“General” for time evolution of quantum state

Question

I am reading a book in which at some point they find the time-evolved wavefunction $\phi_0(\mathbf{r},t)$ from the static $\phi_0(\mathbf{r})$.

They say that "employing the Heisenberg time evolution formalism, one can write the general result as "

$$ \phi_0(\mathbf{r},t) = e^{-3i\pi/4} \left ( \frac{m}{2\pi\hbar t}\right )^{3/2} \int d\mathbf{r}'\phi_0(\mathbf{r'}) \exp \left [ i\frac{m}{2\hbar t}(\mathbf{r} - \mathbf{r'})^2\right ].$$

Needless to say, I do not understand how this is justified. As far as I know the Heisnenberg picture time evolution just expresses the time dependence of the operator as a commutator of that operator with the Hamiltoninan...
Anyone know how to get to this expression?

Answer 1

It should be "everyone", not quite "anyone": isn't the FT of a Gaussian one such too? I'd let you chase after errant normalizations, if you were so inclined. $$ \phi(\vec r, t)= \langle \vec r| e^{-itH_{free}/\hbar} | \phi_0\rangle \\ =\int d\vec r ' \langle \vec r| e^{-\frac{it}{2\hbar m} \hat p ^2}| \vec r'\rangle \langle \vec r '|\phi_0\rangle \\ = \int d\vec r ' \phi_0(\vec r')~~ d\vec p ~ \langle \vec r|\vec p\rangle e^{-\frac{it}{2\hbar m} \vec p ^2}\langle \vec p | \vec r ' \rangle \\ = \int d\vec r ' \phi_0(\vec r')~~ \frac{d\vec p}{(2\pi \hbar)^3} ~ e^{-\frac{it}{2\hbar m} \vec p ^2} e^{i \vec p \cdot (\vec r- \vec r') /\hbar}\\ = \int d\vec r ' \phi_0(\vec r')~~ \left (\frac{ m}{2\pi i \hbar t}\right )^{3/2} ~ \exp {\frac{im (\vec r - \vec r')^2}{2\hbar t}}~~. $$ The penultimate line involves completing the square. Confirm your expectations as $t\to 0$.