Show that if $\mathbf{⟨}\mathbf{h}\mathbf{\mid }\widehat{\mathbf{Q}}\mathbf{h}\mathbf{⟩}\mathbf{=}\mathbf{⟨}\widehat{\mathbf{Q}}\mathbf{h}\mathbf{\mid }\mathbf{h}\mathbf{⟩}$for all functions$\mathbf{h}$(in Hilbert space), then$\mathbf{⟨}\mathbf{f}\mathbf{\mid }\widehat{\mathbf{Q}}\mathbf{g}\mathbf{⟩}\mathbf{=}\mathbf{⟨}\widehat{\mathbf{Q}}\mathbf{f}\mathbf{\mid }\mathbf{g}\mathbf{⟩}$for allrole="math" localid="1655395250670" $\mathit{f}$and$\mathit{g}$(i.e., the two definitions of "Hermitian" -Equations 3.16 and 3.17- are equivalent).

For a Hermitian operator (say Q), the following condition must be satisfied:

${\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{g}}^{*}\mathrm{Qfdx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}{\left(\mathrm{Qg}\right)}^{*}\mathrm{dx}$

For a Hermitian operator (say Q), the following condition must be satisfied:

${\int }_{\mathrm{a}}^{\mathrm{b}}{\mathrm{g}}^{*}\mathrm{Qfdx}={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}{\left(\mathrm{Qg}\right)}^{*}\mathrm{dx}$

This condition can be written as: Using the more compact bracket notation, this condition can be written as:

$⟨g\mid \widehat{Q}f⟩=⟨\widehat{Q}g\mid f⟩$

Start with a less restrictive condition that is:

$⟨\mathrm{h}\mid {\mathrm{Q}}^{\mathrm{h}}⟩=⟨{\mathrm{Q}}^{\mathrm{h}}\mid \mathrm{h}⟩......\left(1\right)$

Let ,$\mathrm{h}=\mathrm{f}+\mathrm{g}$so:

$\begin{array}{c}⟨\mathrm{h}\mid \widehat{\mathrm{Q}}\mathrm{h}⟩=⟨\mathrm{f}+\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}+\widehat{\mathrm{Q}}\mathrm{g}⟩......\left(2\right)\\ =⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩+⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩+⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩+⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩......\left(3\right)\end{array}$

We can also write this equation as:

$\begin{array}{c}⟨\widehat{\mathrm{Q}}\mathrm{h}\mid \mathrm{h}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}+\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}+\mathrm{g}⟩......\left(4\right)\\ =⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{f}⟩+⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{g}⟩+⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{g}⟩+⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}⟩......\left(5\right)\end{array}$

Use assumption $⟨\mathrm{h}\mid \widehat{\mathrm{Q}}\mathrm{h}⟩=⟨\widehat{\mathrm{Q}}\mathrm{h}\mid \mathrm{h}⟩$for all functions, we have $⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{f}⟩$and $⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩=$$⟨Q\widehat{Q}g\mid g⟩$so, equating (2) and (3) and cancel out the common terms, we get:$⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩+⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{g}⟩+⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}⟩$

Now we do the same to above but with $\mathrm{h}=\mathrm{f}+\mathrm{ig}$, so:

$\begin{array}{c}⟨\mathrm{h}\mid \widehat{\mathrm{Qh}}⟩=⟨\mathrm{f}+\mathrm{ig}\mid \widehat{\mathrm{Q}}\mathrm{f}+\mathrm{i}\widehat{\mathrm{Q}}\mathrm{g}⟩\\ =⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩+(-\mathrm{i})\left(\mathrm{i}\right)⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩+\mathrm{i}⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩-\mathrm{i}⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩\\ =⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩+⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩+\mathrm{i}⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩-\mathrm{i}⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩......\left(6\right)\\ ⟨\mathrm{Qh}\mid \mathrm{h}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{f}⟩+⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{g}⟩+\mathrm{i}⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{g}⟩-\mathrm{i}⟨\mathrm{Qg}\mid \mathrm{f}⟩......\left(7\right)\end{array}$

Using the assumption $⟨\mathrm{h}\mid \widehat{\mathrm{Q}}\mathrm{h}⟩=⟨\widehat{\mathrm{Q}}\mathrm{h}\mid \mathrm{h}⟩$for all functions, we have $⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=$$⟨\widehat{\mathrm{Q}}\mid \mathrm{f}⟩$and $⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩=⟨\widehat{\mathrm{Q}}\mid \mathrm{g}⟩$so, equating (2) and (3) and cancel out the common terms, we get:$⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩-⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{g}⟩-⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}⟩......\left(8\right)$

Add equation (4) and (8), we get,

$⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩=⟨\widehat{\mathrm{Q}}\mathrm{f}\mid \mathrm{g}⟩$

If we subtract equation (8) from equation (4), we get,

$⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}⟩$

Thus, the given statement is proved, i.e., $⟨\mathrm{g}\mid \widehat{\mathrm{Q}}\mathrm{f}⟩=⟨\widehat{\mathrm{Q}}\mathrm{g}\mid \mathrm{f}⟩$.