I How does a band pass filter increase the time of arrival of photons?

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A thought experiment explores how a band pass filter affects the time of arrival of photons from a laser pulse. When a pulse is attenuated to single photon intensity, the uncertainty principle allows for a defined time of arrival but not energy. Passing the pulse through a narrow band pass filter seems to randomize the time of arrival while defining energy, suggesting a counterintuitive delay effect. This phenomenon is linked to the concept of "nonlocal collapse," where the filter acts as a measuring apparatus that alters the state of the photons without adhering to light-speed limitations. The discussion highlights the complexities of quantum mechanics and the challenges of intuition in understanding these effects.
Daniel Petka
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If a very short (laser) pulse attenuated to single photon intensity passes through a narrow band pass filter, do the photons get detected at random times?
Here is my thought experiment: Let's say I attenuate a very short laser pulse to single photon intensity. Due to the uncertainty principle, I know the time of arrival of the photons, but not their energy. So let's reverse that by splitting the pulse in its spectral components with a diffraction grating and passing the spectrum through a slit. The photons that pass through the slit will have a defined energy, but not a defined time of arrival, because each new photon arrives on a random spot and so it's also random if it passes through the slit. This seems logical (kind of). But now let's pass the pulse directly through a narrow band pass filter instead. The photons should still appear randomly. Is that true? If yes, it's extremely counter intuitive. It's as if the band pass filter can delay a photon randomly. How does this happen, what's going on here physically? It has to be true because, well, Fourier Transform, but I don't get why.
 
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The filter can the thought of as a "measuring apparatus", which induces a "collapse" of initial superposition (of states with different energies) into a state with well defined energy. The collapse is "nonlocal", in the sense that it does not respect the principle that wave cannot "change" faster than light. As always, the exact physical meaning of "nonlocal collapse" depends on the choice of interpretation (Copenhagen, Bohm, many worlds, ...).
 
Demystifier said:
The filter can the thought of as a "measuring apparatus", which induces a "collapse" of initial superposition (of states with different energies) into a state with well defined energy. The collapse is "nonlocal", in the sense that it does not respect the principle that wave cannot "change" faster than light. As always, the exact physical meaning of "nonlocal collapse" depends on the choice of interpretation (Copenhagen, Bohm, many worlds, ...).
Thank you for the reply! This is probably one of those cases where intuition fails
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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