Thors10 said:
Here's another question for you: If you think one can do QM without intermediate collapse, then how do you calculate probabilities such as ##P(\mathbf{X}(t_1) \in A_1 \wedge \mathbf{X}(t_2) \in A_2)##? In orthodox QM, you would just calculate ##\int \left|(\chi_{A_2}U(t_2-t_1)\chi_{A_1}U(t_1)\psi)(\mathbf{X})\right|^2 \mathrm{d}\mathbf{X}##, but you can't do that without collapse, because the Heisenberg position observables ##\hat{x}(t)## don't commute for different ##t## (not even if they are pointer positions), so one can't perform the calculation at the end of the time evolution. One has to insert the ##\chi##'s at intermediate times.
Counterquestion: How do you measure it? To compare your probability with observation, you have to have information about what was at ##t_1## as well as at ##t_2##. How is this done in QM? Very simple, you measure ##A_1## at ##t_1##, and care about that the measurement result is preserved. So, you start with
##\psi_1(a_1) \psi_2(a_2)\psi(q,0)##. Independent variables defining the states of the two measurement devices and the state of the system, and independent initial wave functions. And physics which makes the interaction between them zero except for the time of the measurements. At ##t_1## this will become a superposition
$$\sum_i c_i \psi^i_1(a_1,t_1) \psi_2(a_2,t_1)\psi_i(q,t_1).$$
After this, the first measurement result will be preserved, thus, no more interaction with all the other things, and trivial evolution of ##a_1## itself. Thus, the future evolution of the ##\psi^i_1(a_1,t)## will be trivial after ##t_1##.
At ##t_2## this will become a superposition
$$\sum_i c_i \psi^i_1(a_1,t_1) \sum_j c_{ij} \psi^j_2(a_2,t_2)\psi_{ij}(q,t_2).$$
This is all yet quantum evolution without collapse, without any ##\chi## inserted. Then comes the final measurement of the positions ##a_1, a_2, q## which gives you ##\rho(a_1, a_2, q, t_2)##. It measures positions of different objects, they commute, no problem.