Does really each and every single photon leave a dot on the screen in the bright area (double slit)?Why does...
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Does really each and every single photon leave a dot on the screen in the bright area (double slit)?
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Does really each and every single photon leave a dot on the screen in the bright area (double slit)?
Why does the photon strike at one or another place on the tape?Do we really know which slit the photon passed through in Afshar's experiment?Single photon double slit experimentDouble slit experiment and single particles. Is the wave function just a mathematical model?Shooting a single photon through a double slitDouble-slit experiment: Difference between observing photon path and interference pattern?Double slit experiment with a single photon streamHow does the light source fire a single photon in the double-slit experimentIs the double slit experiment performed measuring single photons?Why do cylindrical waves create an interference pattern (and sphericals do not)?
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$begingroup$
This is not a duplicate. No other answer has said anything about this specific question.
I have read this question:
Why does the photon strike at one or another place on the tape?
where PhysicsDave says:
All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities. Destructive interference is a violation of conservation of energy, destructive interference is best explained by QM, there is low probability of photons arriving in the dark areas and that is why the dark areas are dark.
Now there are two opinions on this site:
When a single photon is shot, it passes through both slits as a wave and the partial waves of the photon interfere with each other, and create constructive (bright area) or destructive (dark area) interference.
Each and every single photon shot leaves a spot on the screen in the bright areas, there is a low probability for a photon to land in the dark area
These are two different meanings. One says that some photons never reach the screen, and never interact with the screen, that is why we see dark areas.
The other one says, that each and every single photon shot will leave a dot on the screen in the bright areas, and it is just that there is low probability for the photons to arrive in the dark areas, that is why we see dark areas.
Question:
- Which one is right? Does every single photon shot leave a dot on the screen in the bright area?
quantum-mechanics electromagnetism photons double-slit-experiment
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is not a duplicate. No other answer has said anything about this specific question.
I have read this question:
Why does the photon strike at one or another place on the tape?
where PhysicsDave says:
All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities. Destructive interference is a violation of conservation of energy, destructive interference is best explained by QM, there is low probability of photons arriving in the dark areas and that is why the dark areas are dark.
Now there are two opinions on this site:
When a single photon is shot, it passes through both slits as a wave and the partial waves of the photon interfere with each other, and create constructive (bright area) or destructive (dark area) interference.
Each and every single photon shot leaves a spot on the screen in the bright areas, there is a low probability for a photon to land in the dark area
These are two different meanings. One says that some photons never reach the screen, and never interact with the screen, that is why we see dark areas.
The other one says, that each and every single photon shot will leave a dot on the screen in the bright areas, and it is just that there is low probability for the photons to arrive in the dark areas, that is why we see dark areas.
Question:
- Which one is right? Does every single photon shot leave a dot on the screen in the bright area?
quantum-mechanics electromagnetism photons double-slit-experiment
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
$endgroup$
$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
1
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago
add a comment
|
$begingroup$
This is not a duplicate. No other answer has said anything about this specific question.
I have read this question:
Why does the photon strike at one or another place on the tape?
where PhysicsDave says:
All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities. Destructive interference is a violation of conservation of energy, destructive interference is best explained by QM, there is low probability of photons arriving in the dark areas and that is why the dark areas are dark.
Now there are two opinions on this site:
When a single photon is shot, it passes through both slits as a wave and the partial waves of the photon interfere with each other, and create constructive (bright area) or destructive (dark area) interference.
Each and every single photon shot leaves a spot on the screen in the bright areas, there is a low probability for a photon to land in the dark area
These are two different meanings. One says that some photons never reach the screen, and never interact with the screen, that is why we see dark areas.
The other one says, that each and every single photon shot will leave a dot on the screen in the bright areas, and it is just that there is low probability for the photons to arrive in the dark areas, that is why we see dark areas.
Question:
- Which one is right? Does every single photon shot leave a dot on the screen in the bright area?
quantum-mechanics electromagnetism photons double-slit-experiment
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
$endgroup$
This is not a duplicate. No other answer has said anything about this specific question.
I have read this question:
Why does the photon strike at one or another place on the tape?
where PhysicsDave says:
All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities. Destructive interference is a violation of conservation of energy, destructive interference is best explained by QM, there is low probability of photons arriving in the dark areas and that is why the dark areas are dark.
Now there are two opinions on this site:
When a single photon is shot, it passes through both slits as a wave and the partial waves of the photon interfere with each other, and create constructive (bright area) or destructive (dark area) interference.
Each and every single photon shot leaves a spot on the screen in the bright areas, there is a low probability for a photon to land in the dark area
These are two different meanings. One says that some photons never reach the screen, and never interact with the screen, that is why we see dark areas.
The other one says, that each and every single photon shot will leave a dot on the screen in the bright areas, and it is just that there is low probability for the photons to arrive in the dark areas, that is why we see dark areas.
Question:
- Which one is right? Does every single photon shot leave a dot on the screen in the bright area?
quantum-mechanics electromagnetism photons double-slit-experiment
quantum-mechanics electromagnetism photons double-slit-experiment
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
edited 54 mins ago
Árpád Szendrei
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
asked 10 hours ago
Árpád SzendreiÁrpád Szendrei
7,3841 gold badge13 silver badges36 bronze badges
7,3841 gold badge13 silver badges36 bronze badges
$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
1
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago
add a comment
|
$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
1
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago
$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
1
1
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
The photon is a quantum mechanical entity.
Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.
The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.
It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
$endgroup$
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
add a comment
|
$begingroup$
Which one is right?
I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.
Without getting into the real physics (i.e., the math):
- The geometry of the experiment (two slits, and the screen) defines a wave function.
- Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
- Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.
Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
$endgroup$
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The photon is a quantum mechanical entity.
Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.
The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.
It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
$endgroup$
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
add a comment
|
$begingroup$
The photon is a quantum mechanical entity.
Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.
The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.
It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
$endgroup$
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
add a comment
|
$begingroup$
The photon is a quantum mechanical entity.
Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.
The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.
It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
$endgroup$
The photon is a quantum mechanical entity.
Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.
The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$Ψ$ . The $Ψ^*Ψ$ is the probability distribution for the accumulation of photons.
It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
answered 9 hours ago
anna vanna v
169k8 gold badges162 silver badges470 bronze badges
169k8 gold badges162 silver badges470 bronze badges
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
add a comment
|
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
1
1
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Number 1 also dangerously gets close to treating $Psi$ as something physical
$endgroup$
– Aaron Stevens
9 hours ago
add a comment
|
$begingroup$
Which one is right?
I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.
Without getting into the real physics (i.e., the math):
- The geometry of the experiment (two slits, and the screen) defines a wave function.
- Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
- Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.
Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
$endgroup$
add a comment
|
$begingroup$
Which one is right?
I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.
Without getting into the real physics (i.e., the math):
- The geometry of the experiment (two slits, and the screen) defines a wave function.
- Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
- Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.
Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
$endgroup$
add a comment
|
$begingroup$
Which one is right?
I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.
Without getting into the real physics (i.e., the math):
- The geometry of the experiment (two slits, and the screen) defines a wave function.
- Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
- Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.
Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
$endgroup$
Which one is right?
I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.
Without getting into the real physics (i.e., the math):
- The geometry of the experiment (two slits, and the screen) defines a wave function.
- Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
- Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.
Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
edited 9 hours ago
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
answered 9 hours ago
Solomon SlowSolomon Slow
3,27614 silver badges18 bronze badges
3,27614 silver badges18 bronze badges
add a comment
|
add a comment
|
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$begingroup$
It's always possible for a photon to be reflected
$endgroup$
– Ruslan
10 hours ago
$begingroup$
This may help: aapt.scitation.org/doi/full/10.1119/1.4955173
$endgroup$
– J. Manuel
9 hours ago
1
$begingroup$
Excpet in the idealized limit of a diffraction grating there isn't a crisp distinction between "light area" and "dark area"; there is smooth transition between light and dark. It is easier to see in a single slit pattern (and it is why we usually map out the location of the dark fringes for single slit patterns).
$endgroup$
– dmckee♦
9 hours ago
$begingroup$
In the ideal case (no reflection), all photons will interact with the screen as this is required by energy and momentum conservation. The density of interacting points will be proportional to the square of the wave function though.
$endgroup$
– J. Manuel
9 hours ago