Question
Given
$$ \hat H |\psi \rangle = \Big( \frac{1}{2m}\frac{\partial^2 }{\partial x_1^2} + \frac{1}{2m}\frac{\partial^2 }{\partial x_2^2} + k \Big| \prod_i \lim_{t \to T_i} \frac{1}{\langle H \rangle_\psi}\frac{t - T_i}{(x_2 - x_1)} \Big) \Big| | \psi \rangle $$
where:
$$ \langle H \rangle_\psi = \langle \psi |\Big( \frac{1}{2m}\frac{\partial^2 }{\partial x_1^2} + \frac{1}{2m}\frac{\partial^2 }{\partial x_2^2} + k \Big) | \psi \rangle$$
$$ k = \lim_{t \to T_i} \langle \psi |\Big( \frac{1}{2m}\frac{\partial^2 }{\partial x_1^2} + \frac{1}{2m}\frac{\partial^2 }{\partial x_2^2} \Big) | \psi \rangle$$
What happens to time evolution of the wavefunction after a collision $x_2 = x_1$ at $T_j$?
Classical Intuition
Let's say I want to model a gas of $2$ particles where the gas collides at times $T_i$ (including the collisions) -
I use the following Hamiltonian:
$$ H = \frac{1}{2} m \dot x_1^2 + \frac{1}{2} m \dot x_2^2 + k \prod_i \lim_{t \to T_i} f (\frac{t - T_i}{x_2 - x_1} ) $$
Note: $k$ is a parameter which obeys:
$$ k > \frac{1}{2} m \dot x_1^2 + \frac{1}{2} m \dot x_2^2$$
Notice at the time of a collision at when $T_i \to t$ then $ x_2(t) - x_1(t) \to 0$.
One normalise $f$ so that:
$$\lim_{t \to T}H(t) = \lim_{t \to T} \frac{1}{2} m \dot x_1^2 + \frac{1}{2} m \dot x_2^2 + k$$
Hence, the $f$ is:
$$ f (\frac{t - T_i}{x_2 - x_1} ) = \Big| \frac{{t - T_i}}{(\frac{1}{2} m \dot x_1^2 + \frac{1}{2} m \dot x_2^2 + k)({x_2 - x_1})} \Big|$$
Hence, we have:
$$ H = \frac{1}{2} m \dot x_1^2 + \frac{1}{2} m \dot x_2^2 + k \Big| \prod_i \lim_{t \to T_i} \frac{1}{H}\frac{t - T_i}{(x_2 - x_1)} \Big| $$
Hence, upon quantisation in the Schrodinger picture:
$$ \hat H |\psi \rangle = \Big( \frac{1}{2m}\frac{\partial^2 }{\partial x_1^2} + \frac{1}{2m}\frac{\partial^2 }{\partial x_2^2} + k \Big| \prod_i \lim_{t \to T_i} \frac{1}{\langle H \rangle_\psi}\frac{t - T_i}{(x_2 - x_1)} \Big| \Big) | \psi \rangle $$
where:
$$ \langle H \rangle_\psi = \lim_{t \to T_i} \langle \psi |\Big( \frac{1}{2m}\frac{\partial^2 }{\partial x_1^2} + \frac{1}{2m}\frac{\partial^2 }{\partial x_2^2} + k \Big) | \psi \rangle$$