Double slit experiment whith one photon

In summary: ON AVERAGE there is only 1 photon/second emitted and the experiment is only 1foot long - the photon travels through the experiment in 1ns so there is only a 1 in 1billion chance of another photon being emitted in the same nanosecond and being in the experiment at the same time. ) then the odds of detecting another photon are VERY low.
  • #1
alvaros
166
0

Homework Statement



From "http://physicsweb.org/articles/world/15/9/1":

In 1909 Geoffrey Ingram (G I) Taylor conducted an experiment in which he showed that even the feeblest light source - equivalent to "a candle burning at a distance slightly exceeding a mile" - could lead to interference fringes. This led to Dirac's famous statement that "each photon then interferes only with itself".

To be sure that just one photon is producing interference you need to calculate how long the photon last ( t ). Since s = c . t ( space, velocity of light and time ), this will give us a length of the photon.
"length is this experiment" an operative definition.

How can be calculated the power of ligth that must reach the double slit to be shure that there is just a photon at a time ?

Homework Equations



Energy of a photon E = h. nu
Energy of an electromagnetic wave E = 1/2 E . H ( per volume )
Impedance of the empty space E / H = a given number


The Attempt at a Solution



From the tree equations you can derive the volume of a photon, but:
volume = surface . length and we don't know the surface of the photon.
Besides this seems that is a nosense concept ( volume, length, surface ) when applied to a photon.

So my question is

"How was calculated the power of ligth that must reach the double slit to be shure that there is just a photon at a time ? "

I don't understand QM, so I just want an answer to my question.

Thanks in advance.
 
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  • #2
alvaros said:

The Attempt at a Solution



From the tree equations you can derive the volume of a photon, but:
volume = surface . length and we don't know the surface of the photon.
Besides this seems that is a nosense concept ( volume, length, surface ) when applied to a photon.

So my question is

"How was calculated the power of ligth that must reach the double slit to be shure that there is just a photon at a time ? "

You are right, the notions of volume, length, and surface are not applicable to photons. However, you can easily calculate how many photons a light source emits each second. You need to know the light frequency [itex] \nu [/itex], the energy of each photon [itex] E = h \nu [/itex], and the energy emitted by your light source in one second. The latter quantity is called the power of the source. For example, a 100 W(att) lightbulb emits 100 Joules of light energy each second.

You'll see that the lightbulb emits a huge number of photons per second: billions upon billions... In order to reduce the rate of photon emission you can apply filters or let the light to pass through a small aperture (thus blocking a large portion of photons). You'll need to repeat this filtering/blocking procedure several times until you get to the 1 photon/second (or whatever rate you think is appropriate to call "one photon at a time") light source.

Eugene.
 
  • #3
meopemuk:
You'll need to repeat this filtering/blocking procedure several times until you get to the 1 photon/second (or whatever rate you think is appropriate to call "one photon at a time") light source.
I don't know what is the rate appropiate to call "one photon at a time" and this is the relevant answer I want to know.
Thanks.
 
  • #4
alvaros said:
I don't know what is the rate appropiate to call "one photon at a time"
There isn't really an answer to this, it's all a question of probabilites.
If you reduce the power of the lamp so that ON AVERAGE there is only 1 photon/second emitted and the experiment is only 1foot long - the photon travels through the experiment in 1ns so there is only a 1 in 1billion chance of another photon being emitted in the same nanosecond and being in the experiment at the same time.
Are those odds good enough to convince you there is only one photon?
If not reduce the brightness so there is only one photon/hour or 1/day.
 
  • #5
mgb_phys said:
There isn't really an answer to this, it's all a question of probabilites.
If you reduce the power of the lamp so that ON AVERAGE there is only 1 photon/second emitted and the experiment is only 1foot long - the photon travels through the experiment in 1ns so there is only a 1 in 1billion chance of another photon being emitted in the same nanosecond and being in the experiment at the same time.
Are those odds good enough to convince you there is only one photon?
If not reduce the brightness so there is only one photon/hour or 1/day.

Excellent answer! I couldn't say it better.

Eugene.
 
  • #6
all quotes from mgb_phys:

There isn't really an answer to this, it's all a question of probabilites
.
I understand probabilities, let's suppose that the source emits a photon exactly each certain amount of time, like a clock.


If ( ... ) there is only 1 photon/second emitted and the experiment is only 1foot long - the photon travels through the experiment in 1ns
( c = 1 feet/ns ) You are assuming that the length of the photon is 0, why ?
Can you be shure that a photon can't last 2 seconds or 10 minutes, why ?

If not reduce the brightness so there is only one photon/hour or 1/day.
It doesn't seem very scientific.
Are those odds good enough to convince you there is only one photon?
Im convinced that there is just a photon at a time, this is not the question.
 
  • #7
alvaros said:
( c = 1 feet/ns ) You are assuming that the length of the photon is 0, why ?
Can you be shure that a photon can't last 2 seconds or 10 minutes, why ?

From everything we know about photons, they are point-size particles with infinite lifetime.

The fact that they are sizeless particles is supported (among other things) by the observation that each photon interacts with a single AgBr cluster of the photographic plate emulsion. The fact that photons never decay or disappear is supported (among other things) by observations of distant galaxies. However, the infinite lifetime of photons does not imply that their "length" is infinite as well.

Eugene
 
  • #8
I imagine you need to do something like the following for the problem in the OP:

Assume you have an isotropic source of power P, some distance R from the slits. Let the slit area be A - that gives you a total power through the slit. Assume the light is monochromatic at frequency f to find the photon rate, and hence the time between photons arriving at the slits. Finally, assume a distance D from the slits to the screen, and ensure that there must never be more than one photon in this space at any time.
 
  • #9
To all posters: I read carefully your posts and I wanted to discuss each sentence you wrote but I want to focus in the main idea.
Lets suppose you can control exactly the rate of the photons arriving at the double slit ( 1 photon/second ... 1 photon/ns ),
which rate would you choose ? ( to be shure that each photon interferes whith itselt )

meopemuk:
The fact that photons never decay or disappear ...
From this sentence I infer that you never can be shure that there is just a photon at a time. Is there any distribution of probabilities of the "life time" of the photons ? The "life time" depends of the source ( distant galaxies ) ?
 
  • #10
alvaros said:
Let's suppose you can control exactly the rate of the photons arriving at the double slit ( 1 photon/second ... 1 photon/ns ),
which rate would you choose ? ( to be shure that each photon interferes whith itselt )

Suppose that the distance between your proton source and the photographic plate is 1 foot. For a photon it takes 1ns to travel this distance. So, if your source emits 1 photon per 1 ns (let's make it 1 photon per 2ns, to be certain), then you can be sure that when a photon is emitted, the previous photon has already traveled the full length (1 foot), interacted with the photographic plate, and left its mark there. So, the previous photon cannot have any effect on what the present photon will be doing. With this emission rate you can confidently say that each photon behaves on its own, there can be no any photon-photon interaction, and if an intereference pattern is formed in these conditions, then you must conclude that each photon "interferes with itself".


alvaros said:
meopemuk: The fact that photons never decay or disappear ...

From this sentence I infer that you never can be shure that there is just a photon at a time. Is there any distribution of probabilities of the "life time" of the photons ? The "life time" depends of the source ( distant galaxies ) ?

I am not sure I understand your question. I said that the "lifetime" of photons in infinite. This means that once a photon is created (e.g, in an atomic transition, or whatever) it lives forever. It travels with the speed of light until
some other atom absorbs it. The photons emitted by distant galaxies apparently haven't met any absorbers on their way to Earth. If the photons were unstable particles with a finite lifetime t, we would have trouble seeing anything at distances greater than ct from Earth.

Eugene.
 
  • #11
Taking the Heisenberg uncertainty point of view, the "duration" of a photon can be related to it's bandwidth. This is how I'd approach the problem - if you know that your source is a HeNe laser, say, then you can look up it's bandwidth and calculate it's coherence length. This gives you the temporal "duration" of a photon - if the average time separation between photon emissions is longer than this, then you can say that there is very probably only one photon at a time interacting with the slits.
 
  • #12
meopemuk:
Suppose that the distance between your proton source and the photographic plate is 1 foot. For a photon it takes 1ns to travel this distance. So, if your source emits 1 photon per 1 ns (let's make it 1 photon per 2ns, to be certain), then you can be sure that when a photon is emitted, the previous photon has already traveled the full length (1 foot), interacted with the photographic plate, and left its mark there
The experiment doesn't change whit the distance source-double slit, the interference happens when the photon ( or two photons ) arrive at the double slit.
If your argument is true you can't set up this experiment using the photons from stars because you would need years between photon and photon.
JeffKoch:
Taking the Heisenberg uncertainty point of view, the "duration" of a photon can be related to it's bandwidth. This is how I'd approach the problem - if you know that your source is a HeNe laser, say, then you can look up it's bandwidth and calculate it's coherence length. This gives you the temporal "duration" of a photon - if the average time separation between photon emissions is longer than this, then you can say that there is very probably only one photon at a time interacting with the slits.
I don't understand.
What does coherence length mean ? If the bandwith of the laser is 10 Hz, what is the coherence length ?

But I want to further explain my posts.
I want to know how is related the idea of a photon whith the electromagnetic waves. A photon is a piece of electromagnetic wave ( true ? ), so it must start and end, and occupy a certain volume.
From my first post:
Energy of a photon E = h. f
Energy of an electromagnetic wave E = 1/2 E . H ( per volume )
Impedance of the empty space E / H = a given number
From the tree equations you can derive the volume of a photon
What am I missing ?
 
  • #13
alvaros said:
But I want to further explain my posts.
I want to know how is related the idea of a photon whith the electromagnetic waves. A photon is a piece of electromagnetic wave ( true ? ), so it must start and end, and occupy a certain volume.

alvaros,

I think this is a wrong idea. In my opinion, it is more correct to imagine photon as a point particle, and the electromagnetic wave as a (sort of) wavefunction that describes the probability amplitudes of finding this particle in different regions of space.

This situation is very analogous to the QM description of the electron. We consider electron as a point particle, and not as a piece of the De Broglie wavefunction. This is a classic example of the "wave-particle duality" in quantum mechanics.


Eugene.
 
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  • #14
JeffKoch said:
you can look up it's bandwidth and calculate it's coherence length. This gives you the temporal "duration" of a photon
You cannot talk about the coherence length of a single photon. The coherence length is a measure of how long the source gives 'equivalent' photons. Given that the poster is already confused about the size of a photon I don't think it is useful to imagine the coherence length as the length of a photon in an interference experiment.
 
  • #15
mgb_phys said:
You cannot talk about the coherence length of a single photon.

Sure you can, in an average sense. An excellent working model of a photon (whatever that really is - in fact I was asked "What is a photon?" during my dissertation oral exam :eek: I don't think anyone can really answer this.) is a wave with a single oscillation frequency within an envelope function. A Fourier transform gives the bandwidth.

I can't think of any other measure for the "length" of a photon, though I agree that perhaps this will just confuse the poster.

Alvaros, can you clarify what - exactly - the homework question asks? Your reference is to an article, not a homework problem.
 
  • #16
The 'average' of a single photon (smirk)?
I got asked how many constellations there are! Ok I'm an astronomer but my Phd was in building closure phase interferometers.
 
  • #17
mgb_phys said:
The 'average' of a single photon (smirk)?.

More like, "an average photon" from the source.

If you have a better measure for the "length" of a photon, I'm all ears - I can't think of one that doesn't invoke the ghost of Heisenberg. You could consider lifetime instead of length, but it's the same thing.
 
  • #18
Concepts ( like: time, space, coherence length, sizeless particles, wavefunction ) provoke endless discussions. This forum proves it.
I just want to talk about this experiment, if you agree.

1 -The experiment doesn't change with the distance source-double slit ( True / False )

2 -To be shure that we are proving the interference of a single photon with itself we need to reduce the number ( /second ) of photons arriving at the double slit ( T/F )

How can we calculate the highest rate of photons arriving at the double slit in order to prove the interference photon-itself ?

JeffKoch:
Alvaros, can you clarify what - exactly - the homework question asks? Your reference is to an article, not a homework problem.
its not homework, just curiosity.
 
  • #19
I think that a experiment from 100 years ago is easy.
And also its easy to give an answer.

I repeat my last post for your convenience:
1 -The experiment doesn't change with the distance source-double slit ( True / False )

2 -To be shure that we are proving the interference of a single photon with itself we need to reduce the number ( /second ) of photons arriving at the double slit ( T/F )

3 -how can we calculate the highest rate of photons arriving at the double slit in order to prove the interference photon-itself ?

Sorry, I forgot a possible answer: nosense. ( true / false / nosense )
Thanks.
 
  • #20
Hi alvaros,

It is well-established that free photons of visible light do not interact with each other (there is a photon-photon interaction predicted by QED, but it is much too weak to have any effect on interference experiments). So, for the interference picture, it doesn't matter what is the intensity of the light source (the number of photons per second). It could be a super-powerful laser, or it could be a source emitting one photon per year. The accumulated interference picture will be exactly the same. This fact has been proven by experiment.

Eugene.
 
  • #21
meopemuk:
It is well-established that free photons of visible light do not interact with each other (there is a photon-photon interaction predicted by QED, but it is much too weak to have any effect on interference experiments).
I assume this is the answer to my first question:
1 -The experiment doesn't change with the distance source-double slit ( True / False )
True

But
So, for the interference picture, it doesn't matter what is the intensity of the light source (the number of photons per second). It could be a super-powerful laser, or it could be a source emitting one photon per year. The accumulated interference picture will be exactly the same. This fact has been proven by experiment.
says nothing about how must be calculated the highest power to prove:
This led to Dirac's famous statement that "each photon then interferes only with itself
Yes, everybody would agree that one photon per year is enough slow rate, but this is not a satisfactory answer for a physicist.

I infer that there is no way of knowing where is the boundary between interference (photon-another photon) and (photon-itself) and I can't understand why nobody says it clearly.
 
  • #22
alvaros said:
I understand probabilities, let's suppose that the source emits a photon exactly each certain amount of time, like a clock.

But does such a source exist? I don't remember ever reading about one. Are there any photon sources (even idealized, gedankenexperiment-type sources) that are not probabilistic in nature?
 
  • #23
alvaros said:
I infer that there is no way of knowing where is the boundary between interference (photon-another photon) and (photon-itself) and I can't understand why nobody says it clearly.

alvaros,

in physics we have statements of various degrees of plausibility: from "absolutely true" to "absolutely false" and everything in between. I think that the statement "a photon interferes with itself" belongs to the category "most likely to be true". Of course, one can doubt that and imagine (in the case of the source emitting 1 photon per year) that the photon absorbed by the photographic plate 1 year ago has some influence on the photon that we are shooting today. However this assumption is very weird, and not many people would take it seriously. You are welcome to pursue your line of reasoning, but I doubt that it will lead you to any better understanding of physics.

Eugene.
 
  • #24
meopemuk:
in physics we have statements of various degrees of plausibility: from "absolutely true" to "absolutely false" and everything in between. I think that the statement "a photon interferes with itself" belongs to the category "most likely to be true". Of course, one can doubt that and imagine (in the case of the source emitting 1 photon per year) that the photon absorbed by the photographic plate 1 year ago has some influence on the photon that we are shooting today. However this assumption is very weird, and not many people would take it seriously. You are welcome to pursue your line of reasoning, but I doubt that it will lead you to any better understanding of physics.
I understood.

jtbell:
But does such a source exist? I don't remember ever reading about one. Are there any photon sources (even idealized, gedankenexperiment-type sources) that are not probabilistic in nature?
This is not the main question, but thanks for teaching me something.
 
  • #25
alvaros said:
I think that a experiment from 100 years ago is easy. And also its easy to give an answer.

Apparently not. :wink:

alvaros said:
1 -The experiment doesn't change with the distance source-double slit ( True / False ).

If the source is a point source, then it does change because the flux of photons incident on the slit scales as 1/R^2. The source/slit distance only matters to the extent that it affects the number of photons per unit time per unit area at the slit.

alvaros said:
2 -To be shure that we are proving the interference of a single photon with itself we need to reduce the number ( /second ) of photons arriving at the double slit ( T/F ).

True.

alvaros said:
3 -how can we calculate the highest rate of photons arriving at the double slit in order to prove the interference photon-itself ?

As a reasonable first guess, consider a thermal line-emission source. Filter it so that you're almost certain that all the photons hitting your slit are from a single line transition. Measure the power per unit area at your slit plane, and calculate the number of photons incident in the area of the slit per unit time (treat them as photon bullets). Look up the spontaneous emission lifetime of this transition, which is the 1/e time constant of the exponential photon emission probability distribution. Then, attenuate the source emission until the calculated number of photons incident in the area of the slit is 1 per spontaneous emission lifetime. This is the boundary between "multiple hits" and "single hits" - for extra confidence, attenutate the source by another factor of 10 or 100. At that point you're pretty certain that only one photon at a time is incident on the slit.
 
  • #26
JeffKoch:
If the source is a point source, then it does change because the flux of photons incident on the slit scales as 1/R^2. The source/slit distance only matters to the extent that it affects the number of photons per unit time per unit area at the slit.
Ok.
As a reasonable first guess, consider a thermal line-emission source. Filter it so that you're almost certain that all the photons hitting your slit are from a single line transition
Ok, you need monochromatic light.
Measure the power per unit area at your slit plane, and calculate the number of photons incident in the area of the slit per unit time (treat them as photon bullets). Look up the spontaneous emission lifetime of this transition, which is the 1/e time constant of the exponential photon emission probability distribution.
Ok, you are calculating how many photons are arriving at the double slit.
Then, attenuate the source emission until the calculated number of photons incident in the area of the slit is 1 per spontaneous emission lifetime.
1 - I don't understand ( is "lifetime" as the lifetime of U238 ? )
2 - You didnt define "the area of the slit" ( nor the area of the photon )
3 - How many time a photon can interfer with another one ?
If a photon is a point particle this time is 0 ->you don't need to reduce the power per unit area because every photon is interfering with itself.
4 - And I am assuming that the interference is just produced when the photon passes trough the double slit.
 
  • #27
alvaros said:
1 - I don't understand ( is "lifetime" as the lifetime of U238 ? )

The inverse of the Einstein A coefficient. Every excited state has an exponential probability distribution for spontaneously decaying, and you can look up important ones in books or probably just find them by googling. The lifetime relates to the bandwidth, which relates to the "average length" of a photon to the extent this concept isn't nonsense.

alvaros said:
2 - You didnt define "the area of the slit" ( nor the area of the photon )

It's your experiment, so you'll need to know the length and width of the slits. You're overanalyzing the problem by considering the "area of a photon" - it's absorbed over atomic distance scales, so for the present purposes assume it's zero.

alvaros said:
3 - How many time a photon can interfer with another one ? If a photon is a point particle this time is 0 ->you don't need to reduce the power per unit area because every photon is interfering with itself.

I don't understand what you're asking, but isn't the point of your original post to explore how you can prove that each photon interferes only with itself? If you assume you don't need to reduce the power per unit area because every photon interferes with itself, then you've just assumed the answer.

I'm not sure how it's possible to help any further here, good luck calculating!
 
  • #28
I attend your discussion very attentively and I can say that you help me to resolve a many of my dilemmas.
My vision about photons is a slightly different from official science. Reason is very simple. By my opinion, regarding energy aspect, to say that there is some object without a mass, is the same as if we say that there is a wave without a frequency … !? But, though this is not irrelevant fact, this is not reason for my appeal to you.
I must appreciate that I learned much about photons reading your discussion and that is a basical reason to ask you for help.
Please, if you can find some time, give a look at “Heralding single photons from pulsed parametric down-conversion”, T.B. Pittman *, B.C. Jacobs, J.D. Franson. Her is a link:
http://userpages.umbc.edu/~pittmtb1/pdf%20publications/OC%20Heralding%20Efficiency.pdf [Broken]
Regarding Fig/4 (Measured spectrum of the heralded photons), it seems to me that almost all energy content of 780 nm photons is contained within a 10 nm bandwidth because wider-band filter usage did not make noticeable changes in number of average counts. Considering that, one can say that 780 nm photon have bandwidth (spread in frequency) of: df = 7*10^12 Hz, and, coherence time (related to the classical uncertainty principle) is: dt = 1,42*10-13 sec. I think that coherence time is not equal but maybe (somehow, someway) can be related to photon’s “duration” although we defined a photon as point-size particle with infinite lifetime.
By my opinion, notion “photon-duration” can be interpreted quite analogous to diagram showed on fig.4 whereby coinc. counts axis would correspond to photon energy-density-content and wavelength axis should be interpreted as time axis. In this manner, one can say that photon lifetime is infinite, but his “mainly-duration” is time in which spreading energy density is decreased to negligible quantity at concrete path length i.e. coherence length. Means that absolutely all photons energy is implied within space-time infinity. Considering with that, maybe we can say that “mainly-duration” “size” for 780 nm photon is (about) 4,28*10-5 m = 42,8 um ?
What is your opinion?
 
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  • #29
gmarjanovic said:
What is your opinion?

I didn't read the paper, but what you posted is consistent with what I imagine for the "duration" (or length, equivilently) of a photon. Keep in mind, it's just a model though - it's useful in some ways, but can't really capture what a photon "is".
 
  • #30
Yes, You are quite right. We can't really capture what a photon "is".
Yet, I'm very glad because of your accordance with this model though. Quantum world is very "far" from us and all what we can do (for now) is to visualizing a variety of though models. In “our” world we can measure objects radius e.g. However, let me remember you that particle nucleus radius also can’t be “measured” exactly but is defined as length where energy density is decreased around 90 % of energy density in particles center … Considering that fact, I’m addicted to conceive a photon (or any wave-nature object) as (some kind of) “time-ball”, quite and purely analogous to particle as a (some kind of) “space-ball”. …
Your discussion and explanation varied ideas related to photon help me very much in my contemplations about Nature …
Thanks for comment.
 
  • #31
it seem s the solution should be simple, if after exciting a single hydrogen atom's electron to a higher energy state and as it returns a photon is emitted, pass it through the double slit experiment repeatedly.
 
  • #32
The Length of the Photon

Would it be helpful to imagine the photon as a wave thread connection (physical) from the source to where it is absorbed or entangled again? Between the source and destination (entanglement) the photon wave has small amplitude with a huge period to conserve the energy. This long period (threadlike) is perhaps how huge distances can be trans versed with no (little?) loss of total photon energy.
If it entangles with something then it collapses from the source to its source determined wavelength and amplitude again at the destination. But before the entanglement it is stretched out with immensely large period and tiny amplitude not entangled.
Given this idea, the length of a photon is perhaps the distance from the source to the place where it entangles, and only when it entangles does it express its source determined wavelength.
When a double slit is in the photon path, the stretched thread (which is still connected to the source) goes through each slit at a different time. This time difference is related to the small distance between slits. This time amount would perhaps be too small to measure.
With this view the photon does not interact with itself nor does it pass through both slits simultaneously.
Its interaction or entanglement with the final destination is where the characteristic "interference" pattern is seen caused by the rapid oscillation of the photon thread oscillating from one slit to the other. The oscillation is due to the 50% chance that it goes through one slit.
In this view the photon does not leave the source and arrive at entanglement.
The source and destination are connected until it entangles (almost instantaneously). The long period energy then collapses at entanglement with a source determined wavelength (reclaiming the thread energy) and producing the interference pattern because it went through different slits at different times. The long period wave has a vector force toward the destination.
So the length of the photon would be the distance from the source to the destination.
The form of the photon energy would have 2 forms. Source determined wavelength and thread wavelength (huge period with tiny tiny amplitude). The further away the destination, the lower the amplitude of the wave and the longer the wavelength, until entanglement.
If the wave lost any energy into intermediate space, the source determined wavelength upon entanglement would be red shifted.
Maybe the stretched out photons are what “dark” matter is as all space is filled with light. A very small proportion of it becomes entangled with matter. Like sunshine.
This concept becomes more interesting when one considers that the form of photonic energy is probably helical. Between the source and the destination the helix approaches straight line proportions.
Entanglement requires the recoiling of this photon "spring thread". The entangled state being the source determined wavelength.

So perhaps the length of a photon is the distance between the source and the destination. Does this idea remove the enigma? Is there a weakness in this model? Does it affect expanding universe models?
Are these ideas admissible in any way into consideration?
 

1. What is the double slit experiment with one photon?

The double slit experiment with one photon is a thought experiment that demonstrates the wave-particle duality of light. It involves a single photon being fired at a barrier with two slits, and the resulting interference pattern on a screen behind the barrier.

2. How does the double slit experiment with one photon demonstrate wave-particle duality?

The experiment shows that a single photon can behave as both a particle and a wave. When the photon passes through the slits, it creates an interference pattern on the screen, similar to how waves behave. However, when the photons are observed, they behave as individual particles, passing through only one of the slits.

3. Why is the double slit experiment with one photon important in quantum mechanics?

This experiment is important because it challenges our understanding of the nature of light and matter. It shows that particles can exhibit wave-like behavior and that our observations can affect the behavior of particles. This has significant implications in the field of quantum mechanics and our understanding of the fundamental laws of the universe.

4. Can the double slit experiment with one photon be performed with other particles?

Yes, the experiment has been performed with other particles such as electrons, neutrons, and even large molecules. The results are similar to the photon experiment, demonstrating that wave-particle duality is not limited to light but is a fundamental characteristic of particles.

5. What other applications does the double slit experiment with one photon have?

The experiment has been used in various fields, including optics, quantum computing, and cryptography. It has also been used to study the behavior of particles in different environments and to test the validity of quantum mechanics theories.

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