91影视

The harmonic oscillator potential has the form

$\mathit{V}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{蝇}}^{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}$

The value of A$A^2\left(\underset{{}_{-\frac{a}{2}}}{\overset{0}{鈭�}}{\left(\frac{a}{2}+x\right)}^{2}dx+{鈭�}_0^{\frac{a}{2}}{\left(\frac{a}{2}+x\right)}^{2}dx\right)=A^2\left(2鈰�\frac{a^3}{24}\right)=1$

$A=\sqrt{\frac{12}{a^3}}$

The expectation value of kinetic energy

localid="1656042395443" $$\begin{gathered} 鉄�T鉄�=-\frac{{鈩�}^2}{2m}{A}^2{鈭�}_{-a/2}^{a/2}蠄\frac{d^2}{d{x}^2}蠄dx \\ \frac{d}{dx}蠄=\left\{A,\frac{-a}{2}鈮�x鈮�0 \\ -A,\&0鈮�x鈮�\frac{a}{2} \\ 0,Otherwise\right\} \end{gathered}$$

Now delta function,

$\frac{d^2蠄}{d{x}^2}=未\left(x+\frac{a}{2}\right)-2A未\left(x\right)+A未\left(x-\frac{a}{2}\right)$

Now integrate $鉄�T鉄�=\frac{-{鈩�}^2}{2m}鈭�\left[A未\left(x+\frac{a}{2}\right)-2A未\left(x\right)+A未\left(x-\frac{a}{2}\right)\right]蠄\left(x\right)dx$

role="math" localid="1656042956837" $$\begin{gathered} =\frac{-{鈩�}^2}{2m}\left[A蠄(-a/2)-2A蠄\left(0\right)+A蠄(a/2)\right] \\ =\frac{{鈩�}^2{A}^{2}a}{2m} \\ =\frac{6{鈩�}^2}{m{a}^2} \end{gathered}$$

The expectation value of potential energy

$$\begin{gathered} 鉄�\mathrm{V}鉄�=-\mathrm{伪}鈭�\mathrm{未}\left(\mathrm{x}\right){\mathrm{蠄}}^2\left(\mathrm{x}\right) \\ =-\mathrm{伪}\frac{{\mathrm{A}}^2{\mathrm{a}}^2}{4} \\ =\frac{-3\mathrm{伪}}{\mathrm{a}} \end{gathered}$$

The expectation value of Hamiltonian is

$$\begin{gathered} 鉄�H鉄�=鉄�T鉄�+鉄�V鉄� \\ 鉄�H鉄�=\frac{6{鈩�}^2}{m{a}^2}-\frac{-3伪}{a} \end{gathered}$$

Find the value of a.

role="math" localid="1656043876100" $$\begin{gathered} \frac{\mathrm{d}鉄�\mathrm{H}鉄�}{\mathrm{da}}=\frac{-12{鈩�}^2}{{\mathrm{ma}}^3})+\frac{3\mathrm{a}}{{\mathrm{a}}^2} \\ -0 \\ \mathrm{a}=\frac{4{鈩�}^2}{\mathrm{ma}} \\ 鉄�\mathrm{H}鉄�\frac{6{鈩�}^2}{\mathrm{m}}{\left(\frac{\mathrm{尘伪}}{4{鈩�}^2}\right)}^2{\left(\frac{\mathrm{尘伪}}{4{鈩�}^2}\right)}_{\min } \\ =\frac{-3{\mathrm{尘伪}}^2}{8{鈩�}^2} \\ 鉄�\mathrm{H}鉄�{\frac{-3{\mathrm{尘伪}}^2}{8{鈩�}^2}}_{\min } \end{gathered}$$

Thus, the minimum value of the Hamiltonian that approximates the ground state energy is then$<\mathrm{H}{>}_{min}=\frac{-3{\mathrm{尘伪}}^2}{8{鈩�}^2}$