Complementarity with single slit diffraction

[DParlevliet]

Sometimes I see complementarity explained as being not able to measure any wave and particle property at the same time. But it looks like in a standard double slit experiment the detector measures both particles as the interference, so the wave. The same with a single slit. If a photon is detected outside the centre of the detector, this can only be explained by both particle and wave (diffraction) properties. Of course it is not possible to measure all properties of wave and particle at the same time, but it seems to be possible for some.

 

[bhobba]

The theory that explains both is - wait for it - its a biggie - Quantum Mechanics.

This wave particle stuff, ever since Dirac came up with his transformation theory, which is generally what goes by the name of Quantum Mechanics today, about 1927, has been consigned to the dustbin of history.

It not a particle, it's not a wave, its quantum stuff that obeys the laws of QM.

Exactly how does QM explain it - check out:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Thanks
Bill

 

[DParlevliet]

I know, but my question is: does the detector show (some) wave and particle properties at the same time?

 

[UltrafastPED]

I know, but my question is: does the detector show (some) wave and particle properties at the same time?

No. The detector picks up just a point of detection: so it looks like a particle at the detector.

But at the slits it acts like a wave.

PS: I don't care for the "dustbin of history" characterization given above. At the experimental level things appear as particles or waves in their interactions. When I carry out my electron diffraction experiments the electrons operate as waves while passing through the crystal, and as particles when absorbed by the detector. The same with the photons from my laser.

 

[DParlevliet]

But the detector does not only detect the particle, it also detects it at a certain position. This position is determined by the wave. It measures not the whole wave, but the effect of the wave. So anyway shows the presence of the wave (property).

 

[DParlevliet]

Or another question: is it possible to measure the wave (property) of a photon without the particle?