I am doing some revision on theoretical physics, specifically propagator theory. This is talking about how to work out the probability amplitude at some time $t_{f}$ and position $x){f}$, given that we know what happens at $(t,x)$.
Working in natural units, and we can rearrange the Schrodinger equation to get an operator.
$$ i\partial_{t_f} \psi(t_f,x_f) = \hat{H}(t_{f},x_{f})\psi(t_f,x_f) $$ $$ [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\psi(t_f,x_f) = 0 $$
and then we can define an operator $\hat{O}$ as: $$ \hat{O}\psi(t_f,x_f) = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\psi(t_f,x_f) = 0. $$
$\psi$ is arbitrary, therefore we can write $\hat{O}$ as
$$ \hat{O} = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]$$
My question is what I would get if I multiplied $\hat{O}$ with a theta function $\theta(t_{f}-t)$, where $\theta = 1$ for $t_{f} \geq t$ and $\theta = 0$ for $t_{f} < t$
$$\hat{O}\theta(t_{f}-t) = [i\partial_{t_f} - \hat{H}(t_{f},x_{f})]\theta(t_{f}-t)$$
I know that since a theta function only changes gradient at $t_{f} = t$, I will get a Dirac-delta function, but what about when it multiplies with $\hat{H}(t_{f},x_{f})$
$$i\partial_{t_{f}}\theta(t_{f} - t) = i\delta(t_f - t) $$
$$ - \hat{H}(t_{f},x_{f})\theta(t_{f} - t) = ?$$
I want to know this result because solving the propagation equation is done by:
$$ \hat{O}(t_f,x_f)\theta(t_{f} - t)\psi(t_f,x_f) = \int dx [\hat{O}(t_f,x_f) i G(t_f,x_f; t,x)]\psi(t,x) $$
Where G is a Green Function
which is apparently just:
$$i\delta(t_f - t)\psi(t_f,x_f) =\int dx [\hat{O}(t_f,x_f) i G(t_f,x_f; t,x)]\psi(t,x)$$
