I was wondering what the outcome of this experiment would be:
You shoot single photons at a double slit. On their way there you preform a measurement in $x$ so you get the time ($t_0$) the particle moved through the detector. $[\hat{X}, \hat{Y}]=0$ so this measurement did not disturb the spacial wave function in $y$). You then stop the clock once the photon interacts with the detector screen. Because you haven't measured the photons' position (in $y$), the probability of the location it lands on the detector will be described by the intensity of a typical double slit interference pattern. Would you be able to tell what slit the photon passed through by looking at its time-of-flight (time difference of $\delta/c$) between the two paths?
I figure no, you would not. But what would the time measurement read?
Edit (in response to question bellow)
If the final detector was a CCD chip, then the freed electrons from the pixel at the location of interaction were an energy measurement, so therefore you would have an uncertainty in time of arrival (if I understood the time-energy uncertainty relation correctly, $\Delta t \Delta H \geq \hbar/2$ ). However, If you were to do a poor measurement of energy (i.e. measure an interaction, or no interaction) would the uncertainty in time go down? Would you then be able to accurately measure time-of-flight?
Of course to do this you would need an extremely fast digitizer (fs sampling) and be able to measure the particle only in $x$. So this is a hypothetical question.
Edit my first example had a momentum measurement instead of $\hat{X}$. This wasn't necessary and was complicating the problem so I changed the question.