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?

  • 1
    $\begingroup$ Quick Hint : Free particle propagator for time dependent Schrodinger equation in 3D. $\endgroup$
    – Sunyam
    Nov 23, 2018 at 15:10
  • $\begingroup$ @Sunyam 's hint should suffice. It represents the matrix elements of the exponential of the Hamiltonian, the full operator solution of the Schrodinger equation. What more do you want for an initial value problem? $\endgroup$ Nov 23, 2018 at 19:02
  • $\begingroup$ Close duplicate. $\endgroup$ Nov 23, 2018 at 21:20

1 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 $t\to 0$.


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