[proofplan] We compute the [Fourier transform](/page/Fourier%20Transform) of $P\psi = -i\hbar\psi'$ directly from the definition. The only point requiring justification is [integration by parts](/theorems/2098) on the unbounded domain, so we first integrate over $[-R,R]$ and then let $R \to \infty$. The [Schwartz Space](/page/Schwartz%20Space) decay of $\psi$ removes the boundary terms, while the remaining integral converges to the Fourier transform of $\psi$. [/proofplan]

[step:Truncate the Fourier integral to a compact interval] Throughout the proof, $\mathbb C$ denotes the complex field and $\mathcal B(\mathbb R)$ denotes the Borel $\sigma$-algebra on $\mathbb R$. Fix $\psi \in \mathcal S(\mathbb R)$ and $k \in \mathbb R$. Since differentiation preserves the [Schwartz Space](/page/Schwartz%20Space), $\psi' \in \mathcal S(\mathbb R)$. Every Schwartz function on $\mathbb R$ is Lebesgue integrable, because for $N=2$ there is a constant $C_{\psi'} \geq 0$ such that $|\psi'(x)| \leq C_{\psi'}(1+|x|)^{-2}$ for all $x \in \mathbb R$, and $(1+|x|)^{-2} \in L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. Define the measurable function \begin{align*} f_k: \mathbb R \to \mathbb C, \qquad f_k(x)=\psi'(x)e^{-ikx}. \end{align*} Since $|e^{-ikx}|=1$ for all $x \in \mathbb R$, the function $f_k$ belongs to $L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. Therefore \begin{align*} (\mathcal F(P\psi))(k) = \frac{-i\hbar}{\sqrt{2\pi}}\int_{\mathbb R}\psi'(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} For $R > 0$, define the truncated integral $I_R \in \mathbb C$ by \begin{align*} I_R := \int_{[-R,R]} \psi'(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} For each $R > 0$, let $\mathbb{1}_{[-R,R]}: \mathbb R \to \{0,1\}$ denote the indicator function of the interval $[-R,R]$. The functions $\mathbb{1}_{[-R,R]} f_k$ converge pointwise to $f_k$ as $R \to \infty$, and they satisfy $|\mathbb{1}_{[-R,R]} f_k| \leq |f_k|$ with $|f_k| \in L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. By the [Dominated Convergence Theorem](/theorems/4), \begin{align*} \int_{\mathbb R}\psi'(x)e^{-ikx}\,d\mathcal L^1(x) = \lim_{R \to \infty} I_R. \end{align*} [/step]

[step:Integrate by parts on the truncated interval]On the compact interval $[-R,R]$, the functions $x \mapsto \psi(x)$ and $x \mapsto e^{-ikx}$ are continuously differentiable. The [Integration By Parts](/theorems/210) theorem is stated for real-valued functions, and applying it to the real and imaginary parts of these complex-valued functions gives the same formula by linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) over $\mathbb C$. Therefore \begin{align*} I_R = \psi(R)e^{-ikR} - \psi(-R)e^{ikR} + ik\int_{[-R,R]}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*}[/step]

[guided]We integrate by parts only on $[-R,R]$ because the usual one-dimensional integration by parts formula is a compact-interval statement. For fixed $R > 0$, define \begin{align*} u_R: [-R,R] \to \mathbb C, \qquad u_R(x)=\psi(x). \end{align*} Also define \begin{align*} v_R: [-R,R] \to \mathbb C, \qquad v_R(x)=e^{-ikx}. \end{align*} Both maps are continuously differentiable, with $v_R'(x)=-ik e^{-ikx}$. The [Integration By Parts](/theorems/210) theorem is a real-valued compact-interval formula; applying it separately to the real and imaginary parts of $u_R$ and $v_R$, then recombining the four resulting identities by complex linearity of the integral, gives the complex-valued formula. Hence \begin{align*} \int_{[-R,R]} \psi'(x)e^{-ikx}\,d\mathcal L^1(x) = \psi(R)e^{-ikR} - \psi(-R)e^{ikR} - \int_{[-R,R]}\psi(x)(-ik e^{-ikx})\,d\mathcal L^1(x). \end{align*} Rearranging the last term gives \begin{align*} I_R = \psi(R)e^{-ikR} - \psi(-R)e^{ikR} + ik\int_{[-R,R]}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} This is the key calculation: differentiating $\psi$ in physical space has moved onto the exponential, and differentiating $e^{-ikx}$ produces the multiplier $-ik$.[/guided]

[step:Use Schwartz decay to remove the boundary term] Since $\psi \in \mathcal S(\mathbb R)$, the function $x \mapsto x\psi(x)$ is bounded on $\mathbb R$. Hence there exists $M_\psi \geq 0$ such that \begin{align*} |x\psi(x)| \leq M_\psi \end{align*} for all $x \in \mathbb R$. For $R > 0$, \begin{align*} |\psi(R)e^{-ikR}| = |\psi(R)| \leq \frac{M_\psi}{R} \end{align*} and \begin{align*} |\psi(-R)e^{ikR}| = |\psi(-R)| \leq \frac{M_\psi}{R}. \end{align*} Thus \begin{align*} \lim_{R \to \infty}\left(\psi(R)e^{-ikR} - \psi(-R)e^{ikR}\right)=0. \end{align*} [/step]

[step:Pass to the limit and identify the Fourier multiplier] Since $\psi \in \mathcal S(\mathbb R)$, the same Schwartz integrability estimate gives $\psi \in L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. Define the measurable function \begin{align*} g_k: \mathbb R \to \mathbb C, \qquad g_k(x)=\psi(x)e^{-ikx}. \end{align*} Since $|e^{-ikx}|=1$ for all $x \in \mathbb R$, the function $g_k$ belongs to $L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. For each $R > 0$, the functions $\mathbb{1}_{[-R,R]} g_k$ converge pointwise to $g_k$ as $R \to \infty$, and $|\mathbb{1}_{[-R,R]} g_k| \leq |g_k|$ with $|g_k| \in L^1(\mathbb R,\mathcal B(\mathbb R),\mathcal L^1)$. By the [Dominated Convergence Theorem](/theorems/4), \begin{align*} \lim_{R \to \infty}\int_{[-R,R]}\psi(x)e^{-ikx}\,d\mathcal L^1(x) = \int_{\mathbb R}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} Taking limits in the integration-by-parts identity yields \begin{align*} \int_{\mathbb R}\psi'(x)e^{-ikx}\,d\mathcal L^1(x) = ik\int_{\mathbb R}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} Multiplying by $-i\hbar/\sqrt{2\pi}$ gives \begin{align*} (\mathcal F(P\psi))(k) = \frac{-i\hbar}{\sqrt{2\pi}} ik\int_{\mathbb R}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} Since $(-i)(i)=1$, this becomes \begin{align*} (\mathcal F(P\psi))(k) = \hbar k \frac{1}{\sqrt{2\pi}}\int_{\mathbb R}\psi(x)e^{-ikx}\,d\mathcal L^1(x). \end{align*} By the definition of $\hat{\psi}(k)$, \begin{align*} (\mathcal F(P\psi))(k) = \hbar k\,\hat{\psi}(k). \end{align*} This proves the claimed diagonalization formula for every $\psi \in \mathcal S(\mathbb R)$ and every $k \in \mathbb R$. [/step]