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The double-slit experiment is no proof for the existence of single photons. It can be fully explained (at any light intensity) by assuming a classical electric

by fsh 4y ago

The double-slit experiment is no proof for the existence of single photons. It can be fully explained (at any light intensity) by assuming a classical electric field. The "clicks" of a "single-photon detector", such as a photomultiplier, only require the existence of quantized charges (electrons).

kergonath 4y ago
No, not really. If the amplitude of the EM wave is enough to make an electron jump, all the electrons with an available energy level would jump across the whole detector. To have a single detection in a given region of the detector, you need a localised energy transfer. At this point we’re just reinventing photons.
- phkahler 4y ago
  >> At this point we’re just reinventing photons. Or is it another case of "wave function collapse"? Remember that diffraction grating work too, and they rely on the wave nature of light. It is sometimes said (and can be modeled) that light takes every path from source to destination and the diffraction grating causes many paths to cancel out so it appears to go specific ways. Mirror reflection can also be modeled this way, and making periodic scratches will turn it into a diffraction grating. In the end, a photon detection is like a "measurement" that collapses the reflected wave function. Photons are never observed "in flight" they are only detected when they hit something. In between they may as well BE a wave function.
- fsh 4y ago
  This is not how quantum mechanics works. One standard textbook example is to calculate the dynamics of a two-level atom that is perturbed by an oscillating (classical) electric field. The result is that there is a certain probability per unit time for the atom to transition to the excited state. In a solid, the two levels are replaced by a band structure, and the excitation leads to the generation of an electron-hole pair. The calculation then immediately gives quantized absorption events that follow a Poisson distribution. No photons required.

by fsh 4y ago

kergonath 4y ago

No, not really. If the amplitude of the EM wave is enough to make an electron jump, all the electrons with an available energy level would jump across the whole detector. To have a single detection in a given region of the detector, you need a localised energy transfer. At this point we’re just reinventing photons.

phkahler 4y ago
>> At this point we’re just reinventing photons. Or is it another case of "wave function collapse"? Remember that diffraction grating work too, and they rely on the wave nature of light. It is sometimes said (and can be modeled) that light takes every path from source to destination and the diffraction grating causes many paths to cancel out so it appears to go specific ways. Mirror reflection can also be modeled this way, and making periodic scratches will turn it into a diffraction grating. In the end, a photon detection is like a "measurement" that collapses the reflected wave function. Photons are never observed "in flight" they are only detected when they hit something. In between they may as well BE a wave function.
fsh 4y ago
This is not how quantum mechanics works. One standard textbook example is to calculate the dynamics of a two-level atom that is perturbed by an oscillating (classical) electric field. The result is that there is a certain probability per unit time for the atom to transition to the excited state. In a solid, the two levels are replaced by a band structure, and the excitation leads to the generation of an electron-hole pair. The calculation then immediately gives quantized absorption events that follow a Poisson distribution. No photons required.