B Interferometry: Photons Needed for Interference Pattern

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In a two-mirror interferometer, only one photon is needed at a time to create an interference pattern, although many photons must be accumulated to visualize the pattern on a screen. Each photon contributes a single dot, and over time, these dots combine to reveal the interference pattern. This principle also applies to electrons in the double slit experiment, where reducing the current allows for one electron to pass through at a time. Despite the individual contributions, a single particle cannot create a discernible pattern on its own. Thus, while one photon or electron is sufficient for interference, a collection of them is necessary to observe the resulting pattern.
Bob8102
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Number of photons needed to create interference pattern.
If you have, say, a two-mirror interferometer that is producing an interference pattern, how many photons are interfering? Two, or just one?
 
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Just one.
 
You need a lot of photons to see an interference pattern, but you only need one in the apparatus at any time. Each one gives you a dot on your screen - then you wait until enough dots appear for you to see the pattern.
 
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Bob8102 said:
If you have, say, a two-mirror interferometer that is producing an interference pattern, how many photons are interfering? Two, or just one?
Only one at a time is needed. The same is true of, say, electrons in the double slit experiment. The current through the slits can be reduced so low that only one electron at a time is passing through the slits. An interference pattern will still be produced, though you still need to accumulate many electrons at the detector over time to build up the observed pattern. After all, a single particle cannot make a pattern.
 
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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