91影视

The trial wave function has the form:

$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

The harmonic oscillator potential has the form of:

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

(a)

Solve the problem by denoting the ground state as ${蠄}_0$and the first excited state by ${蠄}_1$with energies $E_0$and $E_1$ respectively.

Because any wave function may be expressed as a set of orthonormal eigenfunctions,

role="math" localid="1658305418096" $蠄\left(x\right)=\underset{n=0}{\overset{鈭�}{鈭�}}{c}_n{蠄}_n\left(x\right)$

Where the coefficients $c_n$are given by$c_n=鉄�{蠄}_n鈭�蠄鉄�$

Since we have $鉄�{蠄}_1鈭�蠄鉄�=0$ then, the coefficient of the ground state is :

$鉄�{蠄}_1鈭�蠄鉄�=0$

The Hamiltonian expectation value is now given by:

Therefore, as the equation's second term is positive, we can deduce:$鉄�H鉄�鈮�E_1$

Which is an upper bound on the first eigen state's energy.

(b)

A's normalization

Using Gauss integral as:

${鈭�}_{-鈭�}^{鈭�}{\mathrm{e}}^{-2{\mathrm{bx}}^2}\mathrm{dx}=\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}}$

Perform this integral and get,

$$\begin{gathered} {鈭�}_{-鈭�}^{鈭�}{\mathrm{x}}^2{\mathrm{e}}^{-{\mathrm{bx}}^2}\mathrm{dx}=\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}}\frac{1}{4\mathrm{b}} \\ {\mathrm{A}}^2\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}}\frac{1}{4\mathrm{b}}=1 \\ \mathrm{A}=\sqrt{4\mathrm{b}\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}}} \end{gathered}$$

Find the normalization of T as:

role="math" localid="1658312071939"

This can be solved as follows:

role="math" localid="1658313404784" $$\begin{gathered} \frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\left({\mathrm{xe}}^{-{\mathrm{bx}}^2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(-2{\mathrm{bx}}^2{\mathrm{e}}^{-{\mathrm{bx}}^2}+{\mathrm{e}}^{-{\mathrm{bx}}^2}\right) \\ ={\mathrm{e}}^{-{\mathrm{bx}}^2}\left(4{\mathrm{x}}^3{\mathrm{b}}^2-6\mathrm{xb}\right) \\ =\frac{\mathrm{d}}{\mathrm{db}}\left(\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}}\frac{1}{4\mathrm{b}}\right) \\ =\frac{3}{16{\mathrm{b}}^2}\sqrt{\frac{\mathrm{蟺}}{2\mathrm{b}}} \end{gathered}$$

Next, do the integration for,

${鈭�}_{鈭�}^{鈭�}{x}^4{e}^{-2b{x}^2}dx=\frac{3}{16b^2}\sqrt{\frac{蟺}{2b}}$

Then

Find the value of V as:

role="math" localid="1658310803085"

Find the value of Hamiltonian as:

Derivate with respect to b is:

Then the value of b can be calculated as:

$b=\frac{m蝇}{2鈩�}$

And the minimum energy can be calculated as:

Thus, the best bound on the one-dimensional harmonic oscillator's first excited state $鉄�H鉄�{\frac{3鈩�蝇}{2}}_{min}$.