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I want to show the mathematical equivalence between the Schrödinger:

$$\Psi(x,t)=e^{-it \hat{H}/\hbar} \psi'(x,0), \tag{1}$$

and Green's function (propagator method) :

$$\Psi(x,t)=\int_{-\infty}^{+\infty} K(x, x';t,0)\psi(x',0)~dx', \tag{2}$$

for the case of a free particle; then, in this case, we have $\hat{H}=(\hat{p}^2/2m)=(-i\hbar \partial_{x})^2 /2m$, and $K(x, x';t,0)=\sqrt{\frac{m}{i2\pi \hbar t}}e^{i m \left(x-x' \right)^2/2\hbar t}$. Besides, in Eqs. (1) and (2), $\psi(x',0)$ represents an arbitrary initial wave function. Hence, with this consideration, the Eqs. (1) and (2) becomes

$$\Psi(x,t)=e^{a \partial_{x}^2} \psi'(x,0), \tag{3}$$ $$\Psi(x,t)=\sqrt{\frac{m}{i2\pi \hbar t}}\int_{-\infty}^{+\infty} e^{-\frac{m\left(x-x' \right)^2}{2 i \hbar t}} \psi(x', 0)~ dx'\tag{4},$$ where $a=it \hbar/2m$. Then, the idea is to proof the equivalence between the Eqs. (3) and (4); that is, to proof that both equations gives the same temporal evolution for a free particle. Therefore, we proceed as follows.

We work with Eq. (3) in order to obtain Eq. (4). First, we expand Eq. (3) in a McLaurin series

$$\Psi(x,t)= e^{a \partial_{x}^{2}} \psi'(x,0) =\sum_{k=0}^{\infty} \frac{a^{k}}{k!} \partial_{x}^{2k} \psi'(x,0) \tag{5}$$

Now, apply a Fourier transform on both sides of Eq. (5)

$$F\left[e^{a \partial_{x}^{2}} \psi'(x,0)\right](p_{x})=\sum_{k=0}^{\infty} \frac{a^{k}}{k!} F\left[\partial_{x}^{2k} \psi'(x,0)\right](p_{x}) \\ =\sum_{k=0}^{\infty} \frac{a^{k}}{k!} (i p_{x})^{2k}F\left[\psi'(x,0)\right](p_{x}) \\ = e^{-ap_{x}^2}F\left[ \psi'(x,0)\right](p_{x}), \tag{6} $$

where $p_{x}$ is the conjugate variable of $x$. Besides, in the second line, we have used the derivative property of the Fourier transform: $F\left[\frac{d^{n}f(q)}{dq^{n}} \right](p)=(ip)^{n}F\left[f(q) \right](p)$. It must be noted that the exponential $e^{-a p_{x}^{2}}$ can be expressed by means of a Fourier transform of a function $f(x)$; that is,

$$F\left[f(x) \right](p_{x})=e^{-a p_{x}^{2}}, \tag{7}$$

Using the definition of the inverse Fourier transform: $F^{-1}\left[M(b) \right]=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} N(c) e^{+ibc} dc$, we trivially verify

$$f(x)=\frac{1}{\sqrt{2 a}}e^{-\frac{x^2}{4 a}}. \tag{8}$$

Now, using Eq. (7) in the last line of Eq. (6) we have

$$F\left[e^{a \partial_{x}^{2}} \psi'(x,0)\right](p_{x})=F\left[f(x) \right](p_{x})\cdot F\left[ \psi'(x,0)\right](p_{x}). \tag{9}$$

Take the inverse Fourier transform on both sides of the last equation, that is,

$$\Psi(x,t)=e^{a \partial_{x}^{2}} \psi'(x,0)=F^{-1}\left[F\left[f(x) \right](p_{x})\cdot F\left[ \psi'(x,0)\right](p_{x}) \right], \tag{10}$$

then, take the convolution theorem for a product of Fourier transforms: $f \ast g = F^{-1} \left[F\left[f \right]\cdot F \left[g \right] \right]=\left( \sqrt{2\pi}\right)^{-1}\int_{-\infty}^{\infty} f(\eta) g(x-\eta, 0)~ d\eta$; therefore, the Eq. (10) is

$$\Psi(x,t)=e^{a \partial_{x}^{2}} \psi'(x,0)=f(x) * \psi'(x,0)=\left( \sqrt{2\pi}\right)^{-1}\int_{-\infty}^{\infty} f(\eta) \psi(x-\eta, 0)~ d\eta \\ =\left( \sqrt{2\pi}\right)^{-1}\int_{-\infty}^{\infty} f(x-\eta) \psi(\eta, 0)~ d\eta, \tag{11} $$

in the second line of the last equation we use the commutative property of the convolution theorem. Now, we use Eq. (8) with the explicit definition $a=it \hbar/2m$ in the last line of Eq. (11), obtaining

$$\Psi(x,t)=e^{a \partial_{x}^{2}} \psi'(x,0)=\sqrt{\frac{m}{i2\pi \hbar t}}\int_{-\infty}^{+\infty} e^{-\frac{m\left(x-\eta \right)^2}{2 i \hbar t}} \psi(\eta, 0)~ d\eta, \tag{12}$$

which of course coincides with Eq. (4) (take $x'=\eta$). Then, my question: is my procedure mathematically correct?

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Everything that you wrote seems to be correct for me, but it feels there's a quicker way (mayhaps not the most rigorous one) to obtain the equivalence you're looking for. Suppose that $\psi^{\prime}\left(x, 0\right)$ admits the following Fourier expansion:

$$\psi^{\prime}\left(x, 0\right) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} e^{-ikx} \tilde{\psi}\left(k\right) \ dk$$

Your equation (3) implies:

$$\Psi \left(x, t\right) = e^{a \partial_x^2} \psi^{\prime}\left(x, 0\right) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \left[ \sum_{n=0}^{\infty} \frac{\left(a \partial_x^2\right)^n}{n!} e^{-ikx}\right] \tilde{\psi}\left(k\right) \ dk = $$ $$ = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} e^{-ak^2-ikx} \tilde{\psi}\left(k\right) \ dk \ \ \ (\ast)$$

Where I omitted some steps using the fact that $e^{-ikx}$ is an eigenfunction of the linear operator $\partial_x^2$ with associated eigenvalue $-k^2$ and the fact that the operator $e^{-a\partial_x^2}$ can switch order with the integral sign and act only in the $e^{-ikx}$ term (if you want to be rigorous, some care would need to be taken, such as argumenting for dominated convergence of the integral).

Using the definition of the inverse Fourier transform to obtain $\psi^{\prime}(x^{\prime},0)$ in terms of $\tilde{\psi}(k)$ and completing squares, (*) implies:

$$\Psi(x, t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp \left\{-a\left[k-\frac{i\left(x-x^{\prime}\right)}{2a}\right]^2\right\} \exp \left[-\frac{\left(x-x^{\prime}\right)^2}{4a}\right] \psi^{\prime}\left(x^{\prime},0\right) \ dk \ dx^{\prime}$$

The integral over $k$ can be immediately computed, since it's a Gaussian integral:

$$\Psi(x, t) = \frac{1}{2 \pi}\sqrt{\frac{\pi}{a}} \int_{-\infty}^{\infty} \exp \left[-\frac{\left(x-x^{\prime}\right)^2}{4a}\right] \psi^{\prime}\left(x^{\prime},0\right) \ dx^{\prime}$$

Plugging your definition of $a$ leads immediately to your equation (4). Thus, (3) is equivalent to (4).

Edit: the reason why I assume such a Fourier decomposition for $\psi^{\prime}\left(x, 0\right)$ is the form of the inverse Fourier transform and the integral representation of the Dirac delta function:

$$\delta\left(x-x^{\prime}\right) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-ik(x-x^{\prime})} \ dk$$

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