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$\hat{x} | \Psi \rangle$ en $p$-rep :
\begin{align*}
\langle p | \hat{x} | \Psi \rangle
& = \int \langle p | x \rangle x \Psi(x) \,\mathrm{d}x \\
& = \int {\underbrace{\langle x | p \rangle}_{\clap{\text{\tiny{plane wave}}}}}^* x \Psi(x) \,\textrm{d}x \\
& = \int \Big(\tfrac{1}{\sqrt{2\pi\hbar}}\mathrm{e}^{i\frac{px}{\hbar}}\Big)^* x \Psi(x) \,\mathrm{d}x \\
{\scriptstyle (p=\hbar k)} \quad & = \frac{1}{\sqrt{2\pi\hbar}} \int \mathrm{e}^{-ikx} x \Psi(x) \,\mathrm{d}x  \\
& = \frac{1}{\sqrt{\hbar}} \mathrm{FT}\big[ x\Psi(x) \big] = \frac{1}{\sqrt{\hbar}} \mathrm{i}\frac{\mathrm{d}\hat{\Psi}}{\mathrm{d}k}(k) \\
{\scriptstyle (\tilde{\Psi}(p) = \frac{1}{\sqrt{\hbar}} \hat{\Psi}(k))} \quad & = \frac{1}{\sqrt{\hbar}} \mathrm{i} \hbar\sqrt{\hbar} \frac{\mathrm{d}\tilde{\Psi}}{\mathrm{d}p}(p) \\
\langle p | \hat{x} | \Psi \rangle & = i\hbar \frac{\mathrm{d}\tilde{\Psi}}{\mathrm{d}p}(p)
\end{align*}

Moyenne de l'impulsion :
\begin{align*}
\langle p \rangle_{\!_\Psi}
& = \langle \Psi | \hat{p} | \Psi \rangle \\
& = \langle \Psi | \bigg(\! \int p | p \rangle \langle p | \,\mathrm{d}p \bigg) | \Psi \rangle \\
& = \int \langle \Psi | p \rangle\,p\,\langle p | \Psi \rangle \,\mathrm{d}p \\
& = \int \langle p | \Psi \rangle^*\,p\,\langle p | \Psi \rangle \,\mathrm{d}p \\
& = \int \tilde{\Psi}^*(p)\,p\,\tilde{\Psi}(p) \,\mathrm{d}p \\ \\
\langle p \rangle_{\!_\Psi}
& = \langle \Psi | \hat{p} | \Psi \rangle \\
& = \int \Psi^*(x) \,\hat{p}\, \Psi(x) \,\mathrm{d}x \\
& = -\mathrm{i}\hbar \int \Psi^*(x) \frac{\mathrm{d}\Psi}{\mathrm{d}x}(x) \,\mathrm{d}x
\end{align*}
(derrière cette identitée se cache fondamentalement Parseval-Plancherel et $\mathrm{FT}\big[ \Psi'(x) \big] = k\hat{\Psi}(k)$)

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