2
$\begingroup$

A version of the double slit experiment is done by firing single photons. It is possible that the fired photon could pass through either slit because of its small size. It is also possible that the photon could pass through both slits because of its wave-like nature. If the photon passes through both slits then its energy is spread across the wave and this would apply right up until until it arrives at the screen at which point all the energy is focused again to a single point. I understand that the screen is a macroscopic object which would interfere with the photon. I just find it incredible that the photon does not show up as an energy- smattering. Obviously photons move at their constant speed so I guess the wave-collapse would also occur at that speed. I just don't understand why some of the energy is not lost in the interaction. It makes me think that even though a photon is in wave-form and spread out it must be indivisible (at least under these conditions) ...

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

It's not true that photons have to deposit their entire energy at a single point. Photons can be inelastically scattered, as in Compton scattering. The fact that a certain type of detector, used with photons of a certain energy, may get predominantly full-energy events is not fundamental, it's just a property of that detector and of the way photons of that energy interact with matter.

Obviously photons move at their constant speed so I guess the wave-collapse would also occur at that speed.

No, the collapse of the wavefunction is not a physical process, and it doesn't even exist in most interpretations. It is a feature of the Copenhagen interpretation (CI). In the CI, the collapse is instantaneous.

It makes me think that even though a photon is in wave-form and spread out it must be indivisible (at least under these conditions) ...

It's not indivisible in the sense of not being able to split up its energy. You certainly can, as in Compton scattering.

It's indivisible in the sense that for a fixed frequency, you have a quantum of energy, which is $h\nu$ -- whereas classically, a wave packet can exist with frequency $\nu$ and any energy you like.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Right, so the "h" is constant and the "v" alternating. This made it clear to me that photons can be visible or invisible. I did not realize this before! $\endgroup$
    – Wookie
    Commented Nov 19, 2019 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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