Phy5646/Problem on Variational Method

Problem:- Consider the one-dimensional potential

$\psi (x)=\lambda {\frac {x^{4}}{4}}+\lambda a{\frac {x^{3}}{4}}-\lambda {\frac {a^{2}x^{2}}{8}}$

(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

$\psi (x)=\left({\frac {\beta }{\pi }}\right)^{\frac {1}{4}}e^{\frac {-\beta (x-x_{0})^{2}}{2}}$

where $x_{0}$ 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

$m{\ddot {x}}=-{V}'(x)$ $\rightarrow$ ${V}'(x_{0})=0,{V}''(x_{0})>0$

The extrema of the potential are the solutions of

${V}'(x)={\frac {\lambda }{4}}(4x^{3}+3ax^{2}-a^{2}x)={\frac {\lambda x}{4}}(4x^{2}+3ax-a^{2})=0$

The second derivative of the potential is

${V}''={\frac {\lambda }{4}}(12x^{2}+6ax-a^{2})$

We obtain a maximum at

$x=0,{V}''(0)=-{\frac {\lambda a^{2}}{4}}<0$

and two minima at

$x_{1}=-a,x_{2}={\frac {a}{4}}$

with

${V}''(-a)={\frac {5\lambda a^{2}}{4}}>0,{V}''({\frac {a}{4}})={\frac {5\lambda a^{2}}{16}}>0$

Of these two minima, $x_{1}=-a$ is the global minimum since it corresponds to the lower energy:

$V(-a)=-{\frac {\lambda a^{4}}{8}}<V({\frac {a}{4}})=-{\frac {3\lambda a^{4}}{4^{5}}}$

(b) Consider the trial wave function with $x_{0}=-a$. The expectation value of the kinetic energy, thanks to translational invariance, does not depend on $x_{0}$. It is

$<T>=-{\frac {({\frac {h}{2\pi }})^{2}}{2m}}{\sqrt {\frac {\beta }{\pi }}}\int _{-\infty }^{\infty }dxe^{\frac {-\beta x^{2}}{2}}{(e^{\frac {-\beta x^{2}}{2}})}''={\frac {\beta ({\frac {h}{2\pi }})^{2}}{4m}}$

The expectation value of the potential is

$<V>={\frac {\lambda }{4}}{\sqrt {\frac {\beta }{\pi }}}\int _{-\infty }^{\infty }dxe^{-\beta x^{2}}\left[(x-a)^{4}+a(x-a)^{3}-{\frac {a^{2}}{2}}(x-a)^{2}\right]$

$={\frac {\lambda }{4}}({\frac {3}{4\beta ^{2}}}+{\frac {5a^{2}}{4\beta }}-{\frac {a^{4}}{2}})$

Thus we get,

$E(\beta )={\frac {({\frac {h}{2\pi }})^{2}{}}{4ma^{2}}}\left[\beta a^{2}+{\bar {\lambda }}\left({\frac {3}{4a^{4}\beta ^{2}}}+{\frac {5}{4a^{2}\beta }}-{\frac {1}{2}}\right)\right]$

where we have introduced the dimensionless coupling ${\bar {\lambda }}={\frac {\lambda ma^{6}}{({\frac {h}{2m}})^{2}}}$. Minimizing the energy with respect to $\xi =-\beta a^{2}$, we get the equation

$\xi ^{3}-{\frac {3{\bar {\lambda }}}{2}}-{\frac {5{\bar {\lambda \xi }}}{4}}=0$

For ${\bar {\lambda }}=1$ there is a special exact solution, namely

${\bar {\lambda }}=1\Rightarrow \xi _{0}={\frac {3}{2}},E_{0}={\frac {({\frac {h}{2m}})^{2}}{2ma^{2}}}({\frac {13}{12}})$