The partial derivative $\partial_j f(x)$ of a function $f : \mathbb{R}^n \to \mathbb{C}$ in the $j$-th coordinate direction, defined as the one-variable derivative obtained by varying the $j$-th coordinate while keeping the others fixed.
Instances For
The inclusion $\mathbb{R} \hookrightarrow \mathbb{C}$ has derivative $1$ at every point.
The map $s \mapsto s^2 : \mathbb{R} \to \mathbb{C}$ has derivative $2s$.
Chain rule: $\frac{d}{ds} \exp(a s^2 + b) = \exp(a s^2 + b) \cdot 2as$ for complex constants $a, b$ and a real variable $s$.
Second derivative of $\exp(a s^2 + b)$: differentiating $\exp(a s^2 + b) \cdot 2as$ once more yields $\exp(a s^2 + b) (4 a^2 s^2 + 2 a)$.
Reformulates the second pure partial derivative $\partial_j^2 f$ as an iterated one-variable derivative obtained by varying the $j$-th coordinate.
Closed-form for the second partial derivative $\partial_j^2$ of the Schrödinger phase $\exp(i|y|^2/(2t))$ at $x$, expressed in the $a = i/(2t)$ parametrization.
Closed-form for the spatial Laplacian of the Schrödinger phase $\exp(i|y|^2/(2t))$: $$\Delta_x \exp(i|x|^2/(2t)) = \exp(i|x|^2/(2t))\left(\frac{i n}{t} - \frac{|x|^2}{t^2}\right).$$
The spatial half-Laplacian of the Schrödinger kernel:
$$\tfrac{1}{2}\Delta_x K(t, x) = K(t, x)\left(\frac{i n}{2t} - \frac{|x|^2}{2t^2}\right).$$
The constant prefactor $(2\pi i t)^{-n/2}$ pulls out and the half-Laplacian of the phase is
computed via laplacian_exp_phase.
Time derivative of the Schrödinger kernel multiplied by $i$:
$$i\,\partial_t K(t, x) = K(t, x)\left(-\frac{n i}{2 t} + \frac{|x|^2}{2 t^2}\right).$$
Combined with half_laplacian_schrodinger, this gives that $K$ solves
$i\partial_t K + \tfrac{1}{2}\Delta K = 0$.
The regularized Gaussian replacing the oscillatory kernel: for $\delta > 0$, $g_\delta(t, \xi) = \exp\!\left(-2\pi^2 (\delta + i) t |\xi|^2\right).$ As $\delta \downarrow 0$, $g_\delta \to \hat K(t, \cdot)$ and the Gaussian is integrable, which permits a rigorous Fourier-inversion argument.
Instances For
The local fourierTransformFin (defined via Fin n-indexed dot products) coincides with the
$n$-dimensional Fourier transform CM16.fourierTransformND developed in Class Meeting 16.
Regularization step: writing the oscillatory kernel as the $\delta \downarrow 0$ limit of $g_\delta$ and applying dominated convergence (with bound $|\hat\phi|$), $$\psi(t, x) = (\hat K(t, \cdot)\,\hat\phi)^\vee(x) = \lim_{\delta \downarrow 0} (g_\delta(t, \cdot)\,\hat\phi)^\vee(x).$$
Coercion identity: $(\sum_i (\xi_i)^2) \in \mathbb{C}$ equals $\sum_i (\xi_i : \mathbb{C})^2$.
For $\delta > 0$ and $t > 0$, the regularized Gaussian $g_\delta(t, \cdot)$ is integrable on $\mathbb{R}^n$: the real part of the exponent is $-2\pi^2 \delta t |\xi|^2 < 0$ for $\xi \neq 0$, which gives Gaussian decay.
Joint integrability on $\mathbb{R}^n \times \mathbb{R}^n$ of the integrand
$g_\delta(t, \xi)\, e^{2\pi i \xi \cdot (x-y)}\, \phi(y)$ needed to apply Fubini's theorem in
convolution_via_FT_inversion.
Convolution-via-Fourier-inversion at the regularized level: for any $\delta > 0$, $$(g_\delta(t, \cdot)\,\hat\phi)^\vee(x) = \int g_\delta^\vee(t, x - y)\,\phi(y)\, d^n y.$$ The proof combines the product formula for the inverse Fourier transform with Fubini's theorem to swap the $\xi$ and $y$ integrals.
The closed-form inverse Fourier transform of the regularized Gaussian: $$g_\delta^\vee(t, z) = (2\pi(\delta + i)t)^{-n/2}\, \exp\!\left(\frac{-|z|^2}{2t(\delta + i)} \right).$$ At $\delta = 0$ this formally recovers the Schrödinger kernel $K(t, z)$.
Instances For
The inverse Fourier transform of the regularized Gaussian agrees with the closed-form
expression regularizedGaussianInverseFT. Proved by reducing to the complex Gaussian Fourier
transform formula CM16.fourier_gaussian_complex.
Pointwise convergence of the regularized inverse Fourier transform to the Schrödinger kernel:
for each $z$,
$g_\delta^\vee(t, z) \to K(t, z)$ as $\delta \downarrow 0$. This is the pointwise input to the
dominated convergence argument in dominated_convergence_convolution.
For $\delta > 0$, the integrand $y \mapsto g_\delta^\vee(t, x - y)\,\phi(y)$ is $\mathrm{AEStronglyMeasurable}$ with respect to Lebesgue measure on $\mathbb{R}^n$.
Inverse Fourier transform expressed via the forward transform at $-z$: $f^\vee(z) = \hat f(-z)$. This is the standard symmetry used to reduce inverse-transform estimates to forward-transform results.
Fourier transform of the complex-scaled Gaussian (Proposition 3.0.3 of CM16, in scaledGaussian
form): for $\operatorname{Re} z > 0$ and $\operatorname{Im} z \neq 0$,
$$\widehat{\exp(-\pi z |\cdot|^2)}(\xi) = z^{-n/2}\, \exp(-\pi |\xi|^2 / z).$$
Uniform $L^\infty$ bound on $g_\delta^\vee(t, \cdot)$ in $\delta > 0$ and $z \in \mathbb{R}^n$: there exists $C > 0$ (depending on $n, t$) with $\|g_\delta^\vee(t, z)\| \le C$ for all $\delta > 0$. One may take $C = (2\pi t)^{-n/2}$.
Localized form of the uniform bound: for any compact set $K \subset \mathbb{R}^n$ there exists a uniform constant $C$ and a threshold $\delta_0 > 0$ such that $g_\delta^\vee(t, z)$ is bounded by $C$ for all $\delta \in (0, \delta_0)$ and $z \in K$.
Integrable dominating function for the convolution integrand: for $\phi \in C_c^\infty(\mathbb{R}^n)$ there is an integrable function $\text{bound}(y) = |C| \cdot \|\phi(y)\|$ such that eventually in $\delta \downarrow 0$, $\|g_\delta^\vee(t, x - y)\,\phi(y)\| \le \text{bound}(y)$ almost everywhere in $y$. This is the hypothesis required for dominated convergence under the integral.
Dominated convergence for the convolution: as $\delta \downarrow 0$, $$\int g_\delta^\vee(t, x - y)\,\phi(y)\, d^n y \longrightarrow \int K(t, x - y)\,\phi(y)\, d^n y = (K(t, \cdot) * \phi)(x).$$
Proposition 2.0.1 (Calculation of the fundamental solution $K(t, x)$ for Schrödinger's equation): for $\phi \in C_c^\infty(\mathbb{R}^n)$ and $t > 0$, the function $\psi$ defined by $\hat\psi(t, \xi) = \hat K(t, \xi)\,\hat\phi(\xi)$ admits the convolution representation $$\psi(t, x) = (K(t, \cdot) * \phi)(x) = \int K(t, x - y)\,\phi(y)\, d^n y,$$ where $K(t, x) = (2\pi i t)^{-n/2}\, e^{i|x|^2/(2t)}$. The proof passes through the regularized Gaussian $g_\delta$ and takes the limit $\delta \downarrow 0$.