)\frac{\partial }{\partial x}(-{\alpha }_2{x}^2)e^{-{\alpha }_2{x}^2}\right)& \\ & =\frac{\partial }{\partial x}\left({\alpha }_1{e}^{-{\alpha }_2{x}^2}+({\alpha }_0+{\alpha }_{1}x)(-2{\alpha }_{2}x)e^{-{\alpha }_2{x}^2}\right)& \\ & =\frac{\partial }{\partial x}\left(({\alpha }_1-2{\alpha }_0{\alpha }_{2}x-2{\alpha }_1{\alpha }_2{x}^2)e^{-{\alpha }_2{x}^2}\right)& \text{By the product rule for derivatives}\\ & =(-2{\alpha }_0{\alpha }_2-4{\alpha }_1{\alpha }_{2}x)e^{-{\alpha }_2{x}^2}+({\alpha }_1-2{\alpha }_0{\alpha }_{2}x-2{\alpha }_1{\alpha }_2{x}^2)(-2{\alpha }_{2}x)e^{-{\alpha }_2{x}^2}& \text{Expanding the product on the 2nd term}\\ & =(-2{\alpha }_0{\alpha }_2-4{\alpha }_1{\alpha }_{2}x)e^{-{\alpha }_2{x}^2}+(-2{\alpha }_1{\alpha }_{2}x+4{\alpha }_0{\alpha }_2^2{x}^2+4{\alpha }_1{\alpha }_2^2{x}^3)e^{-{\alpha }_2{x}^2}& \text{Joining the 2 terms}\\ & =(-2{\alpha }_0{\alpha }_2-4{\alpha }_1{\alpha }_{2}x-2{\alpha }_1{\alpha }_{2}x+4{\alpha }_0{\alpha }_2^2{x}^2+4{\alpha }_1{\alpha }_2^2{x}^3)e^{-{\alpha }_2{x}^2}& \text{Cancelling common terms}\\ & =(-2{\alpha }_0{\alpha }_2-6{\alpha }_1{\alpha }_{2}x+4{\alpha }_0{\alpha }_2^2{x}^2+4{\alpha }_1{\alpha }_2^2{x}^3)e^{-{\alpha }_2{x}^2}& \text{Cancelling common terms}\end{array}$$

The above allows us to calculate the numerator of $E_{\alpha }$ via

$$\begin{array}{rll}⟨{\psi }_{\alpha }|H|{\psi }_{\alpha }⟩& =⟨{\psi }_{\alpha }|-\frac{\hslash }{2m}\frac{{\partial }^2}{\partial x^2}+x^2|{\psi }_{\alpha }⟩& \text{By linearity of brakets}\\ & =-\frac{\hslash }{2m}⟨{\psi }_{\alpha }|\frac{{\partial }^2}{\partial x^2}|{\psi }_{\alpha }⟩+⟨{\psi }_{\alpha }|x^2|{\psi }_{\alpha }⟩& \text{Using the integral form}\\ & =-\frac{\hslash }{2m}{\int }_{{\mathbb{R}}^{+}}({\alpha }_0+{\alpha }_{1}x)(-2{\alpha }_0{\alpha }_2-6{\alpha }_1{\alpha }_{2}x+4{\alpha }_0{\alpha }_2^2{x}^2+4{\alpha }_1{\alpha }_2^2{x}^3)e^{-2{\alpha }_2{x}^2}dx+{\int }_{\mathbb{R}}({\alpha }_0+{\alpha }_{1}x{)}^2{x}^2{e}^{-2{\alpha }_2{x}^2}dx& \text{Cancelling out a 2 factor}\\ & =-\frac{\hslash }{m}{\int }_{{\mathbb{R}}^{+}}({\alpha }_0+{\alpha }_{1}x)(-{\alpha }_0{\alpha }_2-3{\alpha }_1{\alpha }_{2}x+2{\alpha }_0{\alpha }_2^2{x}^2+2{\alpha }_1{\alpha }_2^2{x}^3)e^{-2{\alpha }_2{x}^2}dx+{\int }_{\mathbb{R}}({\alpha }_0+{\alpha }_{1}x{)}^2{x}^2{e}^{-2{\alpha }_2{x}^2}dx& \text{Expanding the product}\\ & =-\frac{\hslash }{m}{\int }_{{\mathbb{R}}^{+}}(-{\alpha }_0^2{\alpha }_2-3{\alpha }_0{\alpha }_1{\alpha }_{2}x+2{\alpha }_0^2{\alpha }_2^2{x}^2+2{\alpha }_0{\alpha }_1{\alpha }_2^2{x}^3-{\alpha }_0{\alpha }_1{\alpha }_{2}x-3{\alpha }_1^2{\alpha }_2{x}^2+2{\alpha }_0{\alpha }_1{\alpha }_2^2{x}^3+2{\alpha }_1^2{\alpha }_2^2{x}^4)e^{-2{\alpha }_2{x}^2}dx+{\int }_{\mathbb{R}}({\alpha }_0^2{x}^2+2{\alpha }_0{\alpha }_1{x}^3+{\alpha }_1^2{x}^4)e^{-2{\alpha }_2{x}^2}dx& \text{Removing the xne−2α2x2 terms}\\ & =-\frac{\hslash }{m}{\int }_{{\mathbb{R}}^{+}}(-{\alpha }_0^2{\alpha }_2+2{\alpha }_0^2{\alpha }_2^2{x}^2-3{\alpha }_1^2{\alpha }_2{x}^2+2{\alpha }_1^2{\alpha }_2^2{x}^4)e^{-2{\alpha }_2{x}^2}dx+{\int }_{\mathbb{R}}({\alpha }_0^2{x}^2+{\alpha }_1^2{x}^4)e^{-2{\alpha }_2{x}^2}dx& \text{Splitting the integrals}\\ & =\frac{\hslash }{m}{\alpha }_0^2{\alpha }_2{\int }_{{\mathbb{R}}^{+}}{e}^{-2{\alpha }_2{x}^2}dx+\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_0^2\right){\int }_{{\mathbb{R}}^{+}}{x}^2{e}^{-2{\alpha }_2{x}^2}dx+\left(-\frac{2\hslash {\alpha }_1^2{\alpha }_2^2}{m}+{\alpha }_1^2\right){\int }_{{\mathbb{R}}^{+}}{x}^4{e}^{-2{\alpha }_2{x}^2}dx& \text{Using In,α as defined above}\\ & =\frac{2\hslash }{m}{\alpha }_0^2{\alpha }_2{I}_{0,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_0^2\right)I_{2,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_1^2\right)I_{4,2{\alpha }_2}& \end{array}$$

Using results from the Table of integrals, we know that $\{\begin{array}{rl}{I}_{0,\alpha }& =\sqrt{\frac{\pi }{\alpha }}\\ I_{2k,\alpha }& =\frac{2k-1}{2\alpha }{I}_{2(k-1),\alpha }\end{array}$. This gives us the result

$$\{\begin{array}{rl}{I}_{0,\alpha }& =\sqrt{\frac{\pi }{\alpha }}\\ I_{2,\alpha }& =\frac{1}{\sqrt{2\alpha }}\sqrt{\frac{\pi }{\alpha }}\\ I_{4,\alpha }& =\frac{3}{2\alpha }\sqrt{\frac{\pi }{\alpha }}\end{array}\Rightarrow \{\begin{array}{rl}{I}_{0,\alpha }& =\sqrt{\frac{\pi }{\alpha }}\\ I_{2,\alpha }& =\frac{1}{\sqrt{2\alpha }}{I}_{0,\alpha }\\ I_{4,\alpha }& =\frac{3}{2\alpha }{I}_{0,\alpha }\end{array}$$

We write the Energy as

$$\begin{array}{rll}{E}_{\alpha }& =\frac{⟨{\psi }_{\alpha }|H|{\psi }_{\alpha }⟩}{⟨{\psi }_{\alpha }|{\psi }_{\alpha }⟩}& \text{Using the above results}\\ & =\frac{\frac{2\hslash }{m}{\alpha }_0^2{\alpha }_2{I}_{0,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_0^2\right)I_{2,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_1^2\right)I_{4,2{\alpha }_2}}{2{\alpha }_0^2{I}_{0,2{\alpha }_2}+2{\alpha }_1^2{I}_{2,2{\alpha }_2}}& \text{Expanding the In,2α2 as functions of I0,2α2}\\ & =\frac{\frac{2\hslash }{m}{\alpha }_0^2{\alpha }_2{I}_{0,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_0^2\right)\frac{1}{2\sqrt{{\alpha }_2}}{I}_{0,2{\alpha }_2}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_1^2\right)\frac{3}{4{\alpha }_2}{I}_{0,2{\alpha }_2}}{2{\alpha }_0^2{I}_{0,2{\alpha }_2}+2{\alpha }_1^2\frac{1}{2\sqrt{{\alpha }_2}}{I}_{0,2{\alpha }_2}}& \text{Dividing both numerator and denominator by I0,2α2}\\ & =\frac{\frac{2\hslash }{m}{\alpha }_0^2{\alpha }_2+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_0^2\right)\frac{1}{2\sqrt{{\alpha }_2}}+2\left(-\frac{\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)}{m}+{\alpha }_1^2\right)\frac{3}{4{\alpha }_2}}{2{\alpha }_0^2+2{\alpha }_1^2\frac{1}{2\sqrt{{\alpha }_2}}}& \text{Pulling the m1 outside}\\ & =\frac{1}{m}\frac{2\hslash {\alpha }_0^2{\alpha }_2+2\left(\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)+m{\alpha }_0^2\right)\frac{1}{2\sqrt{{\alpha }_2}}+2\left(-\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)+m{\alpha }_1^2\right)\frac{3}{4{\alpha }_2}}{2{\alpha }_0^2+2{\alpha }_1^2\frac{1}{2\sqrt{{\alpha }_2}}}& \text{Multiplying both numerator and denominator by 2α2}\\ & =\frac{1}{m}\frac{4\hslash {\alpha }_0^2{\alpha }_2^2+2\left(\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)+m{\alpha }_0^2\right)\sqrt{{\alpha }_2}+3\left(-\hslash (2{\alpha }_0^2{\alpha }_2^2-3{\alpha }_1^2{\alpha }_2^2)+m{\alpha }_1^2\right)}{4{\alpha }_0^2{\alpha }_1+2{\alpha }_1^2\sqrt{{\alpha }_2}}& \text{Expanding the parenthesis}\\ & =\frac{1}{m}\frac{4\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}+4\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2}{4{\alpha }_0^2{\alpha }_1+2{\alpha }_1^2\sqrt{{\alpha }_2}}& \text{Simplifying common terms}\\ & =\frac{1}{m}\frac{8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2}{4{\alpha }_0^2{\alpha }_1+2{\alpha }_1^2\sqrt{{\alpha }_2}}& \text{Simplifying common terms}\end{array}$$

In order to find the minimal Energy we need to solve $0=\frac{\partial }{\partial {\alpha }_0}{E}_{\alpha }=\frac{\partial }{\partial {\alpha }_1}{E}_{\alpha }=\frac{\partial }{\partial {\alpha }_2}{E}_{\alpha }$. We will use the quotient rule for differentiation in a specific form: $0={\left(\frac{f}{g}\right)}^{\prime }=\frac{f^{\prime }g-g{f}^{\prime }}{g^2}\Rightarrow 0=f^{\prime }g-g{f}^{\prime }\Rightarrow f^{\prime }g=f{g}^{\prime }$. Applying this to the ratio above, where

$$\{\begin{array}{rl}\frac{\partial }{\partial {\alpha }_0}\left[8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2\right]\left[2{\alpha }_0^2{\alpha }_1+{\alpha }_1^2\sqrt{{\alpha }_2}\right]& =\left[8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2\right]\frac{\partial }{\partial {\alpha }_0}\left[2{\alpha }_0^2{\alpha }_1+{\alpha }_1^2\sqrt{{\alpha }_2}\right]\\ & \Downarrow \\ \left[16\hslash {\alpha }_0{\alpha }_2^2\sqrt{{\alpha }_2}+4m{\alpha }_0-12\hslash {\alpha }_0{\alpha }_2^2\right]\left[2{\alpha }_0^2{\alpha }_1+{\alpha }_1^2\sqrt{{\alpha }_2}\right]& =\left[8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2\right]\left[4{\alpha }_0{\alpha }_1\right]\\ & \Downarrow \text{Dividing both sides by 4}\\ \left[4\hslash {\alpha }_0{\alpha }_2^2\sqrt{{\alpha }_2}-3\hslash {\alpha }_0{\alpha }_2^2\right]\left[2{\alpha }_0^2{\alpha }_1+{\alpha }_1^2\sqrt{{\alpha }_2}\right]& =\left[8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2\right]\left[{\alpha }_0{\alpha }_1\right]\\ & \Downarrow \text{Expanding the product}\\ \left[4\hslash {\alpha }_0{\alpha }_2^2\sqrt{{\alpha }_2}-3\hslash {\alpha }_0{\alpha }_2^2\right]\left[2{\alpha }_0^2{\alpha }_1+{\alpha }_1^2\sqrt{{\alpha }_2}\right]& =\left[8\hslash {\alpha }_0^2{\alpha }_2^2\sqrt{{\alpha }_2}-6\hslash {\alpha }_1^2{\alpha }_2^2\sqrt{{\alpha }_2}+2m{\alpha }_0^2-6\hslash {\alpha }_0^2{\alpha }_2^2+9\hslash {\alpha }_1^2{\alpha }_2^2+3m{\alpha }_1^2\right]\left[{\alpha }_0{\alpha }_1\right]\end{array}$$

We get from $\frac{\partial }{\partial {\alpha }_0}{E}_{\alpha }$ :

• Solve the above for $0=\frac{\partial }{\partial \alpha }{E}_{\alpha }$. We should get a connection to Hermite polynomials.