The Hermitian conjugate (or adjoint) of an operator $\widehat{\mathbf{Q}}$is the operator${\widehat{\mathbf{Q}}}^{\mathbf{†}}$such that

$\mathbf{⟨}\mathbf{f}\mathbf{\mid }\widehat{\mathbf{Q}}\mathbf{g}\mathbf{⟩}\mathbf{=}\mathbf{⟨}{\widehat{\mathbf{Q}}}^{\mathbf{†}}\mathbf{f}\mathbf{\mid }\mathbf{g}\mathbf{⟩}\mathbf{\text{\hspace{1em}(forall}}\mathbf{f}\mathbf{\text{and}}\mathbf{g}\mathbf{\text{).}}$

(A Hermitian operator, then, is equal to its Hermitian conjugate:$\widehat{\mathbf{Q}}\mathbf{=}{\widehat{\mathbf{Q}}}^{\mathbf{†}}$)

(a)Find the Hermitian conjugates of x, i, and$\mathbf{d}\mathbf{/}\mathbf{dx}$.

(b) Construct the Hermitian conjugate of the harmonic oscillator raising operator,${\mathbf{a}}_{\mathbf{+}}$(Equation 2.47).

(c) Show that${\mathbf{\left(}\widehat{\mathbf{Q}}\widehat{\mathbf{R}}\mathbf{\right)}}^{\mathbf{†}}\mathbf{=}{\widehat{\mathbf{R}}}^{\mathbf{†}}{\widehat{\mathbf{Q}}}^{\mathbf{†}}$.

The adjoint of an operator$\widehat{\mathrm{Q}}$ is defined as the operator${\widehat{Q}}^{†}$ such that:$⟨\mathrm{f}\mid \widehat{\mathrm{Q}}\mathrm{g}⟩=⟨{\widehat{\mathrm{Q}}}^{†}\mathrm{f}\mid \mathrm{g}⟩$

(a)

The adjoint of an operator $\widehat{\mathrm{Q}}$is defined as the operator ${\widehat{\mathrm{Q}}}^{†}$such that:

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

For the position operator $\mathrm{x}$, we have:

For the imaginary number $i$, we have:

For the operator $\frac{\mathrm{d}}{\mathrm{dx}}$, we can use integration by parts to find:

role="math" localid="1655396772124"

b)

The Hermitian conjugate of the harmonic oscillator raising operator is

${\mathrm{a}}_{+}=\frac{1}{\sqrt{2\mathrm{\hslash }\mathrm{m\omega }}}(-\mathrm{ip}+\mathrm{m\omega x})$

the momentum$p$ and the position$x$ operators Hermitian, that is ${\mathrm{x}}^{†}=\mathrm{x}$, and ${\mathrm{p}}^{†}=\mathrm{p}$, but the conjugate of the imaginary number is ${\mathrm{i}}^{†}=-\mathrm{i}$, thus :$\begin{array}{l}{\left({\mathrm{a}}_{+}\right)}^{†}=\frac{1}{\sqrt{2\mathrm{\hslash }\mathrm{m\omega }}}(\mathrm{ip}+\mathrm{m\omega x})={\mathrm{a}}_{-}\\ \Rightarrow {\mathrm{a}}_{+}={\mathrm{a}}_{-}\end{array}$

c)

Solving the given equation as,

$⟨f\mid \left(\widehat{Q}\widehat{R}\right)g⟩=⟨{\widehat{Q}}^{†}f\mid \widehat{R}g⟩$

Substituting the calculated conjugates in above equation as,

$\begin{array}{l}=⟨{\widehat{R}}^{†}{\widehat{Q}}^{†}f\mid g⟩\\ =⟨{\left(\widehat{Q}\widehat{R}\right)}^{†}f\mid g⟩\end{array}$

Thus, the we can write: ${\left(\widehat{Q}\widehat{R}\right)}^{†}={\widehat{R}}^{†}{\widehat{Q}}^{†}$.