Question

Find the best bound on ${\mathbf{E}}_{\mathbf{gs}}$for the delta-function potential$\mathit{V}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathit{\alpha }\mathit{\delta }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathbf{}$ using a triangular trial function (Equation 7.10, only centered at the origin). This time a is an adjustable parameter.

Short answer

The minimum value of the Hamiltonian that approximates the ground state energy is then$⟨H⟩{\frac{-3m{\alpha }^2}{8{\hslash }^2}}_{min}$

Step-by-step solution

harmonic oscillator potential

The harmonic oscillator potential has the form

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

Step 2: Find the normalization of A.

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

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

The expectation value of kinetic energy

localid="1656042395443" $$\begin{gathered} ⟨T⟩=-\frac{{\hslash }^2}{2m}{A}^2{\int }_{-a/2}^{a/2}\psi \frac{d^2}{d{x}^2}\psi dx \\ \frac{d}{dx}\psi =\left\{A,\frac{-a}{2}\le x\le 0 \\ -A,\&0\le x\le \frac{a}{2} \\ 0,Otherwise\right\} \end{gathered}$$

Now delta function,

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

Now integrate $⟨T⟩=\frac{-{\hslash }^2}{2m}\int \left[A\delta \left(x+\frac{a}{2}\right)-2A\delta \left(x\right)+A\delta \left(x-\frac{a}{2}\right)\right]\psi \left(x\right)dx$

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

Step 3: Find the Value of V.

The expectation value of potential energy

$$\begin{gathered} ⟨\mathrm{V}⟩=-\mathrm{\alpha }\int \mathrm{\delta }\left(\mathrm{x}\right){\mathrm{\psi }}^2\left(\mathrm{x}\right) \\ =-\mathrm{\alpha }\frac{{\mathrm{A}}^2{\mathrm{a}}^2}{4} \\ =\frac{-3\mathrm{\alpha }}{\mathrm{a}} \end{gathered}$$

Step 4: Find the Hamiltonian value H.

The expectation value of Hamiltonian is

$$\begin{gathered} ⟨H⟩=⟨T⟩+⟨V⟩ \\ ⟨H⟩=\frac{6{\hslash }^2}{m{a}^2}-\frac{-3\alpha }{a} \end{gathered}$$

Find the value of a.

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

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