Undergrad Why Won't Observation in a 2 Slit Experiment Cause 1 Slit Diffraction?

[Paul Colby]

Why would measuring it some other way way prevent diffraction?

Measuring an optical wavelength photon basically removes it from the field. Why would this not change the diffraction?

 

[Flamel]

Measuring an optical wavelength photon basically removes it from the field. Why would this not change the diffraction?

The particle doesn't necessarily need to be a photon. I believe you could also add a pair of polarizers in opposite directions over the slits and use polarization to determine which slit it came from if I'm not mistaken, so i think the photons don't necessarily need to be destroyed to measure their general location.

 

[PeroK]

Why would measuring it some other way way prevent diffraction? Don't quantum eraser experiments show that measurements can be made on a particle and will allow diffraction to occur, so long as it can't be determined which slit the particle came through?

I don't know where you are getting these ideas. The thread seems to be just repeating itself now.

You say "prevent" diffraction, but surely we are talking about a potential diffraction pattern being destroyed by a post-slit measurement?

If you detemine which way a particle went, then you get single slit diffraction (*). If you don't you get a double-slit interference pattern. Are you confusing "diffraction" with "interference pattern"?

(*) This is the case if you measure which way non-intrusively. This is not easy to do. In the case where you have a detector after one slit, that detector will interact with 50% of the electrons. As a result those electons may not form a diffraction pattern because after they diffracted they interacted with the measuring apparatus. In the simplest case the elecrons are captured and never reach the screen; or, they may be knocked off course and form a much more random pattern all over the screen. If electrons come through a slit and are hit by a baseball bat, then no they don't form a neat single slit diffraction pattern.

But, the electrons that aren't measured to come through that slit are indirectly measured to come through the other slit and they should form a neat single-slit diffraction pattern.

In a typical experiment, therefore, where you measure electrons after one slit, you get a combination of a neat single-slit diffraction pattern for the 50% of electrons that came through the unobserved slit and something messier for the 50% of electrons that were observed, depending on the post-slit measurement process.

 

[Paul Colby]

The particle doesn't necessarily need to be a photon.

You might think about how not being a photon actually would change your question. I don't think it does.

I believe you could also add...

You might think about how adding these polarizers completely changes the original two slit experiment. You have different geometry, materials, altered field boundary conditions. For normal light sources the diffraction pattern in the experiment you describe is determined by the classical EM boundary value problem. The photon probability of detection is proportional to the classical energy density at the detector. The discussion about which slit a photon went through is just noise IMO.

 

[vanhees71]

How does the momentum uncertainty generate the minima in the interference pattern? Does it have something to do with the nodes in the infinite square well, and if so why are there no obvious minima when you know which slit the particle traveled through?

The double-slit (or also single slit or more slits) experiment is calculated in the very same way as for classical electromagnetic waves, though it's much simpler, because only one scalar field is involved. What comes out in Fraunhofer approximation is that the interference pattern at the "infinitely far" observation screen is just the Fourier transform of the openings.

If you want of course the more detailed analysis of what happens at finite distances from the slits you need the Green's function. Then you can also study what happens with wave packets (sharp in momentum vs. sharp in position and looking close to the slits and far away etc.). I'm not aware that this has been calculated in any textbook (but I'm sure one can find it somewhere when googling long enough).

 

[vanhees71]

I don't understand that question. In what scenario are we increasing momentum uncertainty and reducing diffraction?

If the beam is not "monochromatic" enough the contrast of your interference pattern goes down of course.

 

[kent davidge]

If the beam is not "monochromatic" enough the contrast of your interference pattern goes down of course.

because then the beam is no longer coherent?

 

[vanhees71]

Yes. It's because the position of the maxima and minima of intensity behind the double slit depend on the momenta (de Broglie wavelenths) of the particles. So if you have a spread in momentum the minima and maxima get "washed out" and you get "less contrast" of your interference pattern.

 

[Dr_Nate]

Yes, I understand that there are probability amplitudes that constructively and destructively interfere. I think what is confusing is how momentum uncertainty apparently also plays a role in the formation of interference patterns.

Momentum uncertainty is related to spatial uncertainty. That's really all the Heisenberg Uncertainty principle says. It isn't about interference patterns. You really should go to the math at this point. $\Delta x$ and $\Delta p$ represent equations. They are standard deviations of the wave function in position space or momentum space. Those two spaces are related by Fourier transforms; another set of equations.

 

[vanhees71]

Sure, it is also about interference patterns.

First of all look at the incoming wave: If you want double-slit interference the wave packet has to be spatially broad enough to cover (more or less with equal intensity) both slits. This implies that the momentum uncertainty must be small enough to get a sufficient width in position space. Note that the usual textbook treatment uses the limit of plane waves, i.e., as generalized momentum eigenstates (which are strictly speaking not represent true states, because they are no square-integrable functions though; so you have to take their interpretation with a grain of salt), i.e., your two slits are "illuminated" with strictly equal intensity since the spatial width goes to infinity.

The same holds behind the slits: If you put your observation screen (nowadays, e.g., a CCD cam or some other "pixel detector") too close the "partial waves" originating from each slit (in the sense of Huygen's principle) have no significant overlap, i.e., there's no interference and thus no refraction pattern though here you get some which-way information. Only if the partial waves overlap at a far enough put onservation screen you get interference and a double-slit interference pattern at the cost of loosing which-way information.

In some sense the uncertainty principle particularly manifests in this paradigmatic example of wave-function interference phenomena at slits and gratings!