In quantum mechanics, the time-evolved state of a one-particle system is given by $\psi(x,t) = e^{-itH_{0}}\psi_{0}(x)$, where $H_{0} = \frac{1}{2}\Delta$ is the free Hamiltonian, $t\in \mathbb{R}$ and $\Delta$ is the usual Laplace operator. I've seen somewhere that both $e^{itH_{0}}$ and $e^{tH_{0}}$ have integral kernels, i.e. \begin{eqnarray} (e^{-itH_{0}}\varphi)(x) = \int_{\mathbb{R}^{d}}K_{0}(x,y,t)\varphi_{0}(y)dy \tag{1}\label{1} \end{eqnarray} and \begin{eqnarray} (e^{-tH_{0}}\varphi)(x) = \int_{\mathbb{R}^{d}}\tilde{K}_{0}(x,y,t)\varphi_{0}(y)dy \tag{2}\label{2} \end{eqnarray} where $K_{0}(x,y,t) = (2\pi i t)^{-d/2}e^{i\frac{(x-y)^{2}}{2t}}$ and $\tilde{K}(x,y,t) = (2\pi t)^{-d/2}e^{\frac{(x-y)^{2}}{2t}}$.
Question: How can I prove (\ref{1}) and (\ref{2})?
Basically we have a PDE $\partial_t \varphi(t,\vec{x}) = \alpha \, \Delta\varphi(t,\vec{x}).$
Taking the Fourier transform in the $\vec{x}$ argument results in the ODE $\partial_t \hat{\varphi}(t,\vec{k}) = -\alpha \, |\vec{k}|^2 \, \hat{\varphi}(t,\vec{k})$ with solution $$ \hat{\varphi}(t,\vec{k}) = e^{-\alpha \, |\vec{k}|^2 t} \hat{\varphi}(0,\vec{k}). $$
The right hand side is a product of two functions of $\vec{k}$ and is therefore the Fourier transform of a convolution: $$ \varphi(t,\vec{x}) = \left(u(t,\cdot)*\varphi(0,\cdot)\right)(\vec{x}), $$ where $\hat{u}(t,\vec{k}) = e^{-\alpha \, |\vec{k}|^2 t}.$
If $\operatorname{Re}\alpha \geq 0$ then $\hat{u}$ can be transformed resulting in $$ u(t,\vec{x}) = (2\pi\alpha t)^{-d/2}e^{-\frac{|\vec{x}|^{2}}{2\alpha t}}. $$
Thus, $$ \varphi(t,\vec{x}) = \int_{\mathbb{R}^d} (2\pi\alpha t)^{-d/2}e^{-\frac{|\vec{x}-\vec{y}|^{2}}{2\alpha t}} \, \varphi(0,\vec{y}) \, d^dy. $$