(a) Prove the following corollary to the variational principle: If $\mathbf{⟨}\mathbf{\psi }\mathbf{\mid }{\mathbf{\psi }}_{\mathbf{gs}}\mathbf{⟩}\mathbf{=}\mathbf{0}$then$\mathbf{áHñ}\mathbf{\ge }{\mathbf{E}}_{\mathbf{fe}}$ where${\mathit{E}}_{\mathbf{f}\mathbf{e}}$ is the energy of the first excited state. Comment: If we can find a trial function that is orthogonal to the exact ground state, we can get an upper bound on the first excited state. In general, it's difficult to be sure that is orthogonal to${\mathbf{\psi }}_{\mathbf{gsi}}$ since (presumably) we don't know the latter. However, if the potential$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}$ is an even function of x, then the ground state is likewise even, and hence any odd trial function will automatically meet the condition for the corollary.

(b) Find the best bound on the first excited state of the one-dimensional harmonic oscillator using the trial function$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Axe}}^{\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}}$

The trial wave function has the form:

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

The harmonic oscillator potential has the form of:

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

(a)

Solve the problem by denoting the ground state as ${\psi }_0$and the first excited state by ${\psi }_1$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" $\psi \left(x\right)=\sum _{n=0}^{\infty }{c}_n{\psi }_n\left(x\right)$

Where the coefficients $c_n$are given by$c_n=⟨{\psi }_n\mid \psi ⟩$

Since we have $⟨{\psi }_1\mid \psi ⟩=0$ then, the coefficient of the ground state is :

$⟨{\psi }_1\mid \psi ⟩=0$

The Hamiltonian expectation value is now given by:

Therefore, as the equation's second term is positive, we can deduce:$⟨H⟩\ge E_1$

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

(b)

A's normalization

Using Gauss integral as:

${\int }_{-\infty }^{\infty }{\mathrm{e}}^{-2{\mathrm{bx}}^2}\mathrm{dx}=\sqrt{\frac{\mathrm{\pi }}{2\mathrm{b}}}$

Perform this integral and get,

$$\begin{gathered} {\int }_{-\infty }^{\infty }{\mathrm{x}}^2{\mathrm{e}}^{-{\mathrm{bx}}^2}\mathrm{dx}=\sqrt{\frac{\mathrm{\pi }}{2\mathrm{b}}}\frac{1}{4\mathrm{b}} \\ {\mathrm{A}}^2\sqrt{\frac{\mathrm{\pi }}{2\mathrm{b}}}\frac{1}{4\mathrm{b}}=1 \\ \mathrm{A}=\sqrt{4\mathrm{b}\sqrt{\frac{\mathrm{\pi }}{2\mathrm{b}}}} \end{gathered}$$

Find the normalization of T as:

role="math" localid="1658312071939"

This can be solved as follows:

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

Next, do the integration for,

${\int }_{\infty }^{\infty }{x}^4{e}^{-2b{x}^2}dx=\frac{3}{16b^2}\sqrt{\frac{\pi }{2b}}$

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:

$b=\frac{m\omega }{2\hslash }$

And the minimum energy can be calculated as:

Thus, the best bound on the one-dimensional harmonic oscillator's first excited state $⟨H⟩{\frac{3\hslash \omega }{2}}_{min}$.