Phy5646/Problem on Variational Method

Problem:- Consider the one-dimensional potential

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (x)=\lambda \frac{x^{4}}{4}+\lambda a\frac{x^{3}}{4}-\lambda \frac{a^{2}x^{2}}{8}}

(a) Find the points of classical equilibrium for a particle of mass m moving under the influence of this potential.

(b) Using the variational method, consider the trial wave function

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (x)=\left ( \frac{\beta }{\pi } \right )^{\frac{1}{4}}e^\frac{{-\beta (x-x_{0})^{2}}}{2}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{0}} is the global minimum found in (a). Evaluate the expectation value of the energy for this wave function and find the equation defining the optimal values of the parameter β, in order to get an estimate of the ground-state energy. Now take a special, but reasonable, value of the coupling constant, , and obtain the corresponding estimate of the ground-state energy.

Solution:-

(a) The classical equilibrium points are the minima of the potential, so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\ddot{x}=-{V}'(x)} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {V}'(x_{0})=0, {V}''(x_{0})> 0}

The extrema of the potential are the solutions of

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {V}'(x)=\frac{\lambda }{4}(4x^{3}+3ax^{2}-a^{2}x)=\frac{\lambda x}{4}(4x^{2}+3ax-a^{2})=0}

The second derivative of the potential is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {V}''=\frac{\lambda }{4}(12x^{2}+6ax-a^{2})}

We obtain a maximum at

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0, {V}''(0)=-\frac{\lambda a^{2}}{4}< 0}

and two minima at

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1}=-a,x_{2}=\frac{a}{4}}

with

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {V}''(-a)=\frac{5\lambda a^{2}}{4}> 0,{V}''(\frac{a}{4})=\frac{5\lambda a^{2}}{16}> 0}

Of these two minima, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{1}=-a} is the global minimum since it corresponds to the lower energy:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(-a)= -\frac{\lambda a^{4}}{8}< V(\frac{a}{4})=-\frac{3\lambda a^{4}}{4^{5}}}

(b) Consider the trial wave function with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{0}=-a} . The expectation value of the kinetic energy, thanks to translational invariance, does not depend on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{0}} . It is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle < T> =-\frac{(\frac{h}{2\pi })^{2}}{2m}\sqrt{\frac{\beta }{\pi }}\int_{-\infty }^{\infty }dxe^{\frac{-\beta x^{2}}{2}}{(e^{\frac{-\beta x^{2}}{2}})}''=\frac{\beta (\frac{h}{2\pi })^{2}}{4m}}

The expectation value of the potential is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle < V>=\frac{\lambda }{4}\sqrt{\frac{\beta }{\pi }}\int_{-\infty }^{\infty } dxe^{-\beta x^{2}}\left [ (x-a)^{4}+a(x-a)^{3}-\frac{a^{2}}{2}(x-a)^{2} \right ]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{\lambda }{4}(\frac{3}{4\beta ^{2}}+\frac{5a^{2}}{4\beta }-\frac{a^{4}}{2})}

Thus we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E(\beta )=\frac{(\frac{h}{2\pi })^2{}}{4ma^{2}}\left [ \beta a^{2}+\bar{\lambda }\left ( \frac{3}{4a^{4}\beta ^{2}}+\frac{5}{4a^{2}\beta }-\frac{1}{2} \right ) \right ]}

where we have introduced the dimensionless coupling Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bar{\lambda }=\frac{\lambda ma^{6}}{(\frac{h}{2m})^{2}}} . Minimizing the energy with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi =-\beta a^{2}} , we get the equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi ^{3}-\frac{3\bar{\lambda }}{2}-\frac{5\bar{\lambda \xi }}{4}=0}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bar{\lambda }=1} there is a special exact solution, namely

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bar{\lambda }=1\Rightarrow \xi _{0}=\frac{3}{2},E_{0}=\frac{(\frac{h}{2m})^{2}}{2ma^{2}}(\frac{13}{12})}