91影视

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.

Write the normalization condition.

$\begin{array}{c}鈭�{\left|蠄\left(x\right)\right|}^{2}dx=1{\left|A\right|}^2{鈭�}_{鈭�a/2}^{蟺/2}{\cos }^2(蟺x/a)dx\\ =1{\left|A\right|}^2{鈭�}_{鈭�a/2}^{a/2}\left[\frac{1+\cos 2\left(\frac{蟺x}{2}\right)}{2}\right]dx\\ =1{\left|A\right|}^2[{鈭�}_0^{a/2}dx+{鈭�}_0^{a/2}\cos \left(\frac{2蟺x}{a}\right)dx]\\ =1{\left|A\right|}^2{[x+\frac{\sin \left(\frac{2蟺x}{a}\right)}{(鈭�\frac{2蟺}{a})}]}_0^{a/2}\end{array}$

Evaluate the above expression further.

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

It is known that$鉄�H鉄�=鉄�T鉄�+鉄�V鉄�$and for a harmonic oscillator,$T=\frac{鈭�{\mathrm{鈩�}}^2}{2m}\frac{d^2}{d{x}^2}$and$V=\frac{1}{2}m{蝇}^2{x}^2$.

Determine the expectation value of.$T$

$\begin{array}{c}鉄�T鉄�=鈭�{蠄}^{*}\left(x\right)\left(\frac{鈭�{\mathrm{鈩�}}^2}{2m}\frac{d^2}{d{x}^2}\right)蠄\left(x\right)dx\\ =\frac{鈭�{\mathrm{鈩�}}^2}{2m}{\left|A\right|}^2鈭�\cos \left(\frac{蟺x}{a}\right)\frac{d}{dx}(鈭�\sin \left(\frac{蟺x}{a}\right))\left(\frac{蟺}{a}\right)dx\\ =\frac{鈭�{\mathrm{鈩�}}^2}{2m}{\left|A\right|}^2(鈭�\frac{{蟺}^2}{a^2}){鈭�}_{鈭�a/2}^{a/2}{\cos }^2\left(\frac{蟺x}{a}\right)dx\\ =\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right){鈭�}_{鈭�伪/2}^{伪/2}{\left|蠄\right|}^{2}dx\\ =\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right)\left(1\right)\\ =\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right)\end{array}$

Determine the expectation value of $V$.

$\begin{array}{c}鉄�V鉄�=\frac{1}{2}m{蝇}^2{A}^2{鈭�}_{\frac{a}{2}}^{\frac{a}{2}}{x}^2{\cos }^2\frac{蟺x}{a}dx\\ ={鈭�}_{\frac{a}{2}}^{\frac{a}{2}}{x}^2{\cos }^2\frac{蟺x}{a}dx\\ ={鈭�}_{\frac{a}{2}}^{\frac{a}{2}}\frac{x^2}{2}(1+\cos \frac{2蟺x}{a})dx\\ ={鈭�}_{\frac{a}{2}}^{\frac{a}{2}}\frac{x^2}{2}dx+\frac{1}{2}{鈭�}_{\frac{a}{2}}^{\frac{a}{2}}{x}^2\cos \frac{2蟺x}{a}dx\\ =I_1+I_2\end{array}$

Simplify the above expression further.

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

$I_2=\frac{1}{2}{鈭�}_{\frac{a}{2}}^{\frac{a}{2}}{x}^2\cos \frac{2蟺x}{a}dx$

Integrate the function by parts,

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

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

$\begin{array}{c}{I}_2={\frac{a{x}^2\sin \frac{2蟺x}{a}}{2蟺}|}_{\frac{a}{2}}^{\frac{a}{2}}鈭�\frac{a}{蟺}{鈭�}_{\frac{a}{2}}^{\frac{a}{2}}x\sin \frac{2蟺x}{a}dx\\ ={\frac{a{x}^2\sin \frac{2蟺x}{a}}{2蟺}|}_{\frac{a}{2}}^{\frac{a}{2}}鈭�\frac{a}{蟺}{鈭�}_{\frac{a}{2}}^{\frac{a}{2}}x\sin \frac{2蟺x}{a}dx\end{array}$

Further simplify,

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

$v=\frac{鈭�a}{2蟺}\cos \frac{2蟺x}{a}$

Simplify the expression.

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

Further evaluate the expression.

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

Write the value of $l$.

$I=\frac{1}{2}m{蝇}^2\left(\frac{2}{a}\right)(\frac{a^3}{24}鈭�\frac{a^3}{4{蟺}^2})$

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

$鉄�H鉄�=鉄�T鉄�+鉄�V鉄�$

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

$\frac{d鉄�H鉄�}{da}=\frac{鈭�{蟺}^2{\mathrm{鈩�}}^2}{m{a}^3}+\frac{m{蝇}^{2}a}{a{蟺}^2}(\frac{{蟺}^2}{6}鈭�1)$

Equate the above expression to 0.

$\begin{array}{c}\frac{d鉄�H鉄�}{da}=0\\ \frac{鈭�{蟺}^2{\mathrm{鈩�}}^2}{m{a}^3}+\frac{m{蝇}^{2}a}{a{蟺}^2}(\frac{{蟺}^2}{6}鈭�1)=0\\ =\frac{m{蝇}^{2}a}{2{蟺}^2}(\frac{{蟺}^2}{6}鈭�1)\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m{a}^3}\end{array}$

Solve to get the value of $a$.

$\begin{array}{c}a=蟺\sqrt{\frac{\mathrm{鈩�}}{m蝇}}{\left(\frac{2}{(\frac{{蟺}^2}{6}鈭�1)}\right)}^{\frac{1}{4}}\\ a^4=\frac{a{蟺}^2{蟺}^2{\mathrm{鈩�}}^2}{m{蝇}^2(\frac{{蟺}^2}{6}鈭�1)m}\\ =\frac{2{蟺}^4{\mathrm{鈩�}}^2}{m^2{蝇}^2(\frac{{蟺}^2}{6}鈭�1)}\\ a=蟺\sqrt{\frac{\mathrm{鈩�}}{m蝇}}{\left(\frac{2}{(\frac{{蟺}^2}{6}鈭�1)}\right)}^{\frac{1}{4}}\end{array}$

Substitute the above value in the expression for$鉄�H鉄�$.

$\begin{array}{c}鉄�T鉄�=\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right)\left(1\right)\\ =\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right)\\ =\frac{\mathrm{鈩�}蝇}{2}\sqrt{\frac{\frac{{蟺}^2}{6}鈭�1}{2}}+\frac{w\mathrm{鈩�}}{4}\sqrt{2}\sqrt{\frac{{蟺}^2}{6}鈭�1}\\ =\frac{1}{2}\mathrm{鈩�}蝇\sqrt{\frac{{蟺}^2}{3}鈭�2}\\ =\frac{1}{2}\mathrm{鈩�}蝇\left(1.136\right)\\ =\frac{1}{2}\mathrm{鈩�}蝇\end{array}$

Thus, the exact energyis $\frac{1}{2}\mathrm{鈩�}蝇$and .${鉄�H鉄�}_{\min }>E_g$

Write the expression for the wave function.

$蠄\left(x\right)=B\sin (蟺x/a)$

Determine the normalization condition.

$\begin{array}{c}鈭�|蠄\left(x\right){|}^{2}dx=1\\ {\left|B\right|}^2{鈭�}_{鈭�a}^a{\sin }^2\left(\frac{蟺x}{a}\right)=1\\ {\left|B\right|}^2{鈭�}_{鈭�a}^a\frac{1鈭�\cos \left(\frac{2蟺x}{a}\right)}{2}=1\\ \frac{{\left|B\right|}^2}{2}{鈭�}_{鈭�}^a(1鈭�\cos \frac{2蟺x}{a})dx=1\end{array}$

Evaluate the above expression further.

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

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$鉄�H鉄�=鉄�T鉄�+鉄�V鉄�$.

Determine the value of$鉄�T鉄�$.

$\begin{array}{c}鉄�T鉄�=\left(\frac{鈭�{\mathrm{鈩�}}^2}{2m}\right)鈭�蠄\left(x\right)\frac{d^2}{d{x}^2}蠄\left(x\right)dx\\ =\left(\frac{鈭�{\mathrm{鈩�}}^2}{2m}\right){\left|B\right|}^2鈭�\sin \left(\frac{蟺x}{a}\right)[鈭�{\left(\frac{蟺}{a}\right)}^2\sin \left(\frac{蟺x}{a}\right)]dx\\ =\left(\frac{鈭�{\mathrm{鈩�}}^2}{2m}\right)(鈭�{\left(\frac{蟺}{a}\right)}^2)|B{|}^2{鈭�}_{鈭�a}^a{\sin }^2\left(\frac{蟺x}{a}\right)dx\\ =\left(\frac{{\mathrm{鈩�}}^2}{2m}\right)\left(\frac{{蟺}^2}{a^2}\right)\left(\frac{1}{a}\right)\left(a\right)\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{2m{a}^2}\end{array}$

Determine the value of$鉄�V鉄�$.

$\begin{array}{c}鉄�V鉄�=鈭�蠄\left(x\right)\left(\frac{1}{2}m{蝇}^2{x}^2\right)蠄\left(x\right)dx\\ ={\left|B\right|}^2\left(\frac{1}{2}\right)m{蝇}^2{鈭�}_{鈭�胃}^a{x}^2{\sin }^2\left(\frac{蟺x}{a}\right)dx\end{array}$

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

,$x=鈭�a$,$y=鈭�蟺$

$x=a$$y=蟺$

Substitute the above values.

$\begin{array}{c}鉄�V鉄�=\frac{m{蝇}^2}{2a}{\left(\frac{a}{蟺}\right)}^3{鈭�}_{鈭�蟺}^{蟺}{y}^2{\sin }^{2}ydy\\ =\frac{m{蝇}^2{a}^2}{2{\left(蟺\right)}^3}{鈭�}_{鈭�蟺}^{蟺}{y}^2\left(\frac{1鈭�\cos 2y}{2}\right)dy\\ =\frac{m{蝇}^2{a}^2}{2{蟺}^3}[{鈭�}_{鈭�蟺}^{蟺}{y}^2鈭�{鈭�}_{鈭�蟺}^{蟺}{y}^2\cos 2ydy]\\ =\frac{m{蝇}^2{a}^2}{2{蟺}^3}{\left[\frac{y^3}{3}\right]}_{鈭�蟺}^{蟺}鈭�{y^2\frac{\sin 2y}{2}|}_{鈭�蟺}^{*}鈭�{鈭�}_{鈭�蟺}^{蟺}\left(2y\right)\left(\frac{\sin 2y}{2}\right)dy\end{array}$

Evaluate the above expression further.

$\begin{array}{c}鉄�V鉄�=\frac{m{蝇}^2{a}^2}{2{蟺}^3}{[\frac{y^3}{6}鈭�(\frac{y^2}{4}鈭�\frac{1}{8})\sin 2y鈭�\frac{y\cos 2y}{4}]}_{鈭�蟺}^{蟺}\\ =\frac{m{蝇}^2{a}^2}{4{蟺}^2}[\frac{2{蟺}^2}{3}\left[\frac{1}{6}\left(2{蟺}^3\right)鈭�\frac{1}{4}(蟺\cos 2蟺+蟺\cos 2蟺)\right]\\ =\frac{m{蝇}^2{a}^2}{2{蟺}^3}[\frac{{蟺}^3}{3}鈭�\frac{蟺}{2}]\end{array}$

Consequently, the minimum Hamiltonian is as follows,

$\begin{array}{c}{鉄�H鉄�}_{\min }=\frac{{蟺}^2{\mathrm{鈩�}}^2}{2m}{\left(\frac{(2{蟺}^3/3)鈭�1}{2}\right)}^{1/2}\frac{m蝇}{\mathrm{鈩�}{蟺}^2}+\frac{m{蝇}^2}{4{蟺}^2}(\frac{2{蟺}^2}{3}鈭�1){蟺}^2\left(\frac{\mathrm{鈩�}}{m蝇}\right){\left(\frac{2}{\left(\frac{2{蟺}^2}{3}\right)鈭�1}\right)}^{1/2}\\ ={\left(\frac{(2{蟺}^3/3)鈭�1}{2}\right)}^{1/2}\left(\frac{\mathrm{鈩�}蝇}{2}\right)+(\frac{2{蟺}^2}{3}鈭�1)\frac{\mathrm{鈩�}蝇}{4}{\left(\frac{2}{(2{蟺}^2/3)鈭�1}\right)}^{1/2}\\ ={\left(\frac{(2{蟺}^2/3)鈭�1}{2}\right)}^{1/2}\left(\frac{\mathrm{鈩�}蝇}{2}\right)[1+(\frac{2{蟺}^2}{3}鈭�1)\frac{1}{2}\frac{2}{(\frac{2{蟺}^2}{3}鈭�1)}]\\ =\frac{\mathrm{鈩�}蝇}{2}{\left[\frac{(2{蟺}^2/3)鈭�1}{2}\right]}^{1/2}\left(2\right)\\ =\frac{3}{2}\mathrm{鈩�}蝇\left[{\left(\frac{(2{蟺}^2/3)鈭�1}{2}\right)}^{1/2}\left(\frac{2}{3}\right)\right]\\ =\left(\frac{3}{2}\right)\mathrm{鈩�}蝇\left(1.857\right)\\ =\left(\frac{3}{2}\right)\mathrm{鈩�}蝇\end{array}$

Thus, the first actual excited state energy is $\frac{3}{2}\mathrm{鈩�}蝇{鉄�H鉄�}_{\min }>\frac{3}{2}\mathrm{鈩�}蝇$.