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?

Double slit experiment

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.

  • 1
    $\begingroup$ More than the fast digitizer, I'm not sure about non-destructively measuring the momentum. $\endgroup$ – zeldredge May 29 '15 at 19:34
  • $\begingroup$ Very true. But that is theoretically possible to do, right? $\endgroup$ – Ben May 29 '15 at 19:45
  • $\begingroup$ I'm actually not sure. You can non-destructively measure photon number in a cavity; I'm not aware of any experimental protocol for doing it in momentum space but neither am I aware of a no-go theorem. Actually I think I've figured out your question, let me type up an actual answer...(edit: bruce got there first, and the uncertainty principle is basically what you want here) $\endgroup$ – zeldredge May 29 '15 at 19:48
  • $\begingroup$ the wave is too large on the picture ? -- hypothetical too: what gives this experiment in an expanding universe and at a large scale ? $\endgroup$ – user46925 May 29 '15 at 20:35
  • $\begingroup$ The problem already starts with the first sentence. We don't "shoot photons". Photons are quantum numbers of a continuous field. Quantum numbers don't propagate like particles do. There is one quantum number where we prepare the field state and there is another one where we measure it. They do not both belong to the same at all times distinguishable intermediate, but they belong to one and the same carrier: the one and only quantum field that permeates the entire universe. How that quantum field behaves can, for the single photon case, be calculated by ordinary single particle QM. $\endgroup$ – CuriousOne May 29 '15 at 21:52

The answer comes from a modified version of Einstein's thought experiments (see here). You cannot measure the time and the energy of the particle simultaneously, if you measure exactly the time you will have an uncertainty in the particles energy (and thus its momentum). This introduces an uncertainty in the position of the wall and or detector or somewhere else (it depends what you measure). The position of the slit will have an uncertainty equal to equal to $h/\Delta p$. If the wall's momentum were known precisely enough to measure the recoil, the slit's position is uncertain enough to disallow a position measurement.

  • 1
    $\begingroup$ Well, we don't have to worry about the position of the slit, I don't think. If you measure the momentum of the photon, you introduce an uncertainty in the spatial wavefunction enough that this "timing" idea won't work. $\endgroup$ – zeldredge May 29 '15 at 19:57
  • $\begingroup$ @zeldredge My thought was that: if you measured momentum perpendicular to the direction of travel you would have an uncertainty in the perpendicular position (lets say $y$). Taking the time would be a measurement in direction of propagation ($x$), however $[x,y]=0$ so the time measurement wouldn't effect the perpendicular position? $\endgroup$ – Ben May 29 '15 at 22:27
  • $\begingroup$ @bruce are you saying that at the detector you are doing a measurement of the energy of the particle so there's an uncertainty at the time of arrival? (now I'm just grasping:) because the energy of the particle doesn't effect the experiment, could you just measure time of arrival? $\endgroup$ – Ben May 29 '15 at 23:15
  • $\begingroup$ @BenH. Not necessarily, you can measure only the time but are then left with uncertainty in the energy (and momentum), or you measure the energy and are left with uncertainty in the time of flight. $\endgroup$ – user66432 May 29 '15 at 23:26
  • $\begingroup$ @bruce please see edit for a followup, it was getting to lengthy for a comment. $\endgroup$ – Ben May 29 '15 at 23:45

I think the answer has to do with the fact that the slit is preforming an energy measurement. We know this because the intensity of the interference pattern is dependent on wavelength:

$$ I(\theta)\propto \cos^2 \left(\frac{\pi d \sin \theta}{\lambda} \right ) \mathrm{sinc}^2\left( \frac{\pi d \sin \theta}{\lambda} \right ) \\ \\ \mathrm{with:} \; E=\frac{hc}{\lambda} $$

So because the slit is preforming an energy measurement, this introduces an uncertainty in time.

$$ \Delta t \cdot \Delta E \ge \frac{\hbar}{2} $$

And therefore you don't know when the photon passed by.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.