Question

(a) Use a trial wave function of the form

$\psi \left(x\right)=\{\begin{array}{ll}Acos(\pi x/a),& \text{if}(-a/2<x<a/2)\\ 0& \text{otherwise}\end{array}$

to obtain a bound on the ground state energy of the one-dimensional harmonic oscillator. What is the "best" value of$a$? Compare${⟨H⟩}_{min}$with the exact energy. Note: This trial function has a "kink" in it (a discontinuous derivative) at$\pm a/2$; do you need to take account of this, as I did in Example 7.3?

(b) Use$\psi \left(x\right)=Bsin(\pi x/a)$on the interval$(-a,a)$to obtain a bound on the first excited state. Compare the exact answer.

Short answer

(a) The exact energy is $\frac{1}{2}\hslash \omega$and${⟨H⟩}_{\min }>E_g$ .

(b) The first actual excited state energy is $\frac{3}{2}\hslash \omega {⟨H⟩}_{\min }>\frac{3}{2}\hslash \omega$.

Step-by-step solution

Step 1:Definition ofthe Hamiltonian function

The Hamiltonian function is a mathematical formulation for describing the rate of change in the state of a dynamic physical system, such as a group of moving particles, The Hamiltonian of a system reveals all of its potential for generating energy.

(a) Determination of the expectation value of the potential energy

Write the normalization condition.

$\begin{array}{c}\int {\left|\psi \left(x\right)\right|}^{2}dx=1{\left|A\right|}^2{\int }_{-a/2}^{\pi /2}{\cos }^2(\pi x/a)dx\\ =1{\left|A\right|}^2{\int }_{-a/2}^{a/2}\left[\frac{1+\cos 2\left(\frac{\pi x}{2}\right)}{2}\right]dx\\ =1{\left|A\right|}^2[{\int }_0^{a/2}dx+{\int }_0^{a/2}\cos \left(\frac{2\pi x}{a}\right)dx]\\ =1{\left|A\right|}^2{[x+\frac{\sin \left(\frac{2\pi x}{a}\right)}{(-\frac{2\pi }{a})}]}_0^{a/2}\end{array}$

Evaluate the above expression further.

$\begin{array}{c}\int {\left|\psi \left(x\right)\right|}^{2}dx=1{\left|A\right|}^2[\frac{a}{2}+0]\\ =1A\\ =\sqrt{\frac{2}{a}}\end{array}$

It is known that$⟨H⟩=⟨T⟩+⟨V⟩$and for a harmonic oscillator,$T=\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$and$V=\frac{1}{2}m{\omega }^2{x}^2$.

Determine the expectation value of.$T$

$\begin{array}{c}⟨T⟩=\int {\psi }^{*}\left(x\right)\left(\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\right)\psi \left(x\right)dx\\ =\frac{-{\hslash }^2}{2m}{\left|A\right|}^2\int \cos \left(\frac{\pi x}{a}\right)\frac{d}{dx}(-\sin \left(\frac{\pi x}{a}\right))\left(\frac{\pi }{a}\right)dx\\ =\frac{-{\hslash }^2}{2m}{\left|A\right|}^2(-\frac{{\pi }^2}{a^2}){\int }_{-a/2}^{a/2}{\cos }^2\left(\frac{\pi x}{a}\right)dx\\ =\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right){\int }_{-\alpha /2}^{\alpha /2}{\left|\psi \right|}^{2}dx\\ =\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right)\left(1\right)\\ =\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right)\end{array}$

Determine the expectation value of $V$.

$\begin{array}{c}⟨V⟩=\frac{1}{2}m{\omega }^2{A}^2{\int }_{\frac{a}{2}}^{\frac{a}{2}}{x}^2{\cos }^2\frac{\pi x}{a}dx\\ ={\int }_{\frac{a}{2}}^{\frac{a}{2}}{x}^2{\cos }^2\frac{\pi x}{a}dx\\ ={\int }_{\frac{a}{2}}^{\frac{a}{2}}\frac{x^2}{2}(1+\cos \frac{2\pi x}{a})dx\\ ={\int }_{\frac{a}{2}}^{\frac{a}{2}}\frac{x^2}{2}dx+\frac{1}{2}{\int }_{\frac{a}{2}}^{\frac{a}{2}}{x}^2\cos \frac{2\pi x}{a}dx\\ =I_1+I_2\end{array}$

Simplify the above expression further.

$\begin{array}{c}{I}_1={\frac{x^3}{6}|}_{\frac{a}{2}}^{\frac{a}{2}}\\ =\frac{a^3}{8\times 6}-\left(\frac{-a^3}{8\times 6}\right)\\ =\frac{a^3}{48}+\frac{a^3}{48}\\ =\frac{a^3}{24}\end{array}$

$I_2=\frac{1}{2}{\int }_{\frac{a}{2}}^{\frac{a}{2}}{x}^2\cos \frac{2\pi x}{a}dx$

Integrate the function by parts,

Assume $u=x^2$,$dv=\cos \frac{2\pi x}{a}dx$ and $du=2xdx$.

$\begin{array}{c}v=\frac{\sin \frac{2\pi x}{a}}{\frac{2\pi }{a}}\\ =\frac{a\sin \frac{2\pi x}{a}}{2\pi }\end{array}$

$\begin{array}{c}{I}_2={\frac{a{x}^2\sin \frac{2\pi x}{a}}{2\pi }|}_{\frac{a}{2}}^{\frac{a}{2}}-\frac{a}{\pi }{\int }_{\frac{a}{2}}^{\frac{a}{2}}x\sin \frac{2\pi x}{a}dx\\ ={\frac{a{x}^2\sin \frac{2\pi x}{a}}{2\pi }|}_{\frac{a}{2}}^{\frac{a}{2}}-\frac{a}{\pi }{\int }_{\frac{a}{2}}^{\frac{a}{2}}x\sin \frac{2\pi x}{a}dx\end{array}$

Further simplify,

Assume $u=x$,$dv=\sin \frac{2\pi x}{a}dx$,$du=dx$

$v=\frac{-a}{2\pi }\cos \frac{2\pi x}{a}$

Simplify the expression.

$\begin{array}{c}{I}_2=\frac{a{x}^2\sin \frac{2\pi x}{a}}{2\pi }-\frac{a}{\pi }(\frac{-a}{2\pi }x\cos \frac{2\pi x}{a}-{\int }_{\frac{a}{2}}^{\frac{a}{2}}\frac{-a}{2\pi }\cos \frac{2\pi x}{a}dx)\\ ={\frac{a{x}^2}{2\pi }\sin \left(\frac{2\pi x}{a}\right)|}_{\frac{a}{2}}^{\frac{a}{2}}+{\frac{a^2}{2{\pi }^2}x\cos \frac{2\pi x}{a}|}_{\frac{a}{2}}^{\frac{a}{2}}-\frac{a^2}{2{\pi }^2}\frac{a}{2\pi }\sin \frac{2\pi x}{a}\\ ={\frac{a^2}{2{\pi }^2}x\cos \left(\frac{2\pi x}{a}\right)|}_{\frac{a}{2}}^{\frac{a}{2}}\\ =\frac{a^2}{2{\pi }^2}\left(\frac{a}{2}\right)(-1)-\frac{a^2}{2{\pi }^2}\left(\frac{-a}{2}\right)(-1)\end{array}$

Further evaluate the expression.

$\begin{array}{c}{I}_2=\frac{-a^3}{4{\pi }^2}-\frac{a^3}{4{\pi }^2}\\ =\frac{1}{2}\left(\frac{-a^3}{2{\pi }^2}\right)\\ =\frac{-a^3}{4{\pi }^2}\end{array}$

Write the value of $l$.

$I=\frac{1}{2}m{\omega }^2\left(\frac{2}{a}\right)(\frac{a^3}{24}-\frac{a^3}{4{\pi }^2})$

Determine the expectation value of the Hamiltonian by substituting the values in.

$⟨H⟩=⟨T⟩+⟨V⟩$

Take the derivative of above expression with respect to $a$.

$\frac{d⟨H⟩}{da}=\frac{-{\pi }^2{\hslash }^2}{m{a}^3}+\frac{m{\omega }^{2}a}{a{\pi }^2}(\frac{{\pi }^2}{6}-1)$

Equate the above expression to 0.

$\begin{array}{c}\frac{d⟨H⟩}{da}=0\\ \frac{-{\pi }^2{\hslash }^2}{m{a}^3}+\frac{m{\omega }^{2}a}{a{\pi }^2}(\frac{{\pi }^2}{6}-1)=0\\ =\frac{m{\omega }^{2}a}{2{\pi }^2}(\frac{{\pi }^2}{6}-1)\\ =\frac{{\pi }^2{\hslash }^2}{m{a}^3}\end{array}$

Solve to get the value of $a$.

$\begin{array}{c}a=\pi \sqrt{\frac{\hslash }{m\omega }}{\left(\frac{2}{(\frac{{\pi }^2}{6}-1)}\right)}^{\frac{1}{4}}\\ a^4=\frac{a{\pi }^2{\pi }^2{\hslash }^2}{m{\omega }^2(\frac{{\pi }^2}{6}-1)m}\\ =\frac{2{\pi }^4{\hslash }^2}{m^2{\omega }^2(\frac{{\pi }^2}{6}-1)}\\ a=\pi \sqrt{\frac{\hslash }{m\omega }}{\left(\frac{2}{(\frac{{\pi }^2}{6}-1)}\right)}^{\frac{1}{4}}\end{array}$

Substitute the above value in the expression for$⟨H⟩$.

$\begin{array}{c}⟨T⟩=\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right)\left(1\right)\\ =\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right)\\ =\frac{\hslash \omega }{2}\sqrt{\frac{\frac{{\pi }^2}{6}-1}{2}}+\frac{w\hslash }{4}\sqrt{2}\sqrt{\frac{{\pi }^2}{6}-1}\\ =\frac{1}{2}\hslash \omega \sqrt{\frac{{\pi }^2}{3}-2}\\ =\frac{1}{2}\hslash \omega \left(1.136\right)\\ =\frac{1}{2}\hslash \omega \end{array}$

Thus, the exact energyis $\frac{1}{2}\hslash \omega$and .${⟨H⟩}_{\min }>E_g$

(b) Determination of a bound on the first excited state

Write the expression for the wave function.

$\psi \left(x\right)=B\sin (\pi x/a)$

Determine the normalization condition.

$\begin{array}{c}\int |\psi \left(x\right){|}^{2}dx=1\\ {\left|B\right|}^2{\int }_{-a}^a{\sin }^2\left(\frac{\pi x}{a}\right)=1\\ {\left|B\right|}^2{\int }_{-a}^a\frac{1-\cos \left(\frac{2\pi x}{a}\right)}{2}=1\\ \frac{{\left|B\right|}^2}{2}{\int }_{-}^a(1-\cos \frac{2\pi x}{a})dx=1\end{array}$

Evaluate the above expression further.

$\begin{array}{c}\frac{{\left|B\right|}^2}{2}[{\left(x\right)}_{-a}^a-{\frac{{\left(\sin \frac{2\pi x}{a}\right)|}^a}{\left(\frac{2\pi }{a}\right)}|}_{-a}]=1\\ {\left|B\right|}^{2}a=1\\ B=\frac{1}{\sqrt{a}}\end{array}$

Determine the bound to the energy of the first excited state since the given state is orthogonal to the ground state.

It is known that$⟨H⟩=⟨T⟩+⟨V⟩$.

Determine the value of$⟨T⟩$.

$\begin{array}{c}⟨T⟩=\left(\frac{-{\hslash }^2}{2m}\right)\int \psi \left(x\right)\frac{d^2}{d{x}^2}\psi \left(x\right)dx\\ =\left(\frac{-{\hslash }^2}{2m}\right){\left|B\right|}^2\int \sin \left(\frac{\pi x}{a}\right)[-{\left(\frac{\pi }{a}\right)}^2\sin \left(\frac{\pi x}{a}\right)]dx\\ =\left(\frac{-{\hslash }^2}{2m}\right)(-{\left(\frac{\pi }{a}\right)}^2)|B{|}^2{\int }_{-a}^a{\sin }^2\left(\frac{\pi x}{a}\right)dx\\ =\left(\frac{{\hslash }^2}{2m}\right)\left(\frac{{\pi }^2}{a^2}\right)\left(\frac{1}{a}\right)\left(a\right)\\ =\frac{{\pi }^2{\hslash }^2}{2m{a}^2}\end{array}$

Determine the value of$⟨V⟩$.

$\begin{array}{c}⟨V⟩=\int \psi \left(x\right)\left(\frac{1}{2}m{\omega }^2{x}^2\right)\psi \left(x\right)dx\\ ={\left|B\right|}^2\left(\frac{1}{2}\right)m{\omega }^2{\int }_{-\theta }^a{x}^2{\sin }^2\left(\frac{\pi x}{a}\right)dx\end{array}$

Applyand,$y=\frac{\pi x}{a}$so$dx=\left(\frac{a}{\pi }\right)dy$the limits are as follows,

,$x=-a$,$y=-\pi$

$x=a$$y=\pi$

Substitute the above values.

$\begin{array}{c}⟨V⟩=\frac{m{\omega }^2}{2a}{\left(\frac{a}{\pi }\right)}^3{\int }_{-\pi }^{\pi }{y}^2{\sin }^{2}ydy\\ =\frac{m{\omega }^2{a}^2}{2{\left(\pi \right)}^3}{\int }_{-\pi }^{\pi }{y}^2\left(\frac{1-\cos 2y}{2}\right)dy\\ =\frac{m{\omega }^2{a}^2}{2{\pi }^3}[{\int }_{-\pi }^{\pi }{y}^2-{\int }_{-\pi }^{\pi }{y}^2\cos 2ydy]\\ =\frac{m{\omega }^2{a}^2}{2{\pi }^3}{\left[\frac{y^3}{3}\right]}_{-\pi }^{\pi }-{y^2\frac{\sin 2y}{2}|}_{-\pi }^{*}-{\int }_{-\pi }^{\pi }\left(2y\right)\left(\frac{\sin 2y}{2}\right)dy\end{array}$

Evaluate the above expression further.

$\begin{array}{c}⟨V⟩=\frac{m{\omega }^2{a}^2}{2{\pi }^3}{[\frac{y^3}{6}-(\frac{y^2}{4}-\frac{1}{8})\sin 2y-\frac{y\cos 2y}{4}]}_{-\pi }^{\pi }\\ =\frac{m{\omega }^2{a}^2}{4{\pi }^2}[\frac{2{\pi }^2}{3}\left[\frac{1}{6}\left(2{\pi }^3\right)-\frac{1}{4}(\pi \cos 2\pi +\pi \cos 2\pi )\right]\\ =\frac{m{\omega }^2{a}^2}{2{\pi }^3}[\frac{{\pi }^3}{3}-\frac{\pi }{2}]\end{array}$

Consequently, the minimum Hamiltonian is as follows,

$\begin{array}{c}{⟨H⟩}_{\min }=\frac{{\pi }^2{\hslash }^2}{2m}{\left(\frac{(2{\pi }^3/3)-1}{2}\right)}^{1/2}\frac{m\omega }{\hslash {\pi }^2}+\frac{m{\omega }^2}{4{\pi }^2}(\frac{2{\pi }^2}{3}-1){\pi }^2\left(\frac{\hslash }{m\omega }\right){\left(\frac{2}{\left(\frac{2{\pi }^2}{3}\right)-1}\right)}^{1/2}\\ ={\left(\frac{(2{\pi }^3/3)-1}{2}\right)}^{1/2}\left(\frac{\hslash \omega }{2}\right)+(\frac{2{\pi }^2}{3}-1)\frac{\hslash \omega }{4}{\left(\frac{2}{(2{\pi }^2/3)-1}\right)}^{1/2}\\ ={\left(\frac{(2{\pi }^2/3)-1}{2}\right)}^{1/2}\left(\frac{\hslash \omega }{2}\right)[1+(\frac{2{\pi }^2}{3}-1)\frac{1}{2}\frac{2}{(\frac{2{\pi }^2}{3}-1)}]\\ =\frac{\hslash \omega }{2}{\left[\frac{(2{\pi }^2/3)-1}{2}\right]}^{1/2}\left(2\right)\\ =\frac{3}{2}\hslash \omega \left[{\left(\frac{(2{\pi }^2/3)-1}{2}\right)}^{1/2}\left(\frac{2}{3}\right)\right]\\ =\left(\frac{3}{2}\right)\hslash \omega \left(1.857\right)\\ =\left(\frac{3}{2}\right)\hslash \omega \end{array}$

Thus, the first actual excited state energy is $\frac{3}{2}\hslash \omega {⟨H⟩}_{\min }>\frac{3}{2}\hslash \omega$.